Friandises Geek Sudoku

Sudoku with red, yellow and blue

Sudoku with red, yellow and blue by Phil Preen

Each row and column must contain the digits 1-9. Digits may not repeat within the outlined regions. Digits may not repeat in cells of the same colour (except white). Digits outside the grid in red are colour sums. The value is the sum of all coloured cells (i.e. non-white) in that row / column. Digits outside the grid in black are sandwich clues. The value is the sum of all digits in that row / column between the 1 & 9.

Normal sudoku rules do not apply to this puzzle. Rows and columns each contain the digits 1 to 9. However, traditional boxes are replaced with regions of various sizes. Regions must not include repeated digits. Also, some regions are colored, and there must be no repeated digits with identical color.

Let’s start with an initial observation regarding rows 7, 8, and 9. Together these rows contains three sets of the digits 1 to 9. They include regions L and M, each containing one set of the digits 1 to 9. Therefore, the remaining cells of rows 7, 8, 9 contain one last set of the digits 1 to 9. These remaining cells represent a hidden 9-cell region with no repeat.

The first digit is easy to get: in column 7, cell r4c7 must be a 3 (only colored cell in the column). Now this gives a 7-9 pair at the left of row 4.

The 35 sandwich-clue on row 7 forces a 7-cell sandwich on row 7 with the crusts at the endpoints.

Therefore, cells r7c1 and r7c9 are a 1-9 pair.

Suppose for a moment that cell r7c1 were a 9.
Cell r4c1 would need to be a 7.
Cell r7c1 would be the 9-crust for a 7-cell sandwich on column 1.
This sandwich could not be 1-cell (digit 7 is not available).
This sandwich could not be 2-cell (there is no room to put the 1 crust).
This sandwich could not be 3-cell (the only way to make 7 with three cells is with the triple 1, 2, 4, and digit 1 is not available).

Therefore, cell r7c1 is a 1, which yields a 9 in cell r7c9.

Now if cell r4c1 were a 7, the would be no place for the 9-crust in column 1.

Therefore, cell r4c1 is a 9, r4c2 is a 7, r4c3 is a 1.

On row 5, the 12 red-hint tells that the four colored cells must be 1, 2, 3, 6, or 1, 2, 4, 5.

Suppose for a moment that r5c2 were a 1-crust.
Then the 9-crust on row 5 would need to be cell r5c4.
Impossible because the 30 sandwich-clue on column 4 requires the sandwich in column 4 to be at least 5 cells (8, 7, 6, 5, 4 add up to 30).

Therefore, digit 1 cannot be in cell r5c2. It cannot be in cell r5c1 either. It must be in either of the remaining red cells r5c5, or r5c6.

The 11 red-clue on row 6 indicates that the red cells on row 6 must be 2, 3, 6, or 2, 4, 5 (digit 1 is no longer available for red cells).
Digit 2 is no longer possible for the red cells on row 5.
Cell r5c2 must be a 2.
And the red cells in row 5 and 6 are a 1, 2, 3, 4, 5, 6 sextuple.

Therefore, the two cells in region A must be a 7-8 pair.
The 7 in cell r4c2 disambiguates the pair.

Also, the 33 sandwich-clue on row 6 indicates that the sandwich in row 6 is 6-cells, with digit 2 missing.
Therefore, cell r6c1 is digit 2.
And the sandwich crusts on row 6 are fixed by the 9 in r7c9.

The 38 red-clue plus the 1 and 2 already present in column 1 fix a 3-4 pair in cells r2c1 and r3c1.

The 7 red-clue on column 6 allows to do some cleanup in region J.

Now consider the 9 sandwich-clue and the 9-crust in column 2.
The 1-crust in column 2 cannot be in region O, since there is already a 1 in the hidden region (by our initial observation).

Therefore the 1-crust must be cell r3c2.

The 7 sandwich-clue on column 1 fixes the 5 in cell r5c1.
And cells r8c1 and r9c1 must be a 6-8 pair.

In the hidden region, digits 2 and 7 must be in column 9.

The remaining cells in the hidden region, r7c2, r8c2, and r9c2 must be 3, 4, 5 in some order.
Now cell r2c2 must be a 6.

The 12 red-clue on row 5 fixes a 1-4 pair in cells r5c5 and r5c6, then a 3-6 pair in cells r6c5 and r6c6.

The 7 sandwich-clue on row 5, and the presence of a 3 in cell r4c7, make it impossible for cell r5c5 to be the 1-crust.
Therefore, the 1-crust on row 5 must be cell r5c6.

On row 5, the 9-crust cannot be in column 9, it must be in column 8.
Therefore, cell r5c8 is digit 9, which forces a 7 in cell r5c7.

The 2 sandwich-clue on row 3 forces a 2 and a 9 next to digit 1.
The 30 sandwich-clue on column 4 forces a 1-crust in cell r9c4 and a 2-3 pair in cells r1c4 and r2c4.

The only place for digit 1 in the yellow region is in region C. And the 1 in cell r7c9 fixes it in cell r1c8.
The 15 sandwich in row 1, without a 7 and an 8, must have its 9-crust in cell r1c3.
The digits in the sandwich must be 2, 3, 4, 6 in some order.

Cell r1c9 must be a 5, which disambiguates region O and adjacent cells.

Finally we can fill some options in region E.

The 9-crust in row 2 must be in cell r2c6.
The 9-crust in row 9 must be in cell r9c7, with a 1-crust just on top of it.
The 1-crust in row 8 must be in cell r8c5, with a 7 between the 1 and the 9.
By sudoku, there must be a 1 in r2c5.

The 9 sandwich-clue on column 3 fixes cell r2c3, which disambiguates cells r8c9 and r9c9.
The 13 sandwich-clue on row 9 fixes a 5-8 pair in cells r9c5 and r9c6.

The 30 red-clue on column 9 helps simplify region E.

The 20 red-clue on row 2 fixes a 5 and an 8 in cells r2c8 and r2c9 respectively.

By sudoku, there is a 2-7 pair in cells r7c4 and r7c5.

The rest of the journey is uneventful.

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