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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
This NYT Pips easy, by Ian Livengood, is a compact 5‑domino layout that solves with clean, linear deductions. Expect a quick solve: the sum‑7 region immediately sets a strong anchor, and a three‑cell equals region of low pips closes the grid without ambiguity.
The medium puzzle, by Rodolfo Kurchan, raises the bar with seven dominoes and interlocking sum constraints. The key bottleneck is a single‑cell sum‑4 that forces a specific high pip, and the greater‑6 pair in the bottom row demands the only two pips high enough—once those are placed, the rest follows naturally.
The hard puzzle, also by Kurchan, is a masterclass in constraint weaving: a sum‑12 region demands two sixes, a sum‑0 region forces a chain of three zeros, and many single‑cell sums and equals regions create a tight logical path. Expect a satisfyingly dense solve where each deduction builds on the last.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: The power of equals
Look for a region where all cells must show the same pip value—this equality constraint can lock down multiple cells at once.
💡 Hint 2: Focus on the bottom‑right cluster
The equals region spanning row 3 col 5, row 4 col 4, and row 4 col 5 ties three cells together. Consider which low pip can appear in the adjacent less‑3 cell above it.
💡 Hint 3: Full answer
Place the [2,2] domino horizontally at [4,4][4,5] to satisfy the equals region with 2s. The other equals at [2,2][2,3] takes [4,4] vertically with 4s. Start the sum‑7 at [0,0][1,0] by putting [3,5] horizontally at [0,0][0,1] (3,5), then [4,2] vertical at [1,0][2,0] (4,2) to fill the less‑3 cell. Finally, the greater‑4 cell at [2,5] and its partner [3,5] use [5,2] vertical (5,2).
💡 Hint 1: Single‑cell sums
Concentrate on sum constraints that cover a single cell—they immediately fix that cell’s pip, giving you a fixed starting point.
💡 Hint 2: The lone sum‑4
The cell at [0,3] is a sum‑4 region all by itself, so it must contain a 4. That 4 forces the domino [0,4] to be placed vertically, putting a 0 in the cell below it at [1,3].
💡 Hint 3: Full answer
Place [0,4] vertical at [0,3][1,3] (4,0). The greater‑6 region at [2,0][2,1] takes [6,5] horizontal. The equals region [2,2][2,3] gets [4,4] vertical. The sum‑6 at [1,2][1,3] uses [6,1] with 6 at [1,2] and 1 at [1,1]. The sum‑6 at [1,4][2,4] takes [3,3] vertical. Finally, place [6,4] vertical at [3,1][3,2] (6,4) and [4,2] horizontal at [3,3][3,4] (2,4).
💡 Hint 1: Extreme targets
Identify regions with extreme sum targets—a sum‑0 region forces every cell inside to be 0, and a sum‑12 region can only be filled with two 6s. These will anchor the whole grid.
💡 Hint 2: Zero cascade and double sixes
The sum‑0 region covers [3,3], [4,3], and [4,4]—three adjacent cells that must all be 0. The sum‑12 region sits at [2,2][2,3] in the same row and needs 6 and 6. Start by placing dominoes that carry 0s and 6s.
💡 Hint 3: Build around the anchors
Use [3,6] vertical at [1,3][2,3] and [2,6] horizontal at [2,1][2,2] to satisfy sum‑12 and the adjacent sum‑2 cell. Then place zero‑bearing dominoes: [5,0] at [3,2][3,3] (5,0) to also satisfy the sum‑5 cell, [0,3] at [4,2][4,3] (3,0) for sum‑3, and [2,0] at [4,4][5,4] (0,2) to complete the sum‑0 chain.
💡 Hint 4: Top row and single‑cell sums
The top row’s sum‑10 at [0,2][0,3] gets [5,5] horizontal. The sum‑3 cell [0,1] is then satisfied by [4,3] horizontal at [0,0][0,1] (4,3). The sum‑10 at [0,0][1,0] forces [4,6] vertical at [1,0][2,0] (6,4). The equals region [0,4][1,4] takes [2,2] vertical, and the less‑3 [1,1][1,2] gets [0,0] horizontal.
💡 Hint 5: Full answer
Place dominoes in this order: [3,6] vertical [1,3][2,3] (3,6); [2,6] horizontal [2,1][2,2] (2,6); [5,0] horizontal [3,2][3,3] (5,0); [0,3] horizontal [4,2][4,3] (3,0); [2,0] horizontal [4,4][5,4] (0,2). Then [5,5] horizontal [0,2][0,3]; [4,3] horizontal [0,0][0,1] (4,3); [4,6] vertical [1,0][2,0] (6,4); [2,2] vertical [0,4][1,4] (2,2); [0,0] horizontal [1,1][1,2]. Next [5,1] vertical [2,4][3,4] (5,1). Finally [2,1] vertical [4,0][3,0] (2,1); [4,0] vertical [5,0][6,0] (4,0); [5,3] horizontal [6,1][6,2] (3,5); [2,4] horizontal [6,3][6,4] (4,2).
🎨 Pips Solver
Jul 31, 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: Secure the sum‑7 anchor
The sum‑7 region at [0,0] and [1,0] can only be achieved with 3 and 4. The only domino containing a 3 is [3,5]; place it horizontally at [0,0][0,1] to put 3 at [0,0] and 5 at [0,1]. This obliges [1,0] to later receive a 4.
