OOOOOO
OO OO
O O
OO OO
OOOO OOOOOO
OOOO ||
SSSSSSSS UUU UUU DDDDDDDDDDD OOOOOOOO KXK KXK UUU UUU VVV VVV AAAAAA NNN NNN IIIIIIIIIIIIII AAAAAA
SSSSSSSSSSSS UUU UUU DDDDDDDDDDDDD OOOOOOOOOOOO KXK KXK UUU UUU VVV VVV AAAAAA NNN NNN IIIIIIIIIIIIII AAAAAA
SSSS SSSS UUU UUU DDD DDDD OOOO OOOO KXK KXK UUU UUU VVV VVV AAA||AAA NNNN NNN IIII AAA AAA
SSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA||AAA NNNNN NNN IIII AAA AAA OOOO
SSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA||AAA NNNNNN NNN IIII AAA AAA OOOO
SSSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA || AAA NNN NNN NNN IIII AAA AAA
SSSSSSS UUU UUU DDD DDD OOO OOO KXKXKKXK UUU UUU VVV VVV AAAAAAAAAA NNN NNN NNN IIII AAAAAAAAAA
SSSSSSSS UUU UUU DDD DDD OOO OOO KXKXKXK UUU UUU VVV VVV AAAAAAAAAA NNN NNN NNN IIII AAAAAAAAAA
SSSSSSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAAA || AAAA NNN NNN NNN IIII AAAA AAA
SSSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA || AAA NNN NNNNNN IIII AAA AAA
SSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA || AAA NNN NNNNN IIII AAA AAA OOOO
SSS UUU UUU DDD DDD OOO OOO KXK KXK UUU UUU VVV VVV AAA || AAA NNN NNNN IIII AAA AAA OOOO
SSSS SSSS UUUU UUUU DDD DDDD OOOO OOOO KXK KXK UUUU UUUU VVV VVV AAA || AAA NNN NNN IIII AAA AAA
SSSSSSSSSSSS UUUUUUUUUUUU DDDDDDDDDDDDD OOOOOOOOOOOO KXK KXK UUUUUUUUUUUU VVVVVV AAA || AAA NNN NNN IIIIIIIIIIIIII AAA AAA
SSSSSSSS UUUUUUUU DDDDDDDDDDD OOOOOOOO KXK KXK UUUUUUUU VVVVVV AAA || AAA NNN NNN IIIIIIIIIIIIII AAA AAA
||
DDDDDDDDDDD IIIIIIIIIIIIII GGGGGGGG IIIIIIIIIIIIII TTTTTTTTTTTTTT SSSSSSSS OOOOOOOO FFFFFFFFFFFFFF
DDDDDDDDDDDDD IIIIIIIIIIIIII GGGGGGGGGGGG IIIIIIIIIIIIII TTTTTTTTTTTTTT SSSSSSSSSSSS OOOOOOOOOOOO FFFFFFFFFFFFFF
DDD DDDD IIII GGGG GGGG IIII TTTT SSSS SSSS OOOO OOOO FFF
DDD DDD IIII GGG IIII TTTT SSS OOO OOO FFF
DDD DDD IIII GGG IIII TTTT SSS OOO OOO FFF
DDD DDD IIII GGG IIII TTTT SSSS OOO OOO FFF
------------------------------------------ DDD DDD IIII GGG IIII TTTT SSSSSSS OOO OOO FFF
\ DDD DDD IIII GGG IIII TTTT SSSSSSSS OOO OOO FFFFFFFFF
--------------------------------------- \ DDD DDD IIII GGG GGGGGG IIII TTTT SSSSSSS OOO OOO FFFFFFFFF
\ \ DDD DDD IIII GGG GGGGGG IIII TTTT SSSS OOO OOO FFF
\ \ DDD DDD IIII GGG GGG IIII TTTT SSS OOO OOO FFF
\ \ DDD DDD IIII GGG GGG IIII TTTT SSS OOO OOO FFF
\ \ DDD DDDD IIII GGGG GGGG IIII TTTT SSSS SSSS OOOO OOOO FFF
\ \ DDDDDDDDDDDDD IIIIIIIIIIIIII GGGGGGGGGGGGG IIIIIIIIIIIIII TTTT SSSSSSSSSSSS OOOOOOOOOOOO FFF
\ \ DDDDDDDDDDD IIIIIIIIIIIIII GGGGGGGG GG IIIIIIIIIIIIII TTTT SSSSSSSS OOOOOOOO FFF
+==================================+ \ \ ||
| P U Z Z L E B Y | \ \ DDDDDDDDDDD EEEEEEEEEEEEEE SSSSSSSS PPPPPPPPPPP AAAAAA IIIIIIIIIIIIII RRRRRRRRRRR
| | \ \ DDDDDDDDDDDDD EEEEEEEEEEEEEE SSSSSSSSSSSS PPPPPPPPPPPPP AAAAAA IIIIIIIIIIIIII RRRRRRRRRRRRR
| S K E P T I C A L M A R I O | \ \ DDD DDDD EEE || SSSS SSSS PPP PPPP AAA AAA IIII RRR RRRR
+==================================+ \ \ DDD DDD EEE\\ || // SSS PPP PPP AAA AAA IIII RRR RRR
| G U I D E B Y | \ \ DDD DDD EEE \\||// SSS PPP PPP AAA AAA IIII RRR RRR
| | \ \ DDD DDD EEE \||/ SSSS PPP PPPP AAA AAA IIII RRR RRRR
| ~-aRjIMikKaL-~ | \ ---------------------------------------------------DDD--------DDD----EEEEEEEEE----------SSSSSSS----------PPPPPPPPPPPPP-------AAAAAAAAAA-----------IIII---------RRRRRRRRRRRR----
+==================================+ \ DDD DDD EEEEEEEEE SSSSSSSS PPPPPPPPPPP AAAAAAAAAA IIII RRRRRRRRRRRR
| F B I D I S C L A I M E R | ------------------------------------------------------DDD--------DDD----EEE---------------------SSSSSSS-----PPP----------------AAA------AAA----------IIII---------RRR--RRR--------
| | DDD DDD EEE SSSS PPP AAA AAA IIII RRR RRR
| THIS GUIDE IS PROVIDED FOR | DDD DDD EEE SSS PPP AAA AAA IIII RRR RRR
| YOUR PERSONAL USE ONLY | DDD DDD EEE SSS PPP AAA AAA IIII RRR RRR
| I CANNOT BE HELD RESPONSIBLE FOR | DDD DDDD EEE SSSS SSSS PPP AAA AAA IIII RRR RRR
| ILLEGAL DOWNLOADING OF FILES | DDDDDDDDDDDDD EEEEEEEEEEEEEE SSSSSSSSSSSS PPP AAA AAA IIIIIIIIIIIIII RRR RRR
+==================================+ DDDDDDDDDDD EEEEEEEEEEEEEE SSSSSSSS PPP AAA AAA IIIIIIIIIIIIII RRR RRR
Sudokuvania: Digits of Despair is a sudoku puzzle hunt that pays tribute to the classic gameplay of action-adventure games like Metroid and Castlevania: exploring a strange new place full of mystery and danger, then gradually coming to master it by defeating enemies and discovering new equipment and techniques to help you in your journey. It uses rules from a variety of common sudoku variants, primarily fog, thermometers, arrows, kropki dots, and German whisper lines, as well as a number of less established rules. It is widely considered very difficult, even for people well-practiced in variant sudoku solving.
The puzzle was set by Skeptical Mario and first uploaded to logic-masters.de on April 5th, 2025. You can find a link to the puzzle on that site, as well as a personal statement from the author.
Sudokuvania: Digits of Despair uses the SudokuPad web-based solving software. Because Sudokuvania makes heavy use of fog mechanics, it is not possible to complete the "world map" as intended without using this software. As of the time of this writing (mid-July 2025), SudokuPad's support for puzzles the size of Sudokuvania continues to be more than a little janky. Please take the following advice (some of which is repeated in the game rules):
In Sudokuvania: Digits of Despair, you are an adventurer exploring the mysterious castle of Sudokuvania. Sudokuvania is organized into a world map for the castle (the puzzle that the link above takes you to) and a set of boss puzzles (independent puzzles that are linked at various points from the world map). The world map can be further broken down into a set of partially-overlapping grids, which are NxN squares with heavy black outlines in which the standard rules of NxN sudoku hold, as if each grid were an independent puzzle. This information can be found as the first entry in the Learned Rules box, in the top-left corner of the world map. As you progress through Sudokuvania, you will discover new rules that will be added to the box. These rules generally hold only on the world map, and the boss puzzles will have their own rulesets.
When you have a cell selected on the world map, the text at the bottom of the puzzle description box (the green box in the top-right of the SudokuPad UI, beneath the puzzle title) will change to talk about what you've selected:
Boss puzzles are associated with specific cells in each of the major grids of the world map. These cells are marked with the "smiling face with horns" emoji (😈) and never appear in the cells that overlap multiple grids. While you can solve boss puzzles immediately once you enter the region and can read the puzzle link in the description box, the instruction you receive after defeating the boss is always to place a specific digit in the boss cell. Therefore, you cannot actually progress on the world map by solving boss puzzles before you reveal its corresponding boss cell.
As you might expect, the boss puzzles in Sudokuvania are often quite challenging, but you should also not underestimate the difficulty of the world map. You will probably spend at least as much time trying to figure out the logic on the world map as you do on the bosses. Don't imagine you'll knock the whole hunt out in an afternoon, or even a day; my own personal solve time was close to 24 hours, spread over an entire weekend in which I fortunately didn't have many other responsibilities. Pace yourself and enjoy the experience. If nothing else, trying to solve while you're mentally fatigued will probably just lead to a lot of silly logic errors and frustration.
Just like in Metroid and Castlevania, you will often find parts of the world map that you cannot solve until you have worked on a different part of the puzzle. In fact, even the 6x6 entrance area cannot be solved until you've completed several other regions. Backtracking is an important part of the experience. Because the world map really can be quite a hard puzzle on its own, of course, you won't always be sure if you're supposed to be backtracking, or if you're just missing something.
This guide is written in a way that I hope will allow you to "incrementally" get yourself unstuck. Text that is not hidden by default (like this) should provide a general outline of what parts of the puzzle to focus on. Additionally, each paragraph in the guide starts with a hint or prompt that is not an outright spoiler for any of the logic. This initial hint will sometimes itself be hidden, if I feel that finding it is an interesting part of the challenge, but you can safely reveal it without worrying that I'll be giving away the logic. This applies only to the first phrase in the paragraph, whether it is hidden or not. If that phrase trails off and leads into some hidden text, the hidden text is very likely to be at least a partial spoiler. Later hidden text in a paragraph generally builds on the conclusions reached earlier in the paragraph. The hidden text does cover every major leap of logic in Sudokuvania, and you will spoil the puzzle if you read it all.
Please don't imagine that I myself solved Sudokuvania in exactly the way I've presented it here. This guide leaves out an awful lot of details, like marking up the possible values of various cells and staring perplexedly at the screen, in favor of just presenting the core logic, which inevitably makes it seem like I found it all rather straightforward. This walkthrough was written during my second solution of the puzzle. While I did solve it on my own, both times from scratch, I found it a very challenging puzzle, and I frequently found myself making mistakes and unsure of how to make progress, even on the second time. I have also frequently re-ordered the logic from the way I originally discovered it, in order to present related conclusions together. My actual soluton was much more scatter-shot and inelegant.
The hiding of spoilers in this guide only works in a browser that supports CSS. If you have JavaScript disabled, you can read a spoiler by highlighting the text. If you have JavaScript enabled, you can also read a spoiler by clicking on it; you can then re-hide it by clicking on it again.
A very minor early spoiler, which you can read once you've revealed the first half of the Courtyard: clues don't always stay within grids. The rules for cells outside of grids are somewhat different, and you should not make deductions that assume that clues work the same way outside of grids as they do within them. Rest assured that the off-grid rules will eventually be explained.
This guide will use several shorthands and pieces of jargon:
| 4 | |||
| 4 | 2 |
Take a moment to familiarize yourself with both the UI and the shorthands I'll be using in this guide:
You can see a bit of a green line in r3c1. This is a clue, but you don't have a rule for it yet, so you have to ignore it for now.
The puzzle here is straightforward sudoku, but you will not be able to complete it yet. Note that you will have to place digits into cells that still have fog, and that's fine. You should end up having filled two of the three cells on the border with the grid to the lower-right (the Courtyard). You can get some interesting pencil-marks in the upper half of the grid, but there's not really anything else you can do here for now. On to the next grid!
Note that we've gained a new rule, Guide THERMO. Read the description in the Learned Rules box, then click on one of the rule cells in that box and read more about it at the bottom of the puzzle description box. If you've done much variant sudoku, you've probably already seen this rule, but it's good to verify that there's nothing special about it in Sudokuvania. Note that cells on a thermometer always "see" the other cells on the thermometer: because the thermometer must strictly increase, digits may never repeat.
Click on one of the cells you've revealed in the new grid, then read the description of the Courtyard in the puzzle description box. You'll see that we also have a link to the boss puzzle. Since we can't see the boss cell yet, we'll put that off until later.
There's only one clue for how to make progress in this grid right now: the thermometer. We can see that there are at least three more cells on the thermometer. There's a 6 in the bulb, and there are only three digits greater than 3, so we should be able to fill all three cells.
That will reveal all of b4. We can't see the bulbs for the new thermometers. There must be a 1 somewhere in b4. Since the 1 must be in c1 in b4, r7c1 cannot also be 1. What does that mean about the value of r6c2? There must also be a 2 somewhere in b4. Since the 2 is also in c1 in b4, what is the least possible value of r6c2 now? r4c3 is the end of a thermometer.
You've revealed your first boss cell!
This is a great introduction to one of the more important ideas from thermometer sudoku. A digit on a thermometer is always different from every other digit on the thermometer. When a thermometer both (1) crosses a box border and (2) shifts over to fill a different row or column, digits on the thermometer in the first box are often tightly constrained in the second. That happens three times in this puzzle:
Let's start with the first. Where do the digits in r2c2 and r3c2 go in b4? They cannot go in c2 because of sudoku, and they cannot go in c1 because of the thermometer. So they must be in c3. But that means at least one of these digits must be in a bulb of one of the horizontal thermometers there. r2c2 and r3c2 cannot be less than 4 because of the vertical thermometer they're on, so whichever horizontal thermometer(s) they go into the bulb of in b4, the end cell in b6 cannot be less than 8.
The same logic applies to r8c8 and r7c8. The digits in these cells cannot go in c8 or c9 in b6, so they must be in c7. Therefore at least one of these digits must be in the end of one of the horizontal thermometers there. r8c8 and r7c8 cannot be greater than 6 because of the vertical thermometer they're on, so whichever horizontal thermometer(s) they go into the end of in b6, the bulb in b4 cannot be more than 2.
Now, keep in mind that you don't know which of the two horizontal thermometers is constrained by r2c2 and r3c2; in the abstract, it could even be the same one as is constrained by r8c8 and r7c8. But in this case, that's not possible because the thermometer constrained by r2c2 and r3c2 must have at least 4 in its bulb, and the thermometer constrained by r8c8 and r7c8 must have at most 2 in its bulb. So these are in fact the two different horizontal thermometers here, even if you don't know which is which.
Can we tighten the ranges on those thermometers? Consider the possibilities for r3c2. r3c2 must be at least 4. If it were exactly 4, then c1 in in b4 would have to be 1/2/3. But we now know that one of the digits in the bulbs in b4 is at most 2, so this is impossible; r3c2 cannot be 4. If it were more than 5, then both it and r2c2 would have to be greater than 5, and since one of those digits is on a thermometer bulb in b4, the end of that thermometer would have to be greater than 9, which is impossible. So r3c2 must be exactly 5, and that 5 must be the digit in the horizontal thermometer bulb in b4, since the digit in r2c2 would be greater and break the thermometer. So one of the horizontal thermometers starting in b4 is 5/6/7/8/9. Similarly, r7c8 cannot be exactly 6 without putting 7/8/9 in c9 in b6, and you know that one of the horizontal thermometers ends with a 9 in b6; and it cannot be less than 5 without breaking the other horizontal thermometer in b6 by forcing its bulb to be less than 1; so r7c8 = 5, and that 5 must in one of the ends of the horizontal thermometers, so that thermometer is 1/2/3/4/5.
It'd be nice to know is which. Does one seem more constrained by the other clues? the puzzle? r9c3 is is at least 4. It must also be on the thermometer that runs along r7 in b8. So if that's the 1/2/3/4/5 thermometer, it must be one of 2/3/4; but then b8 contains both 2 and 3 on both r7 and r8, which is a contradiction. So the thermometer starting in r6c3 must be the 5/6/7/8/9 thermometer, and the one starting in r4c3 must be 1/2/3/4/5. Now let's revisit what we said before: remember that r3c2 and r2c3 must be in c3 in b4. They are both at least 5, so they must be in r5c3 and r6c3, which means one of them is exactly 5, and it must be r3c2. This puts 2/3/4 in c1 in b4. By the same logic, r7c8 must be 5, and c9 in b6 must be 6/7/8. Remember that you can place those exactly because of the thermometer. Where can 1 be in b1? Only c1. Similarly, 9 in b9 can only be in c9. This makes r3c9 and r7c1 naked singles.
Now we can look at the remaining outer thermometers. All the cells in r8 in b8 are at most 5. r7c3 is at most 4, so r8c2 is at most 3. r8c8 is at most 4. That makes five digits that are 5 or less, and 5 can only be on in b8, forcing it into r8c4. r3c7 must be at least 6, so it is 6/7, r2c8 is 7/8, and r1c8 is 8/9. r2c5 is at least 5, so r2c6 is at least 6. r2c4 cannot be less than 7. r2c2 cannot be less than 6. So there are at least five numbers greater than 5, and only r2c5 can be exactly 5, so it is. This forces 5 in the other three boxes.
Finally we can look at b5. If r4c6 is 6 or more, r5c5 must be 8 or more, and then r6c4 is impossible. So it is 2/3. Similarly, if r6c4 is 4 or less, r5c5 must be 2 or less, and then r4c6 is impossible. So it is 7/8. r4c8 can only be 2/3, so the only possible place for 2 in b5 is r4c6, which also gives r4c8 = 3. The only place for 3 now in b5 is r6c6. r8c6 = 1, which makes a 2/4 pair in r8c5 and r8c8, forcing r8c2 = 3 and r7c3 = 4. r9c6 = 9 is a naked single. r9c4 is the only possible cell for a 3 in b8.
The remainder of the puzzle is straightforward. In case you need to refer back to it, the number to place in the boss cell is 3.
You have now unlocked the Sum WHIP. Take a breather to read about the clue in the Learned Rules and the puzzle description. This is just arrow sudoku, but the clarification about what arrows can do in unrevealed fog cells is useful. You've also revealed some arrows in b5 and b6, and the lower arrow overlaps some thermometers.
What digits can be on that lower arrow? What digits can be on the thermometers leading there? They cannot be 4 or 5, and if either were 6+, the corresponding arrow cell would have to be 7+, forcing the arrow sum to be 9, which it cannot be. So r6c4 is 2/3 and r6c7 is 1/2/3. The arrow sum cannot be greater than 8. r5c5 must be at least 4, so r5c6 cannot be greater than 3. Therefore r6c7 must be 1/2, which forces 3 into r6c4. How is the other arrow constrained? The 1 must be on it, but the other digit must be at least 4. In fact, this is the only option that isn't contradicted.
You can now see an arrow and a thermometer in box 8, although you can't see the end of either. There are at least three cells on the arrow. The only way to make three cells sum to a single digit without using 1 is 2/3/4. There is already a 3 in both columns, so there must be a 1, and it must be in either r7c5 or c8c5. It cannot be in r7c5 because that is an intermediate cell on a thermometer. r7c5 must be 2 or 6. If it is 6, that forces a 9 into r8c4; but r8c1 must be 8/9, and that means r8c6 cannot be greater than 7, which breaks the thermometer. With a 1/2 pair on the arrow, we know that the remaining cells must be either a 4/7 or 6/9 pair. r4c4 must be either 6 or 9, so they must be 4/7.
Take a moment to do some sudoku; you should be able to put a digit in r8c3. This will reveal b7, but you can't do much with that information yet. Observe in particular that there's an arrow leading off the grid from r8c1, which must be at least 8; but it's also the end of an arrow, and if the adjacent cell were 9, that arrow wouldn't work. Something odd's going on. Just leave this alone.
The clues in b9 are a little hard to read. There's one an arrow occupying at least r6c9, r7c9, and r8c8, and another occupying at least r9c9, r8c9, and r7c8. You can't see if the unrevealed cells are the circles of the arrows, though. The clues interact in interesting ways; you might find it helpful to mark the possibilities in the four thermometer cells. r7c8 is at least 5, so r8c9 must be 2 or 3, and r7c8 cannot be greater than 7. The only possible place for 9 in r7 is r7c7. If r7c8 were 7, the sum on the arrow would already be 9 without including r9c9, so r9c9 would have be the circle, and it would have to be 9, which is a contradiction. This creates a 5/6 pair in r7. You should be able to do a lot of sudoku now. r6c9 must be 6/8/9, and the two other cells on its arrow sum to at least 4, so it cannot just be part of the sum; it must be the circle. r8c8 must be 3/5/6, but only 5 actually makes a valid sum in r6c9. This should let you fill all but two cells in boxes 4-9; you can mark your options in the other cells, but you can't do anything there yet. Off to the next grid!
Go ahead and read the description of the new area in the puzzle description. Because these two grids are offset by one cell vertically, the box lines can be a little tricky at first. There's a 1 in r3c1 which can see all three cells on this arrow, so its sum must be 9, and the circle must be in r2c4. r2c5, the circle that was just revealed, cannot be greater than 5 because of the thermometer. It has at least two cells on it, neither of which can be 1, so it must also be at least 5. You can fill in all the thermometer cells you can see. The thermometer continues into r3c6, but this cell sees a 1, a 2, and a 3; it must be 4. You should take a moment to fill in what you can, but you're locked out of the rest of this grid for now.
