Pips answers for Tuesday, December 9, 2025

Start with the 2 region at the bottom-right: the only way to make it with these dominoes is 0 + 2.

• the 3 region at the top: 1, 2
• the “=” region at the top-right: 3, 3
• the 2 region at the bottom-right: 0, 2

1. The 2 region at the bottom-right needs a total of exactly 2 — 4 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 2, column 3 and row 3, column 3.
2. The 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 1–3 domino horizontally on row 1, column 2 and row 1, column 3.
3. Row 2, column 2 has only one open neighbor, so its domino must run to row 2, column 1; the 3 region at the top needs a total of exactly 3 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–2 domino horizontally on row 2, column 1 and row 2, column 2.
4. Row 4, column 3 has only one open neighbor, so its domino must run to row 4, column 2; the 2 region at the bottom-right needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–6 domino horizontally on row 4, column 2 and row 4, column 3.

Start with the 9 region at the left: the only way to make it with these dominoes is 4 + 5.

• the 7 region at the top: 1, 1, 5
• the 9 region at the left: 4, 5
• the 4 region at the right: 0, 4
• the 2 region at the bottom-left: 1, 1
• the 5 region at the bottom-right: 2, 3
• the single-cell 0 region at the bottom: 0

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

Start with the 17 region at the left: the only way to make it with these dominoes is 5 + 6 + 6.

• the 14 region at the top: 4, 5, 5
• the “=” region at the top-left: 0, 0, 0
• the single-cell “<3” region at the top-right: 0
• the “=” region at the right: 2, 2, 2, 2, 2
• the 17 region at the left: 5, 6, 6
• the “=” region at the bottom: 4, 4
• the 12 region at the bottom-right: 6, 6
• the single-cell “<4” region at the bottom-left: 3
• the “>8” region at the bottom: 5, 5

1. The 12 region at the bottom-right needs a total of exactly 12 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino horizontally on row 7, column 3 and row 7, column 4.
2. The “=” region at the right 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 2–6 domino vertically on row 5, column 4 and row 6, column 4.
3. The “=” region at the right needs every pip to match (1 of its 5 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 horizontally on row 4, column 3 and row 4, column 4.
4. 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 2, column 4 and row 3, column 4.
5. The “=” region at the right needs every pip to match (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 2–3 domino vertically on row 3, column 2 and row 4, column 2.
6. The 17 region at the left needs a total of exactly 17 — 5 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 3, column 1 and row 4, column 1.
7. The “=” region at the top-left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 1, column 2 and row 2, column 2.
8. Row 2, column 1 has only one open neighbor, so its domino must run to row 1, 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 0–1 domino vertically on row 1, column 1 and row 2, column 1.
9. The 17 region at the left needs a total of exactly 17 (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 4–6 domino horizontally on row 5, column 1 and row 5, column 2.
10. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the 14 region at the top needs a total of exactly 14 (1 of its 3 cells is already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
11. The 17 region at the left needs a total of exactly 17 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–6 domino vertically on row 6, column 1 and row 7, column 1.
12. Row 6, column 2 has only one open neighbor, so its domino must run to row 7, column 2; the “=” region at the bottom needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 4–5 domino vertically on row 6, column 2 and row 7, column 2.