![]() We conclude that R1C5 does not contain a 9. In either case, they cannot sum to 9, so we get a contradiction. 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). Now in the middle box these two digits have to be at positions R4C4 and R4C6, so R4C4 + R4C6 = 9. 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. So every way to write 9 as a sum of 2 digits occurs once in these pairs. Now the following pairs of cells all sum to 9, and cannot use the same digits: 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: Apparently R5C4 and R6C4 do not contain a 9, so one of R4C5 and R6C6 contains a 9: There are only 4 ways to write 9 as a sum of 2 digits, so we get a contradiction. ![]() However, no two of these sums can consist of the same digits: this would cause duplicated digits in either a column or a box. ![]() (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: This implies that also the remaining two cells (R8C4 and R9C5) sum to 9. The total sum of all numbers in the box is 45 = 5*9. Now we note the following about the bottom box: the three arrows together with the 9 in the circle have sum of 4*9. This would force a 9 in R7C6, which in turn causes a 9 in R3C5: Now suppose that either R5C4 or R6C4 contains a 9. So the 9 of the lower box is in either R8C4 or R7C6. 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. These observations will become important soon. In this case all the arrowheads would contain the numbers from 1 to 4, and the circles would contain 6 to 9. ![]() Note that all the digits on the arrowheads would be even in this case, and the digits in the circles would be odd. 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. 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. If you're having trouble getting started, I recommend looking at the R5C5 and R6C7 first. (h/t to Penpa for the basic image used to draw the puzzle) Finding this answer will truly let you Escape from Sudoku! I hope you enjoy! 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. Unlike Kropki Sudoku, adjacent cells may have one of these properties without being indicated.īut what about the escape room?!? The rules above are NOT enough to uniquely determine the Sudoku grid. A black (filled) dot between two cells indicates that the digits in those cells are in a 1:2 ratio.
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