NYT Pips Hints & Answers for May 25, 2026

May 25, 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’s easy grid for today’s NYT Pips is a cozy starting point, almost entirely built from double dominos. You spot the sum‑8 cluster in the upper left immediately; its arithmetic leaves only one possible pair, and the single empty cells offer obvious homes for the one odd domino. It’s a quick, satisfying cascade where each placement unlocks the next without guesswork.

Livengood’s medium puzzle shifts the focus to equals constraints. You begin with a lone “greater than 5” cell that can only be 6, and from there an equals chain pulls you across the top right. The board is a web of two‑cell and four‑cell equal regions, and once you see how the first forced domino triggers the next, the rest falls into a smooth rhythm. It’s a pleasingly tight design that makes you feel clever at every step.

Rodolfo Kurchan’s hard is a sprawling challenge full of sums, equals, less‑thans, and greater‑thans tangled across an irregular grid. A single‑cell sum‑3 at the top gives you your first certain number, then you chase a chain of inferences down the left side where a triple‑equals column awaits. The solve weaves back and forth across the board, rewarding patient solvers who track every constraint. It’s a dense, interconnected puzzle that feels like Kurchan’s signature blend of elegance and toughness.

💡 Progressive Hints

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

💡 Spot the anchor
Look for the region that demands a specific sum across multiple cells. That’s your starting point — think about which domino values could possibly add up to that total.
💡 A perfect pair
The four‑cell sum region in the upper left uses only two distinct pip values. One of those values is already locked by the dominos you have — only one domino carries that number pair.
💡 Full solution (easy)
Place the 4‑4 domino horizontally across [1,0] and [1,1]. Then the 0‑6 domino goes in [0,1] (0) and [0,2] (6). The 0‑0 domino fills [2,1] and [2,2], forcing a 3‑3 in [3,2] and [3,3]. Finally, the 2‑2 completes the sum‑5 region at [2,3] and [2,4].
💡 Greater than what?
Start with the lone cell that demands a value greater than 5. That limits what it can be — and what domino can cover it.
💡 Equal partners
Once you place that high value at [1,5], the equals region next to it forces its neighbor to match. This cascades into a series of forced equalities across the top right.
💡 Full solution (medium)
Place the 6‑5 domino vertically at [1,5] (6) and [2,5] (5). This forces [2,4] to 5, so put the 2‑5 domino horizontally there (2 at [2,3], 5 at [2,4]). The equals region [1,3]/[2,3] forces [1,3] to 2, so place 2‑4 vertically with 2 at [1,3], 4 at [0,3]; matching [0,4] to 4 via equals, and use 3‑4 vertically to put 4 at [0,4], 3 at [0,5]. For the bottom left, the less‑2 cell at [4,0] must be 0 or 1; place 6‑1 vertically with 1 at [4,0], 6 at [3,0], forcing [3,1] to 6 via equals, then place the 3‑6 there. The large equals block all becomes 3 using the 3‑3 and 2‑3 dominos, and the 2‑3 fills [5,0]/[5,1].
💡 Zero in on single‑cell sums
A region with just one cell must satisfy its sum target exactly. That’s your easiest foothold — find it, and you’ll know the pip value for that cell immediately.
💡 Top‑left chain reaction
The single‑cell sum‑3 at [0,2] is 3. It must pair with [0,1] or [0,3]; since [0,1] is part of a sum‑11 region with [0,0], the only way to hit 11 is with a 6 and a 5. That narrows down the domino that covers [0,1] and [0,2].
💡 Follow the sums downward
After placing the 6‑3 in the top row, the sum‑4 region below forces a 0 and 1 via the 5‑4 domino from [0,0]/[1,0]. This opens up the left column’s equals chain.
💡 The equals column emerges
With [2,0]=0 and [3,0]=1, the sum‑11 row at row 3 needs 10 more from [3,1] and [3,2]. The 5‑5 domino fits perfectly, and then the triple‑equals region in the left column locks as 4‑4‑4, using the 4‑4 domino and the 1‑4 domino.
💡 Full solution (hard)
Place the 6‑3 domino at [0,1]=6, [0,2]=3. Then 5‑4 at [0,0]=5, [1,0]=4. Next, 1‑0 at [3,0]=1, [2,0]=0. The sum‑11 on row 3 takes 5‑5 at [3,1]=5, [3,2]=5. The greater‑3 cell [3,5] gets 5 from the 0‑5 domino (0 at [2,5], 5 at [3,5]), and the sum‑1 pair [1,5]/[2,5] uses the 1 at [1,5] from the 1‑3 domino (3 at [0,5]) and the 0 from [2,5] already placed. The left column equals: place 4‑4 at [7,0]=4, [8,0]=4, and 1‑4 at [9,0]=4, [9,1]=1. Complete right side: 6‑2 at [7,4]=6, [7,3]=2; 3‑4 at [8,3]=3, [9,3]=4; 6‑0 at [9,5]=6, [9,4]=0; 1‑2 at [6,3]=1, [5,3]=2; and the remaining equals and sums fill in with 2‑4 at [5,4]=2, [5,5]=4; 5‑3 at [2,3]=5, [1,3]=3; 1‑1 at [6,0]=1, [6,1]=1.

