Pips answers for Saturday, April 25, 2026

Start with the “=” region at the top: all 4 cells must show the same number, and only 1 or 3 appear on enough dominoes to fill it.

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

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

Start with the 6 region at the right: only 0 + 6, or 2 + 4, or 3 + 3 can make it with these dominoes.

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

1. The single-cell 6 region at the left needs a total of exactly 6 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–6 domino vertically on row 2, column 1 and row 3, column 1.
2. The “<4” region at the bottom must total less than 4 — 5 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–4 domino vertically on row 3, column 2 and row 4, column 2.
3. The 6 region at the right needs a total of exactly 6 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 0–3 domino vertically on row 3, column 3 and row 4, column 3.
4. The 6 region at the right needs a total of exactly 6 — nothing else fits. Place the 3–4 domino vertically on row 2, column 4 and row 3, column 4.
5. The 6 region at the right needs a total of exactly 6 (1 of its 2 cells is already filled) — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 2–2 domino horizontally on row 2, column 2 and row 2, column 3.
6. 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.

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

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

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

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