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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, by Ian Livengood, welcomes you with a single-cell less-than-3 region that practically hands you the first domino. From there, a pair of vertical sum-10 regions lock into place with minimal fuss, and a bottom-row equals constraint tidily closes the loop. It’s a smooth solve where each deduction feeds neatly into the next, leaving no guesswork.
Kurchan’s medium puzzle raises the stakes with a top-row sum-10 that stretches across the first two cells, forcing your eye to the available high-pip dominoes. A neighboring less-than-3 cell and a greater-7 vertical duo on the right create a chain reaction — once you slot the [2,6] domino to satisfy both the sum-10 and the less-than, the bottom-row equals region cascades to fill three identical cells, demanding careful domino selection. The board feels interconnected, with each region borrowing from its neighbor’s leftovers.
The hard grid, also by Kurchan, immerses you in a thicket of tiny sum-3 and less/greater constraints that crisscross the board. You’ll find yourself staring at a cluster of cells all demanding minute sums, forcing you to allocate the [1,1] and [1,2] dominoes with surgical precision. A greater-7 region on the right provides a crucial wedge, but the real key is untangling the bottom-left sum-3 chain that links a less-than and a greater-than cell in a tricky vertical alignment. Every placement feels earned, and the puzzle rewards methodical unwinding of its tight logical knots.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Find the Standalone Cell
Look for a single-cell region that restricts the pip value — it’s the only one of its kind and will give you a quick toehold.
💡 Hint 2: Focus on Column 2
The cell at the top, row 0 column 2, must be less than 3. Check which domino can place a pip under 3 there while its partner fits into the sum-4 region just below it.
💡 Hint 3: The Full Solve
Place the [2,3] domino vertically at [0,2]-[1,2] with 2 above. Then the sum-4 region forces [2,2] to be 1 from the [1,5] domino, which goes vertically in column 2 rows 2-3 with 5 below. The left sum-10 needs [4,6] vertically in column 0 rows 2-3 (4 above, 6 below), and the right sum-10 takes [4,5] horizontally across the bottom row with 4 at [4,1] and 5 at [4,2], satisfying the equals region at [4,0]-[4,1] that already holds 4 from the [4,6] domino.
💡 Hint 1: Check the Top-Left Corner
A sum-10 region spans the first two cells of the top row; only a few domino pairs can achieve that total, but you’ll need to accommodate a less-than-3 cell right next door.
💡 Hint 2: Link the Less-Than Cell
The cell at [0,2] is less than 3. The sum-10 region to its left will occupy [0,0] and [0,1], so the only plausible way to satisfy both is to place the [2,6] domino horizontally on [0,1]-[0,2] with 6 on the left and 2 on the right, which simultaneously feeds the sum-10 with the 6.
💡 Hint 3: The Full Solve
After setting the top row with [4,3] vertically in column 0 (4 top), [2,6] horizontally on [0,1]-[0,2], and [4,6] vertically in column 3 (6 top, 4 below) to satisfy the greater-7 region, the bottom-left less-3 region forces [1,2] vertically in column 0 rows 2-3 (1 above, 2 below). The equals region in row 3 demands three 2’s, so place [2,2] horizontally on [3,1]-[3,2] and [3,3] vertically on [2,3]-[3,3]. The remaining [1,5] domino fits on [2,1]-[2,2] with 1 on left, 5 on right.
💡 Hint 1: Look for the Sum-3 Web
The board is dominated by several sum-3 regions — notice the cluster of cells that must total exactly 3 across multiple adjacent cells.
💡 Hint 2: The Top-Left Sum-3 and Greater
The top two cells [0,1] and [0,2] must sum to 3, and the cell at [0,0] must be greater than 3. Since [0,0] will anchor a high pip, the only combination left for the sum-3 is 0 and 3, placed by two different dominoes that also serve other regions.
💡 Hint 3: Central Sum-3 Cluster
Concentrate on the central sum-3 zone at [2,1], [2,2], [3,1]. To fill these three cells with a total of 3 using two dominoes, you’ll need the [1,1] domino covering [2,1]-[2,2] horizontally and the [1,2] domino’s 1 on [3,1] and its 2 on [3,0], which also satisfies the less-than-3 cell there.