If you filled in the possibilities right, you should be able to fill the other overlapping cell between these grids. And now you've unlocked WHISPER dash! Take the opportunity to read the grid description as well as the description of the new rule. This is the common "German whisper line" rule from variant sudoku. There are two general insights you should be able to figure out immediately about these lines: the digits on the line are never 5 and they alternate polarities between "high" (greater than 5) and "low" (less than 5). Note that the clue in this grid doesn't actually tell us anything we don't know, because the thermometer already makes r8c2 at least 6!
There is some interesting logic you can do here to fill a few cells, but you won't make much headway beyond that. For now, you need to notice that a couple of whispers were also revealed back in the previous grid.
The new clues here are quite constrained. Both r4c7 and r4c9 cannot be greater than 3. They also both see the 2 in r4c1, so you should be able to fill them in. The newly-revealed line must have at least one digit less than 5 on it, but both cells see 1, 2, and 3, so it must be a 4/9 pair, which can be placed.
This clue might be tricky to interpret. There is one line between r6c3 and r5c4, and there is a second line between r5c3 and r6c4; the fact that they cross doesn't matter. One of the digits on the first line must be greater than 5. If the high digit is in r6c3, it must be 6, and r7c4 must be 1. If the high digit is in in r7c4, it must be 7, and r6c3 must be 1 because of r4c1. So the second line cannot have a 1 (in either of the two cells you know of), which means the low digit cannot be r6c4: it would have to be at least 2, so r5c3 would have to be at least 7, and it can't. r5c3 therefore must be 3 or 4, and if it were 4, r6c4 would have to be 9. Therefore these cells are 3 and 8. The newly-revealed cell, r4c8, must be 8 by sudoku.
You can see a new kind of clue, a red line, running off the grid. Curiouser and curiouser, but not much to be done with it yet. The other clues that just got revealed are more promising. These two lines meet in r7c8, so r6c7 and r6c9 must have the same "polarity", either high or low. If they were low, r7c8 would have to be 9, which it cannot be. r7c8 must be low. It cannot be 4 because both of the other cells would have to be 9. It cannot be 3 because they would have to be an 8/9 pair, but they both see an 8. It cannot be 2 by sudoku. So it must be 1. r8c8 sees 8 and 9, and it cannot be 6 because it would force a 1 into r9c7, so it must be 7. The 2 in r9c7 constrains 2 in b6 into r5c9, and the 7 in this box must be in r6c9. The 3 in c8 must be in r9c8. And... well that's interesting to learn.
You can do a little more in this grid before you run off to take advantage of the new rule. The 2 and 3 in r6 can only be in b5, which forces r6c7 = 9. The 1 and 6 can only be in r6c2 and r5c3, so r6c1 must be 4/5. That seems to be it, though; about half the grid is still wide open. Back to the factory.
Well, now that we have a knight's move restriction in this grid (and only in this grid), we can make some real progress. Consider the arrow. It cannot have a 1, and it cannot be a 2/3 pair because of r9c3, so it must sum to at least 6. But its digit is constrained in b7: it cannot be on its own arrow, so it must be 6/7. If r8c1 were 7, r6c1 would be forced to be 6, but r9c3 would also be forced to be 2, and it would be impossible to fill the arrow. If r8c2 were 6, r9c3 would be forced to be 2 again, and either possibility for r6c1 leads to a contradiction. r6c1 is interesting. The arrow cannot have a 1, and it cannot be a 2/3 pair because of r9c3, so it must sum to at least 6. But its digit is constrained in b7: it cannot be on its own arrow, and it cannot be in r8c2 by the knight restriction, so it must be in one of the other three cells. And those cells are filled with 1, 2/3, and 5/6. So it must be 6. You should be able to finish the thermometer and b7.
You can now see a relatively long thermometer in b5. r6c4 has at least two digits both less and greater than it, and it cannot be 4, 5, or 6, so it must be 3 or 7. If it were 3, one of the ends of the thermometer would have to be 321, but both of the cells that a 2 could be in are a knight's move from r7c3. So it is 7. That means one of the ends must be 789. If the 9 were in r6c6, there would be nowhere to place 9 in b8. r6c5 must be less than 7 by the thermometer, but it cannot be 6 or 5, so it is 2/3/4. The 3 in b8 must be in c6, so it must be in c5 in b5. r5c3 must be low, but it cannot be 2 or 4, and it cannot be 3 because it would see all three possible positions of 3 in b5. Therefore it is 1. r7c4 is now a naked single 6. The only possible place for 1 in b8 is r8c5. This forces 1 in b5 into r4c6. And now we've found our second boss cell!
This is a very clever exercise in both of those general insights about German whisper lines.
Note that the whisper lines never occupy a complete column. 5 can never be on a whisper line, so it's heavily constrained here. You should be able to place all nine 5s right away: there's only 1 cell that's not on a whisper line in both c2 and c8, which only leaves one possible cell in c5, which restricts c4 and c6 via the knight constraint, which restricts c3 and c7 via the knight constraint, which finally resists c1 and c9.
What can you deduce about where certain digits go on the lines? 4 and 6 cannot be next to two cells on the same line, so the ends of the two long lines form a 4/6 pair, on both lines. The 4 must be next to a 9, and the 6 must be next to a 1, so the next cells in must be a 1/9 pair, again on both lines. Since 3 cannot be at an end, it must be next to the 8 and 9, and similarly 7 must be next to the 1 and 2. So the next cells in form a 3/7 pair, and the middle two cells must be 2/7. From top to bottom, then, each long line is 4/6, 1/9, 3/7, 2/8, 2/8, 3/7, 1/9, 4/6.
Time for some coloring. If you're new to this, shading cells in different colors can be a very useful tool for reasoning about where certain digits must be even when you aren't sure of the exact digits. In this puzzle, it's going to be very useful to reason about high and low digits. You don't know whether the digit in r1c2 is high or low, but you know it's one of those; and you know that the cell in r2c2 is the opposite, and that the cell is r3c2 is the opposite of that, and so on. Use the color tool (in the bottom right of the SudokuPad UI) to shade r1c2 dark gray and r2c2 purple. Every cell you mark gray will have the same high/low polarity as r1c2, and every cell you mark purple will have the opposite. You can use whatever colors you like, of course, but I think these two look nice and are still readable on the light blue background that Skeptical Mario chose for this grid.
You can start by marking all of the cells on the line in c2, alternating down the line. What do you know about c8? r5c2 and r5c8 form a 2/8 pair, so they must have opposite polarity. That should let you mark the whole line in c8. Note that both lines have the same sequence of polarities from top to bottom. This means the lines are actually the exact same sequence of digits top to bottom: e.g. both r7c2 and r8c8 are either 1 and 9, and they have the same polarity, so they are either both 1 or both 9.
What you're going to do now is basically play sudoku without knowing exactly which digits you're using. Essentially, you're exploring two hypotheses at once: the one where r1c2 is 4 (so gray is low and purple is high) and the one where it's 6 (so gray is high and purple is low). Eventually you'll hit a constraint that only works with one or the other, and you'll be able to fill in a lot of actual digits very quickly. In the meantime, you can still reason about them as "abstract digits", because the combination of (1) constraining a cell to either a low or high digit (e.g. it contains either 3 or 7) and (2) giving it a specific color (e.g. it is purple) means that, under each hypothesis, you've narrowed it down to exactly one digit (e.g. if purple is high, it must be 7). So you can just do sudoku with "abstract digits": the purple 3/7 is a different digit from both the purple 4/6 and the gray 3/7.
Let's start by looking at r1c2. This is the gray 4/6, and it's in r1c2 and r2c8, so it must be in r3 in b2. It cannot be in r3c6 because of the knight restriction. It cannot be in r3c5 because it would be in the middle of a line, so both of the cells it's connected to would have to be the purple 1/9. So you can mark r3c4 as the gray 4/6. Parallel logic places the purple 4/6 in r7c6. This puzzle has a lot symmetry, so whenever you figure something out about one digit, check to see if the same applies to its opposite.
The middle row of boxes is very constrained. The gray 3/7 can only be in r5c4 in r5, and the purple 3/7 can only be in r5c6. There is only one remaining cell for those in r6 and r4, respectively. Where does the gray 1/9 go in r6? Only in c5 or c6, and it's a 1/9, so it cannot go in the center of a thermometer. Parallel logic puts the purple 1/9 in r4c4. And of course they must be 1 and 9, respectively, which means purple is high and gray is low. You should be able to place an exact digit into every cell you've colored now.
You should be able to finish the puzzle from here. The digit to place back in the boss cell is 8.
You've revealed the rest of the factory, and you've unlocked a new rule: DOUBLE jump. Go ahead and read about the new rule. If you're used to variant sudoku, this is just a standard Kropki black dot, making one digit the double of the other. This means the lower digit must be 1, 2, 3, or 4, and the higher must be 2, 4, 6, or 8; so a digit adjacent to a dot can never be 5, 7, or 9.
But first, try to make some progress with the digit you just got. The 8 in b8 can only be in c6, so it must be in c5 in b2, which means it must be in r1c5. In b4, it must be in c3, so it must be in c2 in b1, and therefore r2c2. Now it can no longer be in r4c3, so it's in r6c3, which constrains it in b6 as well.
Now we can use our new clue. The black dot tells us that r5c8 must be 4. So the 4 in r6 can only be in r6c5. r5c2 can only be 3, which means you can fill every cell in b5. This makes r6c7 a naked single 3, which makes r8c6 a naked single 9. There is only one remaining cell for 5 in b4. r4c3 can only be 4. r2c3 can only be 7. That constrains 7 in b2 into c6, which means the only cell left for it in c5 is r7c5. And that places the last two 8s in the grid.
The remaining cells in r8 are 2/3/4, but knight constraints force r8c7 to be 4 and r8c9 to be 3. The dotted cells in b3 now cannot have a 3 or a 4, so they must be a 1/2 pair. The rest of the grid is straightforward, and you should have it completely filled.
Hmm. You can see some book clues (📖) in the new grid, but you don't know what they mean. You must be missing something somewhere else.
Did you forget that there was a bit of green line here? You know what that means now. Go ahead and place a 1 in r2c1, and a little sudoku from there will reveal the rest of the grid.
That black dot is interesting. You know that the left cell must be 4/6 and the right cell must be 1/2/3. So it must be a 2/4 or 3/6 pair. But if it's a 2/4 pair, it'll be impossible to place a digit in r2c2. That should let you finish the front gate at last.
Take a moment to read the description of the new grid. It might not look like you can do much here, but that black dot is more constrained that it might seem. It cannot be a 1/2 or 3/6 pair, so it must be 2/4 or 4/8. Both have a 4, and the 4 cannot be in b7.
An eye outside the grid? Another clue you can't do anything with yet. But there are a lot of other cells revealed. The dotted cells in c1 see a 3 and 4, so it must be a 1/2 pair. r3c2 therefore has to be 1/2/4, but it cannot be 1 or 4. 5 cannot be on either a dot or a whisper, so the only possible cell for it in r3 is r3c9. The remaining dots in r3 must separate 3/6 and 4/8 pairs, but the inner two dotted cells are on a whisper line with each other, and of 3/4/6/8, only 3 and 8 are at least 5 apart. The outer two dotted cells are the 4/6, but if r3c7 were 6, r3c8 would have to be 1. This should finish this row and unlock the new rule.
Alright, now you know what that eye means: it's a SCRY-PEEPER. Go ahead and read the details of the rule in the puzzle description; there's an important clarification for how the rule works on the world map. And you can test your understanding by counting the visible cells in r3: the eye can see the 1, 2, 7, 8, and 9, for a total of 5, exactly as the clue says. In variant sudoku, this is generally called a skyscraper rule.
What about that other eye? 6 seems like a lot of cells. What would happen if r6c1 were 9? or 7? The general math for this is, if an eye can see E cells, the Kth cell in that row or column can't be more than 9 - E + K. But you don't really need to memorize that; you can just puzzle it out. r6c1 can't be more than 4 because the eye must see 5 more cells in addition to it. Since it can't be 1, 2, or 4, it must be 3.
7 also seems like a lot of cells. Where can the 9 be in c5? It must be in r7 at the earliest. But if it were in r7, it wouldn't be possible to see the cell in r6, because it would be at most 4, and there are 5 cells before it in the column. So the eye would only be able to see at most 6 cells. That means the 4 cannot be on the whisper segment in this column. What digits can be in first two cells? r1c5 cannot be more than 3, so it must be 1/2. r2c5 must be 1/2/4. If they were both 1/2, there wouldn't be any low digit left that could be on the whisper; so r2c5 must be 4. This means the 3 isn't visible, and the 1/2 on the whisper line can't be visible. So every other cell must be visible. That means that they must be filled with digits in ascending order in the column! r1c5 can still be either 1 or 2, but r4c5 must be 5, r5c5 must be 6, the high digit on the whisper must be 7, r8c5 must be 8, and r9c5 must be 9. Sudoku puts a 7 in r7c1, so the 7 in c5 must be in r6c5.
Back to r6. This 7 is a fairly tight constraint. r6c4 cannot be greater than 7 without breaking the eye, and it cannot be 3, 4, 5, or 6 by sudoku, so it must be 1/2. Since r6c6 must also be 1/2 because of the whisper line, r6c3 is very constrained: it cannot be 1/2/3/4/7 because they are already placed in r6, it cannot be 5 because that is in c3 in b7, and it cannot be 8/9 because of the eye. Therefore, it must be 6. So the eye can see at least the 3, 4, 6, and 7, and it must see two more, which must be the 8 and 9. If the 9 were in r6c7, the eye would only see 5 cells. So it must be in r6c9. Note that you can ignore this clue now: no matter how you fill the remaining cells, the eye will see the 8 and therefore see 6 cells. Note that placing this digit reveals some clues in the new grid. Don't try to use them yet;
Where is the 8 in c8? It cannot be lower than r5, which means the 8 in r6 can only be in r6c7. Does the 5 in r6c8 onstrain the eye in c8? It does: it cannot see both r5c8 and r4c8 anymore. The only possibility would be if the three cells in this box were 5/6/7, and that is impossible because of the 6 in r5c5. But there are only five other cells before the 9. The eye must see all four cells in rows 6-9 and one of the cells in row 4-5. To be visible over the 5 in r6c8, one of those last two cells must be 6/7.
The eye in c7 can see the 8 in r6c7 and the 9 beneath it. There are only five cells above the 8, and it must see exactly four of them. But you've already placed two digits from 5/6/7 in b6, which means one of r4c7 or r5c7 must be less than 4. So the other three cells must be visible, which means the first two cells must be 1 and 2 in that order. A short amount of sudoku later, and you'll find the third boss cell.
This puzzle is quite fun. There are quite a few clues here, and it can be a little intimidating to break in. The opposing eyes are an important constraint, since the numbers are very small. The arrows are also a powerful restriction; and of course there are also some black dots.
Begin with the place where these all interact most strongly. In both b1 and b7, there is a three-cell arrow at the start of a row observed by an eye with a count of 2. An eye always sees the 9, but these arrows cannot have a digit greater than 6. Therefore the eye can see only the first cell in the row and the 9, wherever it appears. The first cell must be the largest digit on the arrow. Therefore it must be 3, 4, 5, or 6; but it cannot be 5 because it is also on a black dot. One of the two dots in c1 must be a 3/6 pair, and the other must be either 2/4 or 4/8, with the 4 sitting on the arrow.
Now consider the arrows alone for a moment. There are five cells on the arrows in both b1 and b7. The circles on the arrows must sum to at least 15, and so neither can be less than 6; but they also cannot sum to more than 18, and this means that at least the 1 and 2 must appear on the arrows. The 1 and 2 in c1 must therefore be in b4: the 1 in r5c1, and the 2 on a dotted pair. Therefore the four dotted digits are 2/3/4/6.
So there are four options for the dotted cells in c1. Do all of them actually work? Recall that the cell must be the largest digit on the arrow. If it is a 3, the other cells must be 1/2, and the circle must be 6. But the 3 must be paired with a 6 for the black dot, which would be in the same row as the circle. Therefore the digits in c1 of the arrows must be 4/6, and the digits in b4 they are paired with are 2/3. The other digits on the arrow with the 6 must be 1/2, and its circle must be 9. Therefore there is a 9 in either r4c4 or r6c4. Where, then, can the 9 be in c6? Only in r2c6. This means the 9 in r3 must be in b3. There are therefore at least six cells to the left of the 9 in r3. If r3c1 is 4, at least one of these cells must have a digit higher than 4 and less than 9, making it visible to the eye. So r3c1 = 6. Similarly, the 9 must be in r3c7, and the 7 and 8 must be in c8 and c9 — in that order, to satisfy the eye to the right.
Where can the 9 go in r7? Because r7c1 = 4, the 9 in r7 cannot be further right than c5, which is also the furthest left it can appear. The cells between the 4 and 9 are 1/2/3 in some order. The 1 must be on the arrow, or else the sum breaks. If r8c1 = 7, then its arrow cells must be 2/5; but r2c1 = 8, and its arrow cells must be 3/5, and there is nowhere for 5 in c1.
It's finally time to consider some of the other arrows. r7c6 must be 5/6/7/8. It cannot be 8, so the 8 in r7 must be in b9. The least the arrow-sum can be is 7, with a 5 and two 1s. But it cannot be 7 or 8 by sudoku, so it must be 9. The other cells on the arrow cannot sum to more than 4, so they must individually be 3 or less. If r8c7 = 3, it must be paired with 6, which forces a 6 onto the arrow and makes the sum at least 10. So it is 1/2.
The arrow starting in r1c8 is pretty constrained. r1c8 cannot be greater than 6. r3c6 cannot be less than 3, so the other digits on this arrow cannot sum to more than 3. Therefore they are both 1 or 2. At least one of them must be 1, so r6c5 and r8c7 cannot both be 1; thus, r7c6 cannot be 7. By the same token, though, they cannot both be 1, because then r6c5 and r8c7 would have to be 3 and 2, respectively. Therefore they sum to 3, r3c6 = 3, and r1c8 = 6.
Where is 8 in b6? It cannot be on the arrows, because the circle cannot be 9. Therefore it is in r5. r5c6 must be at least 5, but it cannot be 8 or 9. It cannot be 5 because neither arrow can be a 2/3 pair. If it is 6, the bottom arrow is 1/5 and the top arrow is 2/4. But a combination of constraints forces r8c7 = 1, and therefore r8c8 = 2, which makes the 2 impossible to place on the top arrow. Hence r5c6 = 7. The 1/2 pair in c7 means that r4c7 and r6c7 are both at least 3. r4c8 and r6c8 are thus both no more than 4. r2c8 and r8c8 are also no more than 4, so these four cells are collectively 1/2/3/4. But the 3 can only be in r6c8. This places digits in a number of other cells.
Think about the other arrow in b7. r8c3 must be at least 5, so r9c2 must be 4 or less; it can only be 2. r2c3 = 3 because there is no way to make this arrow sum without a 3.
Pay attention to the eyes now. In c6, if the bottom eye can see the 6, it can also see the 7 and the 9. This means it will see four cells. Therefore the 6 must be blocked, which it can only be by the 8. The 8 cannot be in r9c6 or else the eye will see only two cells, so r8c6 = 8.
The rest of the puzzle can be solved with ordinary sudoku logic. The digit to place in the boss cell is 2.
Placing the 2 reveals a number of new eyes throughout the world map. You're not quite done with this grid yet, though. The boss digit exposes a number of opportunities for sudoku: one line of reasoning leads to r2c4 = 1, there is only one possible cell for 6 in b8, and there is only one possible cell for 8 in b9.. And there's one that's very easy to miss: r1c1 = 4 in the factory actually sees r7c9 in this grid! That puts the 4 in r7 in r7c8. As a result, r7c9 can only be 3/6, and since r1c9 and r2c9 can only be 3/6/7 and 6/7, respectively, the other three cells in c9 must be 1/2/4.
This should unlock a flurry of sudoku, but you will not be able to finish this grid yet. You should have 5 of the 7 cells bordering the Hall of Illusions filled, with others being a pair: 3, 7, 5, 1/4, 1/4, 9, 6. Do not fill it any other cells in the Hall of Illusions yet; you have a sense of forboding about it.
Go look at the new eyes. Most of them don't appear immediately useful, but one of them deserves a second look.
The new eye here restricts r9c1 quite a bit: it cannot be greater than 6, so the 7 and 9 must now go in r7c1 and r8c1 (in that order, because of r8c8). The whisper line leading into r8c5 forces it to be at least 7, but now it cannot be 7 or 9. This triggers some useful sudoku, most importantly that r7c7 = 8. The 8 in r9 must be in b7, but it cannot be in r9c2 without limiting the eye to only seeing three cells. And this shows us our fourth boss cell!
You've now finally filled the border with the Courtyard, but we still can't finish that grid. Time for the boss.
Be sure to read the rules carefully. The clues outside the grid here are not eyes; they are "numbered room" clues. The digit in the door must go into the nth cell away in the same row or column, where n is the digit in the immediately adjacent cell (its "index cell"). This can be hard to grasp from a description, so it might be more helpful to see a few examples of valid placements according to this rule alone:
The second of these, of course, violates the standard rules of sudoku and so wouldn't be legal in the actual puzzle.
The rules in the puzzle description aren't completely clear about this, but in numbered room sudoku, "nth" is always a relative index, i.e. the cell n cells away from the clue (as opposed to an absolute index that always starts from the top or left side, like I've been using with "r6" or "c7"). For example, these are legal placements for clues on the right side:
Again, the second example violates the standard sudoku rules and wouldn't be legal in the actual puzzle.
