Pips answers for Thursday, August 13, 2026

Start with the “=” region at the bottom-right: all 3 cells must show the same number, and only 0 or 1 appear on enough dominoes to fill it.

• the single-cell “<2” region at the top-left: 0
• the “<2” region at the bottom-left: 0, 0
• the “=” region at the bottom: 6, 6
• the “=” region at the bottom-right: 1, 1, 1

1. Row 2, column 1 has only one open neighbor, so its domino must run to row 2, column 2; the “<2” region at the bottom-left must total less than 2 — nothing else fits. Place the 0–0 domino horizontally on row 2, column 1 and row 2, column 2.
2. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 1–1 domino horizontally on row 2, column 7 and row 2, column 8.
3. The “=” region at the bottom-right needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–5 domino vertically on row 1, column 6 and row 2, column 6.
4. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the single-cell “<2” region at the top-left must total less than 2 — nothing else fits. Place the 0–6 domino vertically on row 1, column 3 and row 2, column 3.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 2, column 5; the “=” region at the bottom needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 3–6 domino horizontally on row 2, column 4 and row 2, column 5.

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

• the “=” region at the top-left: 2, 2
• the 10 region at the top: 3, 3, 4
• the “=” region at the left: 0, 0
• the 10 region at the right: 4, 6
• the 7 region at the center: 3, 4

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

Start with the 15 region at the top: only 3 + 6 + 6, or 4 + 5 + 6, or 5 + 5 + 5 can make it with these dominoes.

• the single-cell “>3” region at the top-left: 4
• the 15 region at the top: 4, 5, 6
• the single-cell “<3” region at the top: 2
• the “<3” region at the top: 0, 2
• the “=” region at the center: 6, 6
• the 15 region at the bottom-left: 4, 5, 6
• the 5 region at the right: 2, 3
• the single-cell “<3” region at the bottom: 1
• the 15 region at the bottom: 3, 6, 6
• the 5 region at the bottom: 0, 2, 3
• the single-cell 5 region at the bottom-right: 5
• the single-cell 3 region at the bottom-left: 3
• the single-cell “<3” region at the bottom-left: 1
• the 5 region at the bottom-right: 2, 3

1. The 15 region at the bottom-left needs a total of exactly 15 — 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 vertically on row 7, column 1 and row 8, column 1.
2. Row 5, column 1 has only one open neighbor, so its domino must run to row 6, column 1; the 15 region at the bottom-left needs a total of exactly 15 (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 4–5 domino vertically on row 5, column 1 and row 6, column 1.
3. Row 5, column 9 has only one open neighbor, so its domino must run to row 6, column 9; the 5 region at the right needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–3 domino vertically on row 5, column 9 and row 6, column 9.
4. The 5 region at the bottom-right needs a total of exactly 5 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 3–5 domino vertically on row 7, column 9 and row 8, column 9.
5. The 15 region at the top needs a total of exactly 15 — 3 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 1, column 2 and row 1, column 3.
6. The 15 region at the bottom needs a total of exactly 15 — nothing else fits. Place the 6–6 domino vertically on row 7, column 4 and row 8, column 4.
7. Row 8, column 3 has only one open neighbor, so its domino must run to row 8, column 2; the 15 region at the bottom needs a total of exactly 15 (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 1–3 domino horizontally on row 8, column 2 and row 8, column 3.
8. The 5 region at the bottom needs a total of exactly 5 — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–2 domino horizontally on row 8, column 7 and row 8, column 8.
9. Row 7, column 6 has only one open neighbor, so its domino must run to row 8, column 6; the 5 region at the bottom needs a total of exactly 5 (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 0–3 domino vertically on row 7, column 6 and row 8, column 6.
10. The 15 region at the top needs a total of exactly 15 (1 of its 3 cells is already filled) — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 2–5 domino vertically on row 1, column 4 and row 2, column 4.
11. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 15 region at the top needs a total of exactly 15 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–4 domino horizontally on row 1, column 5 and row 1, column 6.
12. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the “<3” region at the top must total less than 3 (1 of its 2 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 0–6 domino vertically on row 3, column 4 and row 4, column 4.
13. Row 5, column 4 has only one open neighbor, so its domino must run to row 6, column 4; the “=” region at the center needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino vertically on row 5, column 4 and row 6, column 4.
14. Row 2, column 9 has only one open neighbor, so its domino must run to row 3, column 9 — nothing else fits. Place the 0–0 domino vertically on row 2, column 9 and row 3, column 9.

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