Pips answers for Wednesday, October 15, 2025

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

• the “=” region at the center: 5, 5, 5, 5
• the “=” region at the bottom-left: 2, 2

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

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

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

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

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

• the 8 region at the top-left: 4, 4
• the 2 region at the top-left: 1, 1
• the 4 region at the top-left: 1, 3
• the single-cell “>3” region at the top-right: 6
• the 10 region at the top-left: 5, 5
• the 0 region at the right: 0, 0, 0
• the single-cell 4 region at the left: 4
• the single-cell “>1” region at the left: 2
• the 9 region at the left: 4, 5
• the 2 region at the center: 0, 2
• the 1 region at the bottom: 0, 1
• the “=” region at the bottom-right: 2, 2
• the single-cell 0 region at the right: 0
• the 12 region at the bottom-left: 6, 6
• the single-cell 3 region at the bottom-left: 3
• the “=” region at the bottom-right: 3, 3, 3
• the single-cell 2 region at the bottom-left: 2

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