The full logical solve of the hard Heyawake for September 15 to September 21, 2026, 10 × 10. 75 steps, 6 techniques, no guessing. Step through it, or jump to any step below.
This room needs one more shaded cell and has exactly one open cell, so it is shaded.
Shaded cells never touch by an edge, so the open cells beside r10c2 are white.
Shading r9c1 would cut the white cells into two groups. White cells must all connect, so it is white.
If r9c3 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 r9c3 are white.
Shading r10c4 would cut the white cells into two groups. White cells must all connect, so it is white.
If r10c5 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 r10c5 are white.
If r9c6 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 r9c6 are white.
Shading r10c7 would cut the white cells into two groups. White cells must all connect, so it is white.
If r10c8 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 r10c8 are white.
If r9c9 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 r9c9 are white.
Shading r10c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r8c10 would cut the white cells into two groups. White cells must all connect, 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 r5c3, r5c4 and r5c5.
Shaded cells never touch by an edge, so the open cells beside r5c3 are white.
Shaded cells never touch by an edge, so the open cells beside r5c5 are white.
Suppose r7c3 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r5c7 is shaded. Following rules 1 to 4 from there, a white line would cross two room borders. 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.
Suppose r5c10 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r7c6 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r6c2 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r7c9 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
Suppose r4c4 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r4c2 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 r4c2 are 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.
This room already has its one shaded cell, so its other cells are white.
If r6c4 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 r6c4 are white.
If r6c7 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 r6c7 are white.
If r7c2 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 r7c2 are white.
If r7c5 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 r7c5 are white.
If r7c8 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 r7c8 are white.
If r8c1 were white, a white line would run from one room through the next into a third. So it is shaded.
If r8c4 were white, a white line would run from one room through the next into a third. So it is shaded.
If r8c7 were white, a white line would run from one room through the next into a third. So it is shaded.
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 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.
Shading r6c1 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r5c1 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r3c1 would cut the white cells into two groups. White cells must all connect, so it is white.
If r2c1 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 r2c1 are white.
Shading r6c9 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r1c2 would cut the white cells into two groups. White cells must all connect, so it is 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 r1c5 and r1c6.
Suppose r2c3 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 r2c3 are white.
If r1c4 were white, a white line would run from one room through the next into a third. So it is shaded.
Shading r2c5 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 r1c7 would cut the white cells into two groups. White cells must all connect, so it is white.
Suppose r2c8 is shaded. Following rules 1 to 4 from there, the white cells would split apart. So it is white.
If r2c9 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 r2c9 are white.
If r1c8 were white, a white line would run from one room through the next into a third. So it is shaded.
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.
This room needs one more shaded cell and has exactly one open cell, so it is shaded.
Shaded cells never touch by an edge, so the open cells beside r6c10 are white.
Shading r1c10 would cut the white cells into two groups. White cells must all connect, so it is white.
Shading r3c10 would cut the white cells into two groups. White cells must all connect, so it is white.