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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 constructs today's NYT Pips easy puzzle as a compact 3x5 layout, using two equals regions to create a forced block of identical pips. A pair of single-cell greater-than constraints isolates high values, while a lone sum-2 cell locks the final placement—an exercise in minimal clue economy. The design reads like a domino syllogism, each step an inevitable deduction.
Rodolfo Kurchan's medium grid showcases a clever arithmetic interplay: a sum-10 doubles region immediately forces a pair of fives, radiating constraints through an adjacent sum-7 column and a sum-10 row. The unequal and less-than regions act as elegant checks rather than primary drivers, giving the solve a satisfying 'center-out' flow that rewards solvers who trust the numbers.
The hard puzzle is pure Kurchan minimalism—an 8x4 grid partitioned into 32 mostly single-cell sum targets, each demanding a specific pip. The domino set is curated to include every 0–X pairing, and the architecture forces a beautiful cascade of 0-placements down the board. One solitary greater-than region breaks the symmetry at the end, a final twist in an otherwise relentlessly logical chain.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Look for Equal Regions
The puzzle features two zones where all cells must display identical pip values. Identify these grouped cells—they will be the key to unlocking which dominos go where.
💡 Zoom in on the Top-Left
The two-cell equals region spanning [0,0] and [0,1] demands a domino with matching pips. Meanwhile, the cells directly below it have 'greater than' requirements. Consider which dominoes from the set can satisfy both the equality and those adjacent greater-than conditions.
💡 Complete Easy Solve
Place the [2,2] domino in the top-left equals region, setting 2s at [0,0] and [0,1]. Next, the equals region at [0,2]-[1,2]-[1,3] needs identical pips; use the [4,4] domino, putting 4s at [0,2] and [1,2] (and later [1,3]). The greater-than-4 cell at [1,1] and greater-than-2 at [1,0] force the [5,3] domino: place 5 in [1,1] and 3 in [1,0]. Then the [4,6] domino extends the equals chain at [1,3] (must be 4) and fills the greater-than-4 cell at [1,4] with a 6. Finally, the [3,2] domino goes to the bottom row: 3 in the empty [2,2] and 2 in the sum-2 cell [2,3].
💡 Target the Sum-10 Pair
A two-cell sum region with a high target sits in the middle of the board. Look at your domino list: only one combination can hit that total, forcing those two cells immediately and starting the entire solve from the center outward.
💡 The Vertical Sum-10 Lock
The sum-10 region at [1,1] and [2,1] has only one possible domino: the double-five. Place it to start the chain. Next, the sum-7 column at [1,2], [2,2], [3,2] must be satisfied; the presence of the 5 at [2,1] strongly restricts what can go into [2,2]. Look for a domino that lets the sum-7 work with a low number up top.
💡 Medium Full Solve
Start by putting the [5,5] domino in [1,1] and [2,1]. For the sum-7 column, place the [1,1] domino in [1,2]-[1,3] to give [1,2]=1 and satisfy the less-4 at [1,3]; then place the [3,3] domino in [2,2] and [3,2] to get 3s, making 1+3+3=7. The sum-10 row at [2,3]-[2,5] is filled by the [6,4] domino at [2,3] (6) and [3,3] (4, adhering to less-6) plus the [2,2] domino at [2,4] and [2,5] (2s). The equals pair at [3,0]-[3,1] takes the [6,6] domino, and the unequal pair at [0,4]-[1,4] gets the [2,1] domino with 2 and 1 to be different.
💡 Single-Cell Sums
Virtually every cell has its own sum constraint. This means each cell's pip is predetermined, but you must figure out which dominos cover which pairs. Focus on the smallest targets first—they drastically limit the possibilities.
💡 The Zero Zones
Several cells require a sum of 0, which means they must contain a 0. Look at rows 4, 6, and 7: the 0-cells are [4,2], [4,3], [6,1], [7,1], and [7,3]. All the 0-paired dominoes will need to connect these with specific neighbors that match the surrounding sums.
💡 Pairing the Zero Dominoes
The sum-1 cell at [5,1] is next to a sum-0 at [6,1]; that forces the [0,1] domino into that vertical pair. Similarly, the sum-0 at [7,1] and sum-4 at [7,0] go together with the [0,4] domino. Use these initial placements to gain a foothold in the bottom rows.
💡 Row 4 and the Remaining Zeros
After placing [0,1] and [0,4], tackle [4,3] (sum-0) and [3,3] (sum-3) with the [0,3] domino; then [4,2] (sum-0) and [3,2] (sum-5) with [0,5]. The sum-2 at [7,2] and sum-0 at [7,3] get the [0,2] domino. With all zeros set, the remaining sum constraints will rigidly guide every other pairing.
💡 Hard Complete Grid
Place: [0,1] domino with 0 at [6,1], 1 at [5,1]; [0,4] with 0 at [7,1], 4 at [7,0]; [0,2] with 0 at [7,3], 2 at [7,2]; [0,3] with 0 at [4,3], 3 at [3,3]; [0,5] with 0 at [4,2], 5 at [3,2]. Then sum-1 cells: [1,2]=1 via [1,5] at [1,2] (1) and [1,1] (5); [1,0]=1 via [1,4] at [1,0] (1) and [2,0] (4); [0,0]=1 via [1,2] at [0,0] (1) and [0,1] (2). Continue: [3,0]=1 (empty) and [4,0]=3 via [1,3]; [2,2]=3 and [2,1]=4 via [3,4]; [0,2]=3 and [0,3]=5 via [3,5]; [5,0]=2 and [6,0]=4 via [2,4]; [4,1]=2 and [3,1]=5 via [2,5]; [1,3]=2 and [2,3]=3 via [2,3]; [6,2]=4 and [5,2]=5 via [4,5]; [5,3]=3 and [6,3]=6 via [3,6].
