Pips answers for Friday, July 3, 2026

Start with the 3 region at the bottom-left: the only way to make it with these dominoes is 0 + 3.

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

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

Start with the “=” region at the left: all 3 cells must show the same number, and only 4 or 6 appear on enough dominoes to fill it.

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

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

Start with the 12 region at the left: the only way to make it with these dominoes is 6 + 6.

• the single-cell 3 region at the top: 3
• the 7 region at the top-right: 2, 5
• the “=” region at the top-left: 0, 0
• the 10 region at the top: 5, 5
• the “=” region at the top-right: 0, 0
• the 12 region at the left: 6, 6
• the 2 region at the right: 1, 1
• the 6 region at the left: 3, 3
• the 5 region at the center: 2, 3
• the single-cell 1 region at the right: 1
• the single-cell 3 region at the right: 3
• the single-cell 5 region at the right: 5
• the “=” region at the bottom: 4, 4, 4, 4
• the 12 region at the bottom-left: 6, 6
• the 10 region at the bottom-left: 5, 5

1. Row 6, column 6 has only one open neighbor, so its domino must run to row 6, column 7; the single-cell 1 region at the right needs a total of exactly 1 — nothing else fits. Place the 1–3 domino horizontally on row 6, column 6 and row 6, column 7.
2. The 2 region at the right needs a total of exactly 2 — nothing else fits. Place the 1–5 domino vertically on row 5, column 8 and row 6, column 8.
3. The “=” region at the top-right needs every pip to match — nothing else fits. Place the 0–1 domino vertically on row 3, column 8 and row 4, column 8.
4. The 12 region at the left needs a total of exactly 12 — 2 placements fit here for now, but every other one leaves some region impossible a few moves later. Place the 5–6 domino horizontally on row 4, column 3 and row 4, column 4.
5. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the single-cell 3 region at the top needs a total of exactly 3 — nothing else fits. Place the 3–5 domino vertically on row 2, column 4 and row 3, column 4.
6. The “=” region at the bottom 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 4–4 domino vertically on row 9, column 4 and row 10, column 4.
7. The 12 region at the bottom-left needs a total of exactly 12 — nothing else fits. Place the 4–6 domino horizontally on row 8, column 3 and row 8, column 4.
8. The 10 region at the bottom-left needs a total of exactly 10 — nothing else fits. Place the 5–5 domino vertically on row 9, column 2 and row 10, column 2.
9. The “=” region at the top-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 0–6 domino horizontally on row 4, column 1 and row 4, column 2.
10. The 6 region at the left needs a total of exactly 6 — nothing else fits. Place the 3–6 domino vertically on row 7, column 2 and row 8, column 2.
11. Row 3, column 1 has only one open neighbor, so its domino must run to row 2, column 1; the “=” region at the top-left needs every pip to match (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 0–4 domino vertically on row 2, column 1 and row 3, column 1.
12. The 7 region at the top-right needs a total of exactly 7 — nothing else fits. Place the 0–2 domino horizontally on row 3, column 6 and row 3, column 7.
13. Row 2, column 6 has only one open neighbor, so its domino must run to row 1, column 6; the 7 region at the top-right needs a total of exactly 7 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–5 domino vertically on row 1, column 6 and row 2, column 6.
14. The 5 region at the center needs a total of exactly 5 — nothing else fits. Place the 3–4 domino vertically on row 6, column 4 and row 7, column 4.
15. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the 6 region at the left needs a total of exactly 6 (1 of its 2 cells is already filled) — nothing else fits. Place the 2–3 domino horizontally on row 6, column 2 and row 6, column 3.
16. Row 1, column 7 has only one open neighbor, so its domino must run to row 1, column 8 — nothing else fits. Place the 2–2 domino horizontally on row 1, column 7 and row 1, column 8.