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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
Easy, by Ian Livengood, is a confidence-builder; in today's NYT Pips easy the grid is littered with single-cell sum regions that act like givens. Expect clean deductions with no forks once you trust those tiny anchors.
Medium, from Rodolfo Kurchan, is a step up in pattern but still approachable. A large equals block and several equal pairs dominate the solve; the bottleneck is the vertical equal pair beside an empty cell, and after that the less-than singles submit quickly.
Hard, also by Rodolfo Kurchan, is a dense architecture of sum-0 locks, equal-row chains, and a bottom greater-than region. The early game feels tight, but once the forced anchors start toppling the equals row, the rest opens up.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Look for the gimmes
Scan the grid for small sum regions, especially single-cell sums. They are the only type of constraint that pins a value before you compare domino partners.
💡 Top-left anchors
Start at [0,1] and [1,1]. Those two isolated sum cells lock the [0,2]–[0,1] and [1,1]–[1,0] placements almost instantly. The nearby bottom row sum cells at [2,1], [2,4], and [2,5] then form a neat chain.
💡 Complete placement
Place 4/6 horizontally with 6 at [0,1] and 4 at [0,2]. Place 5/0 horizontally with 5 at [1,1] and 0 at [1,0]. Place 6/5 horizontally with 6 at [2,5] and 5 at [2,4]. Place 3/4 horizontally with 4 at [2,1] and 3 at [2,2]. Finally place 4/2 vertically with 4 at [1,4] and 2 at [0,4].
💡 Read the equals blocks
The solve is controlled by equal-value regions, especially the four-cell block in the upper center. The small less-than cells are useful, but they are not where the bottleneck is.
💡 The empty-side bottleneck
Look at the vertical equal pair at [1,4]–[2,4] and its empty neighbor [2,3]. The need to place the same value in both equal cells rules out the higher pip there. Once that domino is set, the large equals block dictates the [1,3]–[1,4] domino.
💡 Complete placement
Place 4/6 horizontally with 4 at [2,4] and 6 at [2,3]. Place 2/1 vertically with 2 at [1,1] and 1 at [0,1]. Place 4/5 vertically with 4 at [2,1] and 5 at [3,1]. Place 2/3 vertically with 2 at [1,2] and 3 at [0,2]. Place 1/4 vertically with 1 at [1,0] and 4 at [2,0]. Place 2/5 vertically with 2 at [2,2] and 5 at [3,2]. Place 2/4 horizontally with 2 at [1,3] and 4 at [1,4].
💡 Follow the tightest sums
Hunt for one-cell sum regions and equal bands. They are tighter than other constraints because they dictate exact values even before a domino is chosen.
💡 Two sum-0 keys
The single-cell sum-0 regions at [3,2] and [9,4] are the sharpest entry points. [3,2] forces its partner [2,2] into the long equals row, while [9,4] forces [8,4] into the vertical equals stack. Add the single-cell sum-2 at [3,0] to start the same row.
💡 Lock the six-row
With [2,0] and [2,2] matching, the four-cell equals row at [2,0]–[2,3] is forced to the high value. That in turn sets [2,1] and [2,3], which feed the top-left equals band and the sum-9 region via their partners.
💡 Bottom cascade
The equal stack at [6,4]–[8,4] inherits its value from [8,4], and the greater-than-9 pair at [8,2]–[9,2] must be the 6/6 domino. The small sum-2 pair beside [5,3] then forces [5,2] and starts the sum-3 triple.
💡 Complete placement
Place 0/4 vertical with 0 at [9,4] and 4 at [8,4]; 3/2 horizontal with 3 at [5,4] and 2 at [5,3]; 5/5 vertical with 5 at [0,0] and [1,0]; 0/1 vertical with 0 at [5,2] and 1 at [6,2]; 6/6 vertical with 6 at [8,2] and [9,2]; 0/2 vertical with 0 at [0,6] and 2 at [1,6]; 3/3 vertical with 3 at [0,4] and [1,4]; 2/6 vertical with 2 at [3,0] and 6 at [2,0]; 1/5 vertical with 1 at [2,6] and 5 at [3,6]; 2/2 vertical with 2 at [2,4] and [3,4]; 6/5 vertical with 6 at [2,1] and 5 at [1,1]; 4/4 vertical with 4 at [6,4] and [7,4]; 6/3 vertical with 6 at [2,3] and 3 at [1,3]; 1/1 horizontal with 1 at [7,2] and [7,3]; 0/6 vertical with 0 at [3,2] and 6 at [2,2].
🎨 Pips Solver
Sep 7, 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
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
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: Pin the one-cell sum cells
The regions at [0,1], [1,1], [1,4], [2,1], [2,4], and [2,5] each consist of a single cell with a sum target, so each cell must equal its target. This gives immediate fixed values at those six coordinates and is the main deduction engine.
2
Step 2: Place the anchored dominoes
With [0,1] forced to sum 6, the 4/6 domino must lie horizontally at [0,2]–[0,1], putting 4 at [0,2]. With [1,1] forced to sum 5, the 5/0 domino lies horizontally at [1,1]–[1,0], putting 0 at [1,0]. The bottom pair is equally forced: [2,5] takes 6 from the 6/5 domino at [2,4]–[2,5], because [2,4] must be its sum 5.
