The full logical solve of the hard Bridges for September 22 to September 28, 2026, 14 × 14. 36 steps, 5 techniques, no guessing. Step through it, or jump to any step below.
The 4 at row 1, column 1 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 6 at row 1, column 3 and two bridges to the 2 at row 3, column 1.
The 6 at row 1, column 3 needs 6, and its neighbors can take 6 at most. So it takes all of them: two bridges to the 4 at row 1, column 9 and two bridges to the 6 at row 5, column 3.
The 4 at row 2, column 5 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 8 and two bridges to the 4 at row 4, column 5.
The 4 at row 3, column 7 needs 4, and its neighbors can take 4 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 6, column 7.
The 4 at row 4, column 5 needs 2 more, and only the 6 at row 9, column 5 can still take bridges. So it gets two bridges to the 6 at row 9, column 5.
The 2 at row 5, column 1 needs 2 more, and only the 6 at row 5, column 3 can still take bridges. So it gets two bridges to the 6 at row 5, column 3.
The 1 at row 7, column 2 needs 1 more, and only the 2 at row 13, column 2 can still take bridges. So it gets one bridge to the 2 at row 13, column 2.
The 4 at row 8, column 10 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 2 at row 8, column 8 and two bridges to the 6 at row 8, column 12.
The 4 at row 9, column 7 needs 4, and its neighbors can take 4 at most. So it takes all of them: two bridges to the 5 at row 6, column 7 and two bridges to the 6 at row 9, column 5.
The 5 at row 6, column 7 needs 1 more, and only the 3 at row 6, column 9 can still take bridges. So it gets one bridge to the 3 at row 6, column 9.
The 2 at row 13, column 2 needs 1 more, and only the 4 at row 13, column 5 can still take bridges. So it gets one bridge to the 4 at row 13, column 5.
The 3 at row 1, column 14 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 1, column 12 and at least one bridge to the 2 at row 5, column 14.
The 4 at row 1, column 9 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 3, column 9.
The 7 at row 3, column 9 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 4 at row 3, column 12 and at least one bridge to the 3 at row 6, column 9.
The 4 at row 3, 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 6 at row 8, column 12.
The 6 at row 5, column 3 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 5 at row 9, column 3.
The 6 at row 8, 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 13, column 12.
The 5 at row 9, 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 6 at row 9, column 5 and at least one bridge to the 2 at row 12, column 3.
The 4 at row 13, column 5 needs 4, and its neighbors can take 4 at most. So it takes all of them: one bridge to the 6 at row 9, column 5 and two bridges to the 5 at row 13, column 9.
The 5 at row 9, column 3 needs 5, and its neighbors can take 5 at most. So it takes all of them: a second bridge to the 6 at row 5, column 3 and a second bridge to the 2 at row 12, column 3.
The 3 at row 10, column 6 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 10, column 9 and at least one bridge to the 3 at row 12, column 6.
The 2 at row 11, 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 3 at row 13, column 14.
The 3 at row 12, column 6 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 12, column 8.
The 5 at row 13, 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 5 at row 10, column 9 and at least one bridge to the 4 at row 13, column 12.
The 3 at row 13, column 14 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 13, column 12.
The bridge between the 4 at row 8, column 10 and the 6 at row 8, column 12 blocks the way from the 1 at row 6, column 11 to the 2 at row 10, column 11. The 1 at row 6, column 11 needs 1 more, and only the 3 at row 6, column 9 can still take bridges. So it gets one bridge to the 3 at row 6, column 9.
The 7 at row 3, column 9 needs 7, and its neighbors can take 7 at most. So it takes all of them: a second bridge to the 4 at row 1, column 9 and a second bridge to the 4 at row 3, column 12.
The 5 at row 10, 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 10, column 11.
The bridge between the 3 at row 10, column 6 and the 5 at row 10, column 9 blocks the way from the 2 at row 12, column 8 to the 2 at row 8, column 8. The 2 at row 12, column 8 needs 1 more, and only the 3 at row 12, column 6 can still take bridges. So it gets a second bridge to the 3 at row 12, column 6.
The 3 at row 10, column 6 needs 1 more, and only the 5 at row 10, column 9 can still take bridges. So it gets a second bridge to the 5 at row 10, column 9.
The 1 at row 12, column 11 needs 1 more, and only the 2 at row 10, column 11 can still take bridges. So it gets one bridge to the 2 at row 10, column 11.
The 5 at row 10, column 9 needs 1 more, and only the 5 at row 13, column 9 can still take bridges. So it gets a second bridge to the 5 at row 13, column 9.
Suppose the 2 at row 1, column 12 and the 3 at row 1, column 14 shared two bridges. Following the rules from there breaks one, so they don’t. The 2 at row 1, column 12 needs 1 more, and only the 4 at row 3, column 12 can still take bridges. So it gets one bridge to the 4 at row 3, column 12.
The 3 at row 1, column 14 needs 1 more, and only the 2 at row 5, column 14 can still take bridges. So it gets a second bridge to the 2 at row 5, column 14.
The 6 at row 8, column 12 needs 6, and its neighbors can take 6 at most. So it takes all of them: one bridge to the 1 at row 8, column 14 and a second bridge to the 4 at row 13, column 12.
The 2 at row 11, column 14 needs 1 more, and only the 3 at row 13, column 14 can still take bridges. So it gets a second bridge to the 3 at row 13, column 14.