💡 Hint 4: Right-Side Sum-6 Logic
Once the center is locked, the right side’s sum-6 regions become tractable. The [6,2] domino must slot into column 5/4 with the 6 on [3,5] (a single-cell sum-6) and 2 on [3,4], combining with the [5,4] domino’s 4 on [2,4] to hit the vertical sum-6. The [1,6] domino then places 1 on [5,4] and 6 on [4,4] to satisfy the left sum-3 and right sum-6, respectively.
💡 Hint 5: The Full Solve
Start with the top left: [0,4] domino covers [0,0]-[0,1] with 4 and 0; [4,3] domino covers [0,2]-[1,2] with 3 and 4 (3 in [0,2] to complete sum-3). The less-than-3 at [1,0] forces [2,3] domino vertically ([1,0]=2, [2,0]=3). The central sum-3 group: [1,1] domino horizontally at [2,1]-[2,2] (both 1s), and [1,2] domino covers [3,0]-[3,1] with 2 and 1 (2<3 at [3,0], 1 at [3,1]). The right side: [5,4] domino vertically at [1,4]-[2,4] (5 above, 4 below) to start the greater-7 region. [6,2] domino goes to [3,4]-[3,5] (2 left, 6 right) for sum-6. [1,6] domino covers [4,4]-[5,4] (6 above, 1 below) to satisfy sum-6 at [4,4] and sum-3 at [5,4]-[5,5]. [2,5] domino then covers [4,0]-[5,0] (2 above, 5 below) satisfying less-3 and greater-4. The [3,0] domino handles [4,2]-[5,2] (3 above, 0 below) for sum-3. Finally, [1,3] domino covers [1,5]-[1,6] (3 left, 1 right) completing greater-7 and less-2, and [2,4] domino fills [5,5]-[5,6] (2 left, 4 right) to finish the sum-3 on [5,4]-[5,5] and the empty cell. This sequence unwinds the puzzle’s dense constraints.
🎨 Pips Solver
Jul 9, 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
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Step 1: The Less-Than Anchor
The only single-cell region on the easy grid is [0,2] with a less-than-3 constraint. That cell must be 0, 1, or 2. Scanning the domino roster, the [2,3] domino is the only one that can place a low pip there (2) while its partner (3) fits into the sum-4 region below at [1,2]. So place [2,3] vertically with 2 at [0,2] and 3 at [1,2].
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Step 2: Completing the Sum-4
With [1,2]=3, the sum-4 region now requires [2,1]+[2,2] = 4-3 = 1. The only domino left that contains a 1 is [1,5]. Place it vertically so that [2,2]=1 and [3,2]=5, forcing [2,1] to be 0 to keep the sum-4 region total 4.
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Step 3: Left Vertical Sum-10
Now that [2,1] must be 0, the [4,0] domino is the sole remaining piece with a 0, so place it horizontally covering [2,0]-[2,1] with 4 and 0 (4 at [2,0]). The sum-10 region occupying [2,0] and [3,0] therefore needs [3,0] to be 6. The [4,6] domino fits perfectly: place it vertically with 4 at [4,0] and 6 at [3,0] — satisfying the sum-10 and also setting up the equals region below.
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Step 4: Right Sum-10 and the Equals
The equals region at [4,0]-[4,1] now demands [4,1]=4. The right sum-10 region links [3,2] (already 5) and [4,2], so [4,2] must be 5. The only remaining domino is [4,5], which fits horizontally on [4,1]-[4,2] with 4 and 5, completing the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
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Step 1: Top-Row Sum-10 and Less-3
The sum-10 region across [0,0]-[0,1] must use high pips, and the adjacent [0,2] is a less-than-3 single cell. The domino [2,6] can place a 2 at [0,2] while giving a 6 to [0,1], which not only satisfies the less-than but also contributes to the sum-10. Place [2,6] horizontally on [0,1]-[0,2] with 6 left, 2 right.
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Step 2: Completing the Top Row
With [0,1]=6, the sum-10 cell [0,0] must be 4. The [4,3] domino covers [0,0] and the empty cell [1,0] — set it vertically with 4 at top, 3 below. Meanwhile, the greater-7 region at [0,3]-[1,3] needs a high total; the [4,6] domino’s 6+4=10 works, so place it vertically with 6 at [0,3] and 4 at [1,3].
