Pips answers for Monday, September 29, 2025

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

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

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

Start with the 2 region at the left: only 0 + 0 + 2, or 0 + 1 + 1 can make it with these dominoes.

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

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

Start with the “=” region at the bottom: all 7 cells must show the same number, and only 5s appear on enough dominoes to fill it.

• the “=” region at the top: 4, 4, 4, 4
• the 6 region at the top-left: 3, 3
• the 2 region at the top: 1, 1
• the single-cell 0 region at the top: 0
• the “=” region at the center: 3, 3, 3
• the 0 region at the right: 0, 0, 0
• the 3 region at the left: 1, 1, 1
• the “=” region at the bottom: 5, 5, 5, 5, 5, 5, 5
• the single-cell 2 region at the bottom: 2
• the 18 region at the bottom-right: 6, 6, 6

1. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 3; the 6 region at the top-left needs a total of exactly 6 — nothing else fits. Place the 3–3 domino horizontally on row 2, column 2 and row 2, column 3.
2. Row 5, column 7 has only one open neighbor, so its domino must run to row 6, column 7; the 0 region at the 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.
3. The single-cell 2 region at the bottom needs a total of exactly 2 — nothing else fits. Place the 2–5 domino horizontally on row 7, column 4 and row 7, column 5.
4. The “=” region at the bottom needs every pip to match (1 of its 7 cells is already filled) — nothing else fits. Place the 0–5 domino horizontally on row 6, column 5 and row 6, column 6.
5. The 18 region at the bottom-right needs a total of exactly 18 — nothing else fits. Place the 5–6 domino horizontally on row 8, column 4 and row 8, column 5.
6. The “=” region at the bottom needs every pip to match (3 of its 7 cells are already filled) — nothing else fits. Place the 5–5 domino vertically on row 7, column 3 and row 8, column 3.
7. Row 9, column 5 has only one open neighbor, so its domino must run to row 9, column 6; the 18 region at the bottom-right needs a total of exactly 18 (1 of its 3 cells is already filled) — nothing else fits. Place the 6–6 domino horizontally on row 9, column 5 and row 9, column 6.
8. Row 9, column 3 has only one open neighbor, so its domino must run to row 9, column 2; the “=” region at the bottom 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 4–5 domino horizontally on row 9, column 2 and row 9, column 3.
9. The “=” region at the bottom needs every pip to match (6 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 3–5 domino vertically on row 5, column 4 and row 6, column 4.
10. The “=” region at the center needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 4, column 3 and row 4, column 4.
11. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 2 region at the top needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino horizontally on row 3, column 3 and row 3, column 4.
12. The “=” region at the top needs every pip to match — nothing else fits. Place the 3–4 domino vertically on row 3, column 5 and row 4, column 5.
13. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the “=” region at the top needs every pip to match (1 of its 4 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 1, column 4 and row 2, column 4.
14. Row 2, column 5 has only one open neighbor, so its domino must run to row 2, column 6; the “=” region at the top needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 1–4 domino horizontally on row 2, column 5 and row 2, column 6.
15. The 3 region at the left needs a total of exactly 3 — nothing else fits. Place the 1–1 domino horizontally on row 6, column 2 and row 6, column 3.
16. Row 6, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the 3 region at the left needs a total of exactly 3 (2 of its 3 cells are already filled) — nothing else fits. Place the 1–6 domino vertically on row 5, column 1 and row 6, column 1.