Pips answers for Tuesday, August 18, 2026

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

• the single-cell 1 region at the top-left: 1
• the 1 region at the top: 0, 0, 1
• the single-cell 1 region at the center: 1
• the 1 region at the bottom: 0, 0, 1
• the single-cell 2 region at the bottom-left: 2
• the single-cell 6 region at the bottom-right: 6

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

Start with the “=” region at the left: all 3 cells must show the same number, and only 1 or 4 or 5 appear on enough dominoes to fill it.

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

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

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

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

• the single-cell 1 region at the top-left: 1
• the single-cell 1 region at the top: 1
• the “=” region at the top-right: 0, 0
• the single-cell “>1” region at the top-right: 3
• the 11 region at the top-left: 2, 4, 5
• the 11 region at the top: 2, 4, 5
• the single-cell “>1” region at the top-left: 2
• the single-cell “>1” region at the top-right: 3
• the single-cell 1 region at the left: 1
• the “>11” region at the left: 6, 6
• the “=” region at the right: 0, 0
• the single-cell 1 region at the right: 1
• the single-cell 1 region at the left: 1
• the 1 region at the bottom-left: 0, 0, 1
• the 11 region at the bottom-left: 5, 6
• the “>11” region at the bottom: 6, 6
• the single-cell “>1” region at the bottom-right: 2
• the “=” region at the bottom: 5, 5

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

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