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Step 3: Bottom-Left Less-Than
The less-than-3 region covering [2,0]-[2,1] requires both cells to be under 3. The only available domino that can give a 1 and 2 is the [1,2] domino. Place it vertically across [2,0]-[3,0] with 1 above and 2 below. (The 2 at [3,0] also seeds the row-3 equals region.)
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Step 4: The Row-3 Equals Cascade
The equals region at [3,0],[3,1],[3,2] now contains a 2 at [3,0]; therefore all three cells must be 2. The [2,2] domino covers [3,1]-[3,2] horizontally (both 2s), while the final cell [2,3] is part of another equals region — together with [3,3] it must be equal, and the [3,3] domino gives that pair (3,3) placed vertically.
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Step 5: Finishing the Remaining Empty Cell
The empty cell at [2,2] (unconstrained) will receive a 5 from the [1,5] domino, which you place horizontally on [2,1]-[2,2] with 1 left, 5 right — using the 1 already forced at [2,1] from the less-than-3 region. All constraints are satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Top-Left Greater/Sum-3
The cell [0,0] must be greater than 3, and the sum-3 pair [0,1]+[0,2] must total 3. The [0,4] domino’s 0 and 4 are the only way to give [0,0] a high pip (4) while feeding the sum-3 with 0. Place [0,4] horizontally on [0,0]-[0,1] with 4 on [0,0], 0 on [0,1]. The remaining sum-3 component must be 3 at [0,2], so place [4,3] vertically on [0,2]-[1,2] with 3 above, 4 below (which also satisfies the greater-than-3 at [1,2]).
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Step 2: Central Sum-3 Web
The sum-3 region that includes [2,1],[2,2],[3,1] is the puzzle’s heart. With dominos [1,1] and [1,2] remaining, you can achieve 1+1+1 = 3 across the three cells. Place [1,1] horizontally on [2,1]-[2,2] (both 1s), and [1,2] so that its 1 goes to [3,1] and its 2 goes to [3,0] (less-than-3 spot) — so place [1,2] horizontally on [3,0]-[3,1] with 2 on left, 1 on right. The sum-3 cell [2,0] is a single sum-3, so it must be 3. The [2,3] domino fits: place it vertically on [1,0]-[2,0] with 2 on top (less-than-3 at [1,0]) and 3 on bottom (sum-3 at [2,0]). Now the top left is resolved.
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Step 3: Right Side Greater-7 and Sum-6 Pairs
The greater-7 region at [1,4]-[1,5] needs a total >7; the [5,4] domino provides 5+4. Place it vertically on [1,4]-[2,4] with 5 above, 4 below. The sum-6 region [2,4]+[3,4] gets 4+2, so the [6,2] domino must go to [3,4]-[3,5] with 2 at [3,4] and 6 at [3,5] (the single-cell sum-6 at [3,5] is satisfied by that 6).
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Step 4: Bottom-Left Less/Greater and Sum-3
The less-than-3 at [4,0] and greater-than-4 at [5,0] can be met by the [2,5] domino’s 2 and 5. Place it vertically on [4,0]-[5,0] with 2 above, 5 below. The sum-3 vertical pair [4,2]+[5,2] needs 3+0 to total 3, so the [3,0] domino goes there: place it vertically with 3 at [4,2] and 0 at [5,2].
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Step 5: Completing the Right Lower Quadrant
The single sum-6 cell [4,4] must be 6, so the [1,6] domino covers it: place horizontally on [4,4]-[5,4] with 6 above and 1 below, which supplies the needed 1 for the sum-3 region [5,4]+[5,5] (with 1+2). The [2,4] domino then comfortably fills [5,5]-[5,6] with 2 and 4 — 2 to complete the sum-3, and 4 to fill the empty cell at [5,6].
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Step 6: Final Missing Piece
The only cell left unplaced is the less-than-2 at [1,6] and the accompanying [1,5] second part of greater-7. The [1,3] domino fits: place it horizontally on [1,5]-[1,6] with 3 on the left (so [1,5]=3, making greater-7 sum 5+3=8) and 1 on the right (satisfying <2). The puzzle now fully satisfies all constraints.
💡 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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