NYT Pips Hints & Answers for May 22, 2026

May 22, 2026

🚨 SPOILER WARNING

This page contains the final **answer** and the complete **solution** to today's NYT Pips puzzle. If you haven't attempted the puzzle yet and want to try solving it yourself first, now's your chance!

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Want hints instead? Scroll down for progressive clues that won't spoil the fun.

🎲 Today's Puzzle Overview

Ian Livengood sculpts today's NYT Pips easy around a rare three-cell sum region, using its arithmetic tightness to force a single domino orientation and set off a chain reaction through equals and greater‑5 constraints. The result is a compact, clean solve that feels immediate yet satisfying.

For the medium, Livengood layers a triple‑equals block with a lone sum‑3 cell that acts as a key fob. A less‑1 zero corridor and a cascade of sum‑3 pairs create an interlocking structure where each deduction unlocks the next, showcasing his knack for knotting constraints without losing clarity.

Rodolfo Kurchan's hard puzzle centers on a poetic sum‑0 pocket that instantly zeroes three cells, seeding a dense web of sum‑9, sum‑8, and multiple sum‑1 anchors. The design demonstrates Kurchan's signature: a bold, rigid seed that radiates logical pressure across the entire grid, rewarding solvers who trace each thread with a rich, layered resolution.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Spot the Tightest Sum
Look for a region that demands three cells add up to a fixed number. That sort of multi‑cell sum is the gatekeeper.
💡 Top‑Left Alignment
The sum‑6 region spans two vertically adjacent cells and a horizontal neighbor. The vertical pair must share a single domino, and the only domino with a duplicate pip that fits is [3,3].
💡 The Domino Cipher
Place [3,3] so the two 3s fill (0,0) and (1,0); (0,1) becomes 0. The equals region right of it demands a 0‑bearing domino that can match its partner: [5,0] goes on (0,1)-(0,2) with 5 at (0,2). [5,2] completes the pair at (0,3)-(1,3) with 5 and 2. Below, the greater‑5 cells anchor [6,4] on (3,0)-(4,0) and [6,2] on (3,3)-(4,3); finally, [2,3] lands on (4,1)-(4,2) to match the 2 at (4,3).
💡 An Isolated Single Sum
A lone cell with a sum‑3 target forces its pip immediately. Find it — it's the only region that is a singleton sum.
💡 Bottom‑Row Bind
That sum‑3 cell sits at (2,7) and forces a domino with a 3. The only 3‑bearing domino that can legally extend to (2,6) is [3,5]; placing it also fixes the equals pair in the column above to 5.
💡 Full Unfolding
Place [3,5] with 3 at (2,7) and 5 at (2,6). The equals region at (1,6)/(2,6) becomes 5s, so [2,5] covers (1,5)-(1,6) with 5 at (1,6) and 2 at (1,5). The less‑1 region in the lower middle forces [0,0] on (2,3)-(2,4). The triple‑equals block from (0,2) to (1,3) must all be 6: [0,6] vertical at (0,2)-(0,3) (6,0), [2,6] vertical at (1,1)-(1,2) (2,6), and [3,6] horizontal at (1,3)-(1,4) (6,3). [5,4] fills (2,0)-(2,1) for the empty and sum‑7, and [0,1] wraps (0,4)-(0,5) for the less‑1 and final sum‑3.
💡 The Zero‑Sum Command
A region that sums to zero means every cell inside must be zero. Look for it — it's the most rigid constraint on the board.
💡 Vertical Zero Anchor
The sum‑0 region occupies (1,4), (2,3), and (2,4). Two of these are vertically adjacent: (1,4) and (2,4). That forces the [0,0] domino to cover that column.
💡 The Zero Ripple
With (2,4) and (1,4) set to 0, cell (2,3) must also be 0 and sits next to (2,2) — part of a sum‑9 pair. The only domino providing a 0 and a pip that can later sum to 9 with the cell above is [3,0], placing 0 on (2,3) and 3 on (2,2).
💡 Upper‑Left Domino Stack
Now (2,2)=3 demands (1,2)=6 for the sum‑9. The vertical pair (0,2)-(1,2) then becomes a natural [6,6] domino, feeding a sum‑8 with (0,3). That forces (0,3)=2 and (0,4) less than 3 so it becomes 2 from a [2,2] domino. Meanwhile, sum‑1 cells at (5,3) and (7,4) snap in [1,5] and [1,4] dominoes.
💡 Complete Architecture
Place [0,0] on (1,4)-(2,4). [3,0] on (2,2)-(2,3) with 0 on (2,3). [6,6] on (0,2)-(1,2). With (0,2)=6, [2,2] lands on (0,3)-(0,4) (2,2). Sum‑1 at (5,3) takes [1,5] (1 at (5,3), 5 at (5,4)). Sum‑1 at (7,4) takes [1,4] (1 at (7,4), 4 at (7,3)). Sum‑7 left pair (4,0)-(4,1) uses [1,6] (1 at (4,1), 6 at (4,0)). Greater‑10 column 0 takes [6,5] on (1,0)-(2,0). Bottom‑right: sum‑10 pair (5,8)-(6,8) uses [6,3] (6 at (6,8), 3 at (6,7)) and [3,4] on (4,8)-(5,8) (3,4). Sum‑2 triple in col 1 uses [0,2] on (6,1)-(5,1) (0,2) and [1,1] on (7,1)-(8,1) (1,1). Row 1 sum‑6 on (1,6)-(1,7) gets [2,4] (2,4). Final corner sum‑10 (8,7)-(8,8) uses [5,0] on (8,8)-(7,8) (5,0) and [4,5] on (8,6)-(8,7) (4,5). [1,3] fills (4,7)-(4,6) (1,3).