🎨 Pips Solver
Jul 18, 2026
Click a domino to place it on the board. You can also click the board, and the correct domino will appear.
✅ 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: Identify the Double Pip Demand
The equals region at [0,0] and [0,1] forces a domino with identical pips. Among the available doubles, only [2,2] and [4,4] are candidates. Since the other equals cluster will need a larger double, [2,2] fits here, setting 2s in both top-left cells.
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Step 2: Lock the Second Equals Cluster
The larger equals region spans [0,2], [1,2], and [1,3]. It requires identical pips across all three cells. The [4,4] domino must cover two of them—place it horizontally across [0,2] and [1,2], leaving [1,3] to receive a 4 from a later domino.
3
Step 3: Satisfy the Greater-Than Singles
Cell [1,1] demands a pip greater than 4, and [1,0] greater than 2. With 2s and 4s already placed, the only way to achieve this while covering [1,0] and [1,1] as a pair is the [5,3] domino. Put 5 in [1,1] and 3 in [1,0].
4
Step 4: Close Out with Sum and Empty
The remaining two dominos: [4,6] must go to [1,3] and [1,4] because [1,3] belongs to the equals chain and needs a 4, while [1,4] must exceed 4 and gets the 6. Finally, the [3,2] domino fills the bottom row—3 in empty [2,2] and 2 in sum-2 cell [2,3], satisfying that last constraint.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Force the Sum-10 Vertical
The sum-10 region covering [1,1] and [2,1] has only one domino in the set that can hit 10: the double-five [5,5]. Place it vertically, giving both cells a value of 5.
2
Step 2: Untangle the Sum-7 Column
The sum-7 region includes [1,2], [2,2], and [3,2]. With [2,1]=5 adjacent, [2,2] cannot be a high number. To reach a total of 7, [1,2] must be 1 (from the [1,1] domino placed later), and the remaining 6 must be split as 3 and 3 using the [3,3] domino at [2,2] and [3,2].
3
Step 3: Place the Less-4 and Unequal
Cell [1,3] is less than 4, so it naturally takes the 1 from the [1,1] domino, which also covers [1,2] to complete the sum-7 column's 1. The unequal pair at [0,4] and [1,4] then claims the [2,1] domino, placing 2 and 1 to ensure they differ.
4
Step 4: Solve the Sum-10 Row
The horizontal sum-10 region at [2,3], [2,4], [2,5] must add to 10. With [3,3] being less than 6, the [6,4] domino fits perfectly: 6 at [2,3] and 4 at [3,3]. The remaining sum of 4 is achieved by putting 2s from the [2,2] domino into [2,4] and [2,5].
5
Step 5: Finish with the Equals Pair
The only unused region is the equals pair at [3,0] and [3,1], which conveniently matches the last remaining domino [6,6], placing 6s in both cells.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Anchor the 0s via Sum-0 and Sum-1
Cell [6,1] requires a sum of 0, so it must be 0. Its only neighbor in a domino is [5,1], which has a sum-1 constraint and must be 1. The [0,1] domino provides exactly that pair, so place it vertically: 0 at [6,1], 1 at [5,1].
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Step 2: Extend 0s in Row 7
Similar logic applies at the bottom: sum-0 cell [7,1] forces a 0; its neighbor [7,0] demands a sum-4, so they need a 0 and a 4—the [0,4] domino. Then sum-0 cell [7,3] and sum-2 cell [7,2] take the [0,2] domino, placing 0 and 2 accordingly.
3
Step 3: Zero Out Row 4
Two more sum-0 cells sit in row 4: [4,3] pairs with the sum-3 cell [3,3] via the [0,3] domino (0 and 3). [4,2] pairs with the sum-5 cell [3,2] via the [0,5] domino (0 and 5). Now all zero-required cells are filled with their 0-pipped partners.
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Step 4: Assign the Ones
The sum-1 cells are [0,0], [1,0], and [1,2]. [1,2] partners with [1,1] (sum-5) using the [1,5] domino: 1 at [1,2], 5 at [1,1]. [1,0] partners with [2,0] (sum-4) via [1,4]: 1 at [1,0], 4 at [2,0]. [0,0] partners with [0,1] (sum-2) via [1,2]: 1 at [0,0], 2 at [0,1].
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Step 5: Fill the Middle
With the zeros and ones in place, the middle dominoes fall into line. [3,0] (empty) and [4,0] (sum-3) get the [1,3] domino (1 and 3). [2,2] (sum-3) and [2,1] (sum-4) use [3,4] (3 and 4). [0,2] (sum-3) and [0,3] (sum-5) take [3,5] (3 and 5). [5,0] (sum-2) and [6,0] (sum-4) get [2,4] (2 and 4). [4,1] (sum-2) and [3,1] (sum-5) claim [2,5] (2 and 5).
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Step 6: Finish the Right Side
The remaining region pairs: [1,3] (sum-2) and [2,3] (sum-3) take the [2,3] domino (2 and 3). [6,2] (sum-4) and [5,2] (sum-5) take [4,5] (4 and 5). Finally, [5,3] (empty) and [6,3] (greater than 3) must be covered by the lone [3,6] domino: place 3 at [5,3] and 6 at [6,3] to satisfy the greater-than constraint.
💡 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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