Pips answers for Thursday, February 12, 2026

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

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

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

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

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

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

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

Start with the 0 region at the top-right: the only way to make it with these dominoes is 0 + 0.

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

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