The full logical solve of the hard Bridges for September 15 to September 21, 2026, 14 × 14. 33 steps, 6 techniques, no guessing. Step through it, or jump to any step below.
The 3 at row 1, column 2 needs 3, and its neighbors can take 3 at most. So it takes all of them: one bridge to the 1 at row 1, column 4 and two bridges to the 6 at row 3, column 2.
The 4 at row 1, column 14 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 2 at row 1, column 11 and two bridges to the 5 at row 3, column 14.
The 3 at row 1, column 9 needs 3, and its neighbors can take 3 at most. So it takes all of them: one bridge to the 1 at row 1, column 7 and two bridges to the 7 at row 3, column 9.
The 6 at row 3, column 2 needs 6, and its neighbors can take 6 at most. So it takes all of them: two bridges to the 6 at row 3, column 5 and two bridges to the 4 at row 6, column 2.
The 6 at row 3, column 5 needs 6, and its neighbors can take 6 at most. So it takes all of them: two bridges to the 7 at row 3, column 9 and two bridges to the 5 at row 8, column 5.
The 2 at row 5, column 14 needs 2 more, and only the 5 at row 3, column 14 can still take bridges. So it gets two bridges to the 5 at row 3, column 14.
The 7 at row 3, column 9 needs 7, and its neighbors can take 7 at most. So it takes all of them: one bridge to the 5 at row 3, column 14 and two bridges to the 5 at row 6, column 9.
The 2 at row 7, column 7 needs 2 more, and only the 4 at row 10, column 7 can still take bridges. So it gets two bridges to the 4 at row 10, column 7.
The 2 at row 8, column 3 needs 2 more, and only the 5 at row 8, column 5 can still take bridges. So it gets two bridges to the 5 at row 8, column 5.
The 4 at row 11, column 6 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 3 at row 9, column 6 and two bridges to the 2 at row 11, column 8.
The 6 at row 12, column 5 needs 6, and its neighbors can take 6 at most. So it takes all of them: two bridges to the 3 at row 10, column 5, two bridges to the 6 at row 12, column 10, and two bridges to the 2 at row 14, column 5.
The 2 at row 12, column 12 needs 2 more, and only the 6 at row 12, column 10 can still take bridges. So it gets two bridges to the 6 at row 12, column 10.
The 2 at row 4, 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 6, column 13.
The 3 at row 9, column 2 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 4 at row 6, column 2 and at least one bridge to the 2 at row 9, column 4.
The 4 at row 10, column 7 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 4 at row 10, column 9.
The 2 at row 14, column 10 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 12, column 10.
The bridge between the 6 at row 3, column 2 and the 6 at row 3, column 5 blocks the way from the 1 at row 4, column 4 to the 1 at row 1, column 4. The 1 at row 4, column 4 needs 1 more, and only the 2 at row 6, column 4 can still take bridges. So it gets one bridge to the 2 at row 6, column 4.
The bridge between the 6 at row 3, column 5 and the 5 at row 8, column 5 blocks the way from the 1 at row 4, column 6 to the 1 at row 4, column 4. The 1 at row 4, column 6 needs 1 more, and only the 3 at row 6, column 6 can still take bridges. So it gets one bridge to the 3 at row 6, column 6.
The bridge between the 6 at row 3, column 5 and the 5 at row 8, column 5 blocks the way from the 2 at row 6, column 4 to the 3 at row 6, column 6. The 2 at row 6, column 4 needs 1 more, and only the 4 at row 6, column 2 can still take bridges. So it gets one bridge to the 4 at row 6, column 2.
The 3 at row 9, column 2 needs 1 more, and only the 2 at row 9, column 4 can still take bridges. So it gets a second bridge to the 2 at row 9, column 4.
The 3 at row 9, column 6 needs 1 more, and only the 3 at row 6, column 6 can still take bridges. So it gets one bridge to the 3 at row 6, column 6.
The 3 at row 6, column 6 needs 1 more, and only the 5 at row 6, column 9 can still take bridges. So it gets one bridge to the 5 at row 6, column 9.
The 5 at row 6, 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 4 at row 10, column 9.
The bridge between the 3 at row 6, column 6 and the 3 at row 9, column 6 blocks the way from the 5 at row 8, column 5 to the 2 at row 8, column 10. The 5 at row 8, column 5 needs 1 more, and only the 3 at row 10, column 5 can still take bridges. So it gets one bridge to the 3 at row 10, column 5.
The 4 at row 10, column 7 needs 1 more, and only the 4 at row 10, column 9 can still take bridges. So it gets a second bridge to the 4 at row 10, column 9.
The 3 at row 10, column 13 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 8, column 13.
The 2 at row 4, column 13 and the 2 at row 6, column 13 can’t share a double bridge: both would be full and cut off from the rest. The 2 at row 4, column 13 needs 1 more, and only the 1 at row 4, column 10 can still take bridges. So it gets one bridge to the 1 at row 4, column 10.
The 2 at row 8, column 10 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 8, column 13.
Suppose the 5 at row 6, column 9 and the 2 at row 6, column 13 shared no bridge. Following the rules from there breaks one, so they share at least one.
The 4 at row 10, column 9 needs 1 more, and only the 3 at row 10, column 13 can still take bridges. So it gets one bridge to the 3 at row 10, column 13.
The bridge between the 5 at row 6, column 9 and the 2 at row 6, column 13 blocks the way from the 2 at row 8, column 10 to the 1 at row 4, column 10. The 2 at row 8, column 10 needs 1 more, and only the 3 at row 8, column 13 can still take bridges. So it gets a second bridge to the 3 at row 8, column 13.
The 3 at row 10, column 13 needs 1 more, and only the 1 at row 14, column 13 can still take bridges. So it gets one bridge to the 1 at row 14, column 13.
The 6 at row 12, column 10 needs 1 more, and only the 2 at row 14, column 10 can still take bridges. So it gets a second bridge to the 2 at row 14, column 10.