Pips answers for Thursday, December 25, 2025

Start with the 12 region at the left: the only way to make it with these dominoes is 6 + 6.

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

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

Start with the 10 region at the top-right: the only way to make it with these dominoes is 4 + 6.

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

1. The 10 region at the top-right needs a total of exactly 10 — nothing else fits. Place the 6–6 domino vertically on row 1, column 6 and row 2, column 6.
2. Row 3, column 6 has only one open neighbor, so its domino must run to row 4, column 6; the 10 region at the right needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–4 domino vertically on row 3, column 6 and row 4, column 6.
3. The “>3” region at the top must total more than 3 — nothing else fits. Place the 3–4 domino horizontally on row 1, column 4 and row 1, column 5.
4. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 0–5 domino horizontally on row 5, column 3 and row 5, column 4.
5. The “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 1–1 domino horizontally on row 5, column 1 and row 5, column 2.
6. Row 5, column 5 has only one open neighbor, so its domino must run to row 5, column 6; the “=” region at the bottom-right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino horizontally on row 5, column 5 and row 5, column 6.
7. The “<3” region at the top-left must total less than 3 — 4 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–2 domino horizontally on row 1, column 2 and row 1, column 3.
8. Row 1, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “<3” region at the top-left must total less than 3 (1 of its 2 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–3 domino vertically on row 1, column 1 and row 2, column 1.
9. Row 3, column 1 has only one open neighbor, so its domino must run to row 4, column 1; the 5 region at the left needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 3, column 1 and row 4, column 1.

This board has more than one valid layout; the steps follow the official answer.

Start with the 12 region at the center: the only way to make it with these dominoes is 6 + 6.

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