There's an interesting property that's generally true for numbered room clues at opposite ends of the same row or column: when the index cells sum to 10, the clues end up talking about the same cell. If the digits in the clues are different, this has to be impossible. (If the digits in the clues are the same, this is required, but that doesn't happen in this puzzle.)
Anyway, the triple clues are the natural places to start. The three 2 doors on the top side of the grid are aligned with the box borders. This means that, if one of them goes in b8, the other two cannot. There are three clues and three possible boxes, thus one clue per box. As such, one of the digits in r1 in b2 is 1/2/3, one is 4/5/6, and one is 7/8/9. The index cells for these three clues are also a cage with sum 18, which they can only reach if they contain 3, 6, and 9. Accordingly, there must be a 2 in r3 in b2, in r6 in b5, and in r9 in b8. Since the 2 in r6 is in b5, r6c9 must be 4/5/6 because of the 2 door.
As for the three 5 doors on the left side of the grid, the same logic applies except that there's a single degree of freedom: it's 3/6/9 except that one of the digits is one less, i.e. 2/5/8. This ends up being very open-ended in terms of marking the grid, and it's not worthwhile; but it'll be important, and you'll need to keep it in mind.
Consider the 31 cage. The 2 in b5 must be in r6, which puts it in the cage. So the other 4 cells in the cage must sum to 29, making them 5/7/8/9. You know that the 5 in this box is not in r6c4 because this would make the 17 sum in c1 impossible. It also cannot be in r6c5 because then 5 would also have to be in r6c1. So you know that it must be in c6, and therefore one of the digits in the cage in c1 is 6. r9c6 must be 4/5/6 because of the door, but it cannot be 5 by sudoku: the 5 in b5 is already in c6. This also means that r5c6 is not 5, and so r5c1 is not 6. Finally, since neither possible digit in r9c6 is 2, r1c6 cannot be 9.
Can r1c6 be narrowed further? If it were 6, r9c6 would have to be 4. This creates that situation I mentioned above: the clue at the top of the column would make r6c6 = 2, but the clue at the bottom would make r6c6 = 5. So this is impossible, and r1c6 = 3. Therefore r6c9 must be 5/6. This means that r6c1 can no longer be 6, because this would make r6c6 = 5, and r6c9 would be impossible to fill. So r4c1 = 6. By putting a 5 in r4c6, this also puts a 6 in r9c6. This puts a 6 in c3 in b7, since it cannot go in the 12 cage; and that puts a 6 in the 9 cage in b1, making the sum 1/2/6. (There is also a much more interesting way to prove this directly with the 3 clues.)
Consider the 15 cage in b8. To sum to 15 in four distinct digits without 1 requires 2/3/4/6; but r9c6 is 6, so this is impossible, and there must be a 1 in the cage. The 1 in c6 can now only be in r2c6, and the 4 must be in b8. The digits in the 9 cage can be placed by sudoku.
Where is the 3 in c2? The index cell, r1c2, can only be 4/5/7/8. One of the index cells in b4 must be either 2 or 3 to make the cage sum work. If r1c2 is 4 or 5, the index cells in b4 can no longer be 3, and one of them must be 2. But only r5c1 can be 2, and this would make r5c2 = 5. So if r1c2 = 5, there's an immediate contradiction: r5c2 must be both 3 and 5. And if r1c2 = 4, r4c2 = 3; but this means the 12 cage in b7 sees all three of 3, 4, and 5, making it impossible to fill. So r1c2 must be 7/8, putting the 3 into the 12 cage either way and making it a 3/9 pair.
Now consider the 15 cage in b6. By sudoku, there is no 6 in this cage. Again, to sum to 15 in four distinct digits without 1 requires 2/3/4/6; since this is not possible, there must be a 1 in the cage. Therefore the only place for a 1 in r6 is r6c3. As a result, r6c1 cannot be 3; it must be 8/9, and r5c1 must be 2/3. This creates a 2/7/8/9 set in r6. The remaining digits are 3/4/5/6; of these, r6c2 can only be 4. The other three must go in b6. There is only one way to sum 4 distinct to 15 without 3/5/6: 1/2/4/8. The other two cells in b6 must be a 7/9 pair.
What digits remain to be placed in c2? The 2 cannot be in r1c2 or r9c2, so it must go in b4. This forces r5c1 = 3 and r6c1 = 8, as well as the corresponding 5s. There's a lot of sudoku to do now; you should be able to finish b4, as well as b6 outside of the cage.
Now we can look at the 3 in r2 again. It must be 7/8/9, but it cannot be 7/8 because r2c7 sees both 3 and 9. So it must be 9, which sets r2c9 = 3.
Consider the digits 3, 6, and 9. They've already been placed, or at least heavily constrained, in most of the boxes, rows, and columns. One of the exceptions is c8, about which you seem to know almost nothing. 3/6/9 cannot be in r1c8 or r2c8, or in b6. This leaves only four cells for three digits. At least one of 3/6/9 must be in the 14 cage in b9.
It's finally time to look at the last door. The 8 in r8 cannot go in r8c1 or r8c6 by sudoku. It cannot go in the 15 cage in b8 because the cage would have to be 1/2/4/8, and the 4 is known to be in c9 in this box. It cannot go in r8c7 because r8c9 would have to be 3. It cannot go in r8c9 because r8c9 would also have to be 1. If r8c8 = 8, r8c9 = 2, so r9c8 = 4 to make the cage work; but then it becomes impossible to place all of 3/6/9 in c8. Therefore r8c3 = 8, making r8c9 = 7.
The remainder of the puzzle is straightforward. The digit to place in the boss cell is 1.
You just learned two rules, one of which is tied to a quest.
The first rule is CODEX, which you're going to need when you start the Library grid. You won't need this quite yet, but we'll come back to it very soon.
The second rule is collectible coins. You might as well read about this rule now:
You do not need to count the collectible coins as you go. I recommend that you don't. You will not use these counts until Chapter 39, which is a very, very long time from now. You will frequently find yourself backtracking in Sudokuvania, and you will not always be able to fill these cells as you progress, and it will be very easy to lose track of which coins you've counted. Again, I recommend you just ignore the coins until Chapter 39.
But if you really want to keep track of the coins as you go, perhaps because you are obsessively compelled to do so, I definitely recommend that you don't just count them in these boxes. In general in life, whenever you're doing a particularly long or complicated count, you always want to collect, not count. Break the problem down into a set of facts that you can easily verify but also more easily summarize into the count that you'll ultimately need. In this case, instead of writing down how many 5s you've found in the whole map, you should write down what coins you found in each grid. The number of 5s is a fact you can only verify by searching the entire map again looking for coin cells with 5s in them. The set of coins in a grid is a fact you can quickly verify by looking at that specific grid. And if you write that down on a piece of paper like folows, you can easily verify each grid independently and then total up the number of coins in each column:
I'll tell you right now that you only need space for 9 grids. Most of the grids have exactly 3 coins on them, always of different values, and they're never in the border areas. You don't need to assume any of that, but the fact that it turns out to be true will make your bookkeeping a lot easier. But again, you don't actually need to do the count as you go, and you can just ignore the coins until Chapter 39.
Filling the boss cell lets you finish the Quarters grid; this is generally straightforward. 1 is forced into r6c3, which makes r5c4 = 7. The only possible cell for 7 in c6 is r9c6; the whisper line puts a 2 in r8c6. Sudoku will complete the grid except for a few pairs that can be resolved by remembering the eye in r9, which can only be satisfied by putting the 4 in r9c1.
Taking a look around before you move on, it's interesting that you've finished this area with so much fog still uncleared. Perhaps there's something hidden under the fog still. There's also a pair of whisper lines leading off the bottom of the grid that clearly cannot be satisfied by any digit, and a mysterious red line leading off to the right that you're starting to see pop up elsewhere. Very odd. Anyway, off to the library.
Time to read up about the CODEX rule. This is a column-indexing rule, which may be new to you unless you play a lot of variant sudoku (or even if you do; indexing rules are often different from puzzle to puzzle, as in fact we're seeing right now). Much like the doors from the Keymaster boss you just finished, the idea is that you are placing a fixed digit according to the digit in a specific cell. Unlike that rule, however, the digit you are placing is the column number of a cell with the book emoji in it (📖), and the digit in that cell is the column number where it goes. Columns are numbered starting from the left side, just like I've been writing with e.g. "c8". It might help to look at some legal placements:
Unlike the numbered room rule, where certain digits in the index cell can just never work for certain clues, these clues never restrict cells by themselves: any cell in the row could be a legal place to put the column number of the clue cell. It's only in interaction with other clues, or with digits you've already placed, that the rule comes into its own.
Right, so, let's get started. You can see three codex clues. They are in c2, so the digit in the clue is the column number of the 2 in that row. Since r6c1 = 2, r6c2 = 1. That reveals more of the grid, but there's something else to notice first. Just like you saw with the Keymaster, these clues are aligned within a single row of boxes. Since they are placing the 2, they must be placing it in the different boxes. The placement in b5 cannot be in c4, and the placement in b6 cannot be in c7, so you've narrowed the 2 down to four cells in each box.
Now you can check out the clues in b5. The arrows use a lot of cells in this box. The least that a 5-cell set can sum to is 15. The arrows also require a cell outside the box, so their circles must sum even higher than that. They must be at least 16, and r5c7 cannot be greater than 2. It cannot be 2 because that would require the clue in r5c2 to be 7, so it is 1. r4c5 and r4c6 cannot be 2 and still sum to at least 16. r5c2 must be 5/6, and r4c2 must be 8/9. r4c5 and r4c6 must contain a 9, so r4c2 = 8. The arrows sum to 16, so the arrow cells in b5 sum to 15. By the codex clues, r4c8 = 2, and the other two cells in this row and box are 5/6. You can place the 1/3 in this row by sudoku.
Both of these arrows have 1s on them. The other cells are a 2/3/4/5 set; the 7-sum arrow must be 1/2/4, so the 9-sum arrow must be 1/3/5. Since the 2 must be in r5, the 4 must be in either r6c5 or r6c6. Since these cells are both codex clues, the digit in r6c4 must be 5/6. You know that 5 is on the arrows in this box, so it must be 6, and therefore r6c6 = 4. This also determines the 7, 9, 2, and 6 in this box. Placing the 7 and 9 finishes r4 via the codex clues. The remaining codex clues here are 3/5, but r5c6 = 5 would force r5c5 = 6 when we already know it is 2. The 2 in r5 is in r5c5, so r5c2 = 5.
The puzzle suggests looking at these new cells in b8, and so do I. Again, you have two codex cells in the same column and the same row of boxes, so they must be cells in different boxes. They also have the same high/low polarity because of the whisper line. If they were 1/2/3/4, one of them would have to be a 4 or else they'd be in the same box, but r6c6 = 4 already. Therefore they are 6/7/8/9, and one of them must be a 6. r8c7 cannot be 1, so r9c6 cannot be 6; r8c6 = 6 (indexing itself, which is fine), and r9c5 = 1. r9c1 means that the only remaining value for r9c6 is 7, which sets r8c7 = 2. And r9c7 = 6 aaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhh
Well, dust yourself off. You're in a new grid, but there's also a new rule. Go ahead and read about it in the puzzle description box. Every cell has a digit — that must mean outside of grids, too. There are cells you can't see, but surely the line can't have two different endpoints in the same cell. So this line between the library and the maintenance must be continuous, and you can see both ends of it. Go ahead and fill it in as best you can. You should have everything except a pair of cells, the third from each end. And you have one cell in the maintenance grid, it's a 6, and it's next to a black dot. r9c8 = 3. r9c7 is 8/9, but it's also on a black dot, so it must be 8. r9c6 = 4 (by the dot). r9c5 = 9 (by the whisper). Uh oh.
This is a really fun and unique puzzle.
The first rule to learn here is the shading. You're going to have to shade the entire grid, making every cell either a cave or a wall. The cave cells all have to have some orthogonal path to each other. (Orthogonal means passing through the edges between adjacent cells, not through diagonals.) They form one unified group. The wall cells all have to have some orthogonal path to one (or more) of the edges of the grid, but they can be in multiple groups. So this would be a valid shading of a grid:
Rules like this, with different shadings of orthogonally-connected cells, come with an extremely handy little theorem about every 2×2 portion of the grid:
They can never have a checkerboard pattern like this. There must be an orthogonal path between the diagonally-opposite cave cells, but the wall cells must themselves be orthogonally connected to the edges of the grid, which the cave path cannot possibly get around. So if you ever see two cells of the same color across a diagonal from each other, you know at least one of the adjacent has the same color, too.
I suggest using dark grey for wall cells and light grey for cave cells, because I think it looks nice on this puzzle. You might reasonably decide that the contrast isn't good enough, though, especially between the light-grey-and-pink and the default pink background.
The second rule is the akari rule. (This comes from a completely different kind of logic puzzle called "light up" or akari.) Some of the cave cells are going to be "lanterns". Lanterns casts light in all four orthogonal directions, illuminating any cave cells. Light only travels in straight orthogonal lines, never travels along diagonals. Light is also blocked by wall cells and doesn't illuminate cells on the other side. The big constraint in akari is that no two lanterns are ever permitted to illuminate each other, so if two lanterns are in the same row or column, there must be at least one wall cell between them. Since lantern cells are always cave cells, there's a bit of a visual collision when it comes to marking them in the grid. I suggest marking lantern cells by coloring them yellow rather than the light grey of a cave. As long as the cave and lantern cells are both light colors and the walls are a dark color, you probably won't have trouble remembering that lantern cells are also cave cells. With that in mind, this would be a valid placement of lanterns in the grid above:
Note that it's fine if two different lanterns see the same cell; it's only a problem if they see each other. There are many different possible valid placements of lanterns in this grid, at least according to the rules of akari alone.
The third rule, which is special to this puzzle, is that the digit in a lantern cell tells you how many cells it illuminates, including itself. With the placements above, the digits would have to be:
This, of course, interacts with the rules of sudoku, because it means you can never have two lanterns in the same row or column that illuminate the same number of cells. Our lantern placement violates that, so it would not, in fact, be a legal lantern placement under these rules combined. You can fix it by just moving the lantern in r7c1 to r6c1: all the cave cells are still illuminated, but now that lantern contains the digit 3 and doesn't violate sudoku.
The final rule is the cages. Cages in this puzzle are always in wall cells, and the digit in a cage is the number of lanterns in the orthogonally adjacent cells. So if there were a cage in r5c5 in my little example, it would have to be filled with 2. This will be a key rule for breaking into the puzzle.
Start by shading all of the cages as walls. Remember that all of these will have to connected to an edge of the puzzle. This means the digit in a cage can never be 4, because a lantern cell is always a cave cell, and a wall completely surrounded by cave cannot be connected to an edge. But of course it's fine for a cage that's sitting directly on an edge to be surrounded on all its other sides.
The key to breaking into this puzzle is placing lantern cells, which give you a lot of information. The lantern cells that share a row or column are particularly constrained. You have to place 6 lanterns around the cages in r9, in a total of 8 total cells. The gaps between these lanterns are only two cells, and you can never place two lanterns in the same gap because they would illuminate each other. So these two gaps must each contain a cell that does not contain a lantern. Since there are 6 lanterns to place, the 4 cells that are not in the gaps must be lanterns, and the two gaps must contain a lantern.
Recall that each lantern contains the number of cells that it illuminates. If r3c9 were a lantern, it could illuminate no more than 3 cells, but 1/2/3 are impossible by sudoku. Therefore r4c9 cannot be 3, and the 3 must be r7c9. This puts a lantern into r6c9, which means r5c9 cannot have one, so r4c9 = 1 and r1c9 = 2. Be sure you don't try to mark cells that don't have lanterns specifically as cave or wall cells! You don't necessarily know that; all you know is that it isn't a lantern. I found that I could get by without marking non-lantern deductions, but you might consider finding another way to mark them if you want.
There's another place that's restricted in its illumination. r8c9 has a lantern, and it can illuminate at most four cells, but it cannot be 1/2/3 by sudoku. Therefore it is 4. Mark the cells it illuminates as cave. r8c6 cannot be 3 anymore. It must be adjacent to at least one lantern, though. If r8c5 = 3, both of r8c6's remaining adjacent cells would be adjacent to lanterns. Therefore r8c1 = 3.
Don't forget the theorem I explained above. r2c8, r6c8, r7c2, and r9c2 all must be cave cells because they would make the cave disconnected if they were walls. They cannot be lanterns, so you can mark them as ordinary cave. This brings the lanterns in c8 perilously close to illuminating one another; r3c8 and r5c8 must be walls. Therefore r3c9 and r5c9 must also be walls by the checkerboard theorem. This makes r4c6 a cave cell.
These new cave cells make illumination tricky again. You now have two lanterns in c9 that must have a digit of at least 5 but only illuminate in one direction. The digit in r3c7 can only be 1 now: it is surrounded by ordinary cave on two sides and a wall on a third. r3c6 is a lantern. r4c6 is ordinary cave.
You haven't done much with the lanterns in c8 yet. r1c8 cannot be 2, so it must illuminate at least one cell to its left; that's all you can do with this for now. r7c8 cannot be 3/4, so it must also illuminate at least one cell to its left. That means r7c6 cannot be a lantern. r8c6 = 1 and r8c5 = 2. r9c6, r7c5, and r8c4 are lanterns. r7c6 and r8c3 must be walls to prevent lanterns from illuminating each other. r7c4 must be a cave by the checkerboard theorem, so r7c3 must also be a wall. r9c5 must be a wall or the cages will not be connected to an edge. The illumination of r7c8 now cannot be more than 5, so it is 5.
You just added a bunch of new lanterns. The lantern in r9c6 must be at least 3 by sudoku, so it must illuminate at least r9c7 and r9c8, but that means it also illuminates r9c9; it is 4. r7c5 must be at least six, so r5c5, r4c4, and r3c5 must be cave. You already know that r2c5 is cave, so r7c5 is 8/9. Making r4c5 cave means that r4c8 is at least 4, but it cannot be 5 by sudoku, and it cannot be more than 5 because of the wall in r4c3. So it is 4 and r4c4 is a wall. r8c4 must be at least 5 by sudoku, but it can only possibly illuminate 5 cells now, so it is five and all of those cells are cave.
Time to look at the left side of the puzzle for a chanege. If r9c3 were cave, r9c1 would illuminate 4, which breaks sudoku; therefore r9c3 is wall and r9c1 = 2. r8c2 must be at least 6, so you can shade cave up to at least r3c2. This means r4c3 can no longer be 3, so it is 2, with lanterns above and below. The only remaining paths that can connect r4c3 to an edge of the puzzle require r3c4 and r2c4 to be walls. This forces r2c9 = 5.
How constrained is c1 now? r7c1 must be at least 4, so r6c1 and r5c1 must be cave. If r2c1 were a lantern, it would have to be at least 4 by sudoku, but there would also need to be a wall between it and r7c1. The most it could be is 4, but then it would make r7c1 = 4, which would break sudoku. So the lantern is in r1c2.
That creates interesting constraints in r1. r3c2 must be cave by the checkerboard theorem, so r2c2 must be wall to prevent the lanterns from illuminating each other. r1c2's illumination must therefore run strictly to the right, and it must be at least 3. So the only way of connecting the wall in r2c4 to an edge of the puzzle is if r2c3 and r2c1 are also wall. That means r1c5 must be cave, or else that stretch of cave will be disconnected. r1c6 must then be wall to separate the lanterns. r1c2 = 4. r1c8 = 3.
You're very close to finishing the shading. r3c1 must be cave so that r3c3 can illuminate more than 2; r3c3 = 3. r3c6 must be at least 6, which is the most it can be; r5c6 is cave. r7c5 is now known to be 8. r8c2 = 7. If r5c3 = 8, r6c9 is also forced to be 9, and there is no remaining digit that can be placed in r5c9. This still doesn't finish the shading, but the rest of the grid can be solved with sudoku, and you'll shade the last cells along the way.
The digit for the boss cell is 5.
You've unlocked the AKARI LANTERN rule. This is completely useless; there are no akari lanterns on the world map, and in fact the rules box doesn't even bother to list it. But you've also unlocked the DARK rule, and the rules box does cover that; go and read about it in the puzzle description box. Unlike the earlier rules, this one doesn't build out of the rules of the puzzle you just solved; it's just its own new thing. It's a little weird, don't worry about it. Anyway, the rule is just that dark-shaded cells have a lower value than all of the non-dark cells they're adjacent to; that's going to be the key to finishing the maintenance grid. The rule also mentions that fog doesn't clear easily around dark cells, but you can't control the fog clearing anyway, so it's not anything new.
I used to play a lot of greater-than sudoku puzzles, and this rule has a lot of the same characteristics. Generally, you tackle rules like this by thinking about how they rule out digits, keeping in mind that they give you the most information about the smallest and biggest digits that you haven't placed yet. What digits remain to be placed in r9? The 1 cannot go in r9c2 or r9c4 without making it impossible to place a digit in r9c3. The 1 in b8 must go in r7c4 or r7c6 by the same logic. So the 1 in b9 must be in r8, and that means it must be in r8c8. The 2 in b9 cannot in r8 because both of the remaining cells are greater than two other cells in the box.
That puts some interesting constraints on the arrow in b7. It has at least three cells, so these two cells cannot sum to more than 8. These two must be at least 3, but they cannot be 5 because it is in r9 in this box, so they must be a 3/4 pair, and the dark clue forces the 4 into r7c2.
The dark cells in b8 are getting a little tight now. What digits are actually possible? Neither r7c4 or r7c6 can be greater than 6 because they are less than two cells in the box, neither of which is 9. But they also cannot be 2/3/4/5, so they must be 1/6. Because one of them is 6, its adjacent cells must be 7/8. So r9c4 can now only be 2, which places r9c3 = 1. Where is the 2 in b7? Only r8c1 is possible. The arrow cells in b4 are either 1/8 or 2/9.
