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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
Today's NYT Pips easy, constructed by Ian Livengood, is a confidence-builder—two giveaway regions, a sum-0 column and a sum-12 row, guide the entire solve without any branches. You'll move linearly from zeros to sixes to a clean finish. Medium, from Rodolfo Kurchan, has one tight bottleneck: a greater-than-10 region that seems impossible until you realize it forces a 6+5 pair. Find that, and the surrounding sums and less-than regions unlock neatly. Hard, also by Kurchan, is a more intricate beast. It weaves a net of single-cell sum-4 constraints with two large equals regions; the challenge is orchestrating the many 1s and 4s without conflict. Steady deduction will crack it.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Spot the zero-force regions
The most restrictive constraint is a multi-cell region that demands a sum of zero. Identify it and think about which dominos can supply so many blank pips.
💡 Hint 2: The triple-zero column
The rightmost column has three cells that must sum to zero. That forces the double-zero domino and a single-zero domino. Meanwhile, the middle row has two adjacent cells needing a total of 12, which can only be achieved with the two available sixes.
💡 Hint 3: Full answer
Place [0,0] at [1,3]/[2,3] and [1,0] at [4,3]/[3,3] for the zeros. For the sum-12 region, put [6,5] at [2,0]/[1,0] (6 and 5) and [6,3] at [2,1]/[2,2] (6 and 3). The equals pair [0,1]/[0,2] takes [1,1]. Finally, [5,4] goes to [4,0]/[3,0] with 5 and 4.
💡 Hint 1: Focus on the impossible sum
One region asks for a total so high it feels unreasonable. Locate the greater-than constraint and consider the highest possible pip values.
💡 Hint 2: The >10 pair
The region comprising [1,0] and [1,1] must exceed 10. The only way is 6+5, forcing dominos with those numbers. The 6 is paired with a 3, and the 5 with a 1; their partners will lock into adjacent regions.
💡 Hint 3: Full answer
Place [6,3] at [1,0]=6,[2,0]=3 and [1,5] at [0,1]=1,[1,1]=5 to satisfy the >10 and start the chain. Then [3,0] at [1,2]=3,[1,3]=0 meets sum-1; [4,4] at [2,1]=4,[2,2]=4 completes sum-7. [3,1] at [1,4]=3,[1,5]=1 for less-6. Finally, [1,6] at [2,3]=1,[2,4]=6 and [3,4] at [2,5]=3,[2,6]=4.
💡 Hint 1: Single-cell sums are your anchor
Many cells have a sum-4 requirement alone. Find adjacent pairs of these and you can lock in a domino with two 4s. Also, note the large equals region that will demand many copies of the same number.
💡 Hint 2: Zero in and equals region
The cells at [4,3] and [5,3] form a sum-0 region, so they must both be 0. The large equals region spanning [3,2],[3,3],[4,2],[5,2],[6,2] needs five identical pips. Given the available numbers, they'll all be 1.
💡 Hint 3: The [4,4] anchor and equals start
[2,0] and [3,0] are adjacent single-cell sum-4 regions, perfectly taking the [4,4] domino vertically. The five-cell equals region will mostly be filled with 1s, so place [1,1] to cover two of them, and use a domino with a 1 and 0 to start the sum-0 region.
💡 Hint 4: Building the cascade
The sum-4 pair at [0,2]/[0,3] takes [2,2] for 2+2. The sum-1 cell at [2,1] forces a 1 there, so use [1,6] to place 1 at [2,1] and 6 at [2,2] (part of a sum-9). The equals region of three cells at [6,3],[7,2],[7,3] all need to be the same; the remaining numbers point to 3.
💡 Hint 5: Full answer
Place [4,4] at [2,0]/[3,0]; [1,1] at [5,2]/[6,2]; [2,2] at [0,2]/[0,3]; [1,5] at [3,2]/[3,1]; [6,6] at [1,3]/[1,4]; [4,1] at [3,4]/[3,3]; [3,0] at [6,3]/[5,3]; [5,4] at [8,4]/[8,3]; [3,3] at [7,2]/[7,3]; [1,0] at [4,2]/[4,1]; [4,3] at [1,1]/[1,2]; [1,2] at [2,3]/[2,4]; [4,6] at [3,5]/[2,5]; [4,0] at [4,4]/[4,3]; [1,6] at [2,1]/[2,2]; [4,2] at [8,2]/[8,1].
🎨 Pips Solver
Jul 19, 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 sum-0 column
The region consisting of [1,3], [2,3], and [3,3] must sum to 0, forcing every cell to be 0. The [0,0] domino is the only one with two zeros; place it vertically in [1,3] and [2,3]. The third zero can only come from the [1,0] domino, so reserve it for later.
