Pips answers for Friday, December 26, 2025

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

• the 11 region at the top: 5, 6
• the “=” region at the bottom-left: 4, 4
• the single-cell 3 region at the center: 3
• the 3 region at the bottom-right: 1, 2
• the single-cell “<1” region at the bottom-right: 0

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

Start with the “=” region at the center: all 3 cells must show the same number, and only 2 or 3 appear on enough dominoes to fill it.

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

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

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

Start with the “=” region at the left: all 7 cells must show the same number, and only 3s appear on enough dominoes to fill it.

• the 11 region at the top-left: 1, 5, 5
• the “=” region at the top: 2, 2
• the single-cell 6 region at the top-left: 6
• the “=” region at the left: 3, 3, 3, 3, 3, 3, 3
• the single-cell 5 region at the top-left: 5
• the “>10” region at the top-right: 6, 6
• the 2 region at the left: 1, 1
• the single-cell “<4” region at the right: 0
• the “>11” region at the right: 6, 6
• the single-cell 5 region at the bottom-left: 5
• the “=” region at the bottom: 0, 0, 0, 0
• the “=” region at the bottom-left: 1, 1

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