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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, constructed by Ian Livengood, is a confidence-builder — a compact 10-cell grid with two prominent equals regions and a single sum-1 cell that acts as an immediate key. No guesswork required; the path is linear once you spot which domino must satisfy the singleton. This is the kind of breezy solve that rewards pattern recognition.
NYT Pips medium by Rodolfo Kurchan introduces a tighter bottleneck: a sum-1 cell coupled with an adjacent equals region creates a forced domino pair that unlocks a chain of sum constraints across a 4x6 grid. The puzzle feels more spacious, but the logic remains tight — find that initial anchor, and everything cascades into place. One misstep in the early going can lead to dead ends, so it pays to double-check the domino list.
The NYT Pips hard, also from Kurchan, escalates with an 8x4 lattice packed with tiny sum targets — multiple sum-0 and sum-2 singlets on the top row, a sum-5 and sum-7 duo just below, and equals clusters mid-grid. The design forces a snowball effect: zero-placements in the opening row dictate the entire upper half, and from there the rest fills in with satisfying inevitability. It’s a methodical, rewarding workout that tests your ability to track minimal sums across a large field.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Focus on the singleton sum
Spot the single cell with a sum-1 constraint — it can only be satisfied by one specific domino. That same domino also interacts with an equals group nearby. Think about how a zero from that domino could satisfy the equals requirement.
💡 The zero-placed equals trio
The sum-1 cell at [2,2] forces the domino with a 1 and a 0, with the 1 landing on that cell. The neighboring cell [1,2] is part of a three-cell equals region with [0,2] and [0,3], all of which must now be 0. That directly places the double-zero domino horizontally to fill [0,2]-[0,3].
💡 Full solve: the domino map
Place [0,1] vertically: 1 at [2,2], 0 at [1,2]; place [0,0] horizontally at [0,2]-[0,3] (both 0). The large four-cell equals (cells [0,1],[1,1],[2,0],[2,1]) all need the same value — the only remaining double is [2,2], so place it vertically covering [1,1] and [2,1] (both 2). The greater-3 cell at [0,0] must be >3, so use the [2,5] domino horizontally: 5 at [0,0], 2 at [0,1] (matching the equals group). Finally, the empty cell [1,0] takes the 3 from the [2,3] domino placed vertically: 2 at [2,0] (completing the equals group) and 3 at [1,0].
💡 Sum meets equals
Look for a sum-1 region that shares a column with an equals pair. The domino that satisfies the sum-1 will also seed the equals with its other pip. Keep an eye on the available domino values to see which pair can resolve both constraints at once.
💡 Anchor the top-left corner
The sum-1 cell at [0,3] must take the 1 from the [1,0] domino, forcing its 0 into [1,3]. This makes the equals pair [1,3]/[2,3] both 0. Next, the [0,2] domino delivers that 0 to [2,3] and a 2 into [2,2] (which must be >0). Now the sum-4 pair [3,0]+[3,1] becomes the next target — it needs 1+3, drawing in the [1,1] and [3,2] dominoes.
💡 Full medium grid resolution
Place [1,0] vertically: 1 at [0,3], 0 at [1,3]. Place [0,2] horizontally: 2 at [2,2], 0 at [2,3] (satisfying the equals pair). The sum-4 pair forces [1,1] vertically at [2,0]-[3,0] (both 1) and [3,2] vertically at [3,1]-[2,1] (3 at [3,1], 2 at [2,1]) — this also satisfies sum-3 for [2,0]+[2,1]=1+2. The equals pair [3,2]-[3,3] is filled by the [4,4] domino horizontally (both 4). Top right: place [6,5] at [0,5]-[0,4] (6 at [0,5] empty, 5 at [0,4] equals), then [5,4] vertically at [1,4]-[2,4] (5 at [1,4] matching the equals pair, 4 at [2,4] satisfying the less-6 constraint).
💡 Top-row zeroes
The top row contains three sum-0 regions and one sum-2 region. This heavily restricts which dominoes can be placed in that row — you'll need to deploy multiple zero-containing dominoes while delivering exactly a 2 to a specific cell.
