Pips answers for Saturday, December 6, 2025

Start with the 12 region at the top-left: the only way to make it with these dominoes is 6 + 6.

• the 12 region at the top-left: 6, 6
• the single-cell “<5” region at the top: 4
• the 6 region at the top-right: 3, 3
• the single-cell 2 region at the top-left: 2
• the 10 region at the left: 4, 6
• the 5 region at the bottom-left: 0, 5

1. The 12 region at the top-left needs a total of exactly 12 — nothing else fits. Place the 2–6 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 12 region at the top-left needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–6 domino horizontally on row 1, column 2 and row 1, column 3.
3. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the 6 region at the top-right needs a total of exactly 6 — nothing else fits. Place the 3–3 domino vertically on row 1, column 4 and row 2, column 4.
4. The 10 region at the left 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 5–6 domino vertically on row 3, column 1 and row 4, column 1.
5. Row 3, column 2 has only one open neighbor, so its domino must run to row 4, column 2; the 10 region at the left needs a total of exactly 10 (1 of its 2 cells is already filled) — nothing else fits. Place the 4–4 domino vertically on row 3, column 2 and row 4, column 2.
6. Row 5, column 1 has only one open neighbor, so its domino must run to row 5, column 2; the 5 region at the bottom-left needs a total of exactly 5 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–0 domino horizontally on row 5, column 1 and row 5, column 2.

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

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

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

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

• the 1 region at the top-right: 0, 1
• the 10 region at the top-right: 4, 6
• the single-cell “>1” region at the top-left: 6
• the single-cell 2 region at the top: 2
• the 8 region at the top-right: 2, 6
• the single-cell “>1” region at the left: 5
• the 2 region at the right: 1, 1
• the “=” region at the left: 3, 3
• the “=” region at the bottom-left: 0, 0
• the “=” region at the bottom-left: 4, 4, 4, 4
• the “<2” region at the bottom-right: 0, 0

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