Pips answers for Wednesday, September 3, 2025

Start with the 10 region at the center: the only way to make it with these dominoes is 4 + 6.

• the single-cell 4 region at the left: 4
• the 10 region at the center: 4, 6
• the “=” region at the right: 2, 2

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

Start with the “=” region at the bottom-left: all 3 cells must show the same number, and only 0 or 1 appear on enough dominoes to fill it.

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

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

Start with the 7 region at the left — it has the fewest pip combinations on the board, so the dominoes around it are the easiest to pin down.

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

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