Pips answers for Friday, May 1, 2026

Start with the “=” region at the top: all 3 cells must show the same number, and only 1s appear on enough dominoes to fill it.

• the “=” region at the top: 1, 1, 1
• the 10 region at the bottom-left: 4, 6
• the “=” region at the bottom-right: 2, 2
• the “<6” region at the bottom: 0, 5

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

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

• the single-cell “>1” region at the top: 2
• the single-cell “>1” region at the top-right: 2
• the 10 region at the top-right: 5, 5
• the “=” region at the top: 0, 0
• the single-cell 5 region at the left: 5
• the “≠” region at the bottom-left: 2, 3, 4, 6
• the “>8” region at the right: 5, 6
• the 7 region at the bottom-left: 3, 4
• the single-cell “<6” region at the bottom: 1

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

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

Start with the 9 region at the center: the only way to make it with these dominoes is 4 + 5.

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

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