The full logical solve of the hard Fillomino for September 1 to September 7, 2026, 10 × 10. 44 steps, 3 techniques, no guessing. Step through it, or jump to any step below.
This 9 has one cell and needs 8 more. Its only open side is r1c5, so that cell is 9.
This 7 has one cell and needs 6 more. Its only open side is r4c2, so that cell is 7.
This 6 has one cell and needs 5 more. Its only open side is r10c1, so that cell is 6.
This 6 has 2 cells and needs 4 more. Its only open side is r9c1, so that cell is 6.
This 6 has 3 cells and needs 3 more. Its only open side is r8c1, so that cell is 6.
This 7 has one cell and needs 6 more. Its only open side is r9c3, so that cell is 7.
This 4 has one cell and needs 3 more. Its only open side is r10c4, so that cell is 4.
This 2 has one cell and needs 1 more. Its only open side is r9c7, so that cell is 2.
Each other number either has no room to fit around r1c2 or would touch a region of its own size. Only 7 is left.
Try each number that fits in r10c5. 1 and 5 each lead to a broken rule. So it is 4.
This 4 has 3 cells and needs 1 more. Its only open side is r9c4, so that cell is 4.
This 7 has 2 cells and needs 5 more. Its only open side is r8c3, so that cell is 7.
This 7 has 3 cells and needs 4 more. Its only open side is r8c4, so that cell is 7.
This 5 has one cell and needs 4 more. Its only open side is r9c6, so that cell is 5.
Try each number that fits in r7c4. 3, 6 and 9 each lead to a broken rule. So it is 7.
Try each number that fits in r5c4. 1, 7 and 9 each lead to a broken rule. So it is 6.
Try each number that fits in r2c3. A 6 leads to a broken rule: a region grows too big or runs out of room. So it is 7.
Try each number that fits in r6c4. 1, 6 and 9 each lead to a broken rule. So it is 7.
This 9 has one cell and needs 8 more. Its only open side is r6c2, so that cell is 9.
This 7 has 6 cells and needs 1 more. Its only open side is r8c5, so that cell is 7.
This 5 has 3 cells and needs 2 more. Its only open side is r8c6, so that cell is 5.
Each other number either has no room to fit around r7c2 or would touch a region of its own size. Only 9 is left.
This 9 needs 6 more cells, and exactly that many cells around it could join. So r5c2, r6c1, r5c1, r4c1, r3c1 and r2c1 are 9.
Try each number that fits in r9c10. 2, 3 and 7 each lead to a broken rule. So it is 4.
Try each number that fits in r10c10. 1, 2 and 7 each lead to a broken rule. So it is 4.
Try each number that fits in r10c9. A 7 leads to a broken rule: a region grows too big or runs out of room. So it is 4.
This 7 has one cell and needs 6 more. Its only open side is r9c8, so that cell is 7.
This 7 has 2 cells and needs 5 more. Its only open side is r8c8, so that cell is 7.
This 3 has one cell and needs 2 more. Its only open side is r7c9, so that cell is 3.
This 7 needs 4 more cells, and exactly that many cells around it could join. So r7c8, r8c7, r6c8 and r7c7 are 7.
This 2 has one cell and needs 1 more. The only open side it can grow through is r5c7, so that cell is 2.
This 5 has one cell and needs 4 more. Its only open side is r5c9, so that cell is 5.
This 2 has one cell and needs 1 more. Its only open side is r6c10, so that cell is 2.
This 3 has 2 cells and needs 1 more. Its only open side is r7c10, so that cell is 3.
Try each number that fits in r4c6. 1, 3 and 6 each lead to a broken rule. So it is 7.
Try each number that fits in r5c6. 1 and 6 each lead to a broken rule. So it is 7.
This 6 has 4 cells and needs 2 more. Its only open side is r3c4, so that cell is 6.
This 9 needs 7 more cells, and exactly that many cells around it could join. So r1c6, r2c5, r1c7, r3c5, r2c7, r2c8 and r3c7 are 9.
This 7 has one cell and needs 6 more. Its only open side is r1c9, so that cell is 7.
This 7 has 6 cells and needs 1 more. Its only open side is r6c6, so that cell is 7.
This 7 has one cell and needs 6 more. Its only open side is r3c9, so that cell is 7.
Each other number either has no room to fit around r1c10 or would touch a region of its own size. Only 7 is left.
This 7 has 6 cells and needs 1 more. Its only open side is r3c10, so that cell is 7.
This 5 needs 2 more cells, and exactly that many cells around it could join. So r4c10 and r5c10 are 5.