Pips answers for Wednesday, February 11, 2026

Start with the 11 region at the center: the only way to make it with these dominoes is 5 + 6.

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

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

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

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

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

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

Start with the “=” region at the center: all 5 cells must show the same number, and only 3 or 6 appear on enough dominoes to fill it.

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

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