This next leap is quite good. How do the constraints in b8 affect r6? If r7c4 = 1, r6c4 must be at least 6. That forces r6c5 to be at least 7, but it cannot be 9. Meanwhile, r7c6 = 6, so r6c6 must be at least 7, but cannot be 9 because of the arrow. That creates a 6/7/8/9 set in the row, and only r6c3 can be 9. On the other hand, if r7c4 = 6, then r6c4 = 1. So in either case, r6c3 = 9 and r6c2 = 2. The revealed cells force the choice. If r7c4 = 1, there's a 6/7/8 set in r6 in b5, so r5c4 = 9 by the dark rule. The 7/8 in r6c6 would be on the arrow, and the circle of that arrow would have to be in r4c6 because otherwise the sum would have to be at least 10, but there is no longer a digit in b5 greater than r6c6 that can go in it. So r7c4 = 6, r6c4 = 1, r7c6 = 1, and r8c6 = 3.
Back to the arrows. The arrow in c6 has at least two cells on the arrow in that column, but you can never put more than one cell greater than 4 on an arrow, and you've already placed 1/3/4 in c6, so r4c6 must be the circle, and r5c6 = 2 by the dark rule. r6c9 cannot be less than 3, so r8c9 must be at least 6. That forces r8c7 = 4. By the dark rule, r7c7 = 2. r7c9 must now be at least 5, and if it were any greater the arrow sum would be broken, so r8c9 = 9, r7c9 = 5, and r7c9 = 3.
Remember what I said about the general approach to greater-than rules. 1 in b4 can only be r4 by the dark rule. That forces r5c7 = 1. The 2 in b6 can only be in r4c8 or r4c9. It cannot be in r4c9 because the dark rule would force r4c8 = 1. r4c9 is on a whisper line, but now it cannot be a low cell: it sees 1/2/3, and if it were 4, r3c9 would have to be 9, but that is already in r8c9. So r2c9 and r4c9 are a 7/8 pair. r5c9 = 4 is a naked single.
Time to look at b5 again. r5c5 can only be 3/6/7 by sudoku. It cannot be 7 because r6c5 would be impossible to fill by the dark rule. If it were 6, r6c5 = 7, but this makes it impossible to fill the arrow in c6: r6c6 could only be 5, but the circle in r4c6 cannot be 7. So r5c5 = 3, which makes creates a 1/3 pair in r4c1 and r4c2. r4c3 and r4c5 see each and are at least 4, so by the dark rule, r4c4 must be at least 6; but it cannot be 6/7 by sudoku. Since the arrow sum in r4c6 must be at least 7, there is a 7/8/9 set in r4c4, r4c6, and r4c9. The remaining cells in r4 must be 4/5/6. The only possible cell for 9 in b6 is r5c8.
There's an arrow you haven't done much with. r3c4 must be at least 3 by sudoku. It cannot be 3 because the dark rule would force a 1/2 into r2c4, so it is at least 4. r4c5 must be 4/6 by sudoku, but it can no longer be greater than 5 by the arrow sum, so it is 4. r3c4 cannot be greater than 5 by the arrow sum, and if it were 5, the circle would have to be in r3c3, and it would have to contain 9, which breaks sudoku. So r3c4 = 4. This reveals a few more cells and permits some sudoku. r3c7 is the only possible place for 9 in r3. r2c4 = 3 by the dark rule, which also forces r1c7 = 3. r4c7 = 6, so the 5 in b6 must be in r6. This means r6c6 must be at least 6, so r4c6 must be at least 8, creating an 8/9 pair in r4 and forcing r4c9 = 7.
Now think about small digits in b3. r1c9 and r3c9 are a 1/2 pair, and you've placed 3 already. 4 must be in c8, but there are no remaining digits smaller than it, so by the dark rule it must be in r2c8. This forces 4 into r1c3, whicihi reveals a few more cells. 9 in b1 can only be in r1c2 by the dark rule. The rest of the grid is finishable with sudoku.
Okay! You revealed a short roaming line that leads back to the rest of the map, and the AKARI LANTERN has revealed a bunch of cells all around. You can zoom out and repeatedly undo/redo your last few digits to see exactly what changed, but here's the rundown:
Let's try to mop up the unfinished business in the earlier grids before heading back to the Library.
The whisper line in r2 lets you finish this grid. r2c8 must be 1/3, so r2c7 must be at least 6, which is the highest it can be; therefore r2c7 = 6 and r2c8 = 1.
The eye above r1c9 is now looking at a fully-filled-in column, and you can just compute the right value for it. Go ahead and fill it in! It sees the 4, 7, and 9, so has a value of 3.
The new clues off the bottom of the grid are somewhat interesting, but there isn't much you can do without knowing how rules work off-grid. Hmm. Keep it mind for later.
There's clearly no way to fill a digit in the cell on the whisper line beneath r9c4. Hmm indeed.
The only other new thing here is the eye above r1c5. As before, it's looking at a completed column, so you can just compute the right value: it sees the 7, 8, and 9, so it's 9. You can fill that in, and assuming you also filled the eye clue in grid 1... oh hey.
This sure looks like a grid, and if you click on a cell in it, the puzzle description box confirms that it's a 4×4 sudoku. It has no given digits, but it does have four eyes and a ROAMING palindrome line.
Reasoning about the palindrome is the key here. The center of the palindrome is the pair of r4c1 and the cell immediately beneath it, the one with the eye clue. Cells that are an equal distance away from this will be the same. You'll be thinking about which cells are the same or different without knowing exactly which values they have, so this is a great place to use color (or, if you'd prefer, SudokuPad's letter tool, which will let you mark cells A/B/C/D; you can enable it in the gear menu, then hit the The palindrome has length 14, so you get 7 pairs from it. Two of them, of course, are just the 3s at the endpoints. Some of the pairs are more immediately helpful than others, but you can really start anywhere.
Looking at the center: the palindrome is saying that an eye clue and the first cell it sees must be the same. This cannot be 4: the 4 is the highest digit in the grid, so if the first digit is 4, the eye clue must be 1 because no other cells are visible behind it. It also cannot be 3, because the eye clue would have to be 2. And it cannot be 1, because the eye would have to see at least the 1 and the 4. Therefore it is 2.
Looking at the ends: you can't do much with the eye clues, but you can color the cells in r2c1 and r4c3 the same; I picked blue. Sudoku then places the blue digit in r3c2 and r1c4. And the palindrome therefore also places it in the eye beneath r4c2.
The next cells in from blue are definitely a different digit by sudoku, so you can shade it a different color; I picked purple. It must be r2c4 and r4c1 by sudoku, which means it's 2. The last cells on the palindrome within the grid are r2c3 and r4c2. Again, these are different digits from the others by sudoku, so you can color them differently; I picked green. Sudoku also puts them in r1c1 and r3c4. And the last four cells can be orange.
Alright, you've got the pattern of digits set, even if you mostly don't know which is which. How do the eyes constrain the colors? Blue cannot be 4 because putting a 4 in r3c2 would force the eye in c2 to see 2 cells, but this is also a blue cell by the palindrome. Green cannot be 4 because putting a 4 in r1c1 would force the eye in c1 to see 3 cells. Orange being 4 means that the eye in r2 must see 2 cells. If blue were 1, the eye in c3 would see four cells, but the palindrome forces its digit to be 2. So blue is 3 and green is 1.
If you fill all of the cells, including the eyes, you should see the hidden puzzle get checked off in the quest progress box (in the bottom right of the map).
Just a little to mop up here. The new whisper line forces r5c6 = 9, which should fill everything except 2-cell and 3-cell sets in each of the bottom six boxes.
It's been a bit since you've used a codex clue, and you're about to head back to the library, so r8c7 is a nice refresher. You're placing the 7 in this row. Sudoku forces the digit to be 5/7/9, but r8c5 and r8c9 are already filled with a non-7 digit, so r8c7 = 7.
Three pairs remain, but there's no way to resolve them for now. Don't get tempted by the Hall of Illusions; time to hit the books.
Alright, there are two new codex clues in c4, and they're on a whisper line. The codex clues have the same high/low polarity. If they were high, they'd have to be a 7/9 pair by sudoku. r1c5 is adjacent on the whisper line to both, so it would have to be 5 away from a 7, but it cannot be 1 or 2 by sudoku. So the codex clues are low, and the other cells on the whipser are high. This puts 7/8 in r2c3.
Can you constrain that more tightly? A 2 eye is very constrained: it must only be able to see the first cell and the 9. The 9 is in r6c3, and the 8 must be in c3 in b1, so it is definitely visible in front of the 9. Therefore it must be in r1c3, putting r2c3 = 7. This means r1c4 = 2 because of the whisper line. This is a codex cell, so r1c2 = 4. r1c5 must be at least 7, but it cannot be 7/8 by sudoku, so it is 9. r2c4 is 3/4, but it cannot be 3 because it's a codex clue and r2c3 = 7. So it is 4.
There's some sudoku to be done. r3c3 is the last possible cell for 1 in b1. This creates a 1/5 pair in c6 in b2. Back in b1, the open cells in c2 are 2/9, which makes the open cells in c2 in b7 3/7, which can be placed. The 2 in b7 must be in r9c3. That means the 4 in r9 must be in r9c8 or r9c9. The 4 in c7 can only be in r3c7, which puts the 9 in c7 into r2c7.
Aha, the boss cell. Tidy up here with a little sudoku and then head in to find...
Be sure to read the rules, because there's a lot of them. You've got three full columns of codex clues, six killer sudoku cages, and three circled cells that have to contain odd digits. But you've also got to place a snake: a non-branching, orthogonally-connected path through the grid. That's new, so let's spend some time on it.
You're going to want to mark cells for this. I suggest coloring cells that you know the snake doesn't pass through grey, and then you can use yellow for the snake cells. (Green doesn't show up well against the green background.) The rules tell you five things about the snake:
You can make some immediate inferences about the snake from all this. The snake cannot pass through the cage in c1, but there are only two cells above it, so if the snake entered either of them, it would not be able to turn around and would have to dead-end there, which it cannot. Similarly, the snake cannot enter either of the cells beneath the cage in c9. If the snake was in either r5c8 or r5c9, it would have to be a four-cell snake, so it could never visit c1 or c5; so these are non-snake cells, and the snake must also be in r3c9. But the snake does have to enter c1 at some point. If it did not enter r6c1 (that is, go around the cage in r6), it would have to turn around in b7, but this would require at least 7 cells in the box to be snake, violating the only-odds rule. So it does have to pass through r7c1, r6c1, r5c1, and r5c2. If it passed through r4c3, that would be all five odd cells in b4, so both cells in the 11 cage would have to be even, which is not possible. Similarly, if it passed through r9c1, that would require all five odd cells in c1, but the 5 cage also needs one. The snake must leave this corner somehow; you can actually constrain it already, but it's not particularly elegant, so let's leave it for now.
Recall what you learned about codex clues in earlier grids: three codex clues in the same column and box must contain one apiece of 1/2/3, 4/5/6, and 7/8/9. So the only way to fill the cage in b2 is a 6/9 pair. And these are codex clues, so you've constrained the 5 in b2, b3, and b1. The 10 cage must have a 2/8 or 3/7 pair, which means the 5 cannot be in b8 in these two rows; so it must be in r9c4 or r9c5, making r9c5 4/5.
The snake must pass through self-indexing cells. So they cannot be in cages, meaning the 5 cage cannot be 1/4. Making it 2/3 restricts the 1 in both b1 and b4. If r4c1 = 3, r4c3 = 1; but that only leaves three other odd cells for the snake in the box, which is not enough for it to escape. So r4c1 = 2, r4c2 = 1, r3c1 = 3, and r3c3 = 1. Since there is an odd digit in the 11 cage, the snake must pass through the other four odd digits in order to leave the box, so it must pass through r4c2 and r3c2. If it did not pass through the 1 in r3c3, there would only have three odd digits left in b1, so it would have to leave b1 through r2c2 and r2c3, but then its three cells could not include the 5, which must be in r1 in this box. Three cells in a box cannot all be 7/9.
How does the snake you've been building connect to its ends in b6? The end in b1 must connect to r4c9, since otherwise the other end would be cut off. If it doesn't go through r1c5, it must go around the 15 cage the other way. So it would have to occupy at least r3c4 and r3c6, but that would put 6 odd cells in r3. Therefore r1c4, r1c5, and r1c6 must be snake cells; more specifically, they must be 1/3/7, because 5 and 9 are already restricted in b2. This means r2c4 and r3c4 must be even, so they cannot be snake cells. Forcing the snake through r2c3 and r1c3 means there are four snake cells in b1, which means the 5 must be in r1c3. There are already four odd cells in r1, so the snake cannot pass through r1c8 or r1c9. Similarly, there are four odd cells in r3, so the snake cannot pass through r3c6 or r3c7. So it must pass through at least r2c7 and r2c8.
All these restrictions mean something for the codex clues. In b1, the 1 in r1 must be in b2, so r1c1 must be 4/5/6, and it cannot be 5. The 1 in r2 must be in b3, so r2c1 must be 7/8/9, and it must be 8. The 5 you placed earlier tells you two things: r1c5 = 3, and the 10 cage in b8 cannot be 3/7.
You've constrained digits interestingly in r1. r3c2 and r2c3 are 7/9, so the 9 in r1 must be in b9. It cannot be in r1c7 because that would make r1c9 = 7, but the 7 in r1 is already in b2. It cannot be in r1c9 because it would be self-indexing, and the snake cannot pass through this cell. Well, hold on, what does that mean? The 9 in b3 must be in r1c8, which means it sees every possible snake cell in c9 except r4c9. For the snake to pass through the self-indexing 9, then, it must be true that r4c9 = 9. There's also now an odd cell in b1 which isn't on the snake, so there can only be four snake cells in this box, forcing the snake through r2c6 and making that cell 5. This makes r2c5 = 6 by the indexing rule, so r3c5 = 9 by the cage, and finally r3c9 = 5. This makes r2c3 = 9, so r2c9 = 3 by indexing; that must be a snake cell, and the last snake cell here, r2c7, is 7.
There's a cage you've been neglecting. The 11 cage in c9 must be 4/7 by sudoku, and the 7 cannot be in b6 or there'll be two 9s in the box. That places those digits; don't forget the 9s. The 11 cage in r6 now sees a 4 and 9 in the row and a 5 in both of its columns, so it must be 3/8. The snake cells in b4 must be 5/7/9, and r5c2 can only be 9. r7c1 can only be 1. r6c8 must be odd, but it cannot be 1/3/9 by sudoku, and it cannot be 5 because of indexing from the 10 cage in b8. The 13 cage now sees a 7 and 9, so it must be 5/8, and the 5 must be in r4c7. Therefore r4c5 must be 7.
The snake's path is constrained now. The 3 and 7 in b5 are in r4 and cannot be visited by the snake. So the snake can use at most 3 cells in this box; if it enters r6c4, it must exit as quickly as possible by passing through r6c5 and r6c6, but there simply aren't three odd digits left unplaced in r6. Therefore the snake passes through r9c5, as well as r9c4 and r9c6. r9c5 must be 5. At this point, sudoku will finish most of the top two-thirds of the puzzle. The remaining cells in c5 are 1/4, and the 4 cannot be in r6c5. This eventually sets r3c6 = 4, which forces r1c1 = 6. This puts r1c6 = 1 using the codex clue.
After you're done up there, get back to the snake. The snake must move down into r7c8, which can only be 5. That's the last odd digit in c8, so it must turn left into r7c7. A little sudoku will fill in the last of the codex cells; be sure to place the indexed digit appropriately. The only legal place for 7 in r9 is r9c3. You can finally use the circle clue in r8c3: it is forced to be 3. Sudoku forces r8c7 = 8, so the snake must pass into r7c6 and r8c6, using all of the odd digits in b8. Its path through b7 is also completely determined. Sudoku will take care of the last few cells. The boss cell should be filled with 9.
Placing that digit reveals the rest of the library grid, but it also reveals a new quest: Wall Sandwiches. This is optional, and frankly it's not much of a puzzle; on the one hand, you can skip it, and on the other hand, it's not really a burden to do it anyway. The reward for completing it is some extra clues for the next few grids. These extra clues can all easily be ignored, so if you don't want them, you can just pretend they're not there either way. This walkthrough will be explaining how to solve the puzzles without them. It's really up to you.
Anyway, to do Wall Sandwiches, you need to click the quest marker (just below r9c9 in the library) and read the description in the puzzle description box. This will explain the puzzle, and more importantly, it will tell you how to report the answer in a way that will show you the remaining clues. Basically, there are some sandwich icons in the walls outside certain grids; for example, there's a 17 sandwich outside r3 in the Quarters (grid 2). This number is the sum of the cells between two unknown digits in that row or column; the puzzle is to deduce the two digits. Once you've figured them out, you put them, in increasing order, on the appropriate cells in the quest progress box (in the bottom-right of the world map). This will reveal another set of sandwich icons, which you solve the same way. Complete walkthrough:
Back to the normal library grid. Filling the boss cell immediately creates a few sudoku opportunities. It also gives you an arrow, which can be easily solved (it's 1/8 because every other option runs afoul of digits already in b9), and that resolves the rest of the grid.
This is a fun little interlude. The "portal" is a 5×6 grid connecting the library grid to the "elsewhere" grid on the right edge of the map. The portal is not a sudoku grid; it's a completely different kind of puzzle called fillomino. In a fillomino puzzle, you are responsible for dividing the grid into orthogonally-connected regions. Regions of the same size can never touch orthogonally. Every cell in a region is filled with the size of the region. So, for example, this is a valid division of a 4×3 fillomino grid:
This grid is seeded for you by a 1 and a 2... and by the five digits that overlap c9 of the library grid. So e.g. r2c1 is part of a nine-cell region, each cell of which must be filled with 9. You're going to get started here by applying one of the basic ideas of fillomino: find a region that needs to grow but is constrained in an interesting way, then try to grow it one cell at a time. In this case, every region in c1 except the 1 has to grow, and they can all only grow to the right. Now, all of those regions except the 1 and the 2 still need to grow. The 3 region can grow in two directions, so you can ignore for now; but the 6-region and 9-region have no choice but to grow another cell to the right. Now the 6-region has to take its fourth cell in r2c4, but the 9 region can only grow by taking r4c3. This finally tells us how the 3-region has to grow: it must take r5c2.
This next step is pretty neat. Hint: don't forget how you started. c6 of the portal grid overlaps c1 of Elsewhere (grid 7). This is a sudoku grid again, and digits can't repeat in it! This is actually mentioned in the rules for the portal grid, but it's easy to overlook. Because digits can't repeat in c6, there must be 5 different digits, which must be 5 different regions.
Do some counting. There are 5 × 6 = 30 cells in this grid. The regions that overlap c1 account for 2 + 6 + 9 + 3 + 1 = 21 cells. There are also at least two regions in the center of the grid because of the given 1 and 2, and they must be different regions from the c1 regions; so that's 24 cells spoken for. So there are only 6 cells at most that can go into new regions that overlap c6. Only two of the existing regions can reach c6: the 6-region and the 9-region. You will have to add at least 3 new regions that overlap c6. Thos regions must have different sizes, and the least number of cells that can be in three regions of different sizes is 1 + 2 + 3 = 6, which is exactly the number of cells left unaccounted for. To add more than three new regions, you would need at least 10 unaccounted-for cells, which is more cells than there are in the grid. Therefore, the 6-region and 9-region must reach c6, and the other cells there must be part of a 1-region, a 2-region, and a 3-region.
That constrains the regions a lot. The 6-region can only reach c6 by being a horizontal line from r2c1 to r2c6. Don't forget that you've accounted for all 30 cells, so you cannot add any more regions that don't overlap c6. So the only way to fill r1c4 is to put the new 3-region up there. The remaining new regions are a 1-region and a 2-region that must overlap c6, so only the 9-region can be in r5c3 and r5c4. That's 6 of its 9 cells placed. Wherever the new 2-region in c6 is placed, it must extend leftwards into c5. It cannot be in r3 without touching the 2-region in r3c4, which would put two regions of the same size orthogonally adjacent. It cannot be in r4 without making it impossible to place the second cell of the 2-region in r3c4. So it is in r5c5 and r5c6. There is only one way to fill out the remaining regions: the 9-region floods right until it reaches r4c6, and then the new 1-region goes in r3c6.
Finishing the portal reveals most of the next grid. This is an irregular sudoku grid: it follows the normal row and column restrictions of sudoku, but instead of having a regular 3×3 array of 3×3 boxes, it has arbitrary regions of 9 orthogonally-connected cells. These regions still have the same restriction as boxes, though: they must contain all the digits from 1-9, which means digits cannot repeat within the region.
You can see the borders of 7 of the regions. The border between the other two can be easily deduced: the region that contains r5c5 must also contain the 5 cells to its top, bottom, and left, and running right, it reaches 9 cells at r5c8. If you want to draw the line between these regions, you can use SudukuPad's pen tool: click on the gear icon above the number pad, then click on Enable Pen Tool. This adds a fifth entry mode to the right of the number pad; make sure the drop-down box says either "Edge Only" or "Edge and Center", then drag along the edge in the grid to draw a bold edge line there.
I'm going to need to talk about the regions. Fortunately, they're very symmetric and still relatively strongly associated with specific parts of the grid, just like boxes are. So I'm actually just going to use b1 through b9 again, despite them not being, strictly speaking, boxes:
Irregular regions can be very interesting because cells can restrict the placement of their digit in other regions a lot more than they would with 3×3 boxes. Just by looking at the shape of these regions, for example, you know something interesting about the digits in the corners: they're forced into the "risers" of b2 and b8. The digits in r1c1 and r1c9 are the same as the digits in r2c5 and r3c5, and similarly, the digits in r9c1 and r9c9 are the same as the digits in r7c5 and r8c5.
