# Escape from Sudoku!

An entry in Fortnightly Topic Challenge #50: Escape Rooms

I hesitated to enter this in the Escape Room FTC, but I definitely was inspired by the challenge to create this puzzle, and it is, in some sense, an escape room. This puzzle is definitely a Sudoku, with the following indicators:

• Arrows - As in Arrow Sudoku, the numbers along an arrow sum to the number in the circle. Multiple arrows emanating from a single circle each individually add up to the circled number. The colors are simply for clarity; in the middle square, all four arrows proceed straight through the center to the opposite side.
• Dots - As in Kropki Sudoku, a white (unfilled) dot between two cells indicates that the digits in those cells are consecutive. A black (filled) dot between two cells indicates that the digits in those cells are in a 1:2 ratio. Unlike Kropki Sudoku, adjacent cells may have one of these properties without being indicated.

But what about the escape room?!? The rules above are NOT enough to uniquely determine the Sudoku grid. The answer to this puzzle is the completed Sudoku grid for which there is EXACTLY ONE square on the outer border of the grid that can be reached from the center square (grey dot) strictly via an orthogonally connected path consisting of only odd digits. Finding this answer will truly let you Escape from Sudoku! I hope you enjoy! (h/t to Penpa for the basic image used to draw the puzzle)

HINT

This is a tough one, I think. If you're having trouble getting started, I recommend looking at the R5C5 and R6C7 first.

• I assumed this was like an "odd snake" sudoku, where a snake had its head on the center and its tail somewhere on the edge, but that breaks at the first hurdle. The center square MUST be connected to two odd digits if it is odd itself. Which assumption is false: the path doesn't split, the snake's head is on the center, or the center is odd? Mar 4, 2021 at 22:13
• The assumption that there is a single "odd snake" is incorrect; there may be multiple paths of odd digits from the center to an edge square, as well as odd "alcoves" hanging off these paths. But there can only be one such edge square. Hope that helps! Mar 4, 2021 at 22:30
• For anyone who's interested, here's the f-puzzles link I plan to use to solve the puzzle. (Also, thanks @Jeremy !) Mar 4, 2021 at 22:59

## Step-by-step solution:

We start by looking at the

middle box of the puzzle. The center of this box (R5C5) needs to be odd by the escape room condition (actually also without this condition this would need to be the case, but the argument would be more complicated), and it can be at most 5, since the box contains at least 4 digits which are larger. It turns out that if the cell would contain a 3, then the it is impossible to complete the rest of the arrows in this box without duplicates, so R5C5 is either a 1 or a 5.
If it is a 1, the sums of the arrows would be 1+2 = 3, 1+4 = 5, 1+6 = 7, 1+8 = 9. Note that all the digits on the arrowheads would be even in this case, and the digits in the circles would be odd.
If it is a 5, the sums would be 1+5 = 6, 2+5 = 7, 3+5 = 8, 4+5 = 9. In this case all the arrowheads would contain the numbers from 1 to 4, and the circles would contain 6 to 9. These observations will become important soon. Next we note that R9C5 cannot contain a 9: if it did, then R8C5 would contain an 8, and R9C4 + R8C5 would be at least 9, but the 9 is already placed in this box. So the 9 of the lower box is in either R8C4 or R7C6.
Now suppose that either R5C4 or R6C4 contains a 9. This would force a 9 in R7C6, which in turn causes a 9 in R3C5: Now we note the following about the bottom box: the three arrows together with the 9 in the circle have sum of 4*9. The total sum of all numbers in the box is 45 = 5*9. This implies that also the remaining two cells (R8C4 and R9C5) sum to 9. (Similar observations can be applied to each of the boxes with one circle with three arrows, if the circle contains a 9.) In particular, we have the following sums of 2 cells are all equal to 9:
R1C4 + R2C4 = R1C5 + R2C5 = R7C4 + R7C5 = R8C4 + R9C5 = R9C4 + R8C5 = 9
However, no two of these sums can consist of the same digits: this would cause duplicated digits in either a column or a box. There are only 4 ways to write 9 as a sum of 2 digits, so we get a contradiction. Apparently R5C4 and R6C4 do not contain a 9, so one of R4C5 and R6C6 contains a 9: Next we assume that R1C5 contains a 9. In the box on the right, the 9 now has to be in R6C7 (note that it cannot be in R4C8 by the black circle), and this forces the 9 of the middle box to be in R4C5: Now the following pairs of cells all sum to 9, and cannot use the same digits:
R5C2+R5C3 = R4C7+R5C7 = R4C8+R5C9 = R5C8+R4C9 = 9.
So every way to write 9 as a sum of 2 digits occurs once in these pairs. Also R6C8+R6C9 is 9, and its digits are different from each of the last three pairs, so R6C8, R6C9 should contain the same digits as R5C2, R5C3. Now in the middle box these two digits have to be at positions R4C4 and R4C6, so R4C4 + R4C6 = 9. However, both these cells contain arrowheads, and we have already seen that such cells in the middle region are either all even (if R5C5 = 1) or all at most 4 (if R5C5 = 5). In either case, they cannot sum to 9, so we get a contradiction. We conclude that R1C5 does not contain a 9.
So we know that neither R5C4 nor R5C1 contains a 9, so the 9 of row 5 needs to be in R5C9: We finally filled in a digit, just 80 more to go!

