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NYT Pips answers — Wednesday, August 12, 2026

Pips #360 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Ian Livengood · Hard by Rodolfo Kurchan

Stuck on today's New York Times Pips? Below are three progressive hints for each difficulty, then the full solution as a tap-to-reveal board. Hints first — the answer only when you want it.

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Easy Pips answer — August 12, 2026

3×4 board · 5 dominoes · 6 regions · by Ian Livengood

The easy board runs 3 rows by 4 columns: 5 dominoes to place across 6 regions. The mix is 4 sum regions, 1 "less than", 1 free region. The two-cell region at row 1, column 2 adds to 11: the only split is 5 + 6.. After that, the "<2" region at row 3, column 1 leaves almost no room: only 0s and a single 1 can go here.. Rack check: 2 doubles, 1 blank half, 1 six.

Hints

Hint 1 — where to start

The two-cell region at row 1, column 2 adds to 11: the only split is 5 + 6.

Hint 2 — first domino

Place the 2–6 domino with its 2 at row 1, column 3 and its 6 at row 1, column 2.

Hint 3 — second domino

Next, the 0–3 goes at row 3, column 1 → row 2, column 1. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 11 must add up to 11; 5 must add up to 5; <2 must add up to less than 2; 6 must add up to 6.

11
5
5
<2
6
Solution as text
  1. 3–3 — 3 at row 1, column 4, 3 at row 2, column 4
  2. 2–6 — 2 at row 1, column 3, 6 at row 1, column 2
  3. 0–3 — 0 at row 3, column 1, 3 at row 2, column 1
  4. 2–2 — 2 at row 3, column 3, 2 at row 3, column 4
  5. 4–5 — 4 at row 3, column 2, 5 at row 2, column 2

Medium Pips answer — August 12, 2026

3×5 board · 7 dominoes · 7 regions · by Ian Livengood

Medium today is a 3×5 board with 7 dominoes and 7 regions. Rules on the board: 1 sum region, 2 "all equal", 3 "greater than", 1 free region. The way in: the single cell at row 3, column 2 is marked 4, so it can only hold a 4. After that, the "=" region starting at row 1, column 1 has 3 cells that must all match — every 1 in the puzzle should come here.. Rack check: 1 six.

Hints

Hint 1 — where to start

The single cell at row 3, column 2 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.

Hint 2 — first domino

Place the 4–2 domino with its 4 at row 3, column 2 and its 2 at row 3, column 3.

Hint 3 — second domino

Next, the 3–1 goes at row 2, column 1 → row 1, column 1. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: = must all be the same number; >6 must add up to more than 6; >8 must add up to more than 8; 4 must add up to 4.

=
>6
=
>8
>8
4
Solution as text
  1. 4–2 — 4 at row 3, column 2, 2 at row 3, column 3
  2. 1–4 — 1 at row 1, column 3, 4 at row 1, column 4
  3. 3–6 — 3 at row 1, column 5, 6 at row 2, column 5
  4. 2–1 — 2 at row 2, column 2, 1 at row 1, column 2
  5. 5–4 — 5 at row 3, column 4, 4 at row 3, column 5
  6. 5–2 — 5 at row 2, column 4, 2 at row 2, column 3
  7. 3–1 — 3 at row 2, column 1, 1 at row 1, column 1

Hard Pips answer — August 12, 2026

9×9 board · 16 dominoes · 17 regions · by Rodolfo Kurchan

The hard board runs 9 rows by 9 columns: 16 dominoes to place across 17 regions. The mix is 5 sum regions, 4 "all equal", 2 "less than", 5 "greater than", 1 free region. The single cell at row 1, column 2 is marked 3, so it can only hold a 3. After that, the single cell at row 3, column 5 is marked 5, so it can only hold a 5. Rack check: 5 doubles, 7 blank halves, 5 sixes.

Hints

Hint 1 — where to start

The single cell at row 1, column 2 is marked 3, so it can only hold a 3. Find the domino with a 3 that fits there.

Hint 2 — first domino

Place the 0–3 domino with its 0 at row 2, column 2 and its 3 at row 1, column 2.

Hint 3 — second domino

Next, the 0–5 goes at row 3, column 4 → row 3, column 5. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 3 must add up to 3; = must all be the same number; 16 must add up to 16; >3 must add up to more than 3; 5 must add up to 5.

3
=
16
>3
=
5
<3
>0
<3
>3
=
>0
14
12
=
>3
Solution as text
  1. 0–0 — 0 at row 4, column 4, 0 at row 4, column 5
  2. 0–1 — 0 at row 4, column 2, 1 at row 4, column 1
  3. 0–2 — 0 at row 9, column 8, 2 at row 8, column 8
  4. 0–3 — 0 at row 2, column 2, 3 at row 1, column 2
  5. 0–5 — 0 at row 3, column 4, 5 at row 3, column 5
  6. 0–6 — 0 at row 3, column 2, 6 at row 3, column 1
  7. 1–2 — 1 at row 8, column 1, 2 at row 7, column 1
  8. 1–4 — 1 at row 6, column 4, 4 at row 6, column 5
  9. 1–6 — 1 at row 4, column 8, 6 at row 4, column 7
  10. 3–3 — 3 at row 6, column 6, 3 at row 7, column 6
  11. 3–4 — 3 at row 9, column 6, 4 at row 9, column 5
  12. 3–5 — 3 at row 8, column 6, 5 at row 8, column 5
  13. 4–4 — 4 at row 8, column 4, 4 at row 9, column 4
  14. 5–5 — 5 at row 2, column 7, 5 at row 3, column 7
  15. 5–6 — 5 at row 9, column 1, 6 at row 9, column 2
  16. 6–6 — 6 at row 7, column 8, 6 at row 7, column 9

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