That's pretty abstract. Can you apply the same idea to one of the digits you've already filled in? The 1 in r5c1 sees 7 cells in b5. So the 1 in b5 is in either r4c5 or r6c5, which means it can't be in c5 in b2 or b8, so it has to be in r1 in b2 and r9 in b8.
There are a couple of other digits that are nearly as constraining. The 6 in r4c1 and the 9 in r6c1 see five cells in b1 and b7, respectively. Can you do anything with those? If r8c5 was 9, it would prevent 9 from being anywhere in b7. That's something, but it isn't immediately helpful. r3c4 cannot be 6, because r2c5 would have to be 1 because of the whisper line, but we know the 1 in c5 is in b5. So the 6 is in r2 in b1. Where is the 6 in b3, then? It must be r3c6 or r1c9, but r2c6 would one again force r2c5 to be 1. So r1c9 = 6. Now the only place 6 can go in b2 is r3c5. That reveals a bunch of new cells, including a really intriguing red diamond; but you don't see a rule for this clue yet. Put it out of mind for now and get back to what you were doing. The 6 is now really constrained in b6; in fact, it's forced into one of the dotted cells that were just revealed, making them a 3/6 pair.
That creates an interesting constraint, because r7c4 and r7c6 have the same high/low polarity on the whisper line. They cannot be 2/3, so if they were low, they'd have to be 1/4. And that would make r8c5 adjacent to a 5 on the whisper line, requiring it to be 9; but that would make 9 impossible to place in b7. So r7c4 and r7c6 are high, and r8c5 and r6c7 are low. This row is really starting to fill up. The only place for 1 in r7 is now r7c9. And that means the 1 in r3 must be in either r3c4 or r4c6; since they must have the same high/low polarity, that polarity must be low, and the other two cells on this whisper line must be high.
Remember what we figured out about the corners? The digits in the corners must be the same as the risers of b2 and b8. r2c5 (which must be 7/8/9 because of the whisper line) must be the same as r1c1 (which must be 4/5/7/8 by sudoku); therefore they must be 7/8. Since r2c5 can no longer be 9, r3c4 and r3c6 must be a 1/2 pair. 2 in b6 must be in c9. r8c5 cannot be 1 by sudoku, and it cannot be 4 because its whisper-adjacent cells would both have to be 9, so it is 2/3. That digit must also be in one of the bottom corners, and by sudoku it can only be in r9c9; but this cannot be 2, so they are both 3. This makes r7c4 and r7c6 an 8/9 pair because of the whisper line. But more importantly, it reveals the boss cell.
This is a very fun but also very challenging puzzle. Like the Elsewhere grid, the regions are irregular; unlike the Elsewhere grid, you don't know where any of the borders are at the start, and you have to pull off some pretty tricky reasoning to even start applying the main rule that lets you deduce them!
The puzzle has three kinds of clues:
All of these clues connect "adjacent" cells in one way or another: black dots sit between two cells, and the other two clues involve lines between cells. The big, crazy rule of this puzzle is that each of these kinds of clues interacts with regions in one of two ways:
The rules don't saw which clues have which interaction, but they do say that 2 kinds of clues have the first interaction, and the third kind has the second interaction.
The way to start this is by trying to figure out which rules have which interaction. It would be really convenient if one of the clues boxed itself into a corner, like if there was a diamond in r9c9 connected to r8c9 and r9c8; then this would have to have the first interaction because otherwise r9c9 would have to be a 1-cell region. Unfortunately, you can't see anything like that. But you can try to find things like that that involve interactions between clues.
Consider the clues in the bottom left. If black dots have the second interaction, whisper lines must have the first interaction. But this means r8c1's region can have at most 2 cells: r7c1 and r8c2 must be in different regions (possibly the same as each other), and so must r9c2, since it's in the same region as r8c2. So black dots must have the first interaction.
Now consider the clues in the bottom right. If zipper lines use the second interaction, whisper lines must have the first interaction. But then the region containing r8c8 can have at most 3 cells: it is a different region from both the whisper line and r9c8. So zipper lines must also use the first interaction, and whisper lines must use the second. There are some other ways to puzzle through this, but I like this one.
Having figured out the interactions, it's time to start marking out the regions. The best way to do this is by coloring cells. It's tempting to think that you could mark a region by just drawing the border between them, but that doesn't really work in this kind of puzzle because you'll rarely know where exactly the borders are until fairly late the in the puzzle. Instead, you'll have these amorphous regions that you'll be building up as you go along; coloring is a much better way of marking that. To do that, you just need to make sure that regions that might touch are always shaded with different colors. Now, there's a mathematical theorem that says that something like a sudoku grid can always be unambiguously colored with at most four colors. Unfortunately, you need to know exactly which regions touch before you can do that, and you won't know that perfectly as you're solving the puzzle. Don't be afraid of just using a lot of different colors at once; as long as you can tell them apart (both from each other and from the un-colored background), you'll be fine.
I suggest starting coloring by (spoiler only if you're reading ahead and haven't figured out the interactions yet) shading in all of the zipper lines as different regions. Note that, in this specific grid, different zipper lines can never share a region: there's simply no way to connect two different lines orthogonally without creating a region with more than nine cells. It's going to be useful to have some way to refer to the regions here, and unfortunately, they look like they're going to end up much less symmetric and evenly-distributed than the irregular regions in Elsewhere. I'm just going to pick specific colors for them, which means that if you want to follow along, you'll probably have to pick the same colors. Here we're going to start with four regions:
Now don't forget the interactions for the other rules. Black dots also have the first interaction. So r6c4 is Pink, and r5c2 and r8c4 are Yellow. That puts 7 seven cells in Yellow already. So the three cells with black dots around r8c1 must be in a different region; let's make that Red. Whisper lines have the second interaction, so r9c2 must not be Red; it must be Yellow, and to make it connected, so must be r9c3. That completes Yellow. If you learned the Pen Tool before, I suggest drawing a line around it to remind yourself that it is complete. Red must now take r9c1, r6c1, r5c1, and r4c1, and Pink must take r7c4. One last observation of this kind in a very different part of the grid: can you say anything about the whisper line in r1c8? r2c8 must be in a different region from r1c8 and r2c7, so it must be in Dark Grey, or else its region would be trapped in the corner. Therefore r2c9 is Dark Grey as well.
Okay. The clues around Blue look interesting, but you haven't done much with them yet. r7c9 is not Blue because of the whisper line. It cannot be Dark Grey because that region has only two more cells and cannot reach. It cannot be Pink because then r8c7 would be cut off: it can't be Blue by the whisper line, it can't be Yellow or Pink because they'd both be full, and it cannot be a different region because that region could not possibly have nine cells. Therefore this is the start of a new region, Orange; you can also add r6c9. This corner is now getting quite crowded. r8c7 cannot be another new region: that region would have to grow into r5 in order to reach 9 cells, which would cut off both Orange and Blue. So it must be either Pink or Orange. Both of those seem possible for now, though.
This seems like a good time to remember that this isn't just a coloring book. Let's start with the clues overlapping Yellow. The zipper line has two pairs of digits which each add to r7c3. The least sum of four different digits is 10, so r7c3 must be at least half that, i.e. it is at least 5. The cells at the ends of the zipper line sum to r7c3; let's call them the "end-sum cells". The end-sum cells are both in dotted pairs with cells that aren't on the line; let's call them the "paired cells". The digits in those four cells must all be chosen from 1/2/3/4/6/8. The other two cells on the line must also sum to r7c3; let's call the "other-sum cells". The lower of these, at least, must also be 1/2/3/4. Because of the digit restriction, there's actually a relatively small number of possible ways for the end-sum cells to sum to between 5 and 9: 1 + 4 = 5, 2 + 3 = 5, 2 + 4 = 6, 1 + 6 = 7, 3 + 4 = 7, 2 + 6 = 8, 1 + 8 = 9, and 3 + 6 = 9. Each of these options also tightly constrains the paired and other-sum cells, if it doesn't render them impossible to fill. I'll come back to this soon.
These options are directly affected by another set of clues: those in the Red region. r8c2 is constrained in three important ways. First, it is one end of a 1/2/4 or 2/4/8 sequence in the Red dotted cells. Its value dictates the value of both r8c1 and r7c1. Second, it sees most of the cells in Yellow, including both of the paired cells and both of the other-sum cells. Its digit can only go in r7c3 (the center cell), r9c4 (one of the end-sum cells), or r9c3 (entirely off the clues). Third, it is on a whisper line with a cell in Yellow that is not one of the clue cells and so must have a different digit.
Now, consider each of the options for Yellow in turn. You'll need to explore each path for a bit, but if it looks okay after a few steps of logic, pull back and try another. This is a kind of "bifurcation", in other words guess-and-check, which is sometimes looked down on in sudoku solving, but I haven't been able to find a more elegant approach to breaking in. You don't have to search too far, though.
You should have found that only one option actually works after considering the next few steps, so go ahead and put it in. It's the third bullet in the list above. The remaining cells in Yellow are 1/9, but only 9 can go in r9c2 because r8c2 is low. This creates an interesting constraint in the column: the whisper line in r1c2 and r2c2 needs a high digit and a low digit. If r8c2 = 1, the low digit must be 3/4. The high digit would therefore have to 8/9, but it cannot be 9, so it would have to be 8, and the low digit would have to be 3. But this puts a 4/6 pair in the black dot by sudoku, which is not valid. So r8c2 = 4, which puts the 4 in Yellow in r9c4 and makes the dotted pair in c2 3/6. The pair on the whisper line above is therefore 1 and 7/8.
It's finally time to look at Blue. r9c6 and r7c8 are the same high/low polarity, and they're in the same region. If they were both high, the least they could sum to would be 5 + 6 = 11, but you can't put 11 in r9c8. So they must both be low. r7c8 is at least 3 by sudoku, so r8c7 must be at least 8. But it cannot be 8 by sudoku, so it is 9. r7c8 is 3/4, and r9c6 is 2/3, so the sum in r9c8 must be 5/6/7. r8c8 and r9c7 must sum to 7, so they must be 6 or less; by sudoku, they are 1/3/6 and 2/3/5/6, respectively. If r7c8 = 3, r9c6 = 2 and the sum in r9c8 must be 5, but r9c7 would then have to be 6, breaking the sum. So r7c8 = 4, restricting the sum in r9c8 to 6/7. r8c7 cannot be 6 without making the sum at least 8, so it is 1/3. This means that, if r9c7 were 2/3, its sum with r8c7 would be at most 5, but it must be at least 6. So it cannot be 2 or 3 anymore, making it 5/6. That means that r8c7 cannot be 3 without making the sum greater than 8, so it is 1.
You can get back to coloring now. You just learned that r7c8 = 4, which makes r7c9 = 9. r8c7 therefore must be in Pink, because if it were in Orange, that would be two 9s in the region. Ping must connect to r8c7, which means it must occupy at least two more cells in r7 and r8. If this corner was crowded before, it's ready to burst now. Blue and Orange need to add 11 more cells between them. If they can't get enough cells in the bottom-right corner, one of them will need to pass through the only "exit", r5c6. There are 12 unshaded cells behind this exit, but Pink needs two of them, so that's not enough cells. Blue cannot enter r5 without cutting off Orange; it could have at most seven cells. So r5c6 and r5c7 must be Orange. Does Blue need to grow in any particular direction? There must be a 9 somewhere in Blue. The only place that Blue can still reach that can contain a 9 is r6c8; so that is Blue and 9. Orange must therefore pass around it, taking r5c8 and r5c9.
There are now some interesting constraints in c8. The whisper line in r1c8 and r2c8 must contain a high/low pair; the low digit must be 2/3 by sudoku. The zipper line in r3c8 and r4c8 must also contain a low digit or else the sum will be more than 9, but that accounts for all of them, so the other digit must be at least 5. So the sum is at least 7, but it cannot be more than 8 by sudoku. If the larger digit on that zipper line is 6, it must be a 2/6 pair, making a sum of 8. But if Dark Grey already contains all of 2, 6, and 8, there is no way to fill the dotted pair in r2. So the larger digit must be 5. r5c8 must be 7/8 by sudoku. The 6 in c8 cannot be on the whisper line because it would have to be paired with 1, so the only remaining option is r9c8. This makes r9c7 = 5 and r9c6 = 2.
Keep your focus on Dark Grey. The zipper line in c8 constrains the other clues in the region quite a bit. If the sum there is 3 + 5 = 8, the dotted pair must be either 1/2 or 2/4; by sudoku, r2c8 = 2, which means the other zipper sum must be 1/7, which means r2c9 = 4. If the sum there is 2 + 5 = 7, the dotted pair must be either 3/6 or 4/8; that rules 3/4 out of the other zipper sum, so it must be 1/6, and that means that r2c8 = 8 and r2c9 = 4. So regardless, the other zipper sum is 1 and 6/7, and r2c9 = 4. That creates an interesting constraint in the row. If r2c8 = 2, the other dotted sum in r2 would have to be 3/6, but neither digit is at least 5 away from 2. So r2c8 = 8, r3c7 = 7, the zipper line in c8 is 2/5, the zipper line in c9 is 1/6, and r1c8 = 3.
Look left. You already discovered that the whisper line in r1c2 and r2c2 must be a 1/7 or 1/8 pair. r2c2 can now only be 1/7, and it cannot be 7 because r2c1 cannot be 1/2. So it is 1. The dotted pair in r2c5 and r2c6 must now be 3/6, and the 3 must be in r2c6.
How can Dark Grey grow? It has seven cells, and the remaining digits it needs are 3/9. 3/9 are excluded from r1c9 and r3c7 by sudoku, so the only place it can grow is r4c6.
At this point, it makes sense to start shading some new regions, but we have a color problem. There are three colors on the palette that we haven't used: Green, Light Grey, and Grey. The different greys are actually pretty easy to distinguish from each other; the problem is that Light Grey is really hard to distinguish from the background color of light blue, so I don't want to use it on this grid. We won't need to mention the old Yellow region again, so I'm going to have you coloring one of the the new regions Yellow as well. I'll use it for a region that will definitely never touch the old one, so there won't be any ambiguity. That said, if you aannoys you to have two regions the same color, you can use SudokuPad's extended color palette to recolor the old Yellow region to some other color that's easy to tell apart, like Brown. (I suggest recoloring the old region instead of giving the new region one of the extended colors because you won't want to constantly switch palettes while you're coloring.) To do this, just click the rainbow icon beneath the color pad, which will give you access to a second palette of nine more colors. Brown is in the bottom left of this palette. After you've recolored the old region, click the rainbow two more times to return to the original palette.
Alright. The new Yellow region is going to start in r1c9, r1c8, and r1c7. This region already has a 3, so r2c7 must be another new region; start coloring that one Grey. Grey must also contain r2c6 because black dots always separate cells of the same region (remember that?). That means Yellow is forced to grow to the left, into r1c6 and r1c5. Whisper lines always separate cells of different regions, so Yellow must continue growing left, into r1c4. There will be one more new region, Green, but you don't know how to start it yet.
Growth in the bottom-right corner is pretty restricted. There are 8 unshaded cells in this corner, and they can only be taken by Orange, Pink, or Blue. All three of these regions needs 3 more cells. So between them, they will take the 8 unshaded cells in the corner, and then one more cell elsewhere. This puts a very interesting constraint on Orange: because it can only grow at most one cell leftwards from r5c6, the only place it can possibly acquire a 6 is from r6c7 or r7c7. Therefore it must grow into r6c7 at least. In r8, 6 is forced to be in r8c5. This cell must belong to Pink, since Blue already has its 6. This makes r7c4 a naked single 3. 6 in r7 can now only be in c7, and this must be Orange's 6. r8c6 is therefore the only way for Pink to connect to r8c7, and it must be 7. r8c9 = 3. Blue is now boxed in: it needs three cells, and there are exactly three unshaded cells it can claim. r9c5 and r9c9 must be 7/8, so r9c1 = 3.
Try to wrap up Pink. r6c6 is quite constrained. It sees both cells of the 5/8 pair in r7: r7c6 by being in the same column, and r7c5 by being in Pink (and because of the zipper sum). Therefore it can only be 1/4. The dotted pair in Pink cannot be 3/6 or 4/8 by sudoku, so it must be 1/2 or 2/4, making these three cells a 1/2/4 set. The zipper sum in r7c5 therefore cannot be more than 6, so it must be 5. Its line is a 1/4 pair, and r6c4 = 2. So not quite finished, but really close.
There's a lot of sudoku left to do in that row. 7 can only be in r6c1. That means r2c1 can only be 9. The 5 in r2 can only be in r2c4, so the 2 can only be in r2c3, and the 7 must be in r2c5. That makes r9c5 = 8 and r9c9 = 7. Coming back to r6, you can place the 5/6/8 by sudoku. One of these is particularly useful: the 6 in r6c3 cannot be Pink because there's already a 6 in that region. That must be a new region; go ahead and color it Green. And Pink can now only grow to one of three cells; neither r5c4 or r5c6 can be 8, so it must take r7c6, which completes it. Orange must therefore grow into r5c5, which completes it as well.
The unplaced digits in Orange are 1/2/3/4. Of these, r5c9 can only be 2, which kicks off a nice little chain: r1c9 = 8, r1c2 = 7, and r3c4 = 7. Coming back to that row, r5c4 = 9, r5c3 = 8, and r5c1 = 5.
At this point, you're mostly trying to figure out how you can get the right digits in every remaining region. The hardest constraint here is in the top-left, where the remaining whisper line means that different regions have to take different digits. There's a couple ways you can puzzle this out. I suggest starting with Yellow. The 5 in r1 can only go in r1c6. Yellow therefore has a 5 and cannot take r2c4, so it must grow into r1c3. It now only has two cells left to claim, and it must find a 7 somewhere. It cannot reach r2c5 or r3c4, so it must take r1c2. r1c3 is therefore the only 9 it can possibly reach. Yellow can grow into either r1c1 or r2c3. If it grows into r2c3, then r2c3 and r1c1 must be the same digit, because both r1 and Yellow must be a complete set of nine digits, and the other eight digits come from the exact same cells; but r1c1 cannot be 2. Therefore Yellow must grow into r1c1, completing it. Red can only take a 9 in r2c1, so it must grow there, completing it as well. You should be able to complete c1, c2, c8, and c9 with sudoku.
Time to look at Grey. Grey needs an 8, and there's only one place it can get it: r3c2. This means there must be a wall-to-wall connection of Grey cells. The last two incomplete regions, Green and Dark Grey, have cells beneath this connection and cannot possibly have cells above it. So in addition to r3c2, Grey must also take r2c2, r2c3, r2c4, and r2c5. Grey now needs two more digits: 4 and 9. It cannot reach r5c4, so this must belong to Green. It does not need the digits in r4c2, r4c4, or r3c4, so these must also belong to Green. Green must also take r4c3 to be connected. Green now also only needs two more digits: 3 and 4. Green can no longer reach r3c6, so this must belong to Grey; it must be a 4. You can place the rest of the digits by sudoku now, and SudokuPad will tell that you're done, but you're not really done until you've finished coloring the regions. r3c3 = 3, so it must be Green. Green also needs a 2, which must be r3c5. Grey can now only reach the 9 in r3c6, and r4c5 must be Dark Grey.
The digit to place in the boss cell is 4.
Filling the boss cell reveals that zipper lines (excuse me, ZIPLINES) have also been added to the world map. Go ahead and read about the new rule in the puzzle description box. Yes, it's the same rule you just used for the boss puzzle, but the puzzle description box says in a way that's very subtly different:
That is, cells on the line should always contain digits. And you can see in the revealed cells that sometimes that digit is allowed to be 0. Very interesting!
Anyway, you can see an awful lot of pairs of cells on this zipper line; wouldn't it be nice if you knew what the sum was? c8 contains three disjoint pairs of digits that sum to it, so it must be at least 7. b6 has a very useful shape. The 9 in r3 must be in b6, so the sum cannot be 9; it must be 7/8. In b5, the sum must be in either r4c5 or r6c5. r2c5, r4c5, and r6c6 therefore form a 1/7/8 set. This means r7c5 cannot be 7; it must be 4/5. Where is the sum in r3? It must be in b4, either r3c2 or r3c3. But now where can it be in r7? Not in b4 or b6, and it cannot be r7c5 because 7/8 has no overlap with 4/5. Therefore it must be in one of the other two cells, which are an 8/9 pair. It cannot be 9, so it must be 8.
You can do three things with that right away:
Stay there for a moment. There are four remaining digits. The 6 can only be in r6c8 or r8c8. Both of these cells are on the zipper line. So that 6 must be paired with a 2 in either r2c8 or r4c8. Therefore r3c6 = 1 and r3c4 = 2. That means the 1 in c8 must also be in either r6c8 or r8c8, so the 7 must be in either r2c8 or r4c8, and it cannot be in r2c8. Therefore r2c8 = 2, r4c8 = 7, r6c8 = 1, and r8c8 = 6.
There's a lot you can figure out just with sudoku now. Spiralling from r3: r3c7 and r3c9 are a 7/9 pair, so r4c9 and r6c9 are a 2/4 pair. The 4 in r8 must be in b9, so r6c7 cannot be 1/3/4, so it must be 2, which settles r6c9 and r4c9.
There's one last zipper pair in this puzzle. What digits are constrained in c7? The 1 can only be in one of the zipper cells, so they must be a 1/7 pair, but r1c7 cannot be 7. You can finish r3 by sudoku, but something else much more interesting just happened: a new quest right above r1c8, and some new revealed cells way in the bottom-left of the map. You don't need this to finish Elsewhere, but we'll come back to it soon, I promise.
c1 and c9 have some interesting structure: r2c1 and r8c1 are a 5/8 pair, and r2c9 and r8c9 are a 5/9 pair. Meanwhile, r7c4 and r7c6 are an 8/9 pair. If they were 8 and 9 respectively, both r8c1 and r8c9 would have to be 5.