2
Step 2: Cover the less‑3 cell
The cell [2,0] has a less‑3 constraint (value 0, 1, or 2). The [4,2] domino is the only one that can deliver a 2 there while also giving a 4 to [1,0]. Place [4,2] vertically at [1,0][2,0] with 4 above and 2 below, completing the sum‑7 and less‑3 regions.
3
Step 3: Resolve the large equals region
The three‑cell equals region [3,5][4,4][4,5] forces all three cells to hold the same pip. The only domino with identical pips is [2,2]. Place it horizontally at [4,4][4,5], giving both cells a 2 and also satisfying [3,5] through the domino that will cover it.
4
Step 4: Finish the remaining equals and greater‑4
The other equals region at [2,2][2,3] now takes the [4,4] domino vertically (4 and 4). Finally, the greater‑4 cell [2,5] must be >4, so it gets a 5 from the [5,2] domino placed vertically at [2,5][3,5]. The 2 partner falls into [3,5], matching the existing equals value.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Lock the lone sum‑4 cell
The single‑cell sum‑4 region at [0,3] must be exactly 4. The domino [0,4] provides that 4, and the only way to place it without conflict is vertically: 4 at [0,3] and 0 at [1,3].
2
Step 2: Confront the greater‑6 region
The region [2,0][2,1] with a greater‑6 constraint demands the highest possible pips. The only domino that fits is [6,5], placed horizontally with 6 at [2,0] and 5 at [2,1].
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Step 3: Fill the equals pair
The equals region at [2,2][2,3] requires identical values. The [4,4] domino fits perfectly—place it vertically so both cells become 4.
4
Step 4: Top‑row sum‑6 and less‑4
The sum‑6 region [1,2][1,3] now has [1,3]=0, so [1,2] must be 6. The [6,1] domino supplies the 6; place it vertically with 6 at [1,2] and 1 at [1,1] (satisfying the less‑4 constraint on [1,1]). Then the sum‑6 region [1,4][2,4] gets the [3,3] domino vertically, both cells 3.
5
Step 5: Bottom‑row placements
The empty cell [3,1] and the sum‑6 region [3,2][3,3] are resolved with two dominoes. Place [6,4] vertically at [3,1][3,2] (6 and 4) to set the empty cell and part of the sum‑6. Then place [4,2] horizontally at [3,3][3,4] (2 and 4) to complete the sum‑6 and satisfy the less‑6 cell at [3,4].
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Anchor the extreme sums
The sum‑12 region [2,2][2,3] forces two 6s. Place [3,6] vertically at [1,3][2,3] (3,6) to get one 6 at [2,3] and cover the later sum‑3 at [1,3]. The other 6 comes from [2,6] horizontally at [2,1][2,2] (2,6), also satisfying the sum‑2 cell at [2,1]. Simultaneously, the sum‑0 region [3,3][4,3][4,4] demands all zeros. Place [5,0] horizontal at [3,2][3,3] (5,0) to fill the sum‑5 cell [3,2] and give [3,3]=0. Then [0,3] horizontal at [4,2][4,3] (3,0) covers the sum‑3 at [4,2] and [4,3]=0. Finally, [2,0] horizontal at [4,4][5,4] (0,2) completes the sum‑0 chain.
2
Step 2: Top‑row sum‑10 and associated cells
The sum‑10 region [0,2][0,3] requires two 5s; place [5,5] horizontally there. Next, the sum‑3 single cell [0,1] must be 3—use [4,3] horizontally at [0,0][0,1] (4,3), placing 4 at [0,0] and 3 at [0,1]. The sum‑10 region [0,0][1,0] now needs a 6 at [1,0]; the [4,6] domino delivers it vertically at [1,0][2,0] (6,4), which also sets [2,0]=4 for the sum‑5 region below.
3
Step 3: Equals and less‑3
The equals region [0,4][1,4] must have equal pips. The [2,2] domino fits vertically, giving both cells 2. The less‑3 region [1,1][1,2] requires values <3; the [0,0] domino placed horizontally there (0,0) satisfies it cleanly.
4
Step 4: Single‑cell sum‑5 and sum‑1
The cell [2,4] has a sum‑5 constraint, so it must be 5. Place the [5,1] domino vertically at [2,4][3,4] (5,1), which also gives [3,4]=1, fulfilling its sum‑1 requirement.
5
Step 5: Bottom‑row and final horizontal sums
The sum‑5 region [2,0][3,0] currently has [2,0]=4, so [3,0] must be 1. Use [2,1] vertically at [4,0][3,0] (2,1), giving [3,0]=1 and [4,0]=2. The sum‑6 region [4,0][5,0] then needs a 4 at [5,0]; place [4,0] vertical at [5,0][6,0] (4,0), also setting [6,0]=0. The sum‑3 region [6,0][6,1] now has [6,0]=0, so [6,1]=3. Place [5,3] horizontal at [6,1][6,2] (3,5), giving [6,1]=3 and [6,2]=5. Finally, the sum‑9 region [6,2][6,3] gets its remaining 4 from [2,4] horizontal at [6,3][6,4] (4,2), completing 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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