Pips answers for Sunday, April 12, 2026

Start with the 22 region at the top: the only way to make it with these dominoes is 5 + 5 + 6 + 6.

• the 22 region at the top: 5, 5, 6, 6
• the single-cell 4 region at the top-right: 4
• the 4 region at the left: 2, 2
• the “=” region at the center: 4, 4
• the single-cell “>1” region at the right: 2

1. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the 22 region at the top needs a total of exactly 22 — nothing else fits. Place the 6–6 domino horizontally on row 1, column 1 and row 1, column 2.
2. The 22 region at the top needs a total of exactly 22 (2 of its 4 cells are already filled) — nothing else fits. Place the 4–5 domino horizontally on row 1, column 4 and row 1, column 5.
3. Row 1, column 3 has only one open neighbor, so its domino must run to row 2, column 3; the 22 region at the top needs a total of exactly 22 (3 of its 4 cells are already filled) — nothing else fits. Place the 0–5 domino vertically on row 1, column 3 and row 2, column 3.
4. Row 3, column 1 has only one open neighbor, so its domino must run to row 3, column 2; the 4 region at the left needs a total of exactly 4 — nothing else fits. Place the 2–2 domino horizontally on row 3, column 1 and row 3, column 2.
5. The “=” region at the center 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 2–4 domino horizontally on row 3, column 4 and row 3, column 5.
6. Row 3, column 3 has only one open neighbor, so its domino must run to row 4, column 3; the “=” region at the center needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 1–4 domino vertically on row 3, column 3 and row 4, column 3.

Start with the 1 region at the bottom-right: the only way to make it with these dominoes is 0 + 1.

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

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

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

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

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