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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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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Today's NYT Pips easy, set by Ian Livengood, is a compact 10-cell board with a friendly layout. Two sum-10 regions and a sum-3 pair create a clean logical chain; the low pip count guarantees a smooth warm-up with no guesswork. If you spot the single zero domino early, the rest tumbles out in a neat cascade.
Livengood's medium puzzle stretches to 18 cells and leans heavily on equals constraints. The opening bottleneck is a single-cell sum-5 that, coupled with a three-cell equals column, locks the first few dominos. After that, a web of small sums and an equals pair on the right edge make the middle game a satisfying fill.
Rodolfo Kurchan's hard puzzle is a 32-cell monster laced with single-digit sum regions and a towering four-cell equals column. The degree of difficulty is high, but the path is anchored by a cluster of sum-1, sum-3, and sum-5 cells near the bottom that ripple outward. Expect to chain deductions from the single-cell anchors through the equals cascade.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Follow the Tens
Identify the two regions that require a sum of 10; that limits the pip combinations sharply. Also, keep an eye on the tiny less-than-2 cells—they'll accept only 0 or 1.
💡 Hint 2: Column Zero Connection
The sum-10 at [0,0] and [1,0] forces a 6 and a 4. The 6 can only come from one domino, which also places a 1 next door to satisfy a less-2 constraint. That fixes the entire upper-left corner.
💡 Hint 3: Full Layout
Place [1,6] horizontally covering [0,0]=6 and [0,1]=1. Next, [3,4] goes vertically at [1,0]=4 and [2,0]=3, fulfilling the other sum-10 and setting up the sum-3 region. [0,5] then sits horizontally at [3,0]=0 and [3,1]=5 to hit that sum-3. The second sum-10 locks [5,6] across [3,2]=5 and [3,3]=6, and finally [1,3] vertically covers [1,2]=1 and [2,2]=3.
💡 Hint 1: Single Digits and Equals
A one-cell sum region gives a fixed number immediately. Pair that with an entire column that must have identical values, and you'll crack the opening quickly.
💡 Hint 2: The Sum-5 Key
The cell [0,1] is a sum-5 all by itself, so it's a 5. The equals region down the left column ([0,0], [1,0], [2,0]) then forces all three to match. Since [0,0] must pair with [0,1] via a single domino, that domino can only be [4,5], placing a 4 at [0,0] and making the whole column 4.
💡 Hint 3: Complete Assembly
Set [4,5] at [0,0]=4 and [0,1]=5. Then [4,4] vertical at [1,0]=4, [2,0]=4. [2,4] domino goes vertically at [1,8]=2, [0,8]=4 to satisfy sum-8 and equals. [4,0] covers [0,7]=4, [0,6]=0. Fill the middle equals clusters with [6,6] at [2,1]=6, [2,2]=6; [2,6] at [1,2]=6, [0,2]=2; [1,1] at [1,4]=1, [2,4]=1; [4,3] at [1,6]=4, [2,6]=3; and [3,6] at [2,7]=3, [2,8]=6.
💡 Hint 1: Target Single-Cell Sums
Hunt for one-cell regions with tiny sum targets—especially 1, 3, and 5. These cells have forced values and sit very close to each other, giving you a solid foothold.
💡 Hint 2: Bottom-Right Tangle
The cell [5,5] must be 1, and the adjacent [5,6] must be 3; they snap together with domino [3,1]. Neighbouring [5,7] is a sum-5, forcing a 5. It has only one free adjacent partner in [4,7], which must later form a sum-2 with [3,7], so that domino is [1,5].
💡 Hint 3: Left Side Takes Shape
The sum-3 at [1,3] sets that cell to 3. It can only pair with [2,3] below it, so the domino is [5,3] with 5 and 3. That 5 at [2,3] will later feed a sum-10 with [3,3] which will need another 5—watch for that.
💡 Hint 4: The Equals Tower and Double 6s
The four-cell equals region spanning [6,3] to [9,3] will all read 4. The sum-12 at [7,1] and [7,2] demands two 6s, pulling in [6,4] and [6,3] dominos to deliver those 6s while placing 4s into the equals column. The sum-6 at [5,1]+[6,1] then forces a pair of 3s.
💡 Hint 5: Final Hard Answer
Place [3,1] horizontally at [5,5]=1, [5,6]=3. [1,5] vertically at [4,7]=1, [5,7]=5. [5,3] vertically at [2,3]=5, [1,3]=3. [0,2] horizontally at [2,6]=0, [2,5]=2. [0,1] vertically at [2,7]=0, [3,7]=1. [6,4] horizontally at [7,2]=6, [7,3]=4. [6,3] vertically at [7,1]=6, [6,1]=3. [4,4] vertically at [8,3]=4, [9,3]=4. [5,6] horizontally at [3,3]=5, [3,2]=6. [0,6] horizontally at [3,0]=0, [3,1]=6. [0,4] vertically at [2,0]=0, [1,0]=4. [5,5] horizontally at [8,1]=5, [9,1]=5. [3,4] horizontally at [5,3]=3, [6,3]=4. [3,2] horizontally at [5,1]=3, [5,2]=2. [2,2] horizontally at [0,6]=2, [0,7]=2. [5,4] vertically at [1,5]=5, [0,5]=4.
🎨 Pips Solver
Jul 3, 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: The 10‑Sum Corner
The region at [0,0] and [1,0] demands a sum of 10. With pip values capped at 6, the only feasible pair is 6 and 4. Those two cells must hold 6 and 4 in some order.
