Pips answers for Thursday, April 9, 2026

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

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

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

Start with the “=” region at the bottom: all 3 cells must show the same number, and only 0 or 5 appear on enough dominoes to fill it.

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

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

Start with the 3 region at the right: only 0 + 3, or 1 + 2 can make it with these dominoes.

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

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