NYT Pips Hints & Answers for August 25, 2026

Aug 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!

Click here to play today's official NYT Pips game first.

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

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🎲 Today's Puzzle Overview

Today’s NYT Pips easy, from Ian Livengood, feels like a warm-up that still makes you read every cap: two tiny less-than cells act as early signals, and the sum regions channel the remaining dominoes into place without much backtracking. You’ll probably lock the top-left sum pair early, then sweep across the top band as each adjacent cell inherits a forced value.

Rodolfo Kurchan’s medium turns up the architectural pressure. A single greater-than cell and a less-than cell bookend the grid, but the real momentum comes from a chain of equals regions. Once one corner is settled, the matching values ripple upward through neighboring regions, so a modest deduction in the bottom row ends up steering the whole solve.

Kurchan’s hard is the most scenic: a large unequal ring on the right needs six distinct values, while the left side stacks several small sum pockets. Solvers tend to oscillate between the bottom sum strip and the middle sum chain before the ring finally closes. It’s a patience puzzle that rewards following the forced cascade.

💡 Progressive Hints

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

💡 Scan the tiny caps
Look for the one-cell less-than regions first. They create narrow caps that tell you a pip must sit below a low limit, so they act as free early signals before any cross-checking is needed.
💡 Anchor the top-left sum
The two-cell sum in the top-left has only one available domino that can reach its total, which locks [0,0] and [0,1] immediately. Then look at the less-than cell [1,1] and its neighbor [1,2] — the only legal way in gives [1,1] the smallest possible value.
💡 Complete the easy grid
Place [4,5] in [0,0]-[0,1] with 5 in [0,0] and 4 in [0,1]; [6,1] in [1,1]-[1,2] with 1 in [1,1] and 6 in [1,2]; [3,3] in [0,3]-[1,3] with both 3s; [0,1] in [0,4]-[0,5] with 0 in [0,4] and 1 in [0,5]; [6,5] in [1,4]-[1,5] with 6 in [1,4] and 5 in [1,5].
💡 Follow the equals threads
Start by identifying all equals regions and the two one-cell inequality caps. Equals regions force matching pip values, so they form a spine that drives the rest of the grid before the sum regions even enter the picture.
💡 Ignite the bottom-left corner
The less-than cell at [3,0] and the equals pair at [3,1]-[3,2] are the tightest corner. Settling that corner forces [3,0]=1 and [3,1]=6, which then makes [3,2]=6 and sends a 3 up to [2,2] through the adjacent equals region.
💡 Complete the medium grid
Place [1,6] in [3,0]-[3,1] with 1 in [3,0], 6 in [3,1]; [6,3] in [2,2]-[3,2] with 3 in [2,2], 6 in [3,2]; [3,2] in [1,1]-[1,2] with 2 in [1,1], 3 in [1,2]; [4,6] in [0,2]-[0,3] with 4 in [0,2], 6 in [0,3]; [6,6] in [1,3]-[2,3] with both 6; [6,5] in [2,4]-[2,5] with 6 in [2,4], 5 in [2,5]; [0,0] in [0,5]-[1,5] with both 0.
💡 Spot the unequal ring
For hard, begin by noting the large unequal region on the right side — its six cells must all carry different pip values. That ring works like a filter for the endgame, but first let the small sum pockets on the left and bottom show their hand.
💡 Use the bottom sum strip
The bottom sum region covering [6,0]-[6,2] is the best early lock. Once it forces the adjoining [6,3] through the sum pair at [5,3]-[6,3], you can follow the pressure upward into the middle sum regions around [3,3] and [4,3].
💡 Climb the left-center chain
With [6,3] fixed, [5,3] must be 3; that makes the [4,3]-[5,3] domino place 5 at [4,3] and forces [3,3]=5 via the sum-10 region. The 0/5 domino then puts 0 at [3,2], so the sum-4 region at [3,1]-[3,2] requires [3,1]=4, and [3,0]=6 follows with [2,0]=0 below.
💡 Assign the unequal ring
With the left and bottom mostly locked, the unequal ring consumes the last distinctive dominoes. Use the all-different requirement to set [3,6]-[3,7] as 4 and 0, then fill the vertical [4,6]-[5,6] with 3 and 1, and the bottom pair [6,5]-[6,6] with 2 and 5.
💡 Complete the hard grid
Place [1,1] in [6,0]-[6,1] and [6,6] in [6,2]-[6,3]; [5,3] in [4,3]-[5,3] with 5 in [4,3], 3 in [5,3]; [0,5] in [3,2]-[3,3] with 0 in [3,2], 5 in [3,3]; [4,6] in [3,0]-[3,1] with 6 in [3,0], 4 in [3,1]; [0,2] in [1,0]-[2,0] with 2 in [1,0], 0 in [2,0]; [0,6] in [0,0]-[0,1] with 0 in [0,0], 6 in [0,1]; [3,3] in [0,2]-[0,3] with both 3; [0,4] in [3,6]-[3,7] with 4 in [3,6], 0 in [3,7]; [3,1] in [4,6]-[5,6] with 3 in [4,6], 1 in [5,6]; [5,2] in [6,5]-[6,6] with 2 in [6,5], 5 in [6,6].