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Step 3: Fill the remaining sums
The single-cell sum 4 at [1,4] forces the 4/2 domino vertically at [1,4]–[0,4], placing 2 at [0,4]. The single-cell sum 4 at [2,1] forces the 3/4 domino horizontally at [2,1]–[2,2], placing 3 at [2,2], which also satisfies the greater-than-2 constraint.
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Step 4: Verify the grid
All five dominoes are placed: 4/6, 5/0, 6/5, 3/4, and 4/2. Every sum region matches its target, the [2,2] cell is greater than 2, and the remaining empty cells [0,2], [0,4], and [1,0] are correctly filled by their partners.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Resolve the vertical equals bottleneck
Focus on the equal pair [1,4]/[2,4]. If [2,4] were the higher pip of the 4/6 domino, [1,4] would also have to be that value, but no remaining domino on [1,3] can supply the other half. Therefore [2,4] must be 4, the 4/6 domino lies horizontally at [2,3]–[2,4] with 6 at [2,3], and [1,4] is forced to 4.
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Step 2: Set the large equals block
Because [1,4] is 4, the domino covering [1,3]–[1,4] is the 2/4 domino, so [1,3] is 2. Since [1,3] belongs to the four-cell equal region at [1,1],[1,2],[1,3],[2,2], all four cells must be 2.
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Step 3: Use the equal block on surrounding cells
The 2/1 domino must cover [1,1] and [0,1], giving [0,1]=1 and satisfying less-than-3. The 2/3 domino must cover [1,2] and [0,2], giving [0,2]=3 and satisfying less-than-6. The 2/5 domino covers [2,2] and [3,2], giving [3,2]=5.
4
Step 4: Finish the lower-left equal pairs
Since [3,1] and [3,2] are equal, [3,2]=5 forces [3,1]=5. Then the 4/5 domino covers [2,1]–[3,1], placing 4 at [2,1]. The equal pair [2,0]/[2,1] then forces [2,0]=4, and the 1/4 domino at [1,0]–[2,0] puts 1 at [1,0], satisfying its less-than-3 region.
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Step 5: Verify all placements
The seven dominoes are now assigned: 4/6, 2/1, 4/5, 2/3, 1/4, 2/5, and 2/4. Check the equal regions, the less-than singles at [0,1], [0,2], and [1,0], and the empty cell at [2,3] to confirm the grid is consistent.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Use the single-cell sum locks
The one-cell sum-0 at [3,2] forces a 0 there and, via the 0/6 domino, puts a 6 at [2,2]. The one-cell sum-0 at [9,4] forces the 0/4 domino vertically, placing 0 at [9,4] and 4 at [8,4]. The one-cell sum-2 at [3,0] forces the 2/6 domino at [3,0]–[2,0], giving [2,0]=6. The one-cell sum-3 at [5,4] forces the 3/2 domino at [5,4]–[5,3], giving [5,3]=2.
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Step 2: Lock the long equals row
The region [2,0]–[2,3] is an equals constraint. Since [2,0] and [2,2] are both 6, all four cells must be 6. Thus [2,1] and [2,3] are 6. The 6/5 domino at [2,1]–[1,1] gives [1,1]=5, and the 6/3 domino at [2,3]–[1,3] gives [1,3]=3.
3
Step 3: Fill the top-row structures
The top-left equals region [0,0],[1,0],[1,1] all match [1,1]=5, so the 5/5 domino covers [0,0]–[1,0]. The sum-9 region [0,4],[1,3],[1,4] now has [1,3]=3, so all three are 3; the 3/3 domino covers [0,4]–[1,4]. The sum-4 pair [2,4]–[3,4] takes the 2/2 domino.
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Step 4: Resolve the bottom and center cascades
The vertical equals stack [6,4],[7,4],[8,4] is forced to 4 by [8,4], so the 4/4 domino covers [6,4]–[7,4]. The greater-than-9 region at [8,2],[9,2] must be 6 plus 6, so the 6/6 domino covers those two. With [5,3] already 2, the sum-2 pair [5,2]–[5,3] forces [5,2]=0; the 0/1 domino then places 1 at [6,2].
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Step 5: Complete the sum-3 triple and unequal column
The sum-3 region [6,2],[7,2],[7,3] already has [6,2]=1 from the 0/1 domino, so the remaining two cells must sum to 2; the 1/1 domino covers [7,2]–[7,3]. The unequal column [0,6]–[3,6] uses the 0/2 domino at [0,6]–[1,6] and the 1/5 domino at [2,6]–[3,6], yielding distinct values 0,2,1,5.
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Step 6: Verify the full grid
All fifteen dominoes are placed: 0/4, 3/2, 5/5, 0/1, 6/6, 0/2, 3/3, 2/6, 1/5, 2/2, 6/5, 4/4, 6/3, 1/1, and 0/6. Confirm every equals, sum, greater, and unequal region matches its target, and the long domino chain has no overlaps.
💡 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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