Pips answers for Thursday, March 19, 2026

Start with the “=” region at the left: all 3 cells must show the same number, and only 6s appear on enough dominoes to fill it.

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

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

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

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

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

Start with the 15 region at the right: only 3 + 6 + 6, or 4 + 5 + 6 can make it with these dominoes.

• the “=” region at the top-left: 4, 4
• the “=” region at the top-right: 2, 2, 2
• the 10 region at the top-right: 5, 5
• the single-cell 1 region at the left: 1
• the single-cell 0 region at the left: 0
• the single-cell 2 region at the left: 2
• the 15 region at the right: 3, 6, 6
• the single-cell 3 region at the left: 3
• the single-cell 2 region at the center: 2
• the single-cell 4 region at the bottom-left: 4
• the 5 region at the bottom-left: 1, 4
• the “=” region at the bottom-right: 3, 3, 3

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