The full logical solve of the hard Bridges for September 1 to September 7, 2026, 14 × 14. 32 steps, 6 techniques, no guessing. Step through it, or jump to any step below.
The 3 at row 3, column 3 needs 3, and its neighbors can take 3 at most. So it takes all of them: two bridges to the 5 at row 1, column 3 and one bridge to the 1 at row 3, column 5.
The 2 at row 5, column 7 needs 2 more, and only the 7 at row 9, column 7 can still take bridges. So it gets two bridges to the 7 at row 9, column 7.
The 5 at row 6, column 4 needs 5, and its neighbors can take 5 at most. So it takes all of them: two bridges to the 2 at row 6, column 2, two bridges to the 5 at row 6, column 6, and one bridge to the 1 at row 8, column 4.
The 3 at row 9, column 2 needs 3, and its neighbors can take 3 at most. So it takes all of them: two bridges to the 7 at row 9, column 7 and one bridge to the 1 at row 13, column 2.
The 1 at row 11, column 13 needs 1 more, and only the 2 at row 6, column 13 can still take bridges. So it gets one bridge to the 2 at row 6, column 13.
The 2 at row 13, column 11 needs 2 more, and only the 3 at row 13, column 7 can still take bridges. So it gets two bridges to the 3 at row 13, column 7.
The 7 at row 9, column 7 needs 7, and its neighbors can take 7 at most. So it takes all of them: two bridges to the 5 at row 9, column 12 and one bridge to the 3 at row 13, column 7.
The 6 at row 14, column 12 needs 6, and its neighbors can take 6 at most. So it takes all of them: two bridges to the 7 at row 12, column 12, two bridges to the 2 at row 14, column 8, and two bridges to the 2 at row 14, column 14.
The 2 at row 12, column 14 needs 2 more, and only the 7 at row 12, column 12 can still take bridges. So it gets two bridges to the 7 at row 12, column 12.
The 5 at row 1, column 3 needs 5, one less than the 6 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 2 at row 1, column 1 and at least one bridge to the 4 at row 1, column 6.
The 2 at row 1, column 13 needs 2, one less than the 3 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 2 at row 1, column 9.
The 4 at row 1, column 6 needs 4, one less than the 5 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 7 at row 4, column 6.
The 7 at row 4, column 6 needs 7, one less than the 8 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 3 at row 4, column 1, at least one bridge to the 5 at row 4, column 8, and at least one bridge to the 5 at row 6, column 6.
The 3 at row 8, column 1 needs 3, one less than the 4 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 3 at row 4, column 1 and at least one bridge to the 2 at row 12, column 1.
The 5 at row 9, column 12 needs 5, one less than the 6 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 4 at row 4, column 12 and at least one bridge to the 7 at row 12, column 12.
The 7 at row 12, column 12 needs 7, one less than the 8 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 2 at row 12, column 9.
The bridge between the 2 at row 5, column 7 and the 7 at row 9, column 7 blocks the way from the 5 at row 6, column 6 to the 2 at row 6, column 13. The 5 at row 6, column 6 needs 5, and its neighbors can take 5 at most. So it takes all of them: a second bridge to the 7 at row 4, column 6 and one bridge to the 1 at row 8, column 6.
The 2 at row 6, column 13 needs 1 more, and only the 2 at row 1, column 13 can still take bridges. So it gets one bridge to the 2 at row 1, column 13.
The 3 at row 8, column 8 needs 3, one less than the 4 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 5 at row 4, column 8 and at least one bridge to the 2 at row 8, column 10.
The bridge between the 3 at row 8, column 8 and the 2 at row 8, column 10 blocks the way from the 2 at row 1, column 9 to the 2 at row 12, column 9. The 2 at row 1, column 9 needs 1 more, and only the 4 at row 1, column 6 can still take bridges. So it gets one bridge to the 4 at row 1, column 6.
The bridge between the 3 at row 9, column 2 and the 1 at row 13, column 2 blocks the way from the 2 at row 12, column 1 to the 2 at row 12, column 9. The 2 at row 12, column 1 needs 1 more, and only the 3 at row 8, column 1 can still take bridges. So it gets a second bridge to the 3 at row 8, column 1.
The 2 at row 12, column 9 needs 1 more, and only the 7 at row 12, column 12 can still take bridges. So it gets a second bridge to the 7 at row 12, column 12.
The 5 at row 9, column 12 needs 1 more, and only the 4 at row 4, column 12 can still take bridges. So it gets a second bridge to the 4 at row 4, column 12.
A double bridge between the 2 at row 2, column 12 and the 4 at row 4, column 12 would close off a group of 15 full islands, so they share one bridge at most. The 2 at row 2, column 12 needs 2, one less than the 3 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 3 at row 2, column 10.
The 4 at row 4, column 12 needs 4, one less than the 5 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 5 at row 4, column 8.
The bridge between the 5 at row 4, column 8 and the 4 at row 4, column 12 blocks the way from the 2 at row 8, column 10 to the 3 at row 2, column 10. The 2 at row 8, column 10 needs 1 more, and only the 3 at row 8, column 8 can still take bridges. So it gets a second bridge to the 3 at row 8, column 8.
The 3 at row 2, column 10 needs 3, one less than the 4 its neighbors could take. Even if the others give all they can, it still gets at least one bridge to the 2 at row 2, column 8.
Suppose the 2 at row 1, column 1 and the 5 at row 1, column 3 shared only one bridge. Following the rules from there breaks one, so they share two.
The 4 at row 1, column 6 needs 1 more, and only the 7 at row 4, column 6 can still take bridges. So it gets a second bridge to the 7 at row 4, column 6.
The 3 at row 4, column 1 needs 1 more, and only the 7 at row 4, column 6 can still take bridges. So it gets a second bridge to the 7 at row 4, column 6.
The 5 at row 4, column 8 needs 5, and its neighbors can take 5 at most. So it takes all of them: one bridge to the 2 at row 2, column 8 and a second bridge to the 4 at row 4, column 12.
The 3 at row 2, column 10 needs 1 more, and only the 2 at row 2, column 12 can still take bridges. So it gets a second bridge to the 2 at row 2, column 12.