Pips answers for Friday, October 3, 2025

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

• the single-cell 2 region at the left: 2
• the “=” region at the center: 4, 4
• the “=” region at the center: 1, 1, 1, 1
• the single-cell 0 region at the bottom-left: 0

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

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

• the “=” region at the top: 3, 3, 3
• the “=” region at the top-right: 4, 4, 4, 4, 4
• the “=” region at the top-left: 1, 1
• the 0 region at the left: 0, 0
• the 12 region at the bottom-left: 6, 6
• the “=” region at the bottom-right: 2, 2
• the 15 region at the bottom: 5, 5, 5
• the “=” region at the bottom-left: 1, 1

1. The 0 region at the left needs a total of exactly 0 — nothing else fits. Place the 0–6 domino vertically on row 4, column 1 and row 5, column 1.
2. The “=” region at the top-left needs every pip to match — nothing else fits. Place the 0–1 domino vertically on row 2, column 1 and row 3, column 1.
3. The 12 region at the bottom-left needs a total of exactly 12 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–6 domino vertically on row 6, column 1 and row 7, column 1.
4. Row 1, column 1 has only one open neighbor, so its domino must run to row 1, column 2; the “=” region at the top 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 1, column 1 and row 1, column 2.
5. The “=” region at the top-right 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 1–4 domino horizontally on row 2, column 2 and row 2, column 3.
6. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the “=” region at the top needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 3–4 domino horizontally on row 1, column 3 and row 1, column 4.
7. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the “=” region at the top-right needs every pip to match (2 of its 5 cells are already filled) — nothing else fits. Place the 4–4 domino vertically on row 2, column 4 and row 3, column 4.
8. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the “=” region at the top-right needs every pip to match (4 of its 5 cells are already filled) — nothing else fits. Place the 2–4 domino vertically on row 4, column 4 and row 5, column 4.
9. The 15 region at the bottom needs a total of exactly 15 — nothing else fits. Place the 1–5 domino vertically on row 6, column 2 and row 7, column 2.
10. The 15 region at the bottom needs a total of exactly 15 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–5 domino horizontally on row 6, column 3 and row 6, column 4.
11. Row 7, column 3 has only one open neighbor, so its domino must run to row 7, column 4; the 15 region at the bottom needs a total of exactly 15 (2 of its 3 cells are already filled) — nothing else fits. Place the 3–5 domino horizontally on row 7, column 3 and row 7, column 4.

Start with the “>11” region at the bottom-left: the only way to make it with these dominoes is 6 + 6.

• the “>9” region at the top: 5, 5
• the single-cell “<3” region at the center: 1
• the “=” region at the bottom-right: 2, 2, 2, 2, 2
• the “=” region at the bottom-right: 3, 3
• the “>11” region at the bottom-left: 6, 6
• the 7 region at the bottom-left: 2, 5
• the “<1” region at the bottom: 0, 0
• the single-cell “<5” region at the bottom-right: 4
• the 3 region at the bottom: 1, 1, 1

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