NYT Pips Hints & Answers for July 25, 2026

Jul 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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🎲 Today's Puzzle Overview

The easy grid, constructed by Ian Livengood, lays out its initial gifts plainly: a sum-8 region demands a pair of fours, and a solitary less-than-2 cell instantly calls for a 1. From there, an equals triple of 3s takes shape with the remaining double, and a greater-than region (despite its high threshold) snaps the final double-five into place. The solve feels like a quick tour through each constraint type, with almost no ambiguity.

Rodolfo Kurchan’s medium puzzle opens with a single-cell sum‑6 that forces a 6 right away, dragging a 4 next to it to complete an 8‑sum pair. Then the left side unravels through interlocking sum‑3 and sum‑5 constraints, while a tiny sum‑2 at the edge accepts only the double‑1. A cascade of placements fills the remaining cells, leaving just an equals region to cap it off. You’ll feel each deduction click precisely.

Kurchan’s hard entry throws you into large equals regions—a belt of five cells all forced to the same digit, and a three‑cell block at the top that must match. The puzzle’s backbone is the chain of 1s; once you spot the sum‑0 cells handing you zeroes that slot into that chain, the 1‑containing dominoes fall into sequence. Double‑fours secure the top, and then sum‑7, equals‑5, and equals‑3 regions weave together in a satisfying finale. Today’s NYT Pips hard demands patience, but the architecture is deeply rewarding.

💡 Progressive Hints

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

💡 First Glance
The puzzle hands you a tight sum region, a restrictive less-than cell, and an equals cluster. Those three will unlock the whole board.
💡 Narrow the Candidates
The sum‑8 region at [2,0]–[2,1] can only be a double-4 domino. The less‑than‑2 cell at [2,3] must be a 0 or 1, and the only domino with a 1 left after the double‑4 is [1,2].
💡 Full Solve
Place [4,4] horizontally at [2,0][2,1]. Place [1,2] vertically: 1 at the less‑2 cell [2,3], 2 at the empty [3,3]. To satisfy the three‑cell equals region [1,1][1,2][1,3], put [3,3] horizontally on [1,2][1,3] and then [3,6] with 3 at [1,1] and 6 at [0,1]. The remaining region at [2,2][3,2] gets [5,5] vertically.
💡 Spot the Anchors
A lone sum‑6 cell and a tiny sum‑2 region are the grid’s early forcing points—one demands a specific digit, the other a special double.
💡 Follow the Dominoes
The sum‑6 cell at [1,4] must be a 6; domino [4,6] fits, putting its 4 at [1,3]. That forces the sum‑8 region next door to complete with a 4 at [1,2] via [4,2]. The sum‑2 pair at [1,5][2,5] is solved by the [1,1] domino.
💡 Full Solve
Place [4,6] horizontally: 6 at [1,4], 4 at [1,3]. Place [4,2] vertically: 4 at [1,2], 2 at [2,2]. Place [1,1] vertically: 1s at [1,5][2,5]. Place [6,0] with 0 at [1,0], 6 at [2,0]. Then [4,3] puts 3 at [0,0] and 4 at [0,1]; [5,1] puts 5 at [2,1] and 1 at [1,1] (satisfying sum‑5). Finish with [3,3] horizontally on [2,3][2,4].
💡 Equals Everywhere
The hardest puzzle is dominated by equals regions—one five cells long, another three cells, and several smaller ones. Look for a digit that can appear in many dominoes.
💡 Follow the Zeroes
Two sum‑0 cells at [2,1] and [1,6] demand zeroes. They immediately interact with the five‑cell equals chain, forcing the first 1s into place.
💡 Build the 1‑Ladder
The five‑cell equals region (row 1, cols 0‑4) must all be 1. With [1,1] already set to 1 from the zero domino, deploy the double‑1 on [1,3][1,4] and then use [1,4] and [1,6] dominoes to fill the rest.
💡 Crown the Top Equals‑4
The three‑cell equals block at row 0 gets its 4s through the [1,4] domino and the double‑4 [4,4]. The leftmost equals‑6 region below it gets a 6 from [2,6].
💡 Full Solve
Place [0,1] (0 at [2,1], 1 at [1,1]) and [0,4] (0 at [1,6], 4 at [1,5]). Place [1,1] at [1,3][1,4]; [1,4] at [1,2]=1, [0,2]=4; [1,6] at [1,0]=1, [2,0]=6. Place [4,4] at [0,0][0,1]. Place [2,6] with 2 at [4,0] (sum‑2) and 6 at [3,0]. For sum‑7 at [2,2][3,2], use [2,4] (2 at [2,2], 4 at [2,3]) then [3,4] (3 at [2,5], 4 at [2,4]) and [4,6] (4 at [3,3], 6 at [4,3]). Right side: [5,5] at [2,6][2,7]; [3,5] (3 at [4,6], 5 at [3,6]); [3,6] (3 at [4,5], 6 at sum‑6 [3,5]). Finish with [3,3] at [5,5][5,6] for the equals‑3 bottom.

