Pips answers for Tuesday, April 7, 2026

Start with the 3 region at the bottom: the only way to make it with these dominoes is 1 + 1 + 1.

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

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

Start with the 4 region at the top-left: only 0 + 1 + 3, or 0 + 2 + 2, or 1 + 1 + 2 can make it with these dominoes.

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

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

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

• the 4 region at the top-left: 2, 2
• the 6 region at the top: 3, 3
• the 2 region at the top-right: 1, 1
• the 8 region at the top-right: 4, 4
• the 0 region at the left: 0, 0
• the single-cell 1 region at the top-left: 1
• the 10 region at the center: 5, 5
• the single-cell 0 region at the top-right: 0
• the 8 region at the left: 4, 4
• the single-cell 0 region at the center: 0
• the 4 region at the right: 2, 2
• the single-cell 2 region at the bottom-left: 2
• the “=” region at the bottom: 3, 3, 3
• the 9 region at the bottom-right: 4, 5
• the 10 region at the bottom-left: 5, 5
• the 2 region at the bottom: 1, 1
• the single-cell 0 region at the bottom-right: 0

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

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