Pips answers for Friday, September 12, 2025

Start with the 11 region at the top: the only way to make it with these dominoes is 5 + 6.

• the 8 region at the top-left: 2, 6
• the 11 region at the top: 5, 6
• the 0 region at the bottom: 0, 0

1. The 0 region at the bottom needs a total of exactly 0 — nothing else fits. Place the 0–5 domino horizontally on row 2, column 2 and row 2, column 3.
2. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 0 region at the bottom 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 3, column 3 and row 3, column 4.
3. 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 2–6 domino horizontally on row 1, column 1 and row 1, column 2.
4. 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 5–6 domino vertically on row 2, column 1 and row 3, column 1.

Start with the 1 region at the top-left: the only way to make it with these dominoes is 0 + 1.

• the single-cell “<3” region at the top-left: 2
• the single-cell “<3” region at the top-right: 2
• the 1 region at the top-left: 0, 1
• the 0 region at the top-right: 0, 0
• the “=” region at the right: 5, 5
• the single-cell 1 region at the left: 1
• the 8 region at the bottom: 1, 3, 4
• the single-cell “<3” region at the bottom: 2

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

Start with the 30 region at the center: the only way to make it with these dominoes is 6 + 6 + 6 + 6 + 6.

• the single-cell 1 region at the top: 1
• the “≠” region at the top: 2, 3, 4, 5
• the single-cell 4 region at the top-right: 4
• the single-cell 0 region at the top-right: 0
• the 20 region at the center: 5, 5, 5, 5
• the 30 region at the center: 6, 6, 6, 6, 6
• the “=” region at the left: 4, 4
• the single-cell 0 region at the bottom-left: 0
• the single-cell 0 region at the bottom: 0
• the single-cell “<3” region at the bottom: 2

1. Row 1, column 5 has only one open neighbor, so its domino must run to row 2, column 5; the single-cell 1 region at the top needs a total of exactly 1 — nothing else fits. Place the 1–4 domino vertically on row 1, column 5 and row 2, column 5.
2. The single-cell 0 region at the top-right 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 0–3 domino horizontally on row 3, column 5 and row 3, column 6.
3. The “=” region at the 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 0–4 domino vertically on row 6, column 1 and row 7, column 1.
4. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the “=” region at the left needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino horizontally on row 6, column 2 and row 6, column 3.
5. Row 2, column 6 has only one open neighbor, so its domino must run to row 2, column 7; the “≠” region at the top needs every pip to be different (2 of its 4 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 2, column 6 and row 2, column 7.
6. Row 9, column 3 has only one open neighbor, so its domino must run to row 8, column 3; the single-cell 0 region at the bottom needs a total of exactly 0 — nothing else fits. Place the 0–5 domino vertically on row 8, column 3 and row 9, column 3.
7. Row 9, column 5 has only one open neighbor, so its domino must run to row 8, column 5; the single-cell “<3” region at the bottom must total less than 3 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–3 domino vertically on row 8, column 5 and row 9, column 5.
8. The 30 region at the center needs a total of exactly 30 (1 of its 5 cells is already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino horizontally on row 7, column 3 and row 7, column 4.
9. The 30 region at the center needs a total of exactly 30 (2 of its 5 cells are already filled) — nothing else fits. Place the 6–6 domino vertically on row 5, column 4 and row 6, column 4.
10. The 20 region at the center needs a total of exactly 20 — nothing else fits. Place the 5–6 domino vertically on row 6, column 5 and row 7, column 5.
11. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the “≠” region at the top needs every pip to be different (3 of its 4 cells are already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 4 and row 4, column 4.
12. Row 4, column 5 has only one open neighbor, so its domino must run to row 5, column 5; the 20 region at the center needs a total of exactly 20 (2 of its 4 cells are already filled) — nothing else fits. Place the 5–5 domino vertically on row 4, column 5 and row 5, column 5.