Pips answers for Sunday, February 8, 2026

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

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

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

Start with the “=” region at the top-right: all 4 cells must show the same number, and only 4s appear on enough dominoes to fill it.

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

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

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

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

• the 22 region at the top: 5, 5, 6, 6
• the 18 region at the top-right: 6, 6, 6
• the 1 region at the left: 0, 0, 0, 0, 1
• the “=” region at the right: 3, 3
• the “=” region at the right: 4, 4, 4
• the “=” region at the bottom-left: 2, 2
• the “=” region at the bottom-right: 5, 5, 5
• the 11 region at the bottom-left: 5, 6
• the single-cell “>2” region at the bottom-left: 4
• the single-cell “>2” region at the bottom-right: 3

1. The 18 region at the top-right needs a total of exactly 18 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 6–6 domino horizontally on row 1, column 7 and row 1, column 8.
2. The 22 region at the top needs a total of exactly 22 — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–5 domino horizontally on row 1, column 3 and row 1, column 4.
3. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the 22 region at the top needs a total of exactly 22 (2 of its 4 cells are already filled) — nothing else fits. Place the 0–6 domino vertically on row 2, column 3 and row 3, column 3.
4. The “=” region at the right 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 4–4 domino vertically on row 4, column 8 and row 5, column 8.
5. Row 4, column 3 has only one open neighbor, so its domino must run to row 5, column 3; the 1 region at the left needs a total of exactly 1 (1 of its 5 cells is already filled) — 3 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 4, column 3 and row 5, column 3.
6. The “=” region at the bottom-right 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 4–5 domino horizontally on row 6, column 7 and row 6, column 8.
7. The “=” region at the bottom-right needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 3–5 domino horizontally on row 7, column 7 and row 7, column 8.
8. Row 6, column 6 has only one open neighbor, so its domino must run to row 7, column 6; the “=” region at the bottom-right needs every pip to match (2 of its 3 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–5 domino vertically on row 6, column 6 and row 7, column 6.
9. Row 1, column 5 has only one open neighbor, so its domino must run to row 1, column 6; the 22 region at the top needs a total of exactly 22 (3 of its 4 cells are already filled) — 3 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino horizontally on row 1, column 5 and row 1, column 6.
10. Row 2, column 8 has only one open neighbor, so its domino must run to row 3, column 8; the 18 region at the top-right needs a total of exactly 18 (2 of its 3 cells are already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 3–6 domino vertically on row 2, column 8 and row 3, column 8.
11. Row 3, column 7 has only one open neighbor, so its domino must run to row 3, column 6; the “=” region at the right needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–3 domino horizontally on row 3, column 6 and row 3, column 7.
12. The single-cell “>2” region at the bottom-left must total more than 2 — nothing else fits. Place the 4–6 domino horizontally on row 7, column 2 and row 7, column 3.
13. The “=” region at the bottom-left needs every pip to match — nothing else fits. Place the 2–5 domino vertically on row 6, column 1 and row 7, column 1.
14. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, 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 1–2 domino horizontally on row 6, column 2 and row 6, column 3.
15. Row 3, column 4 has only one open neighbor, so its domino must run to row 3, column 5; the 1 region at the left needs a total of exactly 1 (4 of its 5 cells are already filled) — nothing else fits. Place the 0–1 domino horizontally on row 3, column 4 and row 3, column 5.