Pips answers for Tuesday, September 1, 2026

Start with the 11 region at the top: the only way to make it with these dominoes is 5 + 6.

• the single-cell 4 region at the top: 4
• the 11 region at the top: 5, 6
• the “=” region at the center: 2, 2
• the “=” region at the bottom-left: 3, 3
• the 7 region at the bottom-right: 1, 6
• the single-cell “<2” region at the bottom-left: 1
• the single-cell 3 region at the bottom-right: 3

1. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the single-cell 4 region at the top needs a total of exactly 4 — nothing else fits. Place the 4–6 domino vertically on row 1, column 3 and row 2, column 3.
2. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the 11 region at the top needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 3 and row 4, column 3.
3. Row 4, column 2 has only one open neighbor, so its domino must run to row 5, column 2; the “=” region at the center needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino vertically on row 4, column 2 and row 5, column 2.
4. 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 must total less than 2 — nothing else fits. Place the 1–3 domino horizontally on row 6, column 1 and row 6, column 2.
5. The 7 region at the bottom-right needs a total of exactly 7 — nothing else fits. Place the 3–6 domino horizontally on row 6, column 4 and row 6, column 5.
6. Row 5, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 7 region at the bottom-right needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino vertically on row 4, column 4 and row 5, column 4.

Start with the 8 region at the left: only 2 + 6, or 3 + 5 can make it with these dominoes.

• the 6 region at the top: 1, 1, 4
• the single-cell “>4” region at the top-left: 5
• the 8 region at the left: 2, 6
• the 5 region at the center: 2, 3
• the single-cell 2 region at the bottom-left: 2
• the single-cell “>4” region at the bottom-right: 5
• the 4 region at the bottom: 2, 2

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

Start with the 12 region at the top-right: the only way to make it with these dominoes is 6 + 6.

• the “=” region at the top-left: 1, 1, 1, 1, 1
• the “=” region at the top: 2, 2
• the single-cell “>4” region at the top: 5
• the 12 region at the top-right: 6, 6
• the single-cell “>3” region at the top-right: 4
• the “=” region at the top: 3, 3
• the “=” region at the right: 0, 0, 0, 0
• the “=” region at the bottom-left: 4, 4, 4
• the single-cell 4 region at the center: 4
• the “≠” region at the bottom-left: 2, 3
• the 7 region at the bottom-right: 2, 5

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