The full logical solve of the hard Bridges for September 8 to September 14, 2026, 14 × 14. 36 steps, 6 techniques, no guessing. Step through it, or jump to any step below.
The 2 at row 1, column 2 needs 2 more, and only the 3 at row 1, column 5 can still take bridges. So it gets two bridges to the 3 at row 1, column 5.
The 4 at row 2, column 1 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 2 at row 2, column 3 and two bridges to the 4 at row 4, column 1.
The 1 at row 2, column 14 needs 1 more, and only the 3 at row 4, column 14 can still take bridges. So it gets one bridge to the 3 at row 4, column 14.
The 2 at row 5, column 8 needs 2 more, and only the 5 at row 7, column 8 can still take bridges. So it gets two bridges to the 5 at row 7, column 8.
The 1 at row 12, column 8 needs 1 more, and only the 2 at row 10, column 8 can still take bridges. So it gets one bridge to the 2 at row 10, column 8.
The 5 at row 7, column 8 needs 5, and its neighbors can take 5 at most. So it takes all of them: two bridges to the 4 at row 7, column 3 and one bridge to the 2 at row 10, column 8.
The 1 at row 13, column 5 needs 1 more, and only the 2 at row 11, column 5 can still take bridges. So it gets one bridge to the 2 at row 11, column 5.
The 3 at row 14, column 1 needs 3, and its neighbors can take 3 at most. So it takes all of them: one bridge to the 1 at row 11, column 1 and two bridges to the 5 at row 14, column 3.
The 4 at row 4, column 1 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 2 at row 9, column 1.
The 5 at row 4, column 5 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 9 and at least one bridge to the 3 at row 6, column 5.
The 6 at row 4, column 12 needs 6, one less than the 7 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 9, at least one bridge to the 3 at row 4, column 14, and at least one bridge to the 3 at row 6, column 12.
The 4 at row 7, column 3 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 9, column 3.
The 5 at row 14, 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 5 at row 9, column 3 and at least one bridge to the 5 at row 14, column 7.
The 5 at row 14, column 7 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 9, column 7 and at least one bridge to the 5 at row 14, column 9.
The 5 at row 14, column 9 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 11, column 9 and at least one bridge to the 4 at row 14, column 12.
The 2 at row 14, column 14 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 4 at row 14, column 12.
The bridge between the 4 at row 7, column 3 and the 5 at row 7, column 8 blocks the way from the 3 at row 6, column 5 to the 2 at row 9, column 5. The 3 at row 6, column 5 needs 3, and its neighbors can take 3 at most. So it takes all of them: a second bridge to the 5 at row 4, column 5 and one bridge to the 1 at row 6, column 7.
The 3 at row 6, column 12 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 6 at row 9, column 12.
The bridge between the 5 at row 9, column 3 and the 5 at row 14, column 3 blocks the way from the 2 at row 11, column 5 to the 1 at row 11, column 1. The 2 at row 11, column 5 needs 1 more, and only the 2 at row 9, column 5 can still take bridges. So it gets one bridge to the 2 at row 9, column 5.
A bridge between the 3 at row 1, column 5 and the 1 at row 1, column 12 would close off a group of 3 full islands, so it can’t be there. The 3 at row 1, column 5 needs 1 more, and only the 5 at row 4, column 5 can still take bridges. So it gets one bridge to the 5 at row 4, column 5.
The 1 at row 1, column 12 needs 1 more, and only the 6 at row 4, column 12 can still take bridges. So it gets one bridge to the 6 at row 4, column 12.
A double bridge between the 4 at row 4, column 1 and the 2 at row 9, column 1 would close off a group of 4 full islands, so they share one bridge at most. The 4 at row 4, column 1 needs 1 more, and only the 1 at row 4, column 3 can still take bridges. So it gets one bridge to the 1 at row 4, column 3.
The 5 at row 4, column 5 needs 1 more, and only the 4 at row 4, column 9 can still take bridges. So it gets a second bridge to the 4 at row 4, column 9.
The 4 at row 7, column 3 needs 1 more, and only the 5 at row 9, column 3 can still take bridges. So it gets a second bridge to the 5 at row 9, column 3.
The 2 at row 9, column 1 needs 1 more, and only the 5 at row 9, column 3 can still take bridges. So it gets one bridge to the 5 at row 9, column 3.
The bridge between the 5 at row 7, column 8 and the 2 at row 10, column 8 blocks the way from the 3 at row 9, column 9 to the 2 at row 9, column 7. The 3 at row 9, column 9 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 6 at row 9, column 12.
Suppose the 6 at row 4, column 12 and the 3 at row 4, column 14 shared only one bridge. Following the rules from there breaks one, so they share two.
The 2 at row 9, column 14 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 6 at row 9, column 12.
Suppose the 5 at row 9, column 3 and the 2 at row 9, column 5 shared no bridge. Following the rules from there breaks one, so they share at least one.
The 5 at row 14, column 3 needs 1 more, and only the 5 at row 14, column 7 can still take bridges. So it gets a second bridge to the 5 at row 14, column 7.
The bridge between the 4 at row 7, column 3 and the 5 at row 7, column 8 blocks the way from the 2 at row 9, column 7 to the 1 at row 6, column 7. The 2 at row 9, column 7 needs 1 more, and only the 5 at row 14, column 7 can still take bridges. So it gets a second bridge to the 5 at row 14, column 7.
The 5 at row 14, column 9 needs 5, and its neighbors can take 5 at most. So it takes all of them: a second bridge to the 2 at row 11, column 9 and a second bridge to the 4 at row 14, column 12.
The 3 at row 9, column 9 needs 3, and its neighbors can take 3 at most. So it takes all of them: one bridge to the 4 at row 4, column 9 and a second bridge to the 6 at row 9, column 12.
The 6 at row 4, column 12 needs 1 more, and only the 3 at row 6, column 12 can still take bridges. So it gets a second bridge to the 3 at row 6, column 12.
The 6 at row 9, column 12 needs 6, and its neighbors can take 6 at most. So it takes all of them: a second bridge to the 2 at row 9, column 14 and one bridge to the 4 at row 14, column 12.
The 1 at row 11, column 14 needs 1 more, and only the 2 at row 14, column 14 can still take bridges. So it gets one bridge to the 2 at row 14, column 14.