Pips answers for Sunday, May 3, 2026

Start with the “=” region at the bottom-left: all 3 cells must show the same number, and only 0 or 3 appear on enough dominoes to fill it.

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

Start with the “=” region at the left: all 3 cells must show the same number, and only 0 or 2 appear on enough dominoes to fill it.

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

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

Start with the 11 region at the top-right: the only way to make it with these dominoes is 5 + 6.

• the 11 region at the top-right: 5, 6
• the 12 region at the top-right: 4, 4, 4
• the 3 region at the left: 1, 1, 1
• the “≠” region at the center: 0, 1, 2, 3, 4, 5
• the single-cell 2 region at the center: 2
• the 8 region at the left: 4, 4
• the 6 region at the bottom-left: 2, 2, 2
• the single-cell 6 region at the bottom-left: 6

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

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