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NYT Pips answers — Sunday, October 4, 2026

Pips #413 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Rodolfo Kurchan · Hard by Rodolfo Kurchan

Grid Puzzles · Updated October 4, 2026, 07:03 UTC · About these pages

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.

Jump to: Easy · Medium · Hard

Also today: LinkedIn games answers — Zip, Pinpoint, Crossclimb, Mini Sudoku, Patches and Wend.

Easy Pips answer — October 4, 2026

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

Easy is 4 regions on a 3×3 grid, filled by 4 dominoes. You're working with 3 sum regions, 1 "all equal". Another early foothold: two cells at row 1, column 3 total 9. The rack holds 1 double, 1 six.

Hints

Hint 1: where to start

At row 1, column 1 two cells must match. That is a double, or matching halves from two dominoes.

Hint 2: first domino

The 2–2 goes there: 2 on row 1, column 1, 2 on row 2, column 1.

Hint 3: second domino

Next, the 4–2 goes at row 1, column 3 → row 1, column 2. That leaves 2 dominoes, each with far fewer homes.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: = must all be the same number; 9 must add up to 9; 7 must add up to 7.

=
9
7
9
Solution as text
  1. 4–2 — 4 at row 1, column 3, 2 at row 1, column 2
  2. 2–2 — 2 at row 1, column 1, 2 at row 2, column 1
  3. 5–6 — 5 at row 3, column 1, 6 at row 3, column 2
  4. 3–5 — 3 at row 3, column 3, 5 at row 2, column 3

Medium Pips answer — October 4, 2026

5×4 board · 7 dominoes · 9 regions · by Rodolfo Kurchan

Medium is 9 regions on a 5×4 grid, filled by 7 dominoes. By rule type: 3 sum regions, 1 "all equal", 1 "less than", 1 "greater than", 2 free regions. Another early foothold: a one-cell region at row 5, column 3 asks for 2, which fixes that cell outright. Rack check: 1 double, 1 six.

Hints

Hint 1: where to start

The single cell at row 4, column 1 is marked 1, so it can only hold a 1. Find the domino with a 1 that fits there.

Hint 2: first domino

The 5–1 goes there: 5 on row 4, column 2, 1 on row 4, column 1.

Hint 3: second domino

Next, the 2–3 goes at row 5, column 3 → row 4, column 3. That leaves 5 dominoes, each with far fewer homes.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: <3 must add up to less than 3; ≠ must all be different; >3 must add up to more than 3; = must all be the same number; 1 must add up to 1.

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

Hard Pips answer — October 4, 2026

6×6 board · 14 dominoes · 18 regions · by Rodolfo Kurchan

The hard Pips has 28 cells to cover with 14 dominoes, split into 18 regions. Rules on the board: 8 sum regions, 2 "all equal", 2 "less than", 1 "greater than", 3 free regions. The single cell at row 2, column 6 is marked 2, so it can only hold a 2. Keep in mind that the single cell at row 5, column 5 is marked 0, so it can only hold a 0. The rack holds 2 doubles, 4 blank halves, 1 six.

Hints

Hint 1: where to start

Row 2, column 2 is a region of its own with target 2: that cell is a 2. Only dominoes carrying a 2 can reach it.

Hint 2: first domino

First domino: 2–0, covering row 2, column 2 (2) and row 1, column 2 (0).

Hint 3: second domino

Second: 2–3 on row 2, column 6 and row 1, column 6. Check the other 12 against the regions they could still fill.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 3 must add up to 3; 8 must add up to 8; <2 must add up to less than 2; = must all be the same number; 2 must add up to 2.

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