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May 22, 2026

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✅ Final Answer & Complete Solution For Hard Level

The key to solving today's hard puzzle was identifying the placement for the critical dominoes highlighted in the starting grid. Once those were in place, the rest of the puzzle could be solved logically. See the final grid below to compare your solution.

Starting Position & Key First Steps

Pips hint for May 22, 2026 – hard level puzzle grid with critical first placements and strategy

This image shows the initial puzzle grid for the hard level, with a few critical first placements highlighted.

Final Answer: The Solved Grid for Hard Mode

NYT Pips May 22, 2026 hard puzzle full solution grid showing final answer with hints

Compare this final grid with your own solution to see the correct placement of all dominoes.

🔧 Step-by-Step Answer Walkthrough For Easy Level

1
Step 1: Sum‑6 trinity forces the [3,3]
The sum‑6 region at (0,0),(0,1),(1,0) requires three numbers adding to 6. The vertical pair (0,0)-(1,0) must belong to one domino; only domino [3,3] can supply two identical numbers that fit the sum with the remaining cell (0,1).
2
Step 2: Zeroed out and equals kick‑off
Placing [3,3] with 3s at (0,0) and (1,0) forces (0,1) to 0. The equals region at (0,2)-(0,3) demands equal values; (0,1)=0 and the need for equality push the [5,0] domino onto (0,1)-(0,2), setting (0,2)=5.
3
Step 3: Completing the top row
With (0,2)=5, the equals partner (0,3) must also be 5, so domino [5,2] lands on (0,3)-(1,3) with 5 and 2.
4
Step 4: Greater‑5 anchors and bottom equals
The greater‑5 regions at (3,0) and (3,3) require pips >5, so they must be 6. Domino [6,4] spans (3,0)-(4,0) (6 and 4); [6,2] covers (3,3)-(4,3) (6 and 2). The bottom equals region (4,2)-(4,3) forces domino [2,3] onto (4,1)-(4,2) with a 2 matching (4,3)'s 2.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Sum‑3 beacon
The single‑cell sum‑3 region at (2,7) requires a 3. The only domino with a 3 is [3,5]; placing it with the 3 on (2,7) forces the 5 onto (2,6).
2
Step 2: Equals pair cascade
The equals region at (1,6)-(2,6) now has a 5 at (2,6), so (1,6) must also be 5. The [2,5] domino can supply a 5 on (1,6) and a 2 on (1,5), fitting the sum‑3 pair with (0,5).
3
Step 3: Less‑1 zero corridor
The less‑1 region at (2,3)-(2,4) forces both cells to 0, so the [0,0] domino fills them.
4
Step 4: Triple‑equals lock
