Pips answers for Monday, June 15, 2026

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

• the 0 region at the top-left: 0, 0
• the 2 region at the top-left: 1, 1
• the 4 region at the left: 1, 3
• the 6 region at the bottom-left: 2, 4

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

Start with the 9 region at the top-left: only 3 + 6, or 4 + 5 can make it with these dominoes.

• the 9 region at the top-left: 4, 5
• the 8 region at the top-right: 3, 5
• the 8 region at the bottom-left: 2, 6
• the 6 region at the bottom-right: 3, 3
• the single-cell “<3” region at the bottom-left: 0

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

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

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

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