![]() 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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