Pips answers for Saturday, January 3, 2026

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

• the single-cell “>1” region at the top-left: 2
• the “=” region at the top-left: 6, 6
• the 7 region at the top-right: 3, 4
• the single-cell “<1” region at the bottom-left: 0
• the single-cell 5 region at the bottom: 5
• the “=” region at the bottom-right: 4, 4, 4

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

Start with the “=” region at the center: all 4 cells must show the same number, and only 1 or 5 appear on enough dominoes to fill it.

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

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

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

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

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