Pips answers for Friday, February 27, 2026

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

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

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

Start with the 4 region at the top-left: only 1 + 3, or 2 + 2 can make it with these dominoes.

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

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

Start with the “≠” region at the bottom-left: the only way to make it with these dominoes is 0 + 1 + 2 + 3 + 4 + 5 + 6.

• the 6 region at the top-left: 2, 4
• the 0 region at the top: 0, 0, 0, 0
• the 3 region at the top: 1, 2
• the 3 region at the top-right: 1, 2
• the 12 region at the left: 6, 6
• the 6 region at the right: 3, 3
• the “≠” region at the bottom-left: 0, 1, 2, 3, 4, 5, 6
• the 3 region at the bottom: 1, 1, 1
• the single-cell 0 region at the bottom-right: 0

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

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