Pips answers for Thursday, October 9, 2025

Start with the “=” region at the top-left: all 3 cells must show the same number, and only 4 or 6 appear on enough dominoes to fill it.

• the “=” region at the top-left: 6, 6, 6
• the 8 region at the bottom-left: 4, 4
• the 8 region at the bottom-right: 3, 5
• the “=” region at the bottom: 1, 1

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

Start with the “=” region at the center: all 8 cells must show the same number, and only 4s appear on enough dominoes to fill it.

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

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

Start with the “=” region at the right: all 8 cells must show the same number, and only 1s appear on enough dominoes to fill it.

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

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