Pips answers for Tuesday, December 16, 2025

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

• the single-cell 0 region at the top-left: 0
• the 1 region at the left: 0, 1
• the “=” region at the center: 3, 3
• the single-cell “>3” region at the right: 4
• the 8 region at the bottom: 3, 5
• the single-cell 6 region at the bottom-right: 6

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

Start with the “=” region at the top: all 2 cells must show the same number, and only 0 or 1 or 3 or 4 or 5 appear on enough dominoes to fill it.

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

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

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

Start with the “=” region at the center: all 7 cells must show the same number, and only 0s appear on enough dominoes to fill it.

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

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