The full logical solve of the hard Fillomino for September 22 to September 28, 2026, 10 × 10. 36 steps, 3 techniques, no guessing. Step through it, or jump to any step below.
This 2 has one cell and needs 1 more. Its only open side is r8c3, so that cell is 2.
This 4 has one cell and needs 3 more. Its only open side is r9c1, so that cell is 4.
This 7 needs 6 more cells, and exactly that many cells around it could join. So r3c5, r4c6, r3c4, r3c6, r5c6 and r5c7 are 7.
This 6 needs 4 more cells, and exactly that many cells around it could join. So r1c4, r1c5, r1c3 and r1c6 are 6.
This 3 has one cell and needs 2 more. Its only open side is r2c2, so that cell is 3.
This 4 has one cell and needs 3 more. Its only open side is r2c7, so that cell is 4.
This 6 has one cell and needs 5 more. Its only open side is r3c8, so that cell is 6.
This 7 needs 6 more cells, and exactly that many cells around it could join. So r8c10, r10c10, r7c10, r8c9, r10c9 and r8c8 are 7.
This 2 has one cell and needs 1 more. Its only open side is r6c8, so that cell is 2.
This 5 has one cell and needs 4 more. Its only open side is r6c9, so that cell is 5.
This 9 has one cell and needs 8 more. Its only open side is r8c6, so that cell is 9.
This 4 has 3 cells and needs 1 more. Its only open side is r10c7, so that cell is 4.
Try each number that fits in r4c3. A 6 leads to a broken rule: a region grows too big or runs out of room. So it is 3.
This 3 has 2 cells and needs 1 more. Its only open side is r5c4, so that cell is 3.
This 5 has one cell and needs 4 more. Its only open side is r6c5, so that cell is 5.
This 6 needs 5 more cells, and exactly that many cells around it could join. So r5c2, r6c3, r5c1, r6c2 and r6c4 are 6.
This 5 has one cell and needs 4 more. Its only open side is r3c1, so that cell is 5.
This 3 has one cell and needs 2 more. Its only open side is r7c1, so that cell is 3.
This 3 has 2 cells and needs 1 more. Its only open side is r8c1, so that cell is 3.
This 3 has 2 cells and needs 1 more. Its only open side is r7c5, so that cell is 3.
This 5 has 4 cells and needs 1 more. Its only open side is r7c6, so that cell is 5.
This 9 has 2 cells and needs 7 more. Its only open side is r9c6, so that cell is 9.
This 4 has 2 cells and needs 2 more. Its only open side is r9c2, so that cell is 4.
Each other number either has no room to fit around r3c2 or would touch a region of its own size. Only 5 is left.
This 5 has 3 cells and needs 2 more. Its only open side is r2c1, so that cell is 5.
This 5 has 4 cells and needs 1 more. Its only open side is r1c1, so that cell is 5.
Each other number either has no room to fit around r8c2 or would touch a region of its own size. Only 4 is left.
This 9 needs 6 more cells, and exactly that many cells around it could join. So r9c5, r10c6, r10c5, r10c4, r10c3 and r9c3 are 9.
Try each number that fits in r3c10. A 2 leads to a broken rule: a region grows too big or runs out of room. So it is 3.
Try each number that fits in r1c9. 3, 4 and 6 each lead to a broken rule. So it is 2.
Each other number either has no room to fit around r2c10 or would touch a region of its own size. Only 3 is left.
Try each number that fits in r4c8. 2, 4 and 5 each lead to a broken rule. So it is 6.
Try each number that fits in r4c9. 2 and 5 each lead to a broken rule. So it is 6.
This 5 needs 2 more cells, and exactly that many cells around it could join. So r5c9 and r5c10 are 5.
This 6 has 5 cells and needs 1 more. Its only open side is r2c8, so that cell is 6.
This 4 has 3 cells and needs 1 more. Its only open side is r1c8, so that cell is 4.