🎨 Pips Solver

May 25, 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 25, 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 25, 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: Lock the sum‑8 region
The sum‑8 region covers four cells: [0,1], [1,0], [1,1], [2,1]. With dominos 3‑3, 0‑0, 0‑6, 4‑4, 2‑2 available, the only way to get 8 from two equal pairs is 4+4+0+0. That forces the 4‑4 domino to occupy [1,0] and [1,1] horizontally, giving both 4s.
2
Step 2: Fill the sum‑8 zeros
The remaining sum‑8 cells need zeros. [0,1] must pair with [0,2] (a single empty cell), so the 0‑6 domino goes here: 0 in [0,1], 6 in [0,2]. Then [2,1] gets its 0 from the 0‑0 domino, which extends into [2,2].
3
Step 3: Resolve the sum‑3 duo
The sum‑3 region links [2,2] and [3,2]. With [2,2]=0 from the previous step, [3,2] must be 3. The only way to cover [3,2] is with the 3‑3 domino placed horizontally across [3,2] and [3,3] (which belongs to the sum‑5 region).
4
Step 4: Finish with sum‑5
The sum‑5 region is [2,3] and [3,3]. Now [3,3]=3, so [2,3] must be 2. The last unused domino is 2‑2; place it horizontally in [2,3] and the adjacent empty cell [2,4].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Greater‑5 forces a 6
The single cell [1,5] must be >5, so it’s 6. Its neighbors are [0,5] (empty) and [2,5] (in an equals pair with [2,4]). Placing a domino across [1,5] and [2,5] requires a 6 and a value that can equal [2,4]. The 6‑5 domino fits: [1,5]=6, [2,5]=5.
2
Step 2: Equals takes over
[2,5]=5 forces [2,4]=5 via the equals region. The only uncovered neighbor of [2,4] is [2,3] (equals with [1,3]). The 2‑5 domino works perfectly: place it horizontally, [2,3]=2, [2,4]=5.
3
Step 3: Extend the equals chain upward
Now [1,3] must equal [2,3], so it’s 2. It pairs vertically with [0,3] (equals with [0,4]). The 2‑4 domino fits: [1,3]=2, [0,3]=4. Then [0,4]=4, so the [0,4]/[0,5] domino becomes 3‑4 placed vertically, [0,4]=4, [0,5]=3.
4
Step 4: Bottom‑left less‑2 gives a 1
Cell [4,0] is less than 2, so 0 or 1. It pairs with [3,0] (equals with [3,1]). The 6‑1 domino satisfies both: 1 at [4,0], 6 at [3,0]. Then [3,1]=6. The only adjacent cell to [3,1] is [3,2], so the 3‑6 domino goes there: [3,1]=6, [3,2]=3.
5
Step 5: Large equals block and final cleanup
The equals region demands [4,2], [5,1], [5,2] all equal [3,2]=3. Place the 3‑3 domino vertically at [4,2]/[5,2]. The 2‑3 domino then fits horizontally at [5,0]/[5,1] (with the empty [5,0] taking the 2). All cells are now satisfied.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Single‑cell sum‑3 locks [0,2]=3
The sum‑3 region at [0,2] is a lone cell, so its pip must be exactly 3. It can only pair with [0,1] among listed cells. Since [0,1] is part of a sum‑11 region with [0,0], the domino covering them must include 3 and leave 6 for [0,1] (so [0,0] can be 5). The 6‑3 domino is the only fit: place it [0,1]=6, [0,2]=3.
2
Step 2: Top‑row sum‑11 resolves
With [0,1]=6, the sum‑11 region [0,0]/[0,1] forces [0,0]=5. [0,0]’s other neighbor is [1,0] (part of a sum‑4 pair with [2,0]). The 5‑4 domino fits: place it vertically, [0,0]=5, [1,0]=4.
3
Step 3: Sum‑4 and the 1‑0 domino
[1,0] + [2,0] must equal 4, so with [1,0]=4, [2,0]=0. [2,0]’s other neighbor is [3,0] (in the row‑3 sum‑11 region). The only domino giving 0 and pairing with [3,0] is 1‑0. Place it vertically: [3,0]=1, [2,0]=0.
4
Step 4: Row 3 sum‑11 and the 5‑5 domino
The sum‑11 region [3,0],[3,1],[3,2] now has 1, so the remaining two cells must sum to 10 — a pair of 5s. Place the 5‑5 domino horizontally: [3,1]=5, [3,2]=5.
5
Step 5: Left column’s triple‑equals
The equals region [7,0],[8,0],[9,0] all must share the same value. With [9,0] adjacent to [9,1] (sum‑1), the column value must be at least 4 to accommodate remaining dominos. The 4‑4 and 1‑4 domino fit perfectly: first place 4‑4 vertically at [7,0]/[8,0]. Then [9,0] must be 4, so place 1‑4 vertically: [9,0]=4, [9,1]=1 (sum‑1 satisfied).
6
Step 6: Fill the right side and bottom
The greater‑3 cell [3,5] pairs with [2,5] (sum‑1 with [1,5]). Use 0‑5: [2,5]=0, [3,5]=5. Then [1,5] must be 1 for sum‑1, so the 1‑3 domino goes [1,5]=1, [0,5]=3. The greater‑4 at [7,4] must be 6, and it pairs with [7,3] (sum‑2), forcing the 6‑2 domino: [7,4]=6, [7,3]=2. Then [8,3] (sum‑3) gets the 3‑4 domino with [9,3]=4 (from less‑5 region). The sum‑6 at [9,5] forces 6‑0: [9,5]=6, [9,4]=0. Remainder: 1‑2 at [6,3]=1, [5,3]=2; 2‑4 at [5,4]=2, [5,5]=4; 5‑3 at [2,3]=5, [1,3]=3; 1‑1 at [6,0]=1, [6,1]=1.

💡 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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