Christmas tree

Mark Goodliffe and Simon Anthony from cracking the cryptic are doing a fantastic job of making us discover the fascinating world of advanced sudoku.

Here is a treat for the season: normal sudoku rules apply, and identical digits cannot be a chess knight’s move apart. The grey square marks an even digit. Any two cages that share an edge and whose totals have a difference of 1 have a white dot between them, and any two cells that share an edge whose totals have a ratio of 2:1 have a black dot between them; there is a negative constraint so two cages touching each other without a dot do not have totals with a difference of 1 or a ratio of 2:1.

Please try the puzzle here.

Possible solution

Cages A, B, F completely fill box 8, therefore add up to 45. The white dots tell that cage B and cage F must each total to cage A plus or minus 1. The only possibility for A, B and F together to total to a multiple of 3 is if they are x, x-1, x+1, in some order. And subsequently x=15.

Therefore, cage A is 15.

Cage B must be 14 or 16. If B were 16, then cage G would be at least 13, and cages C, D, E respectively at least 15, 14, 13, with a total of 42. Cell r7c7 could be no more than 2, making it impossible to fill cage G.

Therefore, cage B is 14 and cage F is 16.

Cage C must be 13 or 15. If C were 15, the C, D, E would again add up to 42 minimum, making it impossible to fill cage G.

Therefore, cage C is 13. Similarly cage D is 12.

Cage H is either 14 or 16. If it were 16, it would contain the digits 7 and 9. Cage I would be either 17 or 15, thus containing the digits 9 and 8, 9 and 6, or 8 and 7. Impossible by sudoku.

Therefore, cage H is 14, and cage J is 7 (two-cell cage J could not be 28).

Cage K must be 8, with an absolute minimum of 15 for 5-cell J+K, comprised exactly of the digits 1, 2, 3, 4, 5.

Now cage H must contain the digits 8 and 6. The digits 7 and 9 complete box 7.

Now the possibilities for cage I are to be a 13-cage (with digits 9 and 4, or 7 and 6) or a 15-cage (with digits 9 and 6, or 7 and 8). By sudoku, only the first possibility is legit.

Therefore, cage I is 13 with the digits 9 and 4.

By sudoku, cage F cannot contain a 7 or a 9. It must contain a digit 8, otherwise it could not add up to 16.

By sudoku, cell r7c7 in cage G cannot be 7, 8 or 9, and must therefore be 6 for cage G to reach its minimum value of 15.

Conversely, cage F without a 6 now must contain the digits 8, 3, 5 in some order.

Also, now that cage G contains the digits 6 and 9, the only possibility for cage C is the pair of digits 5 and 8.

Cage D cannot contain the digits 5 and 8. The only possibility for a 12-cage is the pair of digits 3 and 9.

The digit 3 no longer being available, we know that cell r7c9 must contain the digit 1, cage E is a 13-cage, and contains the digits 2, 4, 7.

In turn this simplifies the situation of cages J and K.

Finally, this further restricts the possibilities of digits for cages A and B.

By knight move, cell r8c6 cannot be 9. The only possibility for 14-cage B is the triple of digits 7, 1, 6.

The only possibility for 15-cage A is the triple of digits 9, 2, 4.

By sudoku and knight move from cage H, digit 6 cannot be contained in cell r9c5. By knight move from cell r7c7, digit 6 cannot be contained in cell r9c6. Thus, the only place for digit 6 in cage B is in cell r8c6.

Further sudoku and knight move considerations yield quite a few digits.

Cage L must be 15 or 17. If it were 17, cell r6c7 would have to contain a 9 (impossible by sudoku) or an 8 (impossible with the digit 8 one knight move away).

Therefore, cage L is 15. It contains the pair of digits 8 and 7, the 8 above the 7.

Digit 9 in box 5 is now restricted: it must belong to cage M. M being a 14-cage or a 16-cage, the possibilities are either the triple 9, 4, 1, or the triple 9, 6, 1.

By sudoku and knight move, cell r6c5 must be 1. Then cell r5c5 must be 6. Finally cell r4c5 must be 9 and M is indeed a 16-cage.

By sudoku, digit 4 must be in cell r4c6, and 2, 3, 5 complete box 5.

By sudoku and knight move, cage O is restricted to either the pair of digits 3 and 2, or the pair of digits 3 and 5. Now is time to use the negative constraint (there is no dot between cage O and cage M) and eliminate the possibility of a 3 and 5 pair (adding up to 8, i.e. half of cage M).

Therefore, cage O contains the pair of digits 3 and 2.

N must be an 8-cage, containing either the triple of digits 1, 2, 5, or 1, 3, 4, which can be simplified by sudoku and knight move.

By sudoku, cell r2c5 must contain 4, 5 or 8 and the rule set requires it to be even. It must be the digit 8 (there is already a digit 4 a knight move away).

Digits 1 and 7 in cage B can be disambiguated.

We fill the possibilities for the remaining cells in row 6.

Then simplify by sudoku and knight move rules.

Do the same for row 5, for row 4, finally for box 2.

Disambiguate the pair 1 and 3 in cage K, and the pair 2 and 5 in cage J.

The only possibility for cell r1c2 (naked single) is digit 8 – by sudoko and knight move with the nearby 6-7 pair in box 2.

By sudoku, cell r3c8 must also contain an 8.

The only place for a 9 in box 4 is r5c1. By sudoku and knight move, 9 in box 1 must be in cell r2c2.

The 9 in box 3 must be in cell r3c9. This disambiguates the 3-9 pair in cage D.

Fill in the possibilities for row 3. Simplify by sudoku and knight move rules.

The 6-7 pair in column 2 looks at cell r5c3, which therefore must be a 3.

Simplify by sudoku and knight move rules. Use the 4-7 pair in row 5.

Digit 1 in box 1 must be in cell r2c3.

Fill in the possibilities for row 2. Simplify by sudoku and knight move rules.

The 6-7 pair in column 2 looks at cell r2c1, making digit 6 impossible.

This gives a 2-3-4-7 quadruple in row 2.

Disambiguate the 6-7 pair in column 4, then the 6-7 pair in column 3.

Continue with sudoku and knight move rules.

Use the 1-3-4-6 quadruple in row 1.

Box N is now complete with the triple of digits 1, 4, 3.

Only place for digit 6 in row 3 is in cell r3c1.

Only place for digit 6 in column 9 is in cell r1c9.

Complete the puzzle.

Merry Christmas!