Next we focus on R6C7. If this cell would contain at most a 6, then the total sum of all cells on the arrows leaving this cell would be 4*6 = 24, and the sum of the remaining 2 cells in the right most box would need to be 45-24 = 21, which is too large. So R6C7 contains at least a 7, by the same argument this is also the case for R3C5 and R7C6.
Now we look at the parity of the sum of all cells in row 5. This sum is 45, which is odd. Furthermore, we have that R5C1+R5C2+R5C3 = 2*R5C1 is even, and also R5C4+R5C5+R5C6 needs to be even. Since R5C9 = 9 is odd, we find that R5C7+R5C8 needs to be even. So these cells are either both odd or both even.
So how can we use this to find R6C7? Well, if R6C7 would be an 8, then the sums in this box would be 8 = 1+7, 8 = 2+6 and 8 = 3+5, so the last remaining cell R4C8 would be a 4. This implies that R5C8 needs to be 2, and therefore R4C9 = 6: However, by the parity reasoning from before this implies that also R5C7 needs to be even. But there are no remaining even digits we can put there!
We conclude that R6C7 cannot be an 8. The only remaining possibility is that it is a 7. This also allows us to fill in the 8 in the box, and complete one of the arrows. Furthermore the parity constraint gives some more information: Lets look at the bottom box now,

in particular at R7C6. This cell also contains at least a 7. Suppose that it is exactly a 7, then R8C4 and R9C5 would contain the 8 and 9 of the box. This would however force R8C5 to be at least 7, which contradicts the fact that R9C4+R8C5 = 7. So R7C6 is not 7.
Next suppose that R7C6 = 9. In this case we get that R4C5 should also be 9, and R3C4 should be 9: Now we have the following pairs summing to 9:
R5C5+R6C5 = R7C4+R7C5 = R8C4+R9C5 = R9C4+R8C5 = R8C6+R9C6 = 9
We can now make an argument which is very similar to one we have already seen before: The pairs R5C5+R6C5 and R8C6+R9C6 both need to be different from the three other pairs, and therefore we see that R5C5, R6C5 and R8C6, R9C6 contain the same digits. These digits are now also forced to go into R1C4 and R2C4, so R1C4 + R2C4 = 9. This would however imply that R3C5 contains 9 by the green arrow, but this is clearly impossible.
So we conclude that R7C6 is neither a 7 nor a 9, and therefore it is an 8. This implies that the digits in the bottom box which are not on an arrow, are exactly 4 and 9. This allows us to get the following: Now we can use another parity argument, this time using column 5. The sum of the top 6 cells in this column is even, since it contains two complete arrows with circles. Since R9C5 is even and R8C5 is odd, we see that R7C5 needs to be even to make the sum of the whole column odd. This gives the following information: Next we look at row 5 again.

