Pips answers for Thursday, July 23, 2026

Start with the “=” region at the right: all 2 cells must show the same number, and only 1 or 6 appear on enough dominoes to fill it.

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

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

Start with the “<1” region at the top: the only way to make it with these dominoes is 0 + 0 + 0.

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

1. The “>11” region at the bottom must total more than 11 — 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 horizontally on row 6, column 2 and row 6, column 3.
2. The “=” region at the center 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 2–5 domino horizontally on row 3, column 1 and row 3, column 2.
3. The “=” region at the center needs every pip to match (1 of its 4 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 2–3 domino horizontally on row 3, column 5 and row 3, column 6.
4. The “<1” region at the top must total less than 1 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–0 domino horizontally on row 2, column 3 and row 2, column 4.
5. Row 1, column 3 has only one open neighbor, so its domino must run to row 1, column 4; the single-cell “<2” region at the top must total less than 2 — nothing else fits. Place the 0–1 domino horizontally on row 1, column 3 and row 1, column 4.
6. The “>11” region at the bottom must total more than 11 (1 of its 2 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 4–6 domino vertically on row 4, column 3 and row 5, column 3.
7. 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 (2 of its 4 cells are already filled) — nothing else fits. Place the 2–2 domino horizontally on row 3, column 3 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 4 region at the center needs a total of exactly 4 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–4 domino vertically on row 4, column 4 and row 5, column 4.
9. Row 6, column 4 has only one open neighbor, so its domino must run to row 6, column 5; the “=” region at the bottom needs every pip to match (1 of its 2 cells is already filled) — nothing else fits. Place the 2–4 domino horizontally on row 6, column 4 and row 6, column 5.

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

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

• the 13 region at the top-left: 4, 4, 5
• the 22 region at the top-right: 5, 5, 6, 6
• the 11 region at the top-left: 5, 6
• the 1 region at the center: 0, 0, 1
• the 9 region at the top-right: 3, 6
• the 9 region at the right: 4, 5
• the 3 region at the bottom-left: 0, 3
• the 7 region at the bottom-right: 3, 4
• the single-cell “>0” region at the bottom: 6
• the 0 region at the bottom-right: 0, 0, 0
• the “=” region at the bottom-left: 2, 2, 2
• the single-cell 1 region at the bottom-right: 1

1. The 7 region at the bottom-right needs a total of exactly 7 — nothing else fits. Place the 1–4 domino vertically on row 5, column 8 and row 6, column 8.
2. The 0 region at the bottom-right needs a total of exactly 0 — nothing else fits. Place the 0–0 domino horizontally on row 6, column 6 and row 6, column 7.
3. Row 2, column 4 has only one open neighbor, so its domino must run to row 3, column 4; the 1 region at the center needs a total of exactly 1 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 0–1 domino vertically on row 2, column 4 and row 3, column 4.
4. Row 5, column 6 has only one open neighbor, so its domino must run to row 4, column 6; the 0 region at the bottom-right needs a total of exactly 0 (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 0–5 domino vertically on row 4, column 6 and row 5, column 6.
5. The 22 region at the top-right needs a total of exactly 22 — 2 placements fit here for now; this one matches the official answer (this board has more than one valid layout). Place the 4–6 domino vertically on row 2, column 6 and row 3, column 6.
6. Row 4, column 4 has only one open neighbor, so its domino must run to row 5, column 4; the 1 region at the center needs a total of exactly 1 (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 0–6 domino vertically on row 4, column 4 and row 5, column 4.
7. The 3 region at the bottom-left needs a total of exactly 3 — nothing else fits. Place the 3–5 domino vertically on row 3, column 1 and row 4, column 1.
8. The 9 region at the top-right needs a total of exactly 9 — nothing else fits. Place the 3–3 domino vertically on row 3, column 8 and row 4, column 8.
9. Row 5, column 1 has only one open neighbor, so its domino must run to row 6, column 1; the 3 region at the bottom-left needs a total of exactly 3 (1 of its 2 cells is already filled) — nothing else fits. Place the 0–2 domino vertically on row 5, column 1 and row 6, column 1.
10. Row 6, column 2 has only one open neighbor, so its domino must run to row 6, column 3; the “=” region at the bottom-left needs every pip to match (1 of its 3 cells is already filled) — nothing else fits. Place the 2–2 domino horizontally on row 6, column 2 and row 6, column 3.
11. The 22 region at the top-right needs a total of exactly 22 (1 of its 4 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 6–6 domino vertically on row 1, column 8 and row 2, column 8.
12. The 13 region at the top-left needs a total of exactly 13 — nothing else fits. Place the 5–6 domino vertically on row 1, column 1 and row 2, column 1.
13. Row 1, column 2 has only one open neighbor, so its domino must run to row 1, column 3; the 13 region at the top-left needs a total of exactly 13 (1 of its 3 cells is already filled) — nothing else fits. Place the 4–4 domino horizontally on row 1, column 2 and row 1, column 3.
14. Row 1, column 6 has only one open neighbor, so its domino must run to row 1, column 7; the 22 region at the top-right needs a total of exactly 22 (2 of its 4 cells are already filled) — nothing else fits. Place the 5–5 domino horizontally on row 1, column 6 and row 1, column 7.

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