It's time to look into the eyes. Remember how to use these clues. The most obvious restriction is that r9c3 cannot be greater than 5, or else it would be impossible for the eye beneath it to see five cells. By the same logic, r1c3 cannot be greater than 7, r8c3 cannot be greater than 6, and r2c3 cannot be greater than 8. So the only cell that 9 can be in c3 is r4c3, and that means the only cell that 8 can be in is r2c3. Since the bottom eye can no longer see the 8, r9c3 cannot be 5 anymore, and r7c3 cannot be 7.
A lot of sudoku is now unblocked, and that very nearly solves the grid by itself: all that's left are four ambiguous cells in the lower left, which can be resolved by observing that the bottom eye already sees 5 cells: r9c3 (whatever it is), r7c3, r6c3, r5c3, and r4c3. Therefore it must not also see r8c3, and so r9c3 = 2 and r8c3 = 1.
Okay. Now it's time to read that quest just above r1c8 in Elsewhere. This quest is not optional! You can skip the hidden puzzles you find, but you need to complete at least the zipline in order to make progress in the map.
Got it? You can fill cells outside the grids with any digit, including 0, but you can also leave them blank. And the rules all actually leave room for this — as you're about to see, because there's a sort of tutorial that's also been revealed in the bottom-left of the grid, just to the left of the Courtyard (grid 1) and under the Front Gate.
The first tutorial is thermos and arrows. You can leave cells on thermos and arrows blank, but the arrow's circle must have a digit in it, and the filled-in cells on the arrow must sum to that digit. Since there must be a digit in the circle, and it's also on the arrow and therefore must be greater than 8, it must be 9. So the remaining cells on the arrow must sum to 1, i.e. there must be a 1 somewhere on the arrow. You cannot put the 1 anywhere on the thermo because of the 8, so it must be in the corner cell.
The second tutorial is whispers and dots. Dotted cells must contain a digit, so they're exactly like normal. Whisper lines are very different, both because they can contain 0 and because they ignore unfilled cells: two cells are now adjacent for the rule if all the cells between them on the line are unfilled. The 8 must be dotted with a 4, and the 0 must be dotted with a 0. For the upper line, there is nothing you can place between 0 and 8 that is at least 5 away from both, but that's fine, because if you leave the center cell blank, the 0 and 8 are adjacent and are more than 5 apart. On the lower line, 0 and 4 are not 5 apart, so the same trick doesn't work. The only digit 5 apart from 4 is 9, and fortunately that's also more the 5 apart from 0.
The final tutorial is about zippers. The center cell must contain a digit, and equidistant pairs must still sum to that. So at least one cell in each pair must be filled. The center cell is on a thermo that starts with 8, so it must be 9. The remaining thermo cell cannot be filled, because there are no digits between 8 and 9. The cell to the left must therefore also be 9.
That completes the tutorial and reveals most of the zipper line earlier in the puzzle. But before you try that, recall that you've been seeing a bunch of clues that run off the grid all along. Why not see if you can satisfy any of them?
In grid 2, there's a whisper cell between a 1 and a 9. That's clearly unfillable. In grid 1, there's a thermometer cell between a 3 and a 4. That's also clearly unfillable. And there's a whisper cell between a 5 and a 6. That can be filled! It has to be 0. Oh hey, look at that, a hidden puzzle.
Most of this puzzle is a 4×4 sudoku grid, which as usual must be filled with 1/2/3/4. That means only two cells count as outside the grid, and one of them is the arrow circle; keep that in mind.
The first thing to do is to try to constrain the arrow sum. The total sum of the digits in the grid is 40. The arrows contain all but two of those cells, and those cells see each other and therefore can sum to at most 7. So the digits on these arrow sum to at least 33, without even counting the cell off the grid. The circle sum must be 9.
The arrow in c1 sums to 9, so r4c1 = 1. If r1c2 = 1, the arrow ending in r4c2 would already sum to 9 without even counting r1c3. So the 1 in c2 must be in r2c2.
The arrow ending in r4c3 uses the same digits in b2 and b4. These digits cannot sum to more than 4 without breaking the arrow sum, so they must be either 1/2 or 1/3; in either case they have the 1, and so r1c4 = 1 and r3c3 = 1.
In b2, neither r2c4 nor r1c3 can 4 without breaking their respective arrow-sums, so r2c3 = 4. Now r1c2 cannot be 4 because r4c4 would have to be 1 to make the arrow-sum work. So r1c1 = 4. This puts a 4 on the arrow ending in r4c2, but the c2 cells of this arrow cannot sum to more than 7 because r1c3 must be at least 2. Therefore r1c2 = 3 and r1c3 = 2.
The rest of the grid can be filled by sudoku, but that's not quite the end: the in-grid cells on the five-cell arrow only sum to 8, so the cell to the right of the circle must be filled with 1.
You can finally count cells on this zipline to figure out which pairs are equidistant — not quite all the way to the end, but a decent distance. It's a little tedious, though, and you don't actually need to: just fill them in from the right.
The cell to the left of the eye above Elsewhere was revealed to be a 6. So the lower half of that pair must be 2. The 2 is dotted, so the cell to its left must be 1/4. The upper half of that pair must therefore be 4/7, but it's dotted, so it and its pair must both be 4. The cell dotted to the upper 4 must be 2/8. The cell dotted with that must be 1/4. So the lower halves of those pairs must be 0/6/blank and 4/7, respectively. If the left cell on the upper line were 1, the right cell would have to be 2, and they would be paired with 7 and 6 respectively. But the lower halves are on a whisper line, which would be broken. So the left pair is 4 and 4. The only digit that 4 can be adjacent to on a whisper line is 9, and that would immediately break the zipline sum. So the lower half of the right pair must be blank, and the top half must be 8, which works.
Next up are the paired thermos. Note that the upper thermo grows to the left and the lower thermo grows to the right. The circle on the bottom is in a dotted pair, and it must be less than 8, so the upper cell paired with it must also be filled with at most 8. Therefore all of the intermediate cells on both thermos are either unfilled or less than 8. All of these cells that are on the zipper are paired with cells with the same constraint. So all of those cells must be filled, because otherwise they would have to be paired with an 8. The rightmost dotted pair on the bottom thermo must be in increasing order to the right. If it were 0/0, 1/2, or 2/4, it would not be possible to fill the three cells to its left; and if it were 4/8, it would break the thermo immediately. So these cells are 3 and 6 in that order. Since the other cells that are on both the thermo and the zipline must be filled, they must be 0, 1, 2, and 7. The cell dotted with 0 must be 0. The intermediate cell on the lower thermo must be unfilled. Don't forget to fill in all the paired cells. And don't forget that the 2 on the lower line is also on a palindrome line, so you can place the corresponding 2.
Continuing to the left, you'll find some unrevealed cells on the zipline. However, since the gaps are all a single cell and the zipline cannot branch, this does not create any ambiguity about the pairing. The 2 in the eye above the Library must be paired with a 6. The 6 in r1c5 of Maintenance must be paired with a 2 — oh look, another hidden puzzle... but you can come back to that in a moment. Just keep filling in pairs. Eventually you'll reach the 2 in r5c9 of Quarters, and that must be paired with a 6 in r6c9 of the Hall of Illusions. You should see in the Quest Progress box that you've now finished the Fast Travel Zipline, but much more importantly, you should also see a very interesting warning about the Hall of Illusions! You'll come back to this later; it's time to solve the hidden puzzle you just found.
Okay. This is a cute little puzzle. It's a 4x4 sudoku, so all of the cells are going to be filled with the digits 1/2/3/4, and most of the grid is on a zipper line.
Your main tool for solving this is thinking about the zipper pairs. It's going to be extremely useful to color in all the different pairs. Fortunately, there are exactly 9 pairs, so you won't run out of colors, although light grey can be hard to distinguish from the white background.
You could approach this by trying to narrow down the sum, but I have a better suggestion: think about the 2.
The 2 in b2 must be in either r1c3 or r2c3. Both of these cells have pairs within the grid, so 2 must have a complement in 1/2/3/4 that makes the zipper sum. Call this P. If r1c3 = 2, P must be in r4c1; note that this means P cannot also be 2. P must be in b2, but it cannot be one of the two paired cells without putting a second 2 in the box, so it must be in r2c4. So P would also have to be in r1c2 by sudoku. But r1c2 is paired with r4c2, which would have to be 2, but cannot be because it sees r4c4. Therefore r2c3 = 2 and r1c4 = P. P must be in either r2c1 or r2c2 by sudoku, but r2c2 is paired with r3c4, which already sees a 2. So r2c1 = P. 2 must be in either r3c1 or r3c2 by sudoku, but if r3c2 = 2, P could only be in r4c2, which is paired with r1c2, putting a second 2 in the column. So r3c1 = 2. By sudoku, r1c2 = 2, and by pairs, r3c3 = P and r4c2 = P.
Now you can think about the sum. The 4 must be on the line in at least three of the boxes, so the sum must be at least 4, and P cannot be 1. If the sum is 5, P is 3, and the other two digits have to be 1/4. Whichever one is r2c2 is, r3c2 would have be the other by sudoku; but r3c4 would also have to be the other because they are pairs, and 1/4 sum to 5. So the sum is 6 and P = 4. r1c3 is 1/3, but if it were 1, r4c1 would have to be 5, so it is 3. At this point, the rest of the puzzle is straightforward. Don't forget to fill the sum and the four cells on the line outside of the grid. If you've done it all right, you should see a checkbox on the Wolf in the Quest Progress box in the bottom-right of the map.
Finally it's time for the Hall of Illusions. Read the warning at the end of r6: every clue in the grid is lying. The rules of sudoku still apply, but every explicit clue within the grid must work out to be "wrong". Be sure to click on the new rule in the Learned Rules box and read the explanation carefully, though, because "wrong" has a specific meaning for each kind of clue which might not be quite what you'd expect. Please feel free to just skip my unbearably pedantic notes on each of the following rules:
Thanks for putting up with me through all that. Note that, crucially, only the clues entirely inside the Hall of Illusions grid are lying; clues that affect the grid but also involve cells outside of it must still be correct in the conventional sense. This is why r6c9 still has to be 6 to satisfy the zipper line: the zipper line extends outside of the Hall of Illusions, and so r6c9 and its paired cell are required to sum to 8. Perhaps more importantly, the eyes above c3 and beneath c7 are outside of the grid and are still honest about how many cells they can see.
If you completed the Wall Sandwiches side quest, there is a clue above c2 here. You do not need this clue in order to solve the Hall of Illusions, and I will show how to solve the puzzle without it, but feel free to use it in your own solution as your reward for finishing that quest.
If a thermometer is lying, then the cells strictly increase as they move towards the bulb. The bulb in r3c1 is 3, so r2c2 must be 1/2; but r1c1 and r2c1 are a 2/8 pair by sudoku. This reveals a few cells in b5 and a lot of very dense clues. Take a moment to remember what these clues actually mean. Because of the lying arrow clues, r4c4 is not the same as either r3c3 or r5c3. Because of the lying dark clue, r4c4 must be greater than r4c3. And because of the lying index clue, whatever digit is in r4c4 is not the column index of the 4 in r4. All three of the revealed cells in b5 are dark, so they are all greater than the adjacent cells in b4. Therefore none of those cells in b4 can be 9. The 9 in c3 must be in b1, but if it's in r1c3 or r2c3, the eye will be broken (and remember: it's outside the grid, so it's not lying). So r3c3 = 9.
Every square in b5 is dark. Since the dark clues are lying, these cells must all be greater than all of the adjacent non-dark cells. They are all adjacent to at least one non-dark cell except r5c5, so the 1 must be there. What does that tell you about the 2? The two cannot be in any of the cells adjacent to r5c5 because it would have to be adjacent to a 1 outside of b5c5, putting two 1s in that row or column. The 2 cannot be in r4c4 because that would put a 1 in r3c4, which is on a whisper line next to a 9, and that whisper line clue must be lying. The 2 cannot be in r6c4 or r6c6 because that would put a 1 in both r6c and r7, which would break the 1/4 pair in r6c1 and r7c1. (r6c4 would also have broken a couple of thermometers.) Therefore the 2 is in r4c6, and there are 1s in r4c6 and r5c7. This fully reveals the bottom half of the puzzle. And sudoku creates an interesting constraint along these lines: The 1 in r6 can only be in r6c1 or r6c3, but r6c3 is the bulb of a thermometer. So it must be in r6c1, which allows you to finally complete the Ramparts grid.
Consider the whisper line. r3c2 cannot be less than 5 because of the 9 in r3c3; it must be 5/6/7 by sudoku. r3c4 also cannot be less than 5; it must be 5/6/7/8 by sudoku. But if it were 8, r4c4 would have to be 9, which breaks the lying arrow. So it is also 5/6/7. r3c5 cannot be greater than 5 because of the 1 in r3c6. It cannot be 1/3 by sudoku, and it cannot be 2 by the lying arrow from r4c6, so it must be 4/5. But if it were 5, the three cells in b3 would have to be 2/4/8. The 4 in r9c9 would put the 4 in one of the dotted cells, and the other two digits are either half or twice it. So r3c5 must be 4.
How does all that affect b5?
There's a lot to be done in b6 now. r5c8 cannot be greater than 5 because of the whisper line, but it cannot be 1/3/4/5 by sudoku, so it is 2. That means that r6c7, which must be less than 5 by the lying dark clue, can only be 4. The 3/4 in r4 now cannot be in b6; they must be in b4, and r4c8 and r4c9 must be a 5/9 pair. The remaining two cells are a 7/8 pair.
That last bit of logic leads to an important deduction on the other side of the grid. The 9 in b4 must be in c2, but it cannot be in r4c2 because that must be 3/4, and it cannot be in r6c2 because that would break the lying thermometer. So it must be in r5c2. Some simple sudoku finishes the row. A quick deduction finishes another: the 9 in r5c2 means r6c3 must be at least 5, and the only remaining candidate is 8. And to nearly wrap up this entire row of boxes, the 6 in r5c3 means r4c4 cannot be 6 because of the lying arrow. Therefore r4c4 = 8 and r4c5 = 6, and oh! Look at that, the boss cell.
You can do the boss puzzle now, and filling the cell will make the rest of the logic a bit simpler. However, you're not actually stuck quite yet, and you can do pretty much all of the remaining logic without filling the boss cell, just leaving a few cells to be disambiguated. I'm going to take you through that now, but feel free to just come back to this point after the boss puzzle if you'd prefer.
There's something else you unlocked by solving b5. r7c6 must be less than 5 by the lying dark clue, but the only remaining candidate is 3. r7c7 therefore cannot be more than 7, but it cannot be 1/2/3/4 by sudoku, it cannot be 5 because of the lying arrow from r6c6, and it cannot be 6 because of the lying dot with the 3 in r7c6. Therefore it is 7. And that's interesting, because the eye beneath c7 must see 4 cells. The 8 and 9 in c7 cannot be in b9 without breaking the eye clue, so they must be in b3. But the 8 in r3 must also be in b3 by sudoku, so it must be in r3c7. The eye therefore sees the 7, 8, and 9, so the only other digit it can see is the one in r9c7. By sudoku, r9c7 must be 2/5, but it cannot be 2 because then r8c7 would also have to be visible to the eye. r8c7 must be 2/6, but it cannot be 6 without breaking the eye, so it is 2, and the two remaining cells in c7 must be a 6/9 pair. And sudoku puts a 2 in r9c3.
Looking again at your deductions in b6, the 2 in r5c8 means that the 2 in r3 can only be in r3c9. By sudoku, the 7 in r3 can only be in r3c2 by sudoku, which means that the 6 can only be in r3c4, and so r3c8 = 5, which lets you finish b6. r2c3 must now be 4/5, but if it were 4, the eye would not be able to see it, and it must. So r2c3 = 5 and r1c3 = 4. Sudoku should now allow you to finish r4, c7, b4, b6, b7, and r7, leaving just 15 unfilled cells in the grid.
Now you're definitely stuck. What was that about a boss...?
This is a fun and very original puzzle. It's really quite novel; I can't remember having seen these exact rules before elsewhere. They are:
Some digits are more constrained by these rules than others. A grey 5, for example, can be adjacent to any other digit. A grey 9 can only be adjacent to four possible digits: 5/6/7/8. But two digits are even more constrained than that when placed in grey cells: 1, which can only be adjacent to 3/4/5, and 2, which can only be adjacent to 3/5/6. The full set of restrictions hold when either cell is grey, so they also apply when the *other* digit is in a grey cell. You need to remember, though, that when both digits are pink, only a 1:2 ratio is prohibited. As an exercise, go ahead and mark where the 4 can go in b3. It cannot go in r3 by sudoku, and it cannot go in r1c8 or r2c9 because those cells are grey and 4 is too distant from 9. But that last restriction doesn't apply to r2c7 because both cells are pink, so the 4 can go in r1c7, r1c9, and r2c7.
It's time to get started for real, and it's pretty clear that you should start in b5. The 2 in r4c5 is adjacent to three cells in b5, so they must be a 3/5/6 set. Remember that a grey 1 can only be adjacent to 3/4/5. 3/4/5 overlaps very closely with 3/5/6! Wherever the 1 is placed in b5, it can only be adjacent to at most one cell that is not part of the 3/5/6 set, and that cell must be filled with 4. The only position that satisfies this and does not see the 1 in r8c6 is r5c4. So this is 4, and r6c4 = 4. Moreover, r4c4 and r5c5 must be a 3/5 pair, and r4c6 = 6. The other three cells in the box are 7/8/9. If r5c5 were 3, it would have to be adjacent to at least one of 8/9, which are too distant. So r5c5 = 5. There's one last deduction you can make in this box for now: r6c5 is 7/8/9, but it cannot be 9 by distance from 4, and it cannot be 8 by ratio with 2. So it is 7.
Stay focused on those constrained low values. In b8, r7c6 must be 3/4/5, but r6c6 must be 8/9, so r7c6 cannot be 3 by distance, and it cannot be 4 because that is half 8 and too distant from 9. So r7c6 = 5. Also, The 1 in r8c6 cannot be next to a 2 on either side (remember that the 1:2 constraint still applies between two pink cells), so the 2 in c6 must be in b2. And in b2, r3c5 must be 3/6.
There are now a lot of constraints up there. r2c6 cannot be 4 because it would be impossible to place the 2 by ratio. Similarly, the 4 in r2 can now only be in r2c5 or r2c7, so r2c6 also cannot be 2, or it would be impossible to place a 4 in this row. r2c6 has been narrowed down to 3/7. r3c6 can be 2/3/7, but it cannot be 3 because that would force a 6 into r3c5 and create a 1:2 ratio. So it is 2/7. This means r2c6 cannot be 7, because it would force a 2 into r3c6, which would violate distance. So r2c6 = 3. This makes r3c5 = 6, and that rules 1 out of r2c5 by distance, forcing it into r1c5. That disallows 2 from r1c6 by ratio, forcing it into r3c6. 9 must be in r1c4 or r3c4, but it cannot be in the latter by distance. In fact, what digits are allowed in the remaining cells in this box? r2c5 must be 4/8. r2c4 must be 5/7/8, but it cannot be 8 because that would force a 1:2 ratio with r2c5, so it can only be 5/7. r3c4 can also only be 5/7, so these form a 5/7 pair. r1c6 can only be 4/7, but now it cannot be 7, so it must be 4. By sudoku, r2c7 = 4, r2c5 = 8, and r9c6 = 7.
All of those deductions are ready to spill over. The 2 in b3 is ruled out of r1c7 by ratio, and it is ruled out of r1c8 and r2c9 by distance from the 9 in r2c8, so it is forced into r1c9. r1c8 and r2c9 cannot be greater be 6, but they also cannot be less than 5, so they are a 5/6 pair. r3c7 must be 3/5/6 because of the 2 in r3c6, but 5/6 are now spoken for, so it is 3. The possibilities for r1c7 are 1/7/8, but 1 is ruled out by sudoku, and 8 is ruled out by ratio, so it is 7. The remaining two cells are 1/8, but r3c8 cannot be 8 by distance from 3, so it is 1 and r3c9 = 8.
Follow the wave of logic down into b6. The 2 cannot be in r4 or c9 by sudoku, and it cannot be in r5c7 or r6c7 by distance from the 8/9 pair. Meanwhile, the 4 cannot be in c7 or r6 by sudoku, and it cannot be in r4c9 by ratio. If the 2 were in r5c8, then, it would be impossible to place the 4 anywhere in b6. So r6c8 = 2. This forces r6c9 = 3 by sudoku. The 1 in b6 cannot be in c7 or r5 by sudoku, and it cannot be r4c7 or r6c7 by distance, so it must be in r4c9. r4c7 cannot be greater than 7 by distance from r3c7, and it cannot be 1/2/3/4/6/7 by sudoku, so it is 5. r6c7 cannot be greater than 6 by distance from r6c8, and it cannot be 1/2/3/4/5 by sudoku, so it is 6. r5c7 must be 8/9 by sudoku, which creates an 8/9 pair in r5, so r5c8 and r5c9 must be a 4/7 pair, and the 4 cannot be in r5c8 by ratio with the 2 in r6c8. The rest of this box can be finished by sudoku. At this point, the puzzle becomes pretty straightforward as the wave of deductions sweeps around the edge of the grid.
In b9, the 1 is ruled out of r7c7 by distance, so it is in r9c7. The 2 then cannot be in r8c7 by ratio with the 1, so it is in r7c7, and r8c7 = 8. The 4 cannot be in r7c8 or r8c8 by ratio, so it is in r9c8. The two remaining cells in c8 are 3/5/6, but they cannot be 3/6 because they would be in ratio. Since they must include a 3, they are 3/5. Sudoku finishes b3 and then b2. The 5 must be in r8c8 by sudoku. r9c9 is 6/9 by sudoku but cannot be 9 by distance from the 4; it is 6. The remaining digits are a 7/9 pair but cannot yet be placed.
