🚨 SPOILER WARNING
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, constructed by Ian Livengood, is a bite-sized intro. With only five dominoes and a tiny grid, the solve hinges on a sum-0 region that forces two zeros immediately. From there, adjacent sum targets chain cleanly—no forks, no guesswork. Perfect for a quick win. Rodolfo Kurchan's medium puzzle adds one clever trap: a lonely cell hemmed in by a less-than-3 rule. That restriction narrows the possibilities drastically, unlocking a domino that then feeds a series of sum intersections. Expect a brief pause, then smooth sailing. The hard puzzle, also by Kurchan, is a masterclass in interlocking constraints. Three distinct sum-11 regions and two separate equals clusters create a tightly woven net. Finding the right first move among the equals cells is the gateway; after that, the sums cascade in a satisfying domino effect. Expect a longer, more deliberate solve.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Seek the smallest sum
Look for a region whose sum target leaves no choice—a sum of zero means both cells must be zero. This is your key entry point.
💡 Hint 2: Lock the top-left corner
The top-left region sums to 0, forcing [0,0] and [0,1] to both be 0. To supply those zeros, you'll need the two dominos that carry a 0 pip—one of them must also place a 1 nearby.
💡 Hint 3: Full solution
Start with the sum-0 region at [0,0]-[0,1]: both must be 0. Place the [0,1] domino vertically with 0 at [0,0] and 1 at [1,0]. Then the [3,0] domino must go horizontally with 0 at [0,1] and 3 at [0,2]. The sum-2 region at [1,0]-[1,1] now has [1,0]=1, so [1,1] must be 1. Use the [1,1] domino vertically at [1,1]-[2,1] (both 1). The sum-4 region at [2,0]-[2,1] gets [2,1]=1, so [2,0] needs 3—place the [3,4] domino horizontally from [2,0] to [3,0] (3,4). Finally, the sum-6 region at [3,0]-[3,1] has [3,0]=4, so [3,1] must be 2. Fill with the [2,2] domino vertically at [3,1]-[4,1] (both 2).
💡 Hint 1: Follow the lone 'less' rule
Focus on the only 'less-than' restriction—it’s a single cell that can’t be 3 or higher, which narrows the possible domino values sharply.
💡 Hint 2: Corner the low-value cell
The cell at [5,0] must be less than 3. Among the remaining dominos, the [6,0] tile is the only one that can supply a low value there (0) while keeping its high value (6) in an adjacent sum region.
💡 Hint 3: Full solution
Place [6,0] vertically at [4,0]-[5,0] (6 above, 0 below). This fills the sum-8 region in column 0: [4,0]=6 forces [3,0]=2. The [5,2] domino goes horizontally at [2,0]-[3,0] (5,2). The sum-9 region above then needs [1,0]=4, so the [4,4] domino fills [0,0]-[1,0] (both 4). On the right, the sum-8 in column 2 gets the [6,5] domino placed vertically at [0,2]-[1,2] (6,5), so [2,2] must be 3 from [3,3] placed at [2,2]-[3,2] (both 3). Finally, the sum-6 region at [3,2]-[4,2] requires [4,2]=3, so place the [3,4] domino horizontally across [4,2]-[5,2] (3 and 4).
💡 Hint 1: Anchor on the equals regions
The equals regions are your anchor; look for a cluster of cells that must all match, then see which domino values can feed them. The largest one involves four cells and will dictate which domino supplies that value.
💡 Hint 2: Target the four-cell match
The four-cell equals region covers [1,0], [1,1], [2,0], [3,0]. Since all four must be identical, the dominos that touch these cells need to work together—look for a domino with a double pip that can cover two of these cells directly.
💡 Hint 3: Lock in the value 2
The [2,2] domino is ideal: place it vertically at [1,0]-[2,0], giving [1,0]=2 and [2,0]=2. To satisfy the equals region, [1,1] must also be 2, so use the [4,2] domino horizontally at [1,1]-[1,2] (2 and 4). Then [3,0] must be 2, which will come from the [5,2] domino later.
💡 Hint 4: Tackle sum-11 and less-4
The less-4 region in row 0 (cols 4-6) forces all three cells to be 0. Use the [3,0] domino at [1,4]-[0,4] (3,0) to get [0,4]=0, and the [0,0] domino at [0,5]-[0,6] (0,0). For the sum-11 at [2,4]-[3,4], the [5,6] and [0,5] dominoes give 5 and 6. Meanwhile, the equals region’s [3,0]=2 pairs with the [5,2] domino at [4,0]-[3,0] to set up the next sum-11.