The first six cells in this row contain two arrows, and these should be formed by the digits 1, 3, 5, 7, 8 and one of 2 and 6. There are only two ways to make these into two valid sums: 8 = 1+7 and 5 = 2+3, or 8 = 3+5 and 7 = 1+6. The first of these options can however not be used: since R5C5 is a 1 or a 5, the orange arrow in the middle would need to correspond to 8 = 1+7, however, the 7 does not fit in R5C6, since it is blocked by R8C6 and R9C6. We conclude that the arrows in row 5 correspond to 8 = 3+5 and 7 = 1+6.
This implies that R5C7 contains the remaining digit of this row, a 2: Next we want to figure out which sum corresponds to which arrow in row 5. First suppose that the orange arrow in the middle corresponds to 8 = 3+5, like this: But where do the 1 and 6 of the middle box go now? They should be in one arrow, on opposite sides of the central 5. However, the 1 and 6 in R6C8 and R6C9 always block this. We get a contradiction.
We find that the middle arrow corresponds to the sum 7 = 1+6 instead: Now we can make some progress in the middle:

In the middle box there is only one position where 9 = 1+8 arrow can go. This also forces the position of the 9 in the top box, and it forces R3C5 (of which we already know it is at least 7) to be exactly 7: Next the top box can be filled out further, and this also gives a lot of information about the bottom box. Now the middle box can be completed. For the left box, we know for each of its rows which digits it contains. Using this information, there is only one way to satisfy all the arrow conditions. We have completed two boxes, we are really getting somewhere!

Consider R6C8, we know that there is a 1 or 6 here. However, if there were a 1 there, then there would need to be a 2 below it, which is impossible. So we get: There is only one way to complete this arrow in the bottom right. To continue, we look at R3C7.

Suppose that this cell does not contain a 3. The only other place in this column where a 3 might go is at the top, R1C7. This would give us some of the neighboring cells. However, now there are no options remaining for R3C7, so we get a contradiction. We conclude that R3C7 does contain a 3. And with that, we have filled over half of the cells!

Note that there cannot be a 2 part of the kropki clue in the bottom right corner: the 1 and 3 are already used in this box. This leaves two positions for the 2 in this right bottom box. In the top right corner, we see that R2C7 can only be 6 or 8 using the kropki clue, in particular this value is always even. We will also need some escape room logic after this, so lets just mark all odd numbers: Now consider column 8. The top 3 digits in this column cannot all be odd: this would create a path from the center to the boundary, and furthermore the remaining odd number would connect another boundary cell to this path, violating the escape room condition. Since 2 is the only remaining even number in this column, we see that 2 needs to be somewhere in these top 3 cells. This leaves one possibility for the 2 in the lower right corner. Now there are only 2 positions for the 6 and 8 to go in this lower right box. This in turn leaves one position for the 9, which forces the 8 to be next to it. Using this all nines can be found. By finding that R2C7 needs to be an 8, also all fours can be found. The result is as follows: We continue by looking at the top of the puzzle:

The 6 and 8 in the top left can be used to see that there are sixes in R2C4 and R3C9. This yields (with some shading again, since we need escape room rules soon): Next we look where the 2 in row 3 can go. Suppose it goes in R3C8, then the sum of R3C1 and R3C2 would be 1 + 5 = 6. This cannot be extended to 6 or 8 by adding the value of R2C3, so the arrow condition cannot be met. This means that there cannot be a 2 in R3C8.
Suppose the 2 goes in R3C1 instead. Now 3 remaining digits in the top left corner would all need to be odd, which would connect two boundary cells with the center, violating the escape room condition.
The only remaining option is that the 2 goes in R3C2. Now some standard sudoku logic can be used to make some progress on the bottom part of the puzzle. Almost done!

Some more progress can be made at the top of the puzzle: And the final 4 cells are resolved by the arrow rule! And that finishes the solution of this great puzzle!

• You got it...this was a joy to read! I missed the logic where you put the 2 in the bottom-right square, very nice. I hope you enjoyed :-) Mar 5, 2021 at 17:21