Pips answers for Tuesday, February 17, 2026

Start with the 12 region at the top-right: the only way to make it with these dominoes is 6 + 6.

• the “=” region at the top-left: 0, 0
• the 12 region at the top-right: 6, 6
• the single-cell 0 region at the bottom-left: 0
• the “=” region at the bottom-left: 4, 4
• the 2 region at the bottom-right: 0, 2
• the single-cell 5 region at the bottom-right: 5

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

Start with the “=” region at the right: all 3 cells must show the same number, and only 0 or 2 or 6 appear on enough dominoes to fill it.

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

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

Start with the “=” region at the right: all 6 cells must show the same number, and only 5s appear on enough dominoes to fill it.

• the 7 region at the left: 1, 2, 2, 2
• the single-cell “>3” region at the top: 6
• the single-cell “<3” region at the top-right: 2
• the 9 region at the center: 3, 6
• the “=” region at the right: 5, 5, 5, 5, 5, 5
• the 1 region at the center: 0, 0, 1
• the “=” region at the bottom-right: 0, 0, 0
• the 6 region at the bottom-right: 2, 4

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

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