The equals region spanning (0,2),(1,2),(1,3) demands three identical pips. The only way to achieve 6's: [0,6] vertically on (0,2)-(0,3) places a 6 at (0,2); [2,6] vertically on (1,1)-(1,2) places a 6 at (1,2); and [3,6] horizontally on (1,3)-(1,4) places a 6 at (1,3) while solving the sum‑3 at (1,4) with its 3.
5
Step 5: Remaining pieces
With the grid nearly filled, [5,4] goes on (2,0)-(2,1) to satisfy the empty space and the sum‑7 with (1,1), and [0,1] wraps the less‑1 region at (0,3)-(0,4) by placing 0 on (0,4) and 1 on (0,5) to complete the sum‑3.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Zero‑sum pocket
The sum‑0 region at (1,4),(2,3),(2,4) dictates all three cells are 0. The vertical domino [0,0] fits perfectly on (1,4)-(2,4).
2
Step 2: The zero extension
Cell (2,3) must be 0 and pairs horizontally with (2,2). The sum‑9 region linking (1,2) and (2,2) will constrain that neighbor. Among the remaining zero‑bearing dominoes, only [3,0] can place 0 at (2,3) and a plausible pip for (2,2) — a 3 — which forces (1,2)=6 for sum‑9.
3
Step 3: Upper‑left sum‑8 and less‑3
Now (1,2)=6 demands a vertical partner; [6,6] on (0,2)-(1,2) puts a second 6 at (0,2). With (0,2)=6, the sum‑8 region with (0,3) forces (0,3)=2. The less‑3 region at (0,4) then must be less than 3, so a [2,2] domino sits on (0,3)-(0,4) giving both cells 2.
4
Step 4: Sum‑1 anchors
The single‑cell sum‑1 at (5,3) requires a 1. Only [1,5] works, putting 1 at (5,3) and 5 at (5,4) (satisfying greater‑4). The sum‑1 at (7,4) forces a 1; [1,4] places 1 there and 4 at (7,3) (which must be less than 5).
5
Step 5: Left‑side sum‑7 and greater‑10 column
The sum‑7 pair (4,0)-(4,1) resolves with [1,6], giving 1 at (4,1) and 6 at (4,0). The greater‑10 cells (1,0) and (2,0) require 6 and 5 in some order; [6,5] on (1,0)-(2,0) achieves it.
6
Step 6: Bottom‑right sums and final fragments
Sum‑10 column (5,8)-(6,8) uses [6,3] with 6 on (6,8) and 3 on (6,7), then [3,4] on (4,8)-(5,8) with 3 and 4. The sum‑2 column (6,1),(7,1),(8,1) gets [0,2] on (6,1)-(5,1) (0,2) and [1,1] on (7,1)-(8,1) (1,1). Row 1 sum‑6 pair (1,6)-(1,7) takes [2,4] (2+4). Final corner sum‑10 (8,7)-(8,8) uses [5,0] on (8,8)-(7,8) and [4,5] on (8,6)-(8,7). [1,3] fills the remaining (4,7)-(4,6) with 1+3.

💡 Pro Tips for Similar Puzzles

Start with Constraints
Always begin with the most constrained regions - sum regions with small numbers or tight spaces.
Use Equal Regions
Use "equal" regions as anchors - they eliminate many possibilities quickly.
Work Systematically
Let the rules guide your placement rather than guessing randomly.
Double-Check
Verify each region's rules are satisfied before moving to the next.

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