💡 Hint 5: Complete solve
Start with [2,2] vertical at [1,0]-[2,0] (2s). Equals: [4,2] horizontal at [1,1]-[1,2] (2,4); [5,2] vertical at [3,0]-[4,0] (2,5). Less-4 row: [3,0] horizontal at [1,4]-[0,4] (3,0); [0,0] at [0,5]-[0,6] (0,0). Sum-11 at [2,4]-[3,4]: [0,5] horizontal at [2,5]-[2,4] (0,5); [5,6] vertical at [4,4]-[3,4] (5,6). Sum-11 at [4,0]-[4,1]: [0,6] vertical at [4,2]-[4,1] (0,6). Sum-11 at [4,4]-[4,5]: [3,6] vertical at [4,6]-[4,5] (3,6). Bottom equals: [5,1] horizontal at [6,4]-[6,3] (5,1); [1,1] vertical at [6,2]-[7,2] (1,1); [2,1] vertical at [9,2]-[8,2] (2,1).
🎨 Pips Solver
Jun 15, 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: Sum-zero anchor
The region at [0,0] and [0,1] must sum to 0. The only way is for both cells to be 0, locking the first two values.
2
Step 2: Place the [0,1] domino
To give [0,0] a 0 and also provide a 1 for the nearby sum-2 region, place the [0,1] domino vertically so [0,0]=0 and [1,0]=1.
3
Step 3: Complete the sum-2 region
With [1,0]=1, the sum-2 region at [1,0]-[1,1] forces [1,1] to be 1. Place the [1,1] domino vertically at [1,1]-[2,1] to set both cells to 1.
4
Step 4: Cascade the remaining sums
[0,1] still needs 0: place the [3,0] domino horizontally with 0 at [0,1] and 3 at [0,2]. The sum-4 region at [2,0]-[2,1] has [2,1]=1, so [2,0] must be 3; place the [3,4] domino horizontally from [2,0] to [3,0] (3,4). Finally, the sum-6 region at [3,0]-[3,1] has [3,0]=4, so [3,1] needs 2—use the [2,2] domino vertically at [3,1]-[4,1] (both 2).
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Exploit the less-than rule
The cell at [5,0] must be less than 3, so it can only be 0 or 2. The [6,0] domino can place a 0 here, with its 6 forced above it.
2
Step 2: Fix the sum-8 in column 0
Place [6,0] vertically at [4,0]-[5,0] (6 on top, 0 below). The sum-8 region at [3,0]-[4,0] now has [4,0]=6, so [3,0] must be 2.
3
Step 3: Link to the sum-9 above
To supply the 2 at [3,0], place the [5,2] domino horizontally at [2,0]-[3,0] (5 and 2). Now [2,0]=5, feeding the sum-9 region above it.
4
Step 4: Satisfy the upper left
The sum-9 region at [1,0]-[2,0] has [2,0]=5, so [1,0] must be 4. Place the [4,4] domino horizontally at [0,0]-[1,0] to set both cells to 4.
5
Step 5: Resolve the right side
The sum-8 region at [1,2]-[2,2] needs 8. Use the [6,5] domino vertically at [0,2]-[1,2] (6,5), giving [1,2]=5 and forcing [2,2]=3. Place the [3,3] domino vertically at [2,2]-[3,2] (both 3). The sum-6 at [3,2]-[4,2] now has [3,2]=3, so [4,2] must be 3—use the [3,4] domino horizontally at [4,2]-[5,2] (3,4).
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Lock the four-cell equals
The region at [1,0],[1,1],[2,0],[3,0] must all share the same pip. The [2,2] domino can cover [1,0] and [2,0], so place it vertically (2s) to start the equals chain.
2
Step 2: Extend the equals cluster
With [1,0] and [2,0] both 2, [1,1] must also be 2. Use the [4,2] domino horizontally at [1,1]-[1,2] (2,4) to set that. The cell [3,0] will need 2 as well—reserve the [5,2] domino to cover it later.
3
Step 3: Zero out the top row
The less-4 region at [0,4]-[0,6] demands values under 4. Place the [3,0] domino horizontally at [1,4]-[0,4] (3,0) to set [0,4]=0. Then place the [0,0] domino horizontally at [0,5]-[0,6] (0,0) to fill the rest.
4
Step 4: Solve the sum-11 triangles
For the sum-11 at [2,4]-[3,4]: use the [0,5] domino horizontally at [2,5]-[2,4] (0,5) to get [2,4]=5, and the [5,6] domino vertically at [4,4]-[3,4] (5,6) to get [3,4]=6. For the sum-11 at [4,0]-[4,1]: place [5,2] vertically at [3,0]-[4,0] (2,5) to satisfy [3,0]=2 and [4,0]=5, then the [0,6] domino vertically at [4,2]-[4,1] (0,6) supplies [4,1]=6. Finally, the third sum-11 at [4,4]-[4,5] uses [5,6] already giving [4,4]=5, so [4,5]=6 comes from the [3,6] domino vertically at [4,6]-[4,5] (3,6).
5
Step 5: Finish the bottom equals
Cells [6,2],[6,3],[7,2],[8,2] must all be 1. Place the [5,1] domino horizontally at [6,4]-[6,3] (5,1) to give [6,3]=1. Then the [1,1] domino vertically at [6,2]-[7,2] (1,1) covers two cells. Finally, the [2,1] domino vertically at [9,2]-[8,2] (2,1) provides [8,2]=1. All empty cells then fall into place.
💡 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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