💡 Forced placements at the summit
Cells [0,0], [0,1], and [0,3] all require 0; [0,2] needs a 2. The domino that provides a 2 alongside a 0 is [0,2], so place it with 2 at [0,2] and 0 at [0,3]. The other two zeros will come from [0,5] (giving its 0 to [0,0] and 5 to [1,0]) and [0,1] (giving 0 to [0,1] and 1 to [1,1]). This sets the entire top row.
💡 Column zero and the sum-5 lock
Now consider column 0: [1,0] sum-5 (already placed 5 from [0,5]), [2,0] sum-3, [3,0] sum-3, [4,0] empty. The only way to get a 3 into [2,0] and [3,0] is via [0,3] and [3,5] dominoes. The [0,3] domino places 0 at [4,0] (filling the empty) and 3 at [3,0]. Then [3,5] places 3 at [2,0] and 5 at [2,1] (sum-5). This progression also wires into the sum-7 pair at [1,1]+[1,2].
💡 Sum-7 and the equals clusters
With [1,1] already 1 (from [0,1]), the sum-7 at [1,1]+[1,2] demands a 6 at [1,2] — use the [3,6] domino with 6 at [1,2] and 3 at [2,2] (satisfying sum-3 there). Now the equals region at [5,0]-[5,2] forces all three cells to 2, pulling in [2,4], [2,3], and [2,5] dominoes to supply those 2s. The equals at [4,2]-[4,3] gets 5 and 5 from [2,5] and [4,5] dominoes. The bottom rows resolve through the remaining sum targets.
💡 Full hard solution
Start with top row: [0,5] horiz at [0,0]-[1,0] (0/5); [0,2] horiz at [0,2]-[0,3] (2/0); [0,1] horiz at [0,1]-[1,1] (0/1). Column 0: [0,3] vert at [4,0]-[3,0] (0/3); [3,5] horiz at [2,0]-[2,1] (3/5). Sum-7: [3,6] vert at [2,2]-[1,2] (3/6). Equals trio: [2,4] horiz at [5,0]-[6,0] (2/4); [2,3] vert at [5,1]-[6,1] (2/3); [2,5] horiz at [5,2]-[4,2] (2/5). Mid: [1,2] vert at [4,1]-[3,1] (1/2); [3,4] horiz at [3,3]-[3,2] (3/4); [0,4] vert at [2,3]-[1,3] (0/4); [4,5] vert at [5,3]-[4,3] (4/5). Bottom: [1,3] vert at [6,3]-[7,3] (1/3); [1,5] vert at [6,2]-[7,2] (1/5); [1,4] horiz at [7,0]-[7,1] (1/4). All placed.
🎨 Pips Solver
Jul 7, 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: Lock the sum-1 cell
Cell [2,2] must sum to 1, so its pip value must be exactly 1. The only domino containing a 1 is [0,1]. Place it vertically so that 1 lands on [2,2] and 0 on [1,2]. This not only satisfies the sum-1 but also sets [1,2] to 0, matching its equals region neighbors.
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Step 2: Solidify the triple-equals zeroes
With 0 at [1,2], the equals region spanning [0,2], [0,3], and [1,2] forces all three cells to 0. The [0,0] double domino is the only way to cover [0,2] and [0,3] with zeroes — place it horizontally there.
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Step 3: The four-cell equals block
The large equals group at [0,1], [1,1], [2,0], [2,1] must all hold the same number. The remaining dominoes include [2,2], [2,5], [2,3]. The [2,2] double is the only candidate to set a uniform value; place it vertically covering [1,1] and [2,1], assigning 2 to both. The entire block is now locked at 2.
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Step 4: Finish the greater-3 and empty cell
Cell [0,0] has a greater-3 constraint — it needs a pip >3. The [2,5] domino offers a 5, so place it horizontally: 5 at [0,0] and 2 at [0,1] (matching the equals block). The empty cell [1,0] needs coverage; the last domino [2,3] goes vertically: 2 at [2,0] (completing the equals block) and 3 at [1,0].
🔧 Step-by-Step Answer Walkthrough For Medium Level
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Step 1: Sum-1 to the rescue
Cell [0,3] is a sum-1 region, so its value must be 1. The only domino with a 1 is [1,0]. Place it vertically so that 1 sits at [0,3] and 0 at [1,3], satisfying the sum-1.
