Joker Sudoku Solution

Because the clues might contain Joker, we can't trust all of them.

• If you trust all the clues in the cages, a contradiction will be found very quickly.

7 cages are interesting, because, without a Joker, a 4 cell cage must sum to 10 or more. No matter 7 is Joker or not, these 7 cages must sum to a single digit value. Therefore, the 7 cages must have a Joker.

Because there is no repeat digit in a cage, we must have 7=1+2+3+J, where J denotes Joker.

Therefore, 1, 2, 3, and 7 are not Joker.

Now we can trust the "32+1" clue, and those cages must sum to 33, which must contain a Joker.

• 32+1 = J+7+8+9

• Therefore 8 and 9 are not Joker either. Joker is one of 4, 5 and 6.

Now we can also trust 19, 17 cages. And 17 cage must be one of

• 17 = 8 + 9

• 17 = 8 + J

• 17 = J + 9

Putting these together, we should have a board like this.

Now, in row3, where can 7 go?

• It can't be in box 1, because they are occupied by J89, J89, 123J respectively.

• It can't be in box 3, because we have J789 quadruple in 32+1 cage.

• Therefore, 7 must be in one of r3c5, r3c6, which is part of the 5 cage.

Now we have 5 = 7 + ?, and the only explanation is "5 is the Joker"!

Therefore we can't trust 5 cages and 15 cages.

Because 5 is Joker

• We already have 5 in the clues of 15 and 5 cages on row 7

• We already use 5 in 1235 quadruple in box 7

• The only cells 5 can go on row 7 is in the 32+1 cage.

Also, on column 7, we have 5789 quadruple in the bottom, therefore, 5 in column 7 must be in box 9.

• I use blue to marked cells that cannot contain 5.

Where can 1 go in box 9? It can't be in 17 nor 19 cage, so it must be in the 5 cage.

• This also implies that 4 cannot be in 5 cage, otherwise 5 = 1 + 4, and Joker (5) is representing itself in the equation.

19 = 4 + ? + ? + ?, where ? are three of 2, 3, 6, 7, 8. The only combination that works is 2, 4, 6, 7.

• 2 must be included, otherwise 3+4+6+7 = 20 > 19

• 19 = 2 + 4 + ? + ?, where ? + ? = 13, 6 + 7 is the only way.

This also gives us,

• 5 = 1 + 3 (5 is Joker, and it represents 4)

• 7 is locked in r6c7

The only cells 2 can go in row 7 are in the 15 cage.

• 15 = 2 + ?, which implies that "2 + ?" is a two digits value

• Therefore ? must be 9

We can lock the 9 in 32+1 cage (r6c6), and a 89 double in r8c3, r9c3.

Now we have done the most difficult part of the puzzle.