2
Step 2: The sum-12 row
The two adjacent cells [2,0] and [2,1] must sum to 12, meaning each must be 6. The only dominos containing 6 are [6,3] and [6,5]. Place [6,5] vertically with the 6 in [2,0] and the 5 in [1,0] (the empty cell). Place [6,3] horizontally with the 6 in [2,1] and the 3 in [2,2], satisfying the less-4 spot.
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Step 3: The equals pair
Cells [0,1] and [0,2] must be equal. Only the [1,1] domino fits, placing both as 1.
4
Step 4: Finishing the puzzle
Now [3,3] still needs a 0, so finish the [1,0] domino by putting its 0 there and the 1 in the empty [4,3]. Finally, the [5,4] domino goes to the remaining cells: place 5 in empty [4,0] and 4 in [3,0] (which is less-5, so 4 works).
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The >10 region
The region at [1,0] and [1,1] must have a sum greater than 10. The only possible combination is 6+5. The domino [6,3] gives the 6 at [1,0] with its 3 at [2,0], while [1,5] gives the 5 at [1,1] with its 1 at [0,1].
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Step 2: The sum-1 region
Cells [1,3] and [2,3] must sum to 1, so they are 0 and 1 in some order. The [3,0] domino places its 0 in [1,3] and its 3 in [1,2], which will join a sum-7 region with [2,2].
3
Step 3: Completing sum-7 with [4,4]
Now [1,2]=3, so [2,2] must be 4 to reach sum 7. The [4,4] domino fits perfectly horizontally at [2,1] and [2,2], supplying a 4 to each. This also satisfies the sum-7 region on [2,0]/[2,1] because [2,0] already holds the 3 from step 1, making 3+4=7.
4
Step 4: The less-6 pair
Region [1,4] and [1,5] must sum below 6. The [3,1] domino, placed vertically, gives 3 at [1,4] and 1 at [1,5], totaling 4. The 1 at [1,5] also meets the less-6 condition.
5
Step 5: Final greater regions
The remaining cells: [2,3] needs the 1 from the sum-1 region, so place [1,6] horizontally with 1 at [2,3] and 6 at [2,4] (part of a greater-8 pair). The larger region [2,4]/[2,5] now has 6+? and needs >8, so the [3,4] domino goes there with 3 at [2,5] and 4 at [2,6], also satisfying the greater-3 constraint on [2,6].
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Adjacent 4s
Both [2,0] and [3,0] are single-cell regions requiring a sum of exactly 4, meaning each must be 4. They sit vertically adjacent, so the [4,4] domino fits perfectly there, placing 4 in both.
2
Step 2: The sum-0 and a 4
Region [4,3]/[5,3] sums to 0, so both cells are 0. The cell directly above, [4,4], is another sum-4 single. The [4,0] domino places its 4 in [4,4] and its 0 in [4,3], starting the zero pair.
3
Step 3: Anchoring the 5-cell equals
The large equals region (5 cells) must all hold the same number; with many 1s available, they become 1. Place [1,1] at [5,2]/[6,2] to cover two of them. Meanwhile, the less-4 cell at [4,1] forces a 0, so [1,0] goes there with 1 at [4,2] (feeding the equals) and 0 at [4,1].
4
Step 4: Top sum-4 and the sum-1 cell
The sum-4 region [0,2]/[0,3] is perfectly taken by [2,2], giving 2+2. The single sum-1 cell at [2,1] demands a 1; the [1,6] domino fills that with 1 at [2,1] and 6 at [2,2], completing the sum-9 region with the adjacent [1,2] which later gets a 3 from [4,3].
5
Step 5: The second equals region
The three-cell equals region at [6,3],[7,2],[7,3] must all be the same. The [3,3] domino covers [7,2]/[7,3] with 3 and 3, while [3,0] places a 3 in [6,3] and its 0 into [5,3], finishing the sum-0 pair.
6
Step 6: Filling the remaining 4s and 5
Now place the remaining dominos: [1,5] at [3,2]=1,[3,1]=5 (finishing the equals and sum-5 region), [6,6] at [1,3]=6,[1,4]=6, [4,1] at [3,4]=4,[3,3]=1, [4,6] at [3,5]=4,[2,5]=6, [4,3] at [1,2]=3,[1,1]=4 (sum-9 and sum-4), and [5,4] at [8,4]=5,[8,3]=4. Finally, [4,2] at [8,2]=4,[8,1]=2 to satisfy the last sum-4 and less-4 cells.
💡 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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