Pips answers for Wednesday, July 22, 2026

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

• the 10 region at the top-left: 4, 6
• the 4 region at the top-right: 0, 4
• the 11 region at the bottom-left: 5, 6
• the “<4” region at the bottom-right: 0, 3

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

Start with the 6 region at the right: only 0 + 6, or 1 + 5, or 3 + 3 can make it with these dominoes.

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

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

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

• the “=” region at the top: 3, 3
• the single-cell 5 region at the top-right: 5
• the 3 region at the top: 1, 2
• the “=” region at the top: 4, 4, 4
• the single-cell 1 region at the top-right: 1
• the 12 region at the top-right: 6, 6
• the “=” region at the left: 4, 4
• the single-cell 2 region at the top-left: 2
• the “=” region at the left: 3, 3
• the 0 region at the center: 0, 0, 0
• the single-cell 2 region at the top-right: 2
• the 11 region at the left: 5, 6
• the 13 region at the right: 3, 5, 5
• the “=” region at the center: 2, 2
• the single-cell 1 region at the center: 1
• the single-cell 0 region at the bottom: 0
• the single-cell “>4” region at the bottom: 5
• the “=” region at the bottom: 3, 3

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