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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
Today’s NYT Pips easy puzzle, crafted by Ian Livengood, is a compact confidence-builder. A sum-6 region locks in the first domino immediately, sparking a chain of sum and less-than deductions with no ambiguity. Expect a smooth solve — perfect for a warm-up. Rodolfo Kurchan’s medium grid introduces equals regions that cascade elegantly. The puzzle’s bottleneck lies in recognizing which domino fits a zero-equals column, which then dictates placements in the adjacent sum-10 rows. Once that clicks, the rest falls into place without much resistance. The hard puzzle, also by Kurchan, is a masterclass in constraint interplay. A sum-0 region spanning three cells forces a rare double-zero domino, while multiple equals regions weave through the grid — including an equals-2 quartet and an equals-6 column. The solving path is tight; the main bottleneck is untangling the zero-sum and equals relationships around rows 2–3. It’s a satisfyingly tough finish.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Look for the smallest sum
Scan for a tiny sum region — it only has two cells, so the required total immediately points to a specific domino pair. No other domino can match it, giving you a guaranteed starting move.
💡 Top-left lock
The sum-6 region occupies cells [0,0] and [0,1]. Among the available dominos, only the double-three adds up to 6 when placed there. Set that horizontally and the whole top row begins to open.
💡 Full solve chain
Place [3,3] at [0,0]-[0,1]. The less-3 region at [0,3] must be below 3, so [2,4] goes at [0,2]-[0,3] (4 and 2). Sum-11 at [2,1]-[2,2] needs 6+5; use [6,3] at [2,2]-[2,3] (6,3) and put [5,5] at [1,1]-[2,1] (5,5). Finally, [6,4] at [1,0]-[2,0] (4,6) supplies the 4 for sum-9 with the 5 from [5,5].
💡 Follow the equals
Equals constraints force multiple cells to share the same pip. Focus on the column that demands three identical values — this will drastically limit which dominos can touch it.
💡 Column of zeros
The equals column at [1,2], [2,2], [3,2] must be all the same digit. The only number available in enough places is 0, using [0,3] vertically and [4,0] horizontally to deliver the first two zeros. This connection unlocks the whole middle.
💡 Complete medium solution
Set [0,3] at [1,2]-[0,2] (0,3). Equals [0,1]-[0,2] then forces [0,1]=3, so [3,6] goes at [0,0]-[0,1] (6,3); equals [0,0]-[1,0] forces [1,0]=6, placing [6,5] at [1,0]-[2,0] (6,5). Less-3 [0,3] gets 2 from [6,2] at [1,3]-[0,3] (6,2); sum-10 [1,3]-[2,3] gets 4 from [4,0] at [2,2]-[2,3] (0,4). Sum-10 [2,0]-[3,0] needs 5 at [3,0]: use [2,5] at [3,1]-[3,0] (2,5). Finally, [0,2] at [3,2]-[3,3] (0,2) fills the zero column.
💡 Settle zero first
Regions that sum to zero are extremely restrictive. Identify the multi-cell sum-0 area — it will demand that every cell in it is 0, so a double-zero domino is nearly forced here.
💡 Triple zero anchor
The sum-0 region covers [1,0], [2,0], and [2,1]. The only way to get three zeros is to place the [0,0] domino vertically on [1,0]-[2,0], then bring a second zero into [2,1] from an adjacent cell. The [6,0] domino is the ideal candidate, putting 0 at [2,1] and 6 at [2,2].
💡 Build the equals-6 pillar
With [2,2]=6, the equals region at [1,2],[2,2],[3,2] forces those cells to all be 6. Use [6,1] to place 6 at [1,2] and 1 at [1,1]; then the sum-3 region [0,1]-[1,1] makes [0,1]=2 (using [2,4] domino, which also satisfies sum-4 at [0,2]).
💡 Equals-2 chain reaction
The sum-0 single cell at [1,3] must be 0, so [0,2] domino places 0 there and 2 at [2,3]. That activates the equals-2 region: [2,3],[2,4],[3,3],[4,3] all become 2. Use [2,5] to give 2 at [2,4] and 5 at [3,4] (satisfying greater-4), then [6,2] to put 6 at [3,2] and 2 at [3,3].
💡 Full logical path to finish
Sum-2 region [2,5]-[2,6] needs 1+1, so place [1,5] at [2,6]-[2,7] (1,5) and [3,1] at [2,5]-[3,5] (1,3). Equals-5 region [1,8],[2,7],[2,8] forces 5s; use [4,5] at [0,8]-[1,8] (4,5) and [5,0] at [2,8]-[3,8] (5,0). Sum-0 [3,7]-[3,8] gets 0 from [0,3] at [3,7]-[3,6] (0,3). Equals-3 region gets 3 at [4,6] from [5,3] at [4,7]-[4,6] (5,3). Remaining cells fill with [2,3] at [4,3]-[4,2] (2,3), [0,1] at [5,1]-[5,2] (0,1), and [0,4] at [5,5]-[5,6] (0,4).
🎨 Pips Solver
Jul 27, 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: Sum-6 lock
The sum-6 region in cells [0,0] and [0,1] is the tightest constraint. Only the [3,3] domino sums to exactly 6, so it must cover those two cells, giving both a value of 3.
2
Step 2: Less-3 placement
Cell [0,3] must be less than 3. The [2,4] domino can provide a 2 there while placing a 4 in the empty cell [0,2]. No other available domino can satisfy this less-3 region without breaking adjacent empty or sum constraints later, so [2,4] is forced horizontally.
