Pips answers for Monday, February 2, 2026

Start with the 3 region at the left: only 0 + 0 + 3, or 0 + 1 + 2 can make it with these dominoes.

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

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

Start with the 8 region at the bottom: only 2 + 6, or 3 + 5 can make it with these dominoes.

• the “=” region at the top-left: 2, 2
• the “=” region at the left: 1, 1
• the “>8” region at the right: 4, 6
• the “=” region at the bottom-left: 0, 0
• the single-cell “>0” region at the right: 2
• the 8 region at the bottom: 3, 5
• the single-cell “>0” region at the bottom-right: 2

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

Start with the “<2” region at the right: only 0 + 0 + 0 + 0, or 0 + 0 + 0 + 1 can make it with these dominoes.

• the “=” region at the top-left: 3, 3
• the single-cell “>4” region at the top-left: 5
• the 9 region at the top: 2, 2, 2, 3
• the 9 region at the top-right: 3, 6
• the 9 region at the left: 2, 2, 5
• the 8 region at the center: 4, 4
• the “<2” region at the right: 0, 0, 0, 0
• the single-cell 1 region at the center: 1
• the single-cell “<2” region at the bottom-left: 1
• the single-cell 3 region at the bottom-left: 3
• the 9 region at the bottom-left: 4, 5
• the single-cell “<2” region at the bottom: 0
• the 9 region at the bottom-right: 3, 6
• the single-cell 1 region at the bottom-right: 1

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