Pi Sudoku 2020 solution

• By using clues in following order

  • 3 in box 3, box 7

  • 3, 14, 15 clues in box 1

  • 9, 14 clue in box 3

  • Thermometer in the middle

• We can have a board like this.

unique rectangle

• If we put 9 in r2c5, cells r2c4, r2c6, r5c4, r5c6 all become 67 pairs.

  • And none of the clues is interacting with those cells, so we can't distinguish them if 9 is in r2c5, which implies there are two solutions.

  • Therefore r2c6 is 9. (By assuming that this puzzle has a unique solution)

• No matter which version you are solving, the next step are both box 7.

  • This write up is for old version, for new version, some deduction can be further simplified.

• In box 7, yellow cells sum to 15, therefore red cells sum to 13.

• 14 must be 5 + 9 or 6 + 8.

• Can 7 be part of 15 cells? ==> No

  • 15 = 7 + 6 + 2 ==> r7c1 is broken

  • 15 = 7 + 5 + 3 ==> 3 is already used

• Therefore, 13 = 7 + ? + ?

  • 13 = 7 + 5 + 1 = 7 + 4 + 2

• 15 cells

  • 1 + 5 + 9

  • 1 + 6 + 8 ( r7c1 is broken)

  • 2 + 4 + 9

  • 2 + 5 + 8 (14 cells are broken)

  • 4 + 5 + 6 (14 cells are broken)

• 8 can only go to r8c1.

• Using the quadruple in column 3, we can go a little further.

• Now the 25 pair in box 7 is interesting

  • If it is 5, we have 13 = 1 + 5 + 7, therefore r4c3 is 2

  • If it is 2, r4c3 also has to be 2

• Therefore, r4c3 must be 2.

  • Another way to think about this is, if r4c3 is 1, r9c3 has no value.

• If we put 4 in cell r9c3, clue 26 will be broken.

• However, this step is quite unreasonable...

• I don't know if there are other way to solve the puzzle. Please let me know if you found one.

• And clue 26 will be broken again if we put 5 in r7c8.