2
Step 2: Six and a Less‑2 Neighbor
A neighboring less‑2 cell at [0,1] accepts only 0 or 1. The domino [1,6] contains a 6 and a 1; placing it horizontally with the 6 on [0,0] and the 1 on [0,1] satisfies both the sum‑10 (6 at [0,0]) and the less‑2 (1 at [0,1]). This forces [1,0] to be 4 to complete the 10.
3
Step 3: A Zero Completes the Sum‑3
With [1,0]=4, the only domino with a 4 is [3,4]. It must go vertically at [1,0]=4 and [2,0]=3. Now the sum‑3 region at [2,0] and [3,0] forces [3,0] to be 0. The only domino with a 0 is [0,5]; place it horizontally at [3,0]=0 and [3,1]=5.
4
Step 4: The Final Dominoes
The second sum‑10 region at [3,1] and [3,2] now has [3,1]=5, so [3,2] must be 5. Domino [5,6] fits horizontally with 5 at [3,2] and 6 at [3,3], satisfying the greater‑5 constraint. The remaining cells [1,2] (less‑2) and [2,2] (empty) pair vertically via [1,3] with 1 and 3.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The Lone Sum‑5
The single‑cell region at [0,1] is a sum‑5. Since only one cell contributes, its value is forced to 5 immediately.
2
Step 2: The Equals Column Locks to 4
The equals region covers [0,0], [1,0], and [2,0], so those three cells share the same pip. The cell [0,0] must pair with [0,1] in a domino, and with [0,1]=5 the only domino containing a 5 is [4,5]. That places a 4 at [0,0], making the entire column 4.
3
Step 3: Double‑4 Drops In
To supply 4s to the remaining two cells of the equals column, the double‑4 domino [4,4] is placed vertically at [1,0] and [2,0], both reading 4.
4
Step 4: Sum‑8 and More Equals on the Right
The sum‑8 region at [1,8] and [2,8] needs a total of 8. The equals pair [0,7] and [0,8] ties [0,8] to whatever value [0,7] gets. Placing domino [2,4] vertically at [1,8]=2 and [0,8]=4 gives [1,8]=2, so [2,8] must be 6. Domino [3,6] fits vertically at [2,7]=3 and [2,8]=6, and the equals region [2,6]‑[2,7] forces both to 3.
5
Step 5: Finishing the Inner Clusters
[4,0] covers [0,7]=4 and the less‑2 [0,6]=0. The central equals block [1,2]/[2,1]/[2,2] becomes all 6 via [6,6] at [2,1] and [2,2]; [2,6] gives [0,2]=2, [1,2]=6. The equals region [1,4]/[2,4] uses [1,1] vertically (both 1). Finally, [4,3] goes to [1,6]=4 and [2,6]=3, completing the grid.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Tiny Sums Anchor the Bottom
The single‑cell regions [5,5] (sum‑1) and [5,6] (sum‑3) force values 1 and 3. They are adjacent horizontally, so the only domino with 1 and 3, [3,1], must cover both: [5,5]=1, [5,6]=3.
2
Step 2: A 5 Finds Its Partner
The sum‑5 at [5,7] dictates a 5. Its neighbor [5,6] is already taken, so the only free neighbor is [4,7] above. The sum‑2 region [3,7]+[4,7] later forces both to be 1, so the domino here must give 1 at [4,7] and 5 at [5,7]. That is exactly [1,5], placed vertically.
3
Step 3: The Left Sum‑3 and a Pending 10
The sum‑3 at [1,3] forces a 3 there. It pairs vertically with [2,3]; the sum‑10 region [2,3]+[3,3] will later require [3,3]=5, so [2,3] must also be 5. Hence domino [5,3] fits vertically: [1,3]=3, [2,3]=5.
4
Step 4: Zeros and a Sum‑7
The equals region [2,6]‑[2,7] forces both to 0. The sum‑7 at [1,5] and [2,5] will be 5+2. Domino [0,2] goes horizontally at [2,5]=2, [2,6]=0. Then [0,1] placed vertically at [2,7]=0, [3,7]=1 completes the sum‑2 with [4,7]=1.
5
Step 5: The Great Equals Column Meets Double 6s
The four‑cell equals region [6,3]‑[9,3] all become 4. The sum‑12 at [7,1]+[7,2] needs 6+6; domino [6,4] (6+4) rests horizontally at [7,2]=6, [7,3]=4, while [6,3] (6+3) vertical at [7,1]=6, [6,1]=3 supplies the other 6. The sum‑6 region [5,1]+[6,1] then sums to 3+3, so [5,1]=3 and the remaining half of [6,3] is 3. Domino [3,2] is placed at [5,1]=3, [5,2]=2 to satisfy a sum‑5 with [5,3]=3 (from [3,4] later).
6
Step 6: Mopping Up with Symmetry
Finish the equals column with [4,4] vertical at [8,3]=4, [9,3]=4, and [3,4] horizontal at [5,3]=3, [6,3]=4. The sum‑10 at [2,3]/[3,3] uses [5,6] horizontal at [3,3]=5, [3,2]=6, and the sum‑12 at [3,1]/[3,2] gets [0,6] horizontal at [3,0]=0, [3,1]=6. Remaining: [0,4] vertical at [2,0]=0, [1,0]=4; [5,5] horizontal at [8,1]=5, [9,1]=5; [5,4] vertical at [1,5]=5, [0,5]=4; and [2,2] horizontal at [0,6]=2, [0,7]=2 to fill the top row.
💡 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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