Pips answers for Saturday, September 27, 2025

Start with the 6 region at the top: the only way to make it with these dominoes is 2 + 4.

• the 6 region at the top: 2, 4
• the 4 region at the left: 1, 3
• the “=” region at the right: 1, 1

1. 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 1–1 domino vertically on row 3, column 2 and row 4, column 2.
2. The 6 region at the top needs a total of exactly 6 — 4 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–4 domino vertically on row 2, column 1 and row 3, column 1.
3. Row 2, column 2 has only one open neighbor, so its domino must run to row 1, column 2; 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 2–2 domino vertically on row 1, column 2 and row 2, column 2.
4. Row 4, column 1 has only one open neighbor, so its domino must run to row 5, column 1; the 4 region at the left needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino vertically on row 4, column 1 and row 5, column 1.

Start with the 7 region at the bottom-right — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

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

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

Start with the 18 region at the center: the only way to make it with these dominoes is 6 + 6 + 6.

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

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