Pips answers for Thursday, October 16, 2025

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

• the 7 region at the top-left: 1, 6
• the 6 region at the top: 2, 4
• the 12 region at the left: 6, 6
• the “=” region at the bottom-right: 5, 5, 5, 5
• the single-cell “<3” region at the bottom-left: 0

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

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

• the “=” region at the top-left: 1, 1, 1, 1
• the 21 region at the right: 3, 6, 6, 6
• the 6 region at the left: 2, 2, 2
• the “=” region at the bottom-right: 5, 5, 5

1. The “=” region at the top-left needs every pip to match — 4 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 1, column 1 and row 1, column 2.
2. 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 (2 of its 4 cells are already filled) — nothing else fits. Place the 1–3 domino horizontally on row 1, column 3 and row 1, column 4.
3. Row 2, column 1 has only one open neighbor, so its domino must run to row 3, column 1; the “=” region at the top-left needs every pip to match (3 of its 4 cells are already filled) — nothing else fits. Place the 1–2 domino vertically on row 2, column 1 and row 3, column 1.
4. The 21 region at the right needs a total of exactly 21 (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 6–6 domino horizontally on row 2, column 4 and row 2, column 5.
5. The “=” region at the bottom-right needs every pip to match — nothing else fits. Place the 5–6 domino vertically on row 3, column 5 and row 4, column 5.
6. The “=” region at the bottom-right 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 4–5 domino vertically on row 3, column 4 and row 4, column 4.
7. The 6 region at the left needs a total of exactly 6 (1 of its 3 cells is already filled) — nothing else fits. Place the 2–5 domino vertically on row 3, column 3 and row 4, column 3.
8. Row 3, column 2 has only one open neighbor, so its domino must run to row 4, column 2; the 6 region at the left needs a total of exactly 6 (2 of its 3 cells are already filled) — nothing else fits. Place the 2–3 domino vertically on row 3, column 2 and row 4, column 2.

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

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

1. The “=” region at the top needs every pip to match — 3 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–5 domino horizontally on row 1, column 5 and row 1, column 6.
2. The “=” region at the bottom-left needs every pip to match — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–4 domino vertically on row 4, column 3 and row 5, column 3.
3. Row 4, column 1 has only one open neighbor, so its domino must run to row 4, column 2; the single-cell “<4” region at the bottom-left must total less than 4 — nothing else fits. Place the 3–4 domino horizontally on row 4, column 1 and row 4, column 2.
4. The 5 region at the left needs a total of exactly 5 — 6 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–0 domino vertically on row 2, column 2 and row 3, column 2.
5. Row 3, column 3 has only one open neighbor, so its domino must run to row 3, column 4; the 5 region at the left needs a total of exactly 5 (2 of its 5 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–2 domino horizontally on row 3, column 3 and row 3, column 4.
6. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 5 region at the left needs a total of exactly 5 (4 of its 5 cells are already filled) — nothing else fits. Place the 3–5 domino horizontally on row 1, column 2 and row 1, column 3.
7. Row 1, column 4 has only one open neighbor, so its domino must run to row 2, column 4; the “=” region at the top needs every pip to match (2 of its 3 cells are already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 5–6 domino vertically on row 1, column 4 and row 2, column 4.
8. The “=” region at the top needs every pip to match (1 of its 3 cells is already filled) — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 1–6 domino vertically on row 2, column 5 and row 3, column 5.
9. Row 2, column 6 has only one open neighbor, so its domino must run to row 2, column 7; the “=” region at the top needs every pip to match (2 of its 3 cells are already filled) — nothing else fits. Place the 2–6 domino horizontally on row 2, column 6 and row 2, column 7.
10. Row 4, column 5 has only one open neighbor, so its domino must run to row 4, column 4; the 2 region at the center needs a total of exactly 2 (1 of its 2 cells is already filled) — nothing else fits. Place the 1–5 domino horizontally on row 4, column 4 and row 4, column 5.

This board has more than one valid layout; the steps follow the official answer.