Pips answers for Saturday, June 13, 2026

Start with the 11 region at the top-right: the only way to make it with these dominoes is 5 + 6.

• the 7 region at the top-left: 3, 4
• the “=” region at the center: 2, 2
• the 11 region at the top-right: 5, 6
• the single-cell “>5” region at the bottom-left: 6

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

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

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

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

Start with the “=” region at the right: all 5 cells must show the same number, and only 0 or 3 or 4 appear on enough dominoes to fill it.

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

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