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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
Livengood designs today's easy as a compact chain of forced placements. The grid is sparse but every cell is tied to either a large sum block, a small sum pair, or a less-than lock, so the dominos cascade from the lower left. It reads as a clinic in using restrictive singletons to seed a larger sum region.
Livengood's medium is built around repeated equals regions. Three separate equals clusters—including a 3-cell run—act as magnets that eliminate candidate pip values, while a couple of tiny sum singletons provide the initial domino fingerprints. The elegance is in how the equality groups and sum cells interlock without giving away the center too early.
Rodolfo Kurchan's hard raises the architecture considerably. A long zero-sum row splits the middle, flanked by sum-6 trios and a four-cell equals block at the bottom. The design leans on domino orientation through narrow corridors, making this NYT Pips hard feel like a lattice where every zero and equality must be placed just so.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Scan for the tight caps
The most restrictive elements aren't the big sum block; they're the two small less-than regions. Start by asking which cells can be filled at all under those caps, and which dominoes carry the allowed pips.
💡 Use the lower-left less-than cell
The less-than region at [2,0] has only one legal pip and only one adjacent grid cell, so it locks a domino horizontally into the edge. The second less-than region at [3,2] then interacts with the large sum block at [2,2].
💡 Commit the forced chain
Place [6,0] at [2,0]=0 and [2,1]=6. Then [1,5] goes at [3,2]=1 and [2,2]=5. The sum-28 block forces [6,6] onto [1,1]-[1,2], leaving [5,2] to give [2,3]=5 and [1,3]=2. Finish with [3,4] at [0,3]=3 and [0,4]=4.
💡 Read the tiny sum signatures
Look for the singleton sum regions and the smallest multi-cell sum; they are the puzzle's domino fingerprints. Rather than starting with the big equals runs, find the sum target that can only be matched by one available pip in one domino.
💡 Anchor at the bottom-left sum
The sum region at [3,2] forces a very specific low pip, and the only domino containing it must reach toward [3,3]. Because [3,3] is part of an equals pair with [2,3], that fixes the partner pip, and the adjacent singleton [2,4] then locks the next horizontal domino.
💡 Chain the equals and sums
Place [2,5] at [3,2]=2 and [3,3]=5; [4,5] at [2,3]=5 and [2,4]=4. Next, [6,5] goes at [2,0]=6 and [2,1]=5, while [1,6] covers [1,2]=1 and [2,2]=6 to satisfy the sum-11 region. The 6-equals cluster is then satisfied by [6,6] at [1,0]-[1,1] and the already-placed [6,5] at [2,0]. Finish with [5,3] at [0,1]=5 and [0,2]=3, [3,1] at [0,3]=3 and [1,3]=1, and [4,3] at [3,0]=4 and [3,1]=3.
💡 Find the zero-pressure zones
Begin with the strictest sum regions that force every cell to the minimum. Also study the bottom single-cell sum and the mid-board sum trio; together they form a tightening loop.
💡 Anchor the four-cell zero row
The region [3,1]-[3,4] must be all minimum, so the double-minimum domino has to cover a pair inside it. The only smooth fit is the middle pair [3,2]-[3,3]. Then [4,5], a single-cell minimum, pulls on [4,4] and the sum-6 trio [4,2]-[4,4].
💡 Resolve the lower middle band
Place the double-minimum at [3,2]-[3,3]. At [4,5], the minimum cannot pair upward into the equals column, so it pairs left with [4,4]. That forces the companion 2, and the sum-6 trio then requires the double-2 at [4,2]-[4,3].
💡 Unlock the bottom-right equals lattice
The single sum cell at [6,4] must be 6; its only neighbor is [7,4], which is part of a four-cell equals group. The only value that can fill all four cells there is 4, so [6,4]-[7,4] locks, then [4,4] covers [8,3]-[8,4], and [1,4] supplies [8,1]-[8,2].
💡 Complete the full solve
Place the opening anchors exactly: [0,0] at [3,2]-[3,3]; [2,0] at [4,4]=2 and [4,5]=0; [2,2] at [4,2]-[4,3]; [6,4] at [6,4]=6 and [7,4]=4; [4,4] at [8,3]-[8,4]; [1,4] at [8,1]=1 and [8,2]=4; [0,1] at [6,1]=0 and [7,1]=1. Then continue: [6,2] at [0,1]=6 and [0,0]=2; [2,4] at [1,0]=2 and [1,1]=4; [6,0] at [2,1]=6 and [3,1]=0; [2,3] at [2,0]=2 and [3,0]=3; [3,3] at [4,0]-[4,1]; [5,5] at [2,5]-[3,5]; [0,3] at [3,4]=0 and [2,4]=3; [3,5] at [1,4]=3 and [1,5]=5; [6,3] at [0,5]=6 and [0,4]=3.
🎨 Pips Solver
Aug 26, 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: Lock the left edge less-than
The less-than region at [2,0] has only one possible pip, 0. Since [2,0] has just one neighbor in the grid, [2,1], the 6-0 domino is forced horizontally: [2,0]=0 and [2,1]=6. This immediately contributes a high pip to the large sum region.
2
Step 2: Settle the second less-than cell
At [3,2], the less-than cap allows only 0 or 1. With the 6-0 domino already used and [2,2] sitting inside the large sum block, the 1-5 domino must run vertically: [3,2]=1 and [2,2]=5.
