Pips answers for Sunday, August 23, 2026

Start with the “=” region at the top-left: all 2 cells must show the same number, and only 2 or 5 or 6 appear on enough dominoes to fill it.

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

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

Start with the “=” region at the left: all 3 cells must show the same number, and only 0 or 3 or 5 appear on enough dominoes to fill it.

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

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

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

• the 8 region at the top-left: 3, 5
• the 4 region at the top-left: 0, 4
• the 7 region at the top: 1, 6
• the 3 region at the left: 1, 2
• the single-cell 5 region at the center: 5
• the 12 region at the left: 0, 6, 6
• the 12 region at the bottom-left: 0, 6, 6
• the 9 region at the bottom: 1, 4, 4
• the “=” region at the bottom-right: 4, 4
• the single-cell “>2” region at the right: 3
• the single-cell 2 region at the bottom-right: 2

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

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