🎨 Pips Solver

Aug 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 August 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 August 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 top-left sum pair
The two-cell sum region [0,0]-[0,1] has only one domino in the easy set whose pips sum to 9: [4,5]. That places 5 in [0,0] and 4 in [0,1], consuming the top-left corner immediately.
2
Step 2: Use the less-than cell below
Because [0,1] is occupied, the adjacent less-than cell [1,1] must pair horizontally with [1,2]. The less-than cap forces [1,1] to be 1, so the [6,1] domino covers [1,1]-[1,2] with [1,1]=1 and [1,2]=6.
3
Step 3: Complete the second sum region
With [1,2]=6, the sum-9 pair [1,2]-[1,3] requires [1,3]=3. The [3,3] domino therefore runs vertically through [0,3]-[1,3], giving both cells 3.
4
Step 4: Fill the remaining tail
The less-than cell [0,5] is set to 1, so the [0,1] domino covers [0,4]-[0,5] with [0,4]=0. Then the sum region [0,3],[0,4],[1,4] forces [1,4]=6, and the [6,5] domino places 5 in [1,5].

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Settle the bottom-left less-than cell
The less-than cell [3,0] must pair with [3,1]. Using the [1,6] domino is the only clean fit, placing 1 at [3,0] and 6 at [3,1].
2
Step 2: Match the corner equals pair
The equals region [3,1]-[3,2] then forces [3,2]=6. The [6,3] domino covers [2,2]-[3,2], putting 3 at [2,2] and 6 at [3,2].
3
Step 3: Send the 3s upward
The equals region [1,2]-[2,2] now demands [1,2]=3. The [3,2] domino covers [1,1]-[1,2], leaving [1,1]=2.
4
Step 4: Unlock the top-right equals chain
The greater-than cell [0,2] is satisfied by the [4,6] domino, with [0,2]=4 and [0,3]=6. The equals region [0,3]-[1,3] then forces [1,3]=6, which extends through the [6,6] domino to [2,3]=6, and the equals region [2,3]-[2,4] forces [2,4]=6.
5
Step 5: Close the sum column
The final sum region [0,5],[1,5],[2,5] is completed with the double-zero domino at [0,5]-[1,5] and the [6,5] domino at [2,4]-[2,5], placing 5 at [2,5].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the bottom sum strip
The sum-8 strip [6,0]-[6,2] is the tightest opening. The double-1 domino occupies [6,0]-[6,1], and the third cell [6,2] must then be 6 to reach the total. That extends the double-six domino into [6,2]-[6,3], forcing [6,3]=6.
2
Step 2: Climb through the middle sum stack
With [6,3]=6, the sum-9 pair [5,3]-[6,3] forces [5,3]=3. The [5,3] domino then covers [4,3]-[5,3] with [4,3]=5, and the sum-10 pair [3,3]-[4,3] forces [3,3]=5.
3
Step 3: Resolve the left-center chain
With [3,3]=5, the [0,5] domino at [3,2]-[3,3] places 0 at [3,2]. The sum-4 region [3,1]-[3,2] then requires [3,1]=4, so the [4,6] domino covers [3,0]-[3,1] with [3,0]=6. The sum-6 pair [2,0]-[3,0] then gives [2,0]=0.
4
Step 4: Stack the top-left sums
With [2,0]=0, the [0,2] domino at [1,0]-[2,0] places [1,0]=2. The sum-2 pair [0,0]-[1,0] then forces [0,0]=0. The [0,6] domino at [0,0]-[0,1] gives [0,1]=6, and the sum-9 pair [0,1]-[0,2] forces [0,2]=3. The [3,3] domino then fills [0,2]-[0,3] with both 3s.
5
Step 5: Open the unequal ring
For the unequal six-cell ring, all values must be distinct. The [0,4] domino covers [3,6]-[3,7], placing 4 at [3,6] and 0 at [3,7], which supplies two of the required distinct values.
6
Step 6: Close the ring
The remaining ring cells take the [3,1] and [5,2] dominoes. Place [3,1] at [4,6]-[5,6] with [4,6]=3 and [5,6]=1, and [5,2] at [6,5]-[6,6] with [6,5]=2 and [6,6]=5. The ring then holds the complete distinct set 0,1,2,3,4,5.

💡 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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