In b8, c4 must be a 2/6/8 set, but r7c4 cannot be 2/8 by ratio, so it is 6. r8c4 = 2 and r9c4 = 8 by sudoku. The 4 cannot be in r8c5 or r9c5 by ratio, so it is in r7c5. r8c5 cannot be 9 by distance, so it is 3, and r9c5 = 9.
Let the wave skip a box. In b4, r5c3 must be 3/4/5 because of the 1, but it can only be 3 by sudoku. The remaining cells in the row are a 2/6 pair, and the 6 cannot be in r5c2 by ratio. The 4 cannot be in r4c2 by ratio, so it is in r4c1. The remaining cells in r4 are a 7/9 pair, and the 9 cannot be in r4c3 by distance.
Now it's mostly mopping up. The 2 in b1 cannot be in r2c3 by ratio or r2c2 by sudoku, so it is in r2c1; the 1 now cannot be in r2c2 by ratio, so it is in r2c3, and r2c2 = 6. The 3 cannot be in r1c2 by ratio, so it is in r1c1. The 8 cannot be r1c2 by distance from the 3, so r1c2 = 5 and r1c3 = 8. Back in b4, r6c3 = 5 by sudoku, and r6c2 = 8 because it cannot be 1 by ratio. The last box can be completed entirely by sudoku.
You can only see nine cells in this grid, and there are already two cells with coins in them. Furthermore, the puzzle description is talking about "dazzling wealth". You've completed grids 1-8 now, so this seems like the right moment to go count all those coins you found. In fact, I'll save you the effort if you want:
To understand how to use this, go re-read the collectible coins rule. There are N coin cells across the entire map with the digit N in them. You have not found any 1-coins, so there is one remaining 1-coin outstanding. You have found 4 7-coins, but there are seven of them in total, so there are still three 7-coins outstanding. Anyway, to quickly summarize, there are three remaining coins of every value except 1 and 2, of which there are 1 and 2 remaining, respectively. That's 24 outstanding coins in total.
There's no clear way to use that yet. For now, all you can see is some indexing clues. Clear your head and remember that these clues aren't lying anymore! r9c3 should contain the column index of the 3 in its row, and r9c2 = 3, so r9c3 = 2. That reveals a few cells.
This first deduction is fun. As a hint, the indexing clues in this box (b7) are controlling the placement of the 1/2/3 in their rows. As a second hint, what digits are still unplaced in this box? If two of the indexing clues in r7 were 8/9, they would both pointing at the whisper line. But they can't be, because then the adjacent cells would both be 1/2/3! So one of the 8/9 must be in r8c3. That puts a 3 in either r8c8 or r8c9. The 3 in b8 must be in r7. So r7c3 must be 4/5/6; it cannot be 4 by sudoku, so it is 5/6. Of r7c1 and r7c2, one is 8/9 and the other is 5/6. This means r7c5 and r7c6 must be 3 and either 1 or 2, in some order. If r7c3 were 5, r7c5 would be 3, and r7c6 would have to be 1/2; but r7c5 has a dark clue, so it must be less than r7c6. Therefore r7c3 must be 6 and r7c6 = 3.
Placing that last digit reveals a very interesting shape in b5: a ring of dark cells, each with a coin in it. (The money bags (💰) are decorative and have no significance in the puzzle.) You know something about one of these cells: r6c6 has to be less than r7c6, which is 3. So it is 1/2. Actually, though, you know more than that. r7c8 and r7c9 contain coins. One of these cells is 1/2 because of the indexing in b7. r7c5 is also 1/2, and it also contains a coin. That means you've accounted for a 1-coin and a 2-coin, both in r7. There's only one 1-coin across the entire map, so no other coins can be 1. Therefore r6c6 must be 2. And that is the last oustanding 2-coin, so no other coins can be 2, either.
There are several places you can use that; I suggest you start with b5. None of the coins here can be 1. All of the coins here are in dark cells, so they must be less than their non-dark neighbors. If the coin cells must be at least 2, and the cells next to them must be at least 3, then the only space for 1 is r5c5. This reveals some cells in the corner, but put them aside for a moment. r7c5 must now be 2, and don't forget to fill in the clue that indexes it: r7c2 = 5. You've revealed the rest of b8. The 1 in b8 cannot be in r8 or c5 by sudoku. It cannot be in r7 because it is indexed into r7c8 or r7c9. It cannot be in r9c6 because that is a coin cell, and you have already accounted for all the 1-coins. So it is in r9c4. You also know something useful about it in c3. Since r5c3 must be at least 3 now, the cells above and below it must be at least 4. The 1 also cannot be in r1c3 or r2c3 because they are coin cells, so it must be in r3c3. Sudoku means that it must be in c2 in b4, but it cannot be in r4c2 because that would make another 1-coin, so r6c2 = 1. In the same box, the 2 is also constrained: it cannot be in r4c1 or r4c2 because there are no more 2-coins, and it cannot be in r5c2 because that must be greater than r5c3, which is at least 3. So r5c1 = 2.
Come back to c3. The only low digits left in the column are 3/4, so one of those must be on the whisper line; the other digit must be 8/9 so that the diference can be at least 5. r4c3 and r6c3 must be greater than r5c3, so if r5c3 were 5, there would be no space for the other 3/4 digit in this column. Therefore r5c3 must also be 3/4. r4c3 and r6c3 must therefore be a 5/7 pair. Now in c2, there's a black dot, and it already sees a 1 and a 3. So the dotted digits must be either 2/4 or 4/8. In either case, there's a 4 in b1, so the 3/4 digit in r1c3 or r2c3 must be 3, and r5c3 = 4.
It's really hard to ignore all the clues in b5, but there isn't much you can say concrete about it yet. Still, there are some ideas worth exploring. All of the dark cells are next to at least 2 non-dark cells in the box, so none of them can be greater than 7; 8/9 must be in the non-dark cells. Conversely, all of the non-dark cells are next to at least one unfilled dark cell, so none of them can be less than 4; 3 must be in the dark cells. That's about it for now.
The puzzle did reveal this little arrow in b3. What can you do with that? The arrow cells can't be 1/2 because they're both coins, so they must be at least 3/4. You've accounted for two 3-coins already: there's one on the whisper line in b1, and there's another on the dark cells in b5. So if you put a 3 on this arrow, that's all of them. That would also force a 3 into r3 in b2. But there's a 3 in r7c6, and there's a 3 in c4 in b5, so that 3 would have to be in r3c5, which is another coin cell. That would be four 3-coins. So there cannot be a 3 on the arrow in b3. That means the digits on the arrow must be 4/5, and r2c8 = 9. Placing that reveals the rest of the grid. Note that the 3 in b2 still must be in c5. If it were in r2c5, r3c5 would have to be less than that, but it cannot be 1/2 by sudoku. Therefore the 3 in b2 must be in one of the two coin cells, which accounts for the last 3-coin. The 3 in b4 must be in c1, but now it cannot be in r4c1 because that would be a fourth 3-coin. So r6c1 = 3. This forces the 3 in b5 into r4c4. Now the 3 in r5 must be in b6, but it cannot be in r5c7 because that would be a fourth 3-coin, and it cannot be in r5c8 because the dark cell, r5c7, cannot be 1/2 by sudoku. So r5c9 = 3. By sudoku, then, r8c8 = 3. And I know it's been awhile since you thought about this, but that 3 is indexed by r8c3, which must now be 8.
Are there any other coins constrained now? You've placed two 4-coins: one in r5c3, and one in the arrow in b3. Placing the 3 in r4c4 means that the 4 in b5 must now be in a dark cell, because all of the non-dark cells are still adjacent to an unfilled dark cell and therefore must be at least 5. That accounts for all of them. There is a 4 in one of the dotted cells in b1, and there is a 4 on the arrow in b3. So the 4 must be in r3 in b2. It cannot be in the coin cell, r3c5, because that would be a fourth 4-coin. But that means it must be adjacent to it, and r3c5 is also a dark cell. r3c5 cannot be 1/2 by sudoku, so it must be 3. By sudoku, the 4 must be in c5 in b8, so it must be r8c5. And it must be in r7 in b9, but it cannot be on the whisper line with a 1, so it must be in r7c7. This puts a 4 in r6 in b6, so the 4 in b6 can only be in r4c6, and the 4 in b2 must be r3c4.
There's a little mopping up to do now. If you didn't do the indexing before, placing the 9 in r7c1 puts a 1 in r7c9. That constrains the 1 in r4, which cannot be in r4c7 because of the dark cell and so must be in r4c8; the same logic also gives you r4c9 = 2. Sudoku then gives you r8c7 = 2 and r3c8 = 2, which finally settles that the dotted cells in b1 are a 2/4 pair. Also, when you filled in the whisper line in b1, you narrowed 3 in b3 down to r1c7, and now 1 must be in r2c7, and therefore also r1c6.
You've now accounted for all of the 9-coins, so the 9 in b4 must be in r5c2. In b2, it must be in r3c6 by sudoku; since it must now be in c5 in b5, r8c4 = 9.
You can constrain another coin with a little work. Hint: consider r6c4. r6c4 must be at least 5, so the cells around it must be at least 6. r6c3 = 7, r4c3 = 5. r5c4 is at least 6, and r5c6 and r5c8 must be greater than r5c7, which is no less than 5; so the 5 in r5 can only be in r5c7. Similarly, the 5 in r6 can only be in r6c4 because the only other position for it would be r6c5, which must be at least 6. You just placed two 5-coins, and with the 5-coin in the arrow in b3, that's all three accounted for. The 5 in r8 must therefore be in r8c9, the 5 in r9 must be in r9c5, and the others can be placed by sudoku.
Now another coin has become constrained. There's an 8-coin in r8c3, and there's another in r4 of b4. If r9c6 were 8, that would force a 7 into r7c4 and another 8 into r7c8. That'd be four 8-coins, so r9c6 = 7. From there, normal sudoku should very nearly finish the puzzle; if you find yourself with a few cells left along the top and right of the puzzle, just 8 cannot go into a coin cell and you should wrap it up.
There are exactly 24 coin cells in the Golden Temple, so that's all of them; there's no other treasure on the map. You can now fill in the numbers 1 through 9 into the "collectible coins found" box in the Quest Progress box.
There's a boss cell to the right of r5. If you're not quite feeling up to that right now, though, you can take a detour: r4c1 is dotted with a cell off the grid. Dotted cells have to contain a digit, even off the grid, so r4c0 must be a 4. And hey, another hidden puzzle!
This is a pretty quick puzzle; if you get the basic idea right, it really just falls straight into place. It's a 4×4 grid, so it must be filled with the digits 1-4. The lines are thermometers, but you can't see any of the bulbs. Different thermometers only share cells at the bulb. Remember that cells outside of the grid can contain 0 or be blank.
Okay. One of those rules is the real key to unlocking this puzzle: different thermometers can only share a cells when the cell is the bulb. (Unfortunately, this isn't stated directly in Sudokuvania's written rules — it's sort of implied, but not really.) So if you see two thermometer lines going into the same cell, even if you can't see what's happening in that cell, either that cell is the bulb and the lines are different thermometers, or the cell links the two lines into a single thermometer.
In this puzzle, there are two cells that that applies to: to the right of r1c4, and to the left of r4c1. The former joins two 3-three thermometers, so it must be a bulb. The latter connects a 1-cell and a 3-cell thermometer, so it may or may not be a bulb.
Since you know where a bulb is, you can place possible digits. On the upper, r1c4 must be 1/2, r2c3 must be 2/3, and r3c4 must be 3/4. ON the lower, r2c4 must be 1/2, r3c3 must be 2/3, and r4c2 must be 3/4. This creates a 1/2 pair in b2, so r2c3 must be 3. That forces r3c3 to be 2 by sudoku, which forces r2c4 to be 1 by the thermometer, making r1c4 = 2 and r1c3 = 4 by sudoku. r3c4 must be 4 by the thermometer, and r4c3 = 1 and r4c4 = 3 by sudoku. r4c2 = 4 by the thermometer, and r4c1 = 2 by sudoku. r4c3 is 1, so its thermometer must grow as it goes to the left and up. Therefore r3c2 = 3 and r3c1 = 1. The last box can be filled by sudoku.
I really enjoy this puzzle, but it has an extremely funky rule that makes it very difficult to wrap your head around. Don't feel bad if you struggle here.
There are two special rules in this puzzle:
This is the most arithmetic-heavy of all the puzzles in Sudokuvania, and it's going to help if I can talk about that concisely. So let's get some notational issues out of the way:
Okay. Start with some general observations:
With all of that in your toolbox, what value does r5c5 actually have? Focus on the three short green lines to the north, northeast, and east. These lines must all have a single segment: their values sum to ≤14.5, ≤19, and ≤14.5, respectively. If r5c5 = 8, its value is 1, and the other values on the line must sum to exactly 9. For this to be true of the north line, the value of r4c5 must be a multiple of 1. The same logic is true of the east line and r5c6. r4c5, r4c6, and r5c6 also form a purple line, which must itself sum to 10; since the other two cells on this line are multiples of 1, r4c6 must be as well. To have values that multiples of 1, these cells must all contain even digits. Under this hypothetical, the 8 is in r5c5, so these cells must be 2/4/6. But that means these cells have a total value of 6, which breaks the 10 line they share! r5c5 must be 4.
Now think that through again with the new value. r5c5 has a value of 0.5, so r4c5 and r5c6 must not be multiples of 1; they must contain odd digits. The sum of those two values is therefore a multiple of 1. To make the purple 10-sum work, r4c6 must have a value that is also a multiple of 1, so its digit must be even. Continuing outward, r3c7 must be odd in order to produce a 0.5 to balance out r5c5. Now come back to the purple line, because there's an easier way to think about its sum: the values must sum to 10, but the values are all half the digit, so the digits must sum to 20. That rules out 1 from both of the odd cells and 2 from r4c6. You won't use this information for awhile, but it's a great tutorial about how to think about the lines.
Does the value of r5c5 change what you know about about how many segments are on each of the green lines and the values in their intersecting cells?
Now consider how the purple lines might break down into segments. You can use SudokuPad's Pen Tool to mark these segments; just draw lines over the puzzle's lines where you know cells are on the same segment. These purple lines are both quite long, but you can tackle them from each end:
Do any of these segments seem especially constrained? The segment including r3c3 to r3c7 also includes three cells in r2. Those cells all see a 2. Therefore, the smallest set that can fill them is 1/3/4; they have a value sum of ≥8. r3c3 and r3c7 are both odd, so they must be at least a 1/3 pair; that is a value sum of ≥2. That's already a sum of 10, so this must be the entire segment, and those minimal sets must all be correct: the r2 stretch is a 1/3/4 set, r3c3 = 1, and r3c7 = 3. And remember that that segment wasn't the end of the line: there must be a two-cell segment in r4c2 and r4c3, although you can't say much about it yet.
You've finally learned something new about the lines you started with. On the northeast green line, the known digits have a value sum of 2. r4c6 must be an even digit, and it cannot be 2/4, so it must be 6/8. If it were 8, that would be a value of 4, and r2c8 would have to be 4; but there is already a 4 in r2. So r4c6 = 6, which makes r2c8 = 5. For the purple line in b5 to sum to 10, the other values on it must sum to 7, which means the digits must sum to 14; they are a 5/9 pair. If r4c5 = 9, that would put a value sum of 5 on the north line, so r3c5 would have to be 5, which it cannot be by sudoku. So r4c5 = 5 and r5c6 = 9, and by the 10-lines, r3c5 = 7 and r5c7 = 5.
What about the other long segment? The value sum of r5c2 and r5c8 is ≥3, and the value sum of the c1 stretch is ≥6. This segment could stretch to include r9c2, but the digit in that cell would have to be 1, which would violate sudoku. Therefore this segment contains exactly these five cells. Either the endpoints are a 1/5 pair and the middle is a 1/2/4 set, or the endpoints are a 1/7 pair and the middle is a 1/2/3 set. What does all this mean for the remainder of this line? There are four cells, and no cell can sum to 10 by itself. So this is either a single 4-cell segment or two 2-cell segments. The three non-intersection cells all see a 1, so they must contain at least 2/3/4, for a value sum of ≥9. The intersection cell, r9c5, also cannot contain 1, so it must have a value of at least 1. But to achieve that, it would have to contain a 2, which would violate sudoku with the 2/3/4 set. So this stretch of the line must be two 2-cell segments.
That tells you something about the south green line. You know that the segment containing r5c5 must also contain r7c6. So if the line were broken into two segments, the outermost two cells would have to be a segment of their own. r8c5 is not an intersection cell, so the digit in it plus the value of r9c5 would have to be 10. But r9c5 is already in a two-cell segment with r9c4, which is also not an intersection cell; whatever value r9c5 has, r9c4 and r8c5 would end up needing the same digit. So the south green line must be a single segment.
The next line to consider is the west green line. r4c3 must be even, but it cannot be 6 by sudoku, and it cannot be 8 because that would put a 6 in r4c2, which would also violate sudoku. So it is 2/4. r5c2 must be odd, and it can only be 1/7 by sudoku. If it were 7, that would put a value sum of 4 on the line; but if r4c3 were 2, r5c4 would have to be 5, which it cannot be, and if r4c3 were 4, r5c4 would have to be 4, which it cannot be. So r5c2 = 1. That puts a value sum of 1 on the line; since the value of r4c3 is 1 or 2, the digit in r5c4 must be 7/8.
Some digits are constrained in b5. On the southwest line, the value sum of the endpoints is ≥3, so r6c4 cannot be greater than 6; it must be 1/2/3. On the southern line, the least possible value sum of the four other cells is ≥3, and 7 is ruled out by sudoku; r6c5 must be 1/2/3. On the southeast line, one of the cells in c7 must be odd, and it must be at least 7 by sudoku; that means the value sum of the line other than r6c6 must be ≥1 + ≥3.5 + ≥1 + ≥1, which is ≥6. So r6c6 is also 1/2/3. Those three deductions mean that r4c4 and r5c4 are a 7/8 pair. But if r5c4 were 7, r4c3 would be 4, so r4c2 would have to be 8; that would make 8 impossible to place in b5. Therefore, r5c4 = 8, r4c3 = 2, r4c2 = 9, and r4c4 = 7. This in turn makes r3c2 = 2.
Sudoku tells you something interesting now. The 1/2/3 set in r6 means that r6c1 must be 4. This settles the ambiguity on this segment that you found earlier: the stretch in c1 must be 1/2/4, and the endpoints must be a 1/5 pair (the 5 being in r8c2). Sudoku puts (in b1) r1c3 = 4, (in b4) r6c3 = 5, and (in b6) r5c8 = 2 Considering the southwest green line again, the endpoints have a value sum of 3, so the others sum to 7; since r6c4 is 1/2/3, r7c3 must be 4/5/6; but it can only be 6, and so r6c4 can only be 1. The only possible way to fill the 2-cell segment in b7 with a 10-sum now is 3/7. Sudoku places them exactly and also gives r7c2 = 4.
You've nearly finished using the information on the lines down here; consider the intersection in r9c5. This cell must be 2/6/8, but if it were 6, r9c4 would have to be 7, which is forbidden by sudoku. If it were 2, r6c5 would be forced to be 3, and r2c5 would be forced to be 1. The least value that could be placed in r8c5, then, would be 6, and the value sum on the line would be ≥10.5 without even including r7c6. So r9c5 = 8, r9c4 = 6, and by sudoku, r9c1 = 9 and r8c3 = 8, which then gives r2c3 = 9 and r5c3 = 7. Returning to the line, the value sum from r5c5, r6c5, and r9c5 is already ≥6.5, sor8c5 cannot be more than 3. This creates a 1/2/3 set in the column.
Time for some sudoku. In c5, r7c5 = 9 and r1c5 = 6. In r3, 4 and 6 are forced into b3; this gives r3c6 = 8 and r3c4 = 9. In c6, the 7 must be on the purple line in b8; but if it were in r7c6, there would be a value sum of 8 on the green line, and it would not be possible to fill r6c5 and r8c5. Therefore r8c6 = 7.
Now you can finally do something with this end of this purple line. Recall that exactly one of the two intersection cells on the southeast green line must be odd. By sudoku, that must be 7/9. That cell cannot be r8c7 because that value (≥3.5) plus the 7 in r8c6 is already more than 10. So the odd intersection must be r7c7, which also cannot be 9; thus r7c7 = 7. By sudoku, r2c7 = 8, which in turn completes b1, b4, and r5. The purple line in b6 cannot take all three cells without exceeding 10, so there must be a 2-cell segment in r4c8 and r5c8; therefore r4c8 = 8. The next segment starts with r6c8 and r7c7; the least value that r6c8 can have is 6, which gives a value sum of 9.5 with just those two cells. Therefore r6c8 = 6, and the segment must continune only into r7c6, which must be 1. The last segment is just r8c6 and r8c7, so r8c7 = 6. On the southeast green line, the current value sum is 8, so r6c6 = 2.
The rest of the puzzle can be solved with sudoku. The boss clue is 6.
Did you think you were done? Well, you're actually pretty close now, I promise. Just one more grid...
To get there, though, you have to solve this little thing. Remember all of the outside the boxes rules?
Got it? You can trace the path of the thermometer around the puzzle. Consider the interesting cells you visit. Ignore all the cells that don't involve any other clues; there's no reason you would ever need to fill them. Just write the interesting clues down in order.
That gives: (top-left whisper) (bottom-left whisper) (center-right whisper / dotted) (right circle) (left circle) (center-left whisper / dotted) (bottom right whisper) (top right whisper)
What's the critical sequence there? The dotted cells and circles must all be filled. So there are at least two digits between the dotted digits. They are either 3/6 or 4/8.
These cells are also connected by the whisper line. Neither pair is 5 apart, so you need at least one digit separating them. Since one of the digits is a 3/4, the adjacent digit to that must be 8/9. That is not 5 away from 6/8, so you also need a low digit to separate those. So the four corner spots on the whisper line must also all be filled.
