Pips answers for Sunday, April 26, 2026

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

• the 9 region at the center: 4, 5
• the “=” region at the bottom-left: 6, 6, 6
• the 2 region at the right: 0, 2
• the “=” region at the bottom-right: 1, 1

1. Row 3, column 1 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 — 2 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 3, column 1 and row 4, column 1.
2. Row 4, column 2 has only one open neighbor, so its domino must run to row 4, column 3; the “=” region at the bottom-left needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 1–6 domino horizontally on row 4, column 2 and row 4, column 3.
3. The 9 region at the center needs a total of exactly 9 — nothing else fits. Place the 2–4 domino vertically on row 2, column 3 and row 3, column 3.
4. Row 2, column 2 has only one open neighbor, so its domino must run to row 1, column 2; the 9 region at the center needs a total of exactly 9 (1 of its 2 cells is already filled) — nothing else fits. Place the 5–5 domino vertically on row 1, column 2 and row 2, column 2.
5. Row 3, column 4 has only one open neighbor, so its domino must run to row 4, column 4; the 2 region at the right needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–1 domino vertically on row 3, column 4 and row 4, column 4.

Start with the 10 region at the top-right: only 4 + 6, or 5 + 5 can make it with these dominoes.

• the 6 region at the top-left: 2, 4
• the 10 region at the top-right: 5, 5
• the single-cell 6 region at the center: 6
• the 3 region at the right: 1, 2
• the 6 region at the left: 3, 3
• the 6 region at the right: 1, 5
• the 6 region at the bottom: 1, 1, 4

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

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

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

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

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

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