Pips answers for Tuesday, August 19, 2025

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

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

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

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

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

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

Start with the “=” region at the bottom: all 6 cells must show the same number, and only 1s appear on enough dominoes to fill it.

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

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

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