Pips answers for Tuesday, August 4, 2026

Start with the 4 region at the right: the only way to make it with these dominoes is 1 + 1 + 1 + 1.

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

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

Start with the 0 region at the bottom-left: the only way to make it with these dominoes is 0 + 0 + 0.

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

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

Start with the 4 region at the left: only 0 + 4, or 1 + 3 can make it with these dominoes.

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

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

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