3
Step 3: Feed the big sum region
The sum-28 region already has [2,1]=6 and [2,2]=5, leaving 17 across [1,1], [1,2], and [2,3]. The 6-6 domino is the only way to provide 12 of that from the two top cells, so it goes at [1,1]=6 and [1,2]=6. The remaining [2,3] must be 5, so the 5-2 domino runs vertically with [2,3]=5 and [1,3]=2.
4
Step 4: Finish the sum-5 corner
With [1,3]=2, the sum-5 region needs [0,3]=3. The remaining 3-4 domino therefore fills [0,3]=3 and the empty cell [0,4]=4, completing the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Start with the tiny sum singleton
The sum-2 region at [3,2] forces a pip of 2. The only available domino with a 2 is the 2-5 domino, and its partner must go to [3,3] because the other neighboring options either don't exist or break the nearby equals region. Thus [3,2]=2 and [3,3]=5.
2
Step 2: Lock the adjacent equals-and-sum pair
Since [2,3] is in an equals region with [3,3], it must also be 5. The cell [2,4] is a one-cell sum-4 region and is only adjacent to [2,3], so the 4-5 domino must lie horizontally with [2,3]=5 and [2,4]=4.
3
Step 3: Solve the 6-equals cluster and sum 11
The three-cell equals region containing [1,0], [1,1], and [2,0] needs the same large pip. The 6-6 domino covers [1,0]-[1,1], and the 6-5 domino puts its 6 at [2,0], with its 5 at [2,1]. That gives [2,1]=5, so the sum-11 pair [2,1]-[2,2] requires [2,2]=6, which comes from the 1-6 domino at [1,2]=1 and [2,2]=6.
4
Step 4: Finish the top equals runs
The greater-than cell [0,1] requires 5, and the equals pair [0,2]-[0,3] requires both to be 3. The 5-3 domino fills [0,1]=5 and [0,2]=3; the 3-1 domino then fills [0,3]=3 and [1,3]=1, which also satisfies the equals pair [1,2]-[1,3] because [1,2] is already 1.
5
Step 5: Fill the bottom singletons
The one-cell sum-4 region at [3,0] needs a 4. The 4-3 domino sits horizontally with [3,0]=4 and the empty cell [3,1]=3. The rest of the grid is already settled by the earlier equalities and sums, so this confirms the completed placement.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Anchor the zero row's double
The four-cell sum-0 region [3,1]-[3,4] forces all four cells to be 0. The only double-0 domino is needed to cover two adjacent cells, and the middle pair [3,2]-[3,3] is the anchor because the outer cells can later connect to [2,1] and [2,4]. Place it: [3,2]=0 and [3,3]=0.
2
Step 2: Resolve the lower zero singleton
At [4,5], the singleton sum-0 forces a 0. Its upward neighbor [3,5] belongs to an equals column that will be 5, not available on any zero-carrying domino, so [4,5] must pair with [4,4]. The 2-0 domino puts [4,5]=0 and [4,4]=2. Now the sum-6 region [4,2],[4,3],[4,4] has 2 at [4,4], so [4,2]+[4,3]=4. The 2-2 domino supplies the only two pips that sum to 4 here, so place it at [4,2]=2 and [4,3]=2.
3
Step 3: Set the bottom-right equals block
The singleton sum-6 at [6,4] forces a 6. Its only neighbor [7,4] belongs to the four-cell equals block [7,4],[8,2],[8,3],[8,4]. The only value that can fill all four cells is 4, so [7,4]=4. Thus the 6-4 domino runs vertically with [6,4]=6 and [7,4]=4; the 4-4 domino fills [8,3]-[8,4]; and the 1-4 domino fills [8,1]=1 and [8,2]=4.
4
Step 4: Use the bottom-left less-than
The less-than cell at [6,1] is capped below 2, and its only neighbor [7,1] must equal [8,1], which is already 1. Therefore [6,1] must be 0, and the 0-1 domino covers [6,1]=0 and [7,1]=1.
5
Step 5: Climb the left-side sums
The singleton sum-6 at [0,1] forces a 6. Because [1,1] is needed for the sum-10 region, [0,1] pairs with [0,0]; the 6-2 domino gives [0,1]=6 and [0,0]=2. The left-column sum-6 triplet [0,0],[1,0],[2,0] now has [0,0]=2, so [1,0] and [2,0] must also be 2. The 2-4 domino goes [1,0]=2 and [1,1]=4; the 2-3 domino goes [2,0]=2 and [3,0]=3.
6
Step 6: Finish the remaining chains
The sum-10 pair [1,1]-[2,1] takes [1,1]=4 and [2,1]=6, so the 6-0 domino covers [2,1]=6 and [3,1]=0. The equals region [3,0],[4,0],[4,1] needs 3s, so the 3-3 domino goes at [4,0]-[4,1]. The equals column [1,5],[2,5],[3,5] needs 5s; put the 5-5 domino at [2,5]-[3,5] and the 3-5 domino at [1,4]=3 and [1,5]=5. The sum-9 triplet [0,4],[1,4],[2,4] needs 3s, so the 0-3 domino fills [3,4]=0 and [2,4]=3, and the 6-3 domino fills [0,5]=6 and [0,4]=3. This completes the grid.
💡 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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