🎨 Pips Solver

Jul 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 July 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 July 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 cells at [2,0] and [2,1] must sum to 8; the only available domino with two 4s is [4,4]. Place it horizontally so that both become 4.
2
Step 2: Claim the Less‑Than‑2 Cell
[2,3] is restricted to a pip less than 2 (0 or 1). The only unused domino with a 1 is [1,2]; put the 1 there and the 2 into the empty cell at [3,3] (vertical placement).
3
Step 3: Triple the 3s
The equals region over [1,1], [1,2], and [1,3] requires three identical pips. Use [3,3] horizontally on [1,2] and [1,3] to supply two 3s. Then place [3,6] with the 3 at [1,1] and the 6 at the only remaining empty cell [0,1].
4
Step 4: Finish with the High‑Value Duo
The region covering [2,2] and [3,2] can only be satisfied by the leftover [5,5] domino. Place it vertically: [2,2]=5, [3,2]=5. The grid is solved.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Nail the Sum‑6 Cell
The single-cell region at [1,4] demands a 6. The [4,6] domino can place its 6 there and its 4 at [1,3]. Place it horizontally.
2
Step 2: Complete the Sum‑8 Pair
The sum‑8 region at [1,2] and [1,3] now has a 4 at [1,3], forcing [1,2] to be 4. Use the [4,2] domino vertically, putting 4 at [1,2] and 2 at the less‑than‑3 cell [2,2].
3
Step 3: Tuck in the Sum‑2 Region
The sum‑2 region at [1,5] and [2,5] requires a total of 2. The only workable domino is the double‑1 [1,1]; place it vertically so both become 1.
4
Step 4: Distribute 0 and 6 on the Left
Place the [6,0] domino with 0 at [1,0] and 6 at [2,0] (the empty cell). This gives the upper‑left sum‑3 region [0,0]/[1,0] a current sum of 0, so [0,0] must become 3.
5
Step 5: Fill the Remaining Sums and Equals
To supply the 3 at [0,0], place [4,3] with 3 at [0,0] and 4 at [0,1]. Now the sum‑5 region [0,1]/[1,1] has a 4 at [0,1], so [1,1] must be 1; the [5,1] domino supplies 5 at [2,1] and 1 at [1,1]. Finally, the equals region at [2,3] and [2,4] is filled by placing [3,3] horizontally.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Plant the Zeroes
The sum‑0 single‑cell regions at [2,1] and [1,6] must be 0. Place [0,1] with 0 at [2,1] and 1 at [1,1]; place [0,4] with 0 at [1,6] and 4 at [1,5].
2
Step 2: Weave the Five‑1s Chain
The five‑cell equals region in row 1 (columns 0‑4) must all be 1. With [1,1]=1 set, place [1,1] double‑1 on [1,3] and [1,4]. Then use [1,4] to put 1 at [1,2] and 4 at [0,2] (the rightmost cell of the top equals‑4). Finally, place [1,6] with 1 at [1,0] and 6 at [2,0]; now all five positions are 1.
3
Step 3: Crown the Top Equals‑4
The three‑cell equals region on row 0 (cols 0‑2) already has [0,2]=4. To make all three 4, place [4,4] horizontally over [0,0] and [0,1].
4
Step 4: Anchor the Bottom‑Left Equals‑6
The equals region at [2,0] and [3,0] needs both to be 6. [2,0] is already 6; use [2,6] to supply 6 at [3,0] and the 2 at [4,0] (satisfying the sum‑2 cell). Place it vertically.
5
Step 5: Solve the Sum‑7 and Equals‑4 Mid‑Section
The sum‑7 region [2,2] and [3,2] needs 2 and 5. Place [2,4] with 2 at [2,2] and 4 at [2,3] (part of the equals‑4 region). Then place [3,4] with 3 at [2,5] and 4 at [2,4]; this completes sum‑7 (4+3) and the equals‑4 region now needs [3,3]=4. Place [4,6] with 4 at [3,3] and 6 at [4,3] (one cell of the bottom‑right equals‑6).
6
Step 6: Finish with the Right‑Side Equals and Sum‑6
The equals‑5 region [2,6],[2,7],[3,6] must be all 5. Place [5,5] on [2,6] and [2,7] horizontally. Then [3,6] gets 5 from [3,5], placing its 3 at [4,6]. The single‑cell sum‑6 at [3,5] must be 6, so place [3,6] with 3 at [4,5] and 6 at [3,5]. Finally, place [3,3] on [5,5] and [5,6] horizontally to complete the bottom equals‑3 region.

💡 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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