The full logical solve of the hard Heyawake for September 29 to October 5, 2026, 10 × 10. 73 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.
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 r6c7, r7c7 and r8c7.
Shaded cells never touch by an edge, so the open cells beside r6c7 are white.
Shaded cells never touch by an edge, so the open cells beside r8c7 are white.
Suppose r6c5 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r8c5 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r8c9 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r8c10 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 r8c10 are white.
Suppose r9c8 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 r9c8 are white.
Suppose r9c6 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is 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.
Suppose r9c3 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is 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.
Suppose r8c1 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r8c3 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 r8c3 are white.
Suppose r7c6 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 r7c6 are white.
If r5c5 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 r5c5 are white.
This room already has its one shaded cell, so its other cells are 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.
Shading r7c8 would cut the white cells into two groups. White cells must all connect, so it is 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.
Shading r6c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Suppose r7c4 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 r7c4 are white.
If r6c3 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 r6c3 are white.
If r5c2 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 r5c2 are white.
Shading r7c2 would cut the white cells into two groups. White cells must all connect, so it is white.
If r7c1 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 r7c1 are white.
Suppose r3c2 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r3c1 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 r3c1 are white.
Suppose r4c3 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 r4c3 are 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.
If r2c3 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 r2c3 are white.
If r2c6 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 r2c6 are white.
If r3c7 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 r3c7 are white.
If r4c6 were white, a white line would run from one room through the next into a third. So it is shaded.
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.
Shading r1c7 would cut the white cells into two groups. White cells must all connect, so it is white.
Suppose r1c4 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r1c5 were white, a white line would run from one room through the next into a third. So it is shaded.
Suppose r2c10 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is 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.
If r4c9 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r5c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r10c9 would cut the white cells into two groups. White cells must all connect, so it is white.
This room needs one more shaded cell and has exactly one open cell, so it is shaded.
Shading r10c7 would cut the white cells into two groups. White cells must all connect, so it is white.
If r10c6 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r10c4 would cut the white cells into two groups. White cells must all connect, so it is white.
If r10c3 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r10c1 would cut the white cells into two groups. White cells must all connect, so it is white.