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Step 3: Sum-11 forced
The sum-11 region [2,1] and [2,2] needs a pair that adds to 11; the only possibility is 5 and 6. The [6,3] domino supplies the 6, so it goes at [2,2]-[2,3] with 6 on [2,2] and 3 on [2,3] (which is <4, satisfying the less-4 region). This forces [2,1] to be 5, so the [5,5] domino will cover [1,1] and [2,1] to provide a 5 there.
4
Step 4: Sum-9 completion
With [2,1] filled by a 5 from the [5,5] domino, the sum-9 region [1,0]-[1,1] now has a 5 in [1,1]. It needs a 4 in [1,0] to sum to 9. The [6,4] domino places 4 at [1,0] and the remaining 6 in empty cell [2,0], completing the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Equals zero column
The equals region at column 2 ([1,2],[2,2],[3,2]) forces all three cells to share the same pip. The only digit that appears on enough dominos to cover three adjacent cells is 0. Place [0,3] vertically at [1,2] (0) and [0,2] (3). Then place [4,0] horizontally so it covers [2,2] (0) and [2,3] (4). This sets two zeros in the column, and the third zero will come later.
2
Step 2: Equals top rows
With [0,2]=3, the equals region [0,1]-[0,2] forces [0,1] to also be 3. The [3,6] domino naturally fits here, placed horizontally with 3 at [0,1] and 6 at [0,0]. Then the equals region [0,0]-[1,0] demands [1,0]=6, so the [6,5] domino goes vertically with 6 at [1,0] and 5 at [2,0].
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Step 3: Less-3 and sum-10 middle
The less-3 region at [0,3] needs a value below 3. The [6,2] domino placed vertically puts 2 at [0,3] and 6 at [1,3]. This automatically creates a sum-10 region at [1,3]-[2,3] where 6 (from [6,2]) plus 4 (already placed at [2,3] from the [4,0] domino) equals 10.
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Step 4: Bottom sum-10
The sum-10 region [2,0]-[3,0] already has a 5 in [2,0] from the [6,5] domino. So [3,0] must be 5. The [2,5] domino placed vertically puts 5 at [3,0] and 2 at [3,1] (which is an empty cell, so no conflict).
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Step 5: Closing the zero column
All that remains are cells [3,2] and [3,3]. The equals column still needs a zero at [3,2]; the last unused domino [0,2] fits vertically, giving 0 at [3,2] and 2 at [3,3]. All constraints are now satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Triple-zero foundation
The sum-0 region covering [1,0], [2,0], and [2,1] forces every cell inside to be zero. The [0,0] domino is the only double-zero, so place it vertically at [1,0] and [2,0]. Now [2,1] must also be 0; the [6,0] domino placed horizontally gives 0 at [2,1] and 6 at [2,2] to start the next chain.
2
Step 2: Equals-6 pillar and top sums
Cell [2,2] is 6, so the equals region [1,2],[2,2],[3,2] forces 6s throughout. Place [6,1] vertically: 6 at [1,2] and 1 at [1,1]. The sum-3 region [0,1]-[1,1] now has 1+?=3, so [0,1] must be 2. The [2,4] domino placed horizontally gives 2 at [0,1] and 4 at [0,2], perfectly hitting the sum-4 region at [0,2].
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Step 3: Zero single and equals-2 cascade
The sum-0 single cell at [1,3] requires a 0. Place [0,2] horizontally: 0 at [1,3] and 2 at [2,3]. This triggers the equals-2 region [2,3],[2,4],[3,3],[4,3]: all must be 2. Use [2,5] to put 2 at [2,4] and 5 at [3,4] (satisfying the greater-4 region). Then use [6,2] vertically to place 6 at [3,2] (completing the equals-6 pillar) and 2 at [3,3].
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Step 4: Sum-2 pair and equals-5 web
The sum-2 region [2,5]-[2,6] requires a total of 2, achievable only with 1+1. Place [1,5] at [2,6]-[2,7] (1,5) and [3,1] at [2,5]-[3,5] (1,3). Now the equals region [1,8],[2,7],[2,8] demands all cells be 5, because [2,7] is already 5. Place [4,5] at [0,8]-[1,8] (4,5) and [5,0] at [2,8]-[3,8] (5,0).
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Step 5: Bottom right sum-0 and equals-3
The sum-0 region [3,7]-[3,8] already has a 0 at [3,8] from the [5,0] domino, so [3,7] must be 0. Place [0,3] horizontally: 0 at [3,7] and 3 at [3,6]. Now the equals-3 region [3,5],[3,6],[4,6] has 3 at [3,5] (from [3,1]) and 3 at [3,6]; thus [4,6] needs 3. Use [5,3] horizontally with 5 at [4,7] (greater-4 satisfied) and 3 at [4,6].
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Step 6: Final corners and sums
Remaining cells: the equals-2 region still needs [4,3] filled as 2, so [2,3] domino goes at [4,3]-[4,2] (2,3) to also satisfy the sum-4 region [4,2]-[5,2] which now needs a 1 at [5,2] (3+1=4). Place [0,1] at [5,1]-[5,2] (0,1) to fulfill the sum-0 at [5,1] and the 1 for sum-4. Finally, the empty cell [5,5] and sum-4 [5,6] take [0,4] at [5,5]-[5,6] (0,4).
💡 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.
💬 Community Discussion
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