The full logical solve of the hard Heyawake for September 8 to September 14, 2026, 10 × 10. 69 steps, 6 techniques, no guessing. Step through it, or jump to any step below.
A 0 room has no shaded cells, so every cell in it is white.
A 0 room has no shaded cells, so every cell in it is white.
Suppose r3c2 is shaded. Following rules 1 to 4 from there, a white line would cross two room borders. So it is white.
Suppose r3c3 is shaded. Following rules 1 to 4 from there, a white line would cross two room borders. So it is white.
If r3c4 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r3c4 are white.
Try every way to fit one shaded cell into the open cells of the 1 room without breaking a rule. All 2 ways that work agree on r5c2 and r5c3.
If r5c4 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r5c4 are white.
Suppose r2c1 is white. Following rules 1 to 4 from there, the white cells would split apart. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r2c1 are white.
If r1c5 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r1c5 are white.
This room already has its one shaded cell, so its other cells are white.
If r2c8 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r2c8 are white.
If r1c9 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r1c9 are white.
Shading r2c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Suppose r5c6 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r7c1 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r8c1 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r8c1 are white.
Suppose r7c4 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Try every way to fit 2 shaded cells into the open cells of the 2 room without breaking a rule. Only one way works, and it settles r8c4, r9c4 and r10c4.
Shaded cells never touch by an edge, so the open cells beside r8c4 are white.
Shaded cells never touch by an edge, so the open cells beside r10c4 are white.
Suppose r4c2 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
This room needs one more shaded cell and has exactly one open cell, so it is shaded.
If r6c2 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r6c2 are white.
If r6c5 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r6c5 are white.
If r7c3 were white, a white line would run from one room through the next into a third. So it is shaded.
If r7c6 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r7c6 are white.
If r9c2 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r9c2 are white.
If r9c5 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r9c5 are white.
Shading r10c1 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r4c5 would cut the white cells into two groups. White cells must all connect, so it is white.
If r4c6 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r4c6 are white.
If r3c7 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r10c6 would cut the white cells into two groups. White cells must all connect, so it is white.
If r10c7 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r10c7 are white.
Suppose r3c9 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r3c10 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r3c10 are white.
Suppose r4c8 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r4c9 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r4c9 are white.
Shading r5c10 would cut the white cells into two groups. White cells must all connect, so it is white.
If r5c8 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r5c8 are white.
If r6c7 were white, a white line would run from one room through the next into a third. So it is shaded.
Suppose r8c7 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r8c8 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r8c8 are white.
If r7c9 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r7c9 are white.
If r6c10 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r8c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Suppose r9c9 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r9c10 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r10c10 were white, a white line would run from one room through the next into a third. So it is shaded.
Shaded cells never touch by an edge, so the open cells beside r10c10 are white.