2
Step 2: Zero equals and the [0,2] domino
Cells [1,3] and [2,3] are an equals pair, forcing [2,3] to also be 0. The [0,2] domino can give 0 to [2,3] and 2 to [2,2] (which only needs to be >0). Place [0,2] horizontally: 2 at [2,2], 0 at [2,3].
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Step 3: Cracking the sum-4 pair
The sum-4 pair [3,0]+[3,1] is the next bottleneck. To sum to 4 with available dominoes, the only viable split is 1+3. Place [1,1] vertically at [2,0]-[3,0] (both 1) and [3,2] vertically at [3,1]-[2,1] (3 at [3,1], 2 at [2,1]). Incidentally, this also satisfies the sum-3 pair [2,0]+[2,1]=1+2.
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Step 4: Equals at [3,2]-[3,3]
Cells [3,2] and [3,3] form an equals pair. The [4,4] double remains — place it horizontally covering both, making each cell 4.
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Step 5: Wrap the top right
The equals pair [0,4] and [1,4] must match. The [6,5] domino puts 5 at [0,4] and 6 at the empty cell [0,5] horizontally. Then [5,4] goes vertically at [1,4]-[2,4]: 5 at [1,4] (matching equals) and 4 at [2,4] (satisfying the less-6 constraint).
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Top-row zeroes and the sum-2
The top row cells [0,0] (sum-0), [0,1] (sum-0), [0,2] (sum-2), [0,3] (sum-0) force direct pip assignments: 0, 0, 2, 0. The [0,2] domino must deliver the 2 paired with a 0, so place it with 2 at [0,2] and 0 at [0,3]. Place [0,5] horizontally at [0,0]-[1,0] (0 at [0,0], 5 at [1,0]) and [0,1] horizontally at [0,1]-[1,1] (0 at [0,1], 1 at [1,1]).
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Step 2: Column zero and the sum-3 chain
Column 0 now has [1,0]=5, [2,0] sum-3, [3,0] sum-3, [4,0] empty. Place [0,3] vertically: 3 at [3,0], 0 at [4,0] (empty). Then place [3,5] horizontally at [2,0]-[2,1]: 3 at [2,0] (sum-3) and 5 at [2,1] (sum-5).
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Step 3: The sum-7 at [1,1]-[1,2]
With [1,1]=1 already, the sum-7 pair [1,1]+[1,2] demands a 6 at [1,2]. Place [3,6] vertically: 6 at [1,2] and 3 at [2,2] — satisfying the sum-3 cell at [2,2].
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Step 4: Row 5 equals and the snowball
The equals region [5,0],[5,1],[5,2] forces all three to the same value. Place [2,4] horizontally at [5,0]-[6,0] (2/4), [2,3] vertically at [5,1]-[6,1] (2/3), and [2,5] horizontally at [5,2]-[4,2] (2/5). This locks the equals trio to 2 and feeds values into rows 4 and 6.
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Step 5: Mid-grid sums and the remaining pairs
Place [1,2] vertically at [4,1]-[3,1] (1 at [4,1] for sum-1, 2 at [3,1] for sum-2). Place [4,5] vertically at [5,3]-[4,3] (4 at [5,3] for sum-4, 5 at [4,3] completing the equals pair [4,2]-[4,3]). Place [3,4] horizontally at [3,3]-[3,2] (3 at [3,3] sum-3, 4 at [3,2] sum-4). Place [0,4] vertically at [2,3]-[1,3] (0 at [2,3] sum-0, 4 at [1,3] sum-4).
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Step 6: Bottom rows wrap-up
Place [1,5] vertically at [6,2]-[7,2] (1 at [6,2] sum-1, 5 at [7,2] sum-5). Place [1,3] vertically at [6,3]-[7,3] (1 at [6,3] sum-1, 3 at [7,3] sum-3). Finally, place [1,4] horizontally at [7,0]-[7,1] (1 at [7,0] sum-1, 4 at [7,1] to complete the sum-7 pair with the 3 already at [6,1]).
💡 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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