Which of the pairs would be more constrained? If the pair is 4/8, both cells next to the 4 must be 9. But the thermometer passes through both of these cells, so this cannot possibly work. The pair must be 3/6. The 3 is in the center-right, the 6 in the center-left. The cells next to the 3 must be 8/9. The thermometer visits the bottom-right first, then the top-right. The cells next to the 6 must be 0/1. The thermometer visits the top-left first, then the bottom-left. The right arrow circle is 4, the left arrow circle is 5. Don't forget to fill in the arrows themselves.
And now you've popped through to the final puzzle!
This is a great puzzle. It is very difficult. It is very easy to mess up, and it is even easier to think you've messed up. Take it easy, take it slow.
Okay. Revealing this grid also filled in the last rule in the Learned Rules box. Go ahead and click on that to read the big new rule for this puzzle. The gist is that this puzzle is an 11×11 grid, and you have to put nine non-overlapping 3×3 boxes into it. The normal rules of sudoku apply in all nine boxes and all 11 rows and columns, except that rows and columns don't necessarily contain all nine digits (but digits still cannot repeat within one). The puzzle also features ziplines, whisper lines, arrows, indexing clues, dotted pairs, eyes, and a thermometer. It's a lot.
Rows and columns are numbered from the top-left corner of the grid. The top-left corner is r1c1; the bottom-right corner is r11c11. Numberings don't change just because cells aren't in boxes; they are absolute in the grid. The rules never say this, but this also applies to indexing clues: if r3c6 (the northern indexing clue) contains 7, then r3c7 must be in a box and must contain 6. The indexing does not change just because some cells in that row are not in boxes. Note that this implies that these clues can never index into c10 or c11.
As you're solving this puzzle, you will need to mark your best knowledge of where the boxes are. When you've fully positioned a box, you can use the Pen Tool to draw a thick square around the box. While you're still positioning boxes, however, I recommend using color to mark cells that you know are in a particular box. Adjacent boxes should use different colors so that you can easily tell them apart. You will also want to color in cells that you know are not in any box, just to make it clear that you can ignore them. I will use dark grey for out-of-boxes cells and blue, green, and orange for different boxes. (Normally, I do not suggest using dark grey for coloring because it can be hard to read digits against it, but in this puzzle, there are never digits in these out-of-boxes cells, so it's fine.) When you finish a box, you should outline its border. When you've got the whole grid determined, you can remove all the coloring from the boxes so that you can use colors for normal sudoku purposes.
Your first step is fairly clear: you should try to figure out where some of the boxes go. In order to get started, it will help to identify cells that must be in boxes. There are two ways to do this: the first is some general abstract reasoning about positioning 3×3 boxes inside an 11×11 grid, and the second is by applying the specific rules of Sudokuvania.
In either case, you now know that the boxes must include at least the nine key cells r3c3, r3c6, r3c9, r6c3, r6c6, r6c9, r9c3, r9c6, and r9c9. Moreover, those nine boxes must non-overlapping, because a 3×3 box in that includes r3c3 cannot possibly reach to any of the other key cells. So these boxes are still basically arranged into a loose 3×3 pattern, and we can still unambiguously number them b1, b2, etc. based on which of the key cells each occupies. I will color the boxes like this: blue in the corners, green on the edges, and orange in the center. Go ahead and color the key cells now.
One of the clues imposes a very straightforward constraint on the boxes: the arrow pointing to the south. b8 must fill at least three cells on this arrow. If more than three cells on this arrow were in boxes, the arrow sum would have to be at least 10, and it would not be possible to satisfy. So b5 cannot also include any boxes on this arrow; it must stretch upwards to include r4c6 and r5c6. Similarly, b2 must include r1c6 and r2c6. Those boxes must also grow to the left or right (or both). The northwest arrow says that r5c5 contains the same digit as r6c6. b5 therefore cannot include r5c5, which means that must be the left edge of the box, and it must stretch into c7 and c8. You can now draw the outline of this box. b5 is now forced to stretch rightwards into c11.
There's a clue you were just using that you can now use in a different way: the northwest arrow. r5c5 must contain the same value as r6c6, which means it must be filled. Either b1 or b4 can reach r5c5, but if b1 reached it, it would be impossible for any box to take the other dotted cell in r4c2. So b4 must stretch up two cells to the right and at least one cell up. If b4 stretched two cells up, into r4, it would create the same problem with r4c2 that you just avoided with b1; so it must also stretch into r7. You can draw its outline now. Go ahead and shade the six cells to the left dark grey, though, since they cannot be used by any boxes. The dotted pair in r4 must be in b1, which is also forced to stretch upwards into r2. That's only two columns, though, so it's not quite finished.
Some cells on zipper lines now cannot be in boxes; what does that mean? At least one of the cells in each zipper pair must always be taken, and if one cell is known to be empty, the other must be the same digit as the sum. Above, r2c5 must be in a box and contain the same value as r3c3. b1 cannot reach it; b2 must stretch into c5. Also, r2c4 must also be in a box, because otherwise r3c3 and r4c2 would be the same digit, and they are both in b2. You don't know which box yet, though. Below, r10c5 must be in a box and contain the same digit as r9c9. Because it is the same digit, b7 cannot include it; it must be in b8. If r8c2 was not in a box, then r10c4 would have to be in a box and contain the same digit as r9c9. But that's the same digit that's known to be in r10c5, which is in the same row. So r8c2 must be in a box, and that box must be b7, which must stretch up to r8. As above, you also know that r10c4 must be in a box or else b7 would contain a duplicate digit, but you don't know which box yet.
There are three boxes you haven't done anything with yet. It's tricky to reason about them directly, but there's something you know because of their neighbors: because b2 and b8 have stretched at least one column to the left, they cannot take r2c8 or r10c8 anymore; if those cells are to be in boxes, they must be in b3 and b9, respectively. At least one of r2c8 and r4c10 must be taken. b3 cannot take both of them because of the placement of b5. If b3 takes r2c8 and r4c10 is empty, r2c8 and r3c9 will be the same digit, which will therefore be repeated in b3. So r4c10 cannot be empty; b6 will have to stretch up to take r4c10. If b3 does not take r2c8, it will be empty, and so r4c10 will have to have the same digit as r3c9. Again, this cannot be in b3, so b6 must stretch up to take r4c10. In either case, then, b6 must stretch up to r4. You can draw its outline now.
One last box about which you've learned nothing. The dotted pair in r8 must be in a box, and that can only be b9 now. That puts r8c10 within b9; if r10c8 is not also in a box, r8c10 will necessarily repeat with r9c9. But only b9 can include r10c8. Stretching to include it completes that box. You can shade the cells above, below, and to the right of b9 dark grey. Since r7c10 must be empty, r10c7 cannot be empty; it must be in b8. That pulls b7 over into c4 so that r10c4 can be in a box and b7 doesn't end up with a duplicate of r9c3; that box is now complete as well.
CHECKPOINT 1. You're going to have to start thinking about digits to make progress from here. Here's what the grid should look like:
To break in, focus on b4. r6c3 is an indexing clue, but it's also on a whisper line. The whisper line is adjacent to two cells that see each other, so r6c3 cannot be 4/6 because those cells would have to be the same. Because it is an indexing clue, it gives the column index of the 3 in this row. It cannot be 1/2 because r6c1 and r6c2 are outside of boxes and therefore cannot contain digits. If it were 3, the adjacent cells on the whisper line would have to be 8/9. There is an eye clue to the right of r7. No other boxes but b4 are in this row; b8 can still grow, but it cannot grow above r8 for multiple reasons. For this eye to see three cells, the digits in the bottom row of b4 must all increase as you move to the left. So if r6c3 were 3, that would put an 8/9 in r7c4, but a 9 would immediately break the eye, and an 8 would require the 9 to be in both r7c3 and r5c4. So r6c3 cannot be 3. Similarly, if r6c3 were 7, r7c4 would have to be 1/2, but a 1 would immediately break the eye, and a 2 would force both r7c5 and r5c4 to be 1. So r6c3 must be 8/9.
That has interesting consequences for b8. r7c4 must be low, but it cannot be 1. Therefore the next cell on the whisper line must be at least 7. If b8 stretched into r11, that next cell would be r9c6. r9c6 is on an arrow, together with two other cells in b8. So if r9c6 were 7, the arrow sum would have to be at least 10, which is impossible. b8 must therefore stretch upward so that the next cell on the whisper line is r8c5 instead. You can finally outline that box. r9c6 must 1/2/3/4, but it cannot be 4 because then r7c5 and r7c7 would both have to be 9. It also cannot be 1 because it is an indexing clue, and r9c1 is not in a box. It must be 2/3, and so the 6 in r9 must be in r9c2 or r9c3. Back to the line. r8c7 must be at at least 7, and the adjacent cell on the whisper line is r6c9, which must be 1/2/3/4. r8c7 cannot be 1/2 because it is an indexing clue, and r6c1 and r6c2 are not in a box. Therefore it is 3/4, and so the 9 in r6 must be in r6c3 or r6c4.
That is useful information, because it means that the arrow sum cannot be 9. That means 6 can no longer be on the south arrow. Now the 6 in b8 must be in either r10c5 or r10c7. r10c7 is paired with an empty cell on the zipper line, so if it were 6, r9c9 would also have to be 6. But you've already figured out that the 6 in r9 is in r9c2 or r9c3. So r10c5 = 6. Since r10c5 is paired with an empty cell on the zipper line, r9c3 must also be 6. That means the indexing clue, r9c6, must be 3. It also means that r6c4 = 6, because that's the only remaining spot for a 6 in b4. That forces the 9 in r6 into r6c3. And since that cell has an indexing clue, r6c9 = 3.
One of the clues is now much more constrained. The arrow sum has a three-cell arrow in b8. It must be at least 6, but it cannot be 6 or 9 by sudoku. It cannot be 7 because a set of three cells summing to 7 must be 1/2/4; they cannot include the 3 in r9c6. So r6c6 = 8, and the south arrow cells are 1/3/4. The northwest arrow means that r5c5 = 8. By sudoku and the whisper line constraint, r8c5 = 9 and r8c7 = 8.
You may have noticed something a little surprising in that logic, when you constrained the 6 in b4 using 6s that you'd placed in b7 and b8. It's not usually possible to do that with digits in a single row or column of boxes, but the normal row and column constraints of sudoku can play out very differently when the boxes are offset from each other. That digit I just mentioned created a very powerful constraint: there is now a 6 in c3, c4, and c5, which means that you cannot place any other box that would span exactly those columns. However, b1 and b2 already overlap in these columns, so that doesn't directly restrict anything in the grid. You also have the same situation with the 8s in c5, c6, and c7. In this case, that's immediately useful: those columns are one of the two possible positions for b2. Therefore b2 cannot stretch right into c7 and must instead stretch left into b4. That in turn forces b1 to also stretch left into c1. You can draw the borders of both of those boxes. What about the last? If b3 did not take c7 or c8, then both r2c7 and r2c8 would be empty. r4c10 and r5c10 would therefore both be paired with an empty cell on the zipper line and have to be the same digit. b3 must stretch to include at least c7.
There's a more subtle example of this working horizontally. Consider r7. There are three digits are in r7, all in b4. Those digits must also be somewhere in r5 and r6, for the simple reason that both of those rows include nine cells in boxes and therefore must include all nine digits. They cannot in b4, because they're already in r7 in b4. So they must be in b5 and b6. That means that those digits cannot be in r4 in those boxes. Therefore the three digits in r7 in b4 are exact the same set as the three digits in r4 of b1. There is a dotted pair in r4 in b1. This pair cannot be 4/8 or 3/6, because the 6 and 8 are not in r7 in b4. So the pair must be either 1/2 or 2/4. In either case, there is a 2, and so there must be a 2 in r7 of b4. The 2 cannot be in r7c3 because it would not be visible to the eye, which would mean the eye only sees 2 cells. The 2 cannot be in r7c5 because the arrow clue would put a 6 in r8c4, and there is already a 6 in b7. Therefore the 2 must be in r7c4. For this cell to be visible to the eye, r7c5 must be 1. The arrow then makes r8c4 = 7.
A lot of digits have gotten constrained in b2 now. The 1 must be in c4 because there are already 1s in c5 and c6. The 2 must be in c6 because it cannot be in either of the other two boxes: b8 has 1/3/4 in this column, and b5 has an 8 and a thermometer with two lower digits. The 3 must be in c5 because it cannot be in either b4 or b8 by sudoku. The 4 can't be in c6 but otherwise isn't constrained. The 5 isn't yet meaningfully constrained. The 6 must be in c6 because there are already 6s in c4 and c5. The 7 can't be in c4 but otherwise isn't restricted. The 8 must be in c4 because there are already 8s in c5 and c6. And the 9 can't be in c5 but otherwise isn't constrained.
One of those restrictions has very significant consequences, starting in b7. The paired cells on the zipper line must add to 6, so they are either 2/4 or 1/5. r10c4 now cannot be 1/2 by sudoku, so it must be 4/5, and r8c2 is 1/2. There is a dotted pair in r8, over in b9. This pair cannot be 4/8 because there is an 8 already in r8. It cannot be 1/2 because it would be impossible to place a digit in r8c2. And it cannot be 2/4 then both r8c2 and r8c6 would have to be 1. Therefore, it is a 3/6 pair, and sudoku says that r8c9 = 6 and r8c10 = 3. Now consider the zipper sum. r10c7 is 2/5/7, and it is paired with an empty cell on the zipper line, so it has the same value as r9c9. But the zipper sum must be greater than 3 now, so these cells are actually restricted to 5/7. r10c8, the pair of the 3 on the zipper line in b9, must therefore be 2/4.
There are another couple of "spillover" deductions you can make from those restrictions in b2. In c5, the only cell available for the 2 is now r9c5, which also makes r9c7 = 5. And in c4, the only cell available for the 3 is now r5c4.
It's time to think about the thermometer. When working with thermometers, it's often useful to try to propagate a maximum or minimum constraint along the thermometer. The end of the thermometer, r4c6, can be at most 9. r5c6 can therefore be at most 8, but it cannot be 8 because of r6c6, so it is at most 7. r6c7 is therefore at most 6, but it cannot be 5 or 6 by sudoku. So it is at most 4. r6c8 is therefore at most 3, but it cannot be 3 because of sudoku, so it is at most 2, which is to say, 1/2. This restricts what's possible in b9. If r10c8 were 2, r10c7 and r9c9 would have to be 5. This would make r10c4 = 4, r10c6 = 1, and r8c8 = 4. But now r6c8 = 1 as well, and there is no possible digit that can be placed in r8c8. So r10c8 = 4, which makes r10c7 and r9c9 = 7. Sudoku and the zipper line should let you completely fill r8.
Swing around into b1. The 2 in the dotted pair must now be in r4c2. r4c3 must be 1/4. Remember that this digit must be in r7 in b4. There isn't a 4 already in r7, so if r4c3 were 4, r7c3 would also have to be 4, breaking sudoku. So r4c3 = 1. Now, what can this zipper sum actually be? The 2 is paired with r2c4, which can only be 1/4/8/9, but 8/9 would make the sum greater than 9, and 4 would make the sum 6, which is ruled out of r3c3 by sudoku. So r2c4 = 1 and r3c3 = 3. r2c5 is paired with an empty cell on the zipper line and so must also be 3. Sudoku then puts the 3 in b7 into r10c2.
While you're in b1, see if you can do anything with the eye clue. The 9 cannot be in r2c1 or r2c2 without breaking the eye. It cannot be in r2c3 by sudoku. It cannot be in r4c1 because this digit must also be in r7c3. So in fact the 9 must be in r3 in b1.
It's somewhat astonishing that c3 still can't be placed exactly. There are some clues that you haven't done much with up to now, but it's hard to see how to make use of a whisper line without at least knowing which positions are low and which are high. Maybe the zipper line can help with that. r4c10 cannot be 1/2/3 by sudoku, soit is at least 4, and since it is paired with a cell in a box, the zipper sum must be at least 5. It cannot be 5 because r3c9 is on a whisper line, so it must be at least 6. r3c9 cannot be 6/7/9 by sudoku, so it must be 8. By sudoku, r3c4 = 4, r1c4 = 8, and you can completely fill b7. But more importantly, r5c10 cannot be 8, which means that r2c7 must be in a box and thus that b3 must stretch left into c7.
Back to the whisper line. r4c8 must be at most 3, but it cannot be 1/2 by sudoku, so it is 3. r5c7 must be at least 8, but it cannot be 8 by sudoku, so it is 9. r5c6 must be at least 3 by the thermometer, but it cannot be 3/4 by sudoku, so it is at least 5. r4c6 must be at least 6, then, but it cannot be 6/8/9 by sudoku, so it is 7. That means r5c6 must be 5/6 by the thermometer, and it also cannot be 6 by sudoku, so it is 5. The 5/7 must be in c5 in b2. You can now completely fill b4 by sudoku; don't forget that r4c1 has the same value as r7c3. This also finishes the thermometer, r9, and the rest of b5.
In b6, r4c9 can only be 9 by sudoku, which lets you finish b9. r4c10 cannot be 8 because its paired cell, r2c8, would have to be empty; therefore it is 6 and r2c8 = 2. r5c10 can only be 2/4, but if it were 4, r2c7 would have to be 4, and it can be by sudoku; therefore r5c10 = 2 and r2c7 = 6. The eye above c10 forces r6c10 = 7.
In b1, sudoku puts r2c3 = 4. The remaining digits in r2 are 5/7/8, but sudoku forces the 5 into r2c9. r2c1 and r2c2 must be 7 and 8, respectively, so that the eye can see three cells. The remaining digits in r3 are a 6/9 pair, which forces r3c6 = 2 by sudoku, and that sets r3c2 = 6 by indexing.
The rest of the cells can be filled by sudoku. And the whole map is revealed! Congratulations, you've finished!
Well, you haven't actually triggered the "puzzle fininshed" screen in SudokuPad yet. If you peek down at the Quest Progress box, you'll see that there's one last quest to complete: Solve the Title.
Now, if you look at the title, you might think that you have no idea how to actually solve this. You might even be afraid that it's some kind of fiendish mathematical cryptogram.
It is not. It is very silly.
If that doesn't bring up the game over screen, check the Quest Progress box again to see what you've missed. Make sure you've put the numbers into the Wall Sandwiches boxes and the Collectible Coins Found cells.
My thanks to Skeptical Mario for a wonderful puzzle hunt, to Sven Neumann for SudokuPad, and to the folks at Cracking the Cryptic for introducing me to a world of puzzles beyond dreary machine-generated killer sudokus.
Sudokuvania: Digits of Despair guide v1.1. Written by John McCall from July–October 2025, with occasional clarifications since.Chapter 11. Front Gate (grid 0)
Chapter 12. Battlements (grid 4)
Chapter 13. Skysitter (Battlements boss)
Chapter 14. Battlements (grid 4)
Chapter 15. Quarters (grid 2)
Chapter 16. Keymaster (Quarters boss)
5 3 5
5 5 5
5 8 5 5 3 5
5 5 5
5 8 5 Chapter 17. Quarters (grid 2)
Grid 1 2 3 4 5 6 7 8 9 1 X 2 X 3 X X X Chapter 18. Library (grid 5)
4 1
7 4
4 Chapter 19. Maintenance (grid 6)
Chapter 20. Prism Break (Maintenance boss)
3 4 2 2 5 5 7 3 4 2 2 Chapter 21. Maintenance (grid 6)
Chapter 22. Courtyard (grid 1)
Chapter 23. Quarters (grid 2)
Chapter 24. Serpent (hidden puzzle 2)
9/A button in any of the number-entry modes).
Chapter 25. Battlements (grid 4)
Chapter 26. Library (grid 5)
Chapter 27. Book Constrictor (Library boss)
Otherwise it would touch itself diagonally.Chapter 28. Library (grid 5)
3 8 in the quest progress box.4 5 in the next line of the quest progress box.2 9 in the last line of the quest progress box.Chapter 29. Portal (grid X)
1 4 4 2 4 4 3 2 2 2 3 3 Chapter 30. Elsewhere (grid 7)
Chapter 31. Enlightningment (Elsewhere boss)
Chapter 32. Elsewhere (grid 8)
Along a red line, cells an equal distance from the center cell of the line contain digits that sum to the center cell.
Chapter 33. Outside the boxes: Tutorial
Chapter 34. Dancer (hidden puzzle 1)
Chapter 35. Outside the boxes: Zipline
Chapter 36. Wolf (hidden puzzle 3)
Chapter 37. Hall of Illusions (grid 8)
(1)-4-2-2 would make a thermometer clue false as a whole, but it isn't valid under the puzzle's rule. The puzzle's rule is almost right if you think of thermometers not as a single clue, but as a set of clues between the adjacent pairs of cells on the thermometer. In that case, the logical negation of the clues would require the thermometer to increase as it approaches the bulb, but it would allow non-strict increases, which the puzzle's rule does not. Regardless, none of this matters for any of the actual thermometer clues in the Hall of Illusions, which are always two cells that see each other.1-7-3 would make a whisper line clue false as a whole, but it is invalid under the puzzle's rule. The puzzle's rule is right if you think of whisper line clues as a collection of clues between the pairs of adjacent cells on the line.
Chapter 38. Negative Zone (Hall of Illusions boss)
Chapter 39. Golden Temple (grid 9)
Grid 1 2 3 4 5 6 7 8 9 1 X X X 2 X X X 3 X X X 4 X X X 5 X X X 6 X X X 7 X X X 8 Chapter 40. Spider (hidden puzzle 4)
Chapter 41. Sacrifission Lambda (Golden Temple boss)
Chapter 42. Secret Passage (grid Z)
Chapter 43. Throne Room (grid 10)
Chapter 44. Title
Closing notes