Pips answers for Sunday, November 2, 2025

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

• the 12 region at the top: 6, 6
• the 10 region at the left: 0, 5, 5
• the 2 region at the right: 1, 1
• the single-cell 2 region at the right: 2

1. The 2 region at the right needs a total of exactly 2 — nothing else fits. Place the 1–2 domino horizontally on row 2, column 4 and row 2, column 5.
2. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 2 region at the right needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino vertically on row 3, column 4 and row 4, column 4.
3. The 10 region at the 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 0–6 domino horizontally on row 2, column 2 and row 2, column 3.
4. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 12 region at the top needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino horizontally on row 1, column 3 and row 1, column 4.
5. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the 10 region at the left needs a total of exactly 10 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 3, column 1 and row 3, column 2.

Start with the “<4” region at the center: the only way to make it with these dominoes is 1 + 1 + 1.

• the 10 region at the left: 4, 6
• the “<4” region at the center: 1, 1, 1
• the single-cell “>4” region at the top-right: 6
• the 11 region at the center: 5, 6
• the 11 region at the right: 5, 6
• the single-cell 2 region at the bottom-left: 2

1. The single-cell “>4” region at the top-right must total more than 4 — nothing else fits. Place the 6–6 domino vertically on row 2, column 4 and row 3, column 4.
2. The single-cell 2 region at the bottom-left needs a total of exactly 2 — nothing else fits. Place the 2–4 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 2, column 2; the 10 region at the left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino horizontally on row 2, column 1 and row 2, column 2.
4. The 11 region at the center needs a total of exactly 11 — nothing else fits. Place the 1–5 domino horizontally on row 3, column 2 and row 3, column 3.
5. Row 2, column 3 has only one open neighbor, so its domino must run to row 1, column 3; the “<4” region at the center must total less than 4 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–4 domino vertically on row 1, column 3 and row 2, column 3.
6. Row 4, column 2 has only one open neighbor, so its domino must run to row 5, column 2; the 11 region at the center needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino vertically on row 4, column 2 and row 5, column 2.
7. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 11 region at the right needs a total of exactly 11 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–5 domino vertically on row 4, column 4 and row 5, column 4.

Start with the “=” region at the center: all 7 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the single-cell 0 region at the top: 0
• the 8 region at the top: 4, 4
• the “=” region at the center: 1, 1, 1, 1, 1, 1, 1
• the single-cell 3 region at the top-right: 3
• the 6 region at the left: 3, 3
• the 8 region at the right: 2, 6
• the single-cell 0 region at the left: 0
• the 0 region at the center: 0, 0, 0
• the 10 region at the center: 5, 5
• the single-cell 2 region at the bottom: 2
• the 12 region at the bottom: 6, 6
• the 8 region at the bottom: 4, 4
• the single-cell “>4” region at the bottom-left: 5
• the single-cell “>4” region at the bottom-right: 6

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