Pips answers for Tuesday, August 26, 2025

Start with the “=” region at the top-left: all 2 cells must show the same number, and only 1 or 2 or 6 appear on enough dominoes to fill it.

• the “=” region at the top-left: 1, 1
• the single-cell 4 region at the left: 4
• the single-cell 6 region at the center: 6
• the single-cell 2 region at the right: 2

1. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell 4 region at the left needs a total of exactly 4 — nothing else fits. Place the 1–4 domino horizontally on row 2, column 1 and row 2, column 2.
2. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the “=” region at the top-left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–2 domino horizontally on row 1, column 2 and row 1, column 3.
3. The single-cell 2 region at the right needs a total of exactly 2 — nothing else fits. Place the 2–6 domino horizontally on row 2, column 4 and row 2, column 5.
4. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the single-cell 6 region at the center needs a total of exactly 6 — nothing else fits. Place the 0–6 domino vertically on row 2, column 3 and row 3, column 3.

Start with the 1 region at the left: the only way to make it with these dominoes is 0 + 1.

• the 10 region at the top-left: 5, 5
• the 1 region at the left: 0, 1
• the 12 region at the right: 6, 6
• the 8 region at the bottom-right: 4, 4
• the “=” region at the bottom-left: 2, 2

1. The “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 2–2 domino vertically on row 7, column 1 and row 8, column 1.
2. The 10 region at the top-left needs a total of exactly 10 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 1–5 domino vertically on row 3, column 1 and row 4, column 1.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, column 1; the 10 region at the top-left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 3–5 domino vertically on row 1, column 1 and row 2, column 1.
4. The 1 region at the left needs a total of exactly 1 (1 of its 2 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–3 domino vertically on row 5, column 1 and row 6, column 1.
5. The 12 region at the right needs a total of exactly 12 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino horizontally on row 5, column 2 and row 5, column 3.
6. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the 12 region at the right needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 5, column 4 and row 6, column 4.
7. Row 7, column 4 has only one open neighbor, so its domino must run to row 8, column 4; the 8 region at the bottom-right needs a total of exactly 8 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 7, column 4 and row 8, column 4.

Start with the 10 region at the top-left: the only way to make it with these dominoes is 4 + 6.

• the 10 region at the top-left: 4, 6
• the 1 region at the top-left: 0, 1
• the “=” region at the left: 3, 3, 3
• the 12 region at the center: 6, 6
• the “≠” region at the center: 2, 4, 5
• the “=” region at the bottom-right: 3, 3, 3
• the single-cell 4 region at the bottom-right: 4
• the single-cell “<2” region at the bottom: 1
• the single-cell 6 region at the bottom: 6
• the single-cell 0 region at the bottom-right: 0

1. The “=” region at the left needs every pip to match — nothing else fits. Place the 3–3 domino vertically on row 4, column 2 and row 5, column 2.
2. The “=” region at the left needs every pip to match (2 of its 3 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–6 domino horizontally on row 3, column 2 and row 3, column 3.
3. The “=” region at the bottom-right needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–3 domino horizontally on row 7, column 5 and row 7, column 6.
4. The “=” region at the bottom-right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–4 domino horizontally on row 6, column 5 and row 6, column 6.
5. The “≠” region at the center needs every pip to be different — nothing else fits. Place the 2–3 domino vertically on row 4, column 5 and row 5, column 5.
6. Row 7, column 3 has only one open neighbor, so its domino must run to row 7, column 4; the single-cell “<2” region at the bottom must total less than 2 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino horizontally on row 7, column 3 and row 7, column 4.
7. The “≠” region at the center needs every pip to be different (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–6 domino horizontally on row 4, column 3 and row 4, column 4.
8. The 10 region at the top-left needs a total of exactly 10 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–4 domino horizontally on row 1, column 1 and row 1, column 2.
9. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the 10 region at the top-left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 domino horizontally on row 2, column 1 and row 2, column 2.
10. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the “≠” region at the center needs every pip to be different (2 of its 3 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 3, column 4 and row 3, column 5.

This board has more than one valid layout; the steps follow the official answer.