Pips answers for Thursday, April 30, 2026

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

• the single-cell 4 region at the top-left: 4
• the 4 region at the top: 1, 1, 2
• the single-cell 4 region at the left: 4
• the single-cell 4 region at the bottom-left: 4
• the single-cell “>4” region at the bottom: 5
• the 4 region at the bottom-right: 2, 2

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

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

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

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

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

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

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

1. The single-cell 5 region at the top-left needs a total of exactly 5 — nothing else fits. Place the 2–5 domino vertically on row 1, column 1 and row 2, column 1.
2. The single-cell 4 region at the bottom-left needs a total of exactly 4 — nothing else fits. Place the 0–4 domino vertically on row 4, column 1 and row 5, column 1.
3. The single-cell 4 region at the bottom-left needs a total of exactly 4 — nothing else fits. Place the 3–4 domino horizontally on row 5, column 2 and row 5, column 3.
4. The single-cell 3 region at the bottom needs a total of exactly 3 — nothing else fits. Place the 0–3 domino horizontally on row 5, column 4 and row 5, column 5.
5. The single-cell 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 2–3 domino horizontally on row 1, column 3 and row 1, column 4.
6. Row 1, column 2 has only one open neighbor, so its domino must run to row 2, column 2; the single-cell 4 region at the top-left needs a total of exactly 4 — nothing else fits. Place the 4–5 domino vertically on row 1, column 2 and row 2, column 2.
7. The single-cell 4 region at the top-right 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 2–4 domino horizontally on row 1, column 5 and row 1, column 6.
8. 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 0–2 domino vertically on row 2, column 6 and row 3, column 6.
9. The single-cell 2 region at the bottom-right needs a total of exactly 2 — nothing else fits. Place the 1–2 domino vertically on row 3, column 5 and row 4, column 5.
10. The single-cell 0 region at the top needs a total of exactly 0 — nothing else fits. Place the 0–5 domino horizontally on row 2, column 4 and row 2, column 5.
11. The single-cell 0 region at the bottom-left needs a total of exactly 0 — nothing else fits. Place the 0–1 domino horizontally on row 4, column 2 and row 4, column 3.
12. Row 4, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the single-cell 4 region at the bottom needs a total of exactly 4 — nothing else fits. Place the 1–4 domino vertically on row 3, column 4 and row 4, column 4.
13. Row 4, column 6 has only one open neighbor, so its domino must run to row 5, column 6; the single-cell “<3” region at the bottom-right must total less than 3 — nothing else fits. Place the 1–5 domino vertically on row 4, column 6 and row 5, column 6.
14. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the single-cell 5 region at the left needs a total of exactly 5 — nothing else fits. Place the 3–5 domino horizontally on row 3, column 1 and row 3, column 2.
15. Row 2, column 3 has only one open neighbor, so its domino must run to row 3, column 3; the single-cell 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 1–3 domino vertically on row 2, column 3 and row 3, column 3.