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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
Ian Livengood’s easy puzzle opens with a friendly 3x6 grid where two greater-5 cells immediately lock in the [6,5] and [6,6] dominoes. From there, the sum-8 columns cascade downward, and you quickly fill the bottom row with a neat pairing of doubles—[2,2] and [3,3]. This NYT Pips easy feels less like a puzzle and more like a warm-up, with every constraint pointing directly to the next piece.
The medium grid expands to 7 rows, and Livengood introduces equals and less-than constraints alongside the sums. You’re forced to think about how dominos bridge regions, especially the long vertical sum-6 column on the left and the duo of equals regions anchoring the right side. The single less-2 cell in column 2 narrows choices dramatically, but the real satisfaction comes from untangling the connected equals pairs that form a chain of matching pips.
Rodolfo Kurchan’s hard puzzle is an entirely different beast: a 5x6 grid where every cell in the top three rows has its own dedicated sum target, and only two cells in the bottom row are left unpinned. This dense field of single-cell sums forces you to match each domino pair precisely to fulfill the tiny regions, leaving little room for deduction—it’s more of a matching game. Yet the experience is strangely meditative, as you systematically assign the [0,1] through [4,5] dominoes and watch the grid resolve without a single ambiguity.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Scan for the most restrictive regions
The grid’s two ‘greater than 5’ cells in the top row are the tightest constraints—they accept only a single pip value, so they immediately dictate which domino halves belong there. The column sums below them then dictate how those dominos must be oriented.
💡 Hint 2: Lock down the top-row greater cells
The greater-5 cells sit at [0,0] and [0,5]. They force you to use the two dominos that contain a 6: [6,5] and [6,6]. The orientation matters because the sum-8 region in column 0 (cells [1,0] and [2,0]) needs a partner that adds to 8 with the 5 or the 6, and the same logic applies to column 5.
💡 Hint 3: The complete solution chain
Place [6,5] with 6 at [0,0] and 5 at [1,0]; then the [3,3] double covers [2,0]=3 and [2,1]=3 to satisfy sum-8 (5+3) and sum-5 (3+2). In the right column, [6,6] lands with both 6s at [0,5] and [1,5]; then the [2,2] double fills [2,4]=2 and [2,5]=2 to complete sum-8 (6+2). Finally, the [4,2] domino slips into the remaining sum-6 region with [2,3]=4 and [2,4]=2.
💡 Hint 1: Follow the equals and less-than clues
This grid mixes equals constraints (forcing two cells to match pips) with a less-than-2 cell that only accepts a 0 or 1. Start by examining the less-2 cell and the equals pairs on the right side—they create a ripple of forced values across several regions.
💡 Hint 2: Zero in on the less-2 cell and the sum-2 region
Cell [0,2] is the only ‘less 2’ region, so it must be 1. That forces the domino covering it to also place a 0 in the sum-2 region at [0,1]. The sum-2 region then needs a 2 in [1,1], which in turn pulls a domino with a 0 into the sum-6 column starting at [2,1]. Meanwhile, the equals constraints on columns 1 and 2 at rows 5–6 and 3–4 dictate matching values that will unlock the bottom rows.
💡 Hint 3: The complete solution chain
Place [1,0] horizontally with 1 at [0,2] and 0 at [0,1]. For the sum-2 pair, [0,2] covers [2,1]=0 and [1,1]=2. The sum-6 column [2,1]‑[4,1] gets [6,2] with 2 at [3,1] and 6 at [3,2]; the equals region above gets the other 6 from [6,3] (6 at [4,2], 3 at [5,2]). Bottom equals pairs: [4,5] gives 4 at [4,1] and 5 at [5,1]; [1,3] gives 1 at [6,3] and 3 at [6,2]. Finally, [3,5] places 3 at [6,0] and 5 at [6,1].
💡 Hint 1: Conquer the single-sum landscape
Nearly every cell in the top three rows has a tiny sum region all to itself, meaning each cell’s pip is exactly its target. Look for the cells that demand the rarest values—the sum‑0 cells—and trace which dominos can supply those zeroes while satisfying the larger sum next door.
💡 Hint 2: Start with the top-row zeroes
The top row alone contains a sum‑0 cell at [0,0] and a sum‑0 cell further right. [0,0] forces a 0 there, and the adjacent [1,0] sums to 4, so the domino covering that pair must be [0,4]. Nearby, [1,3] is a sum‑0 cell paired with [0,3] (sum 5), locking in [0,5]. Use these two early placements to unlock the rest of the zero‑bearing dominos.
💡 Hint 3: Chase the remaining zeroes
After placing [0,4] and [0,5], the grid still has three sum‑0 cells: [2,2], [2,5], and [3,0]. [2,2] connects to [2,3] (sum 1), demanding the [0,1] domino. [3,0] connects to [2,0] (sum 2), forcing [0,2]. [2,5] is linked to the equals pair [1,4]=[1,5]; those cells are both 3, so the domino covering [2,5] must give it a 0 and place the matching 3 in [1,5], using [0,3].
💡 Hint 4: Fill the top-row gaps and the bottom rows
With all zero‑bearing dominos placed, top row cells [0,1] (sum 2), [0,2] (sum 1), [0,4] (sum 4), and [0,5] (sum 5) remain. [0,1]‑[1,1] need a 2‑4 domino, so [2,4] slides in there. [0,2]‑[1,2] take [1,3] for 1‑3. The equals region [1,4]=[1,5]=3 is already fixed, and its domino [3,4] fills 3‑4 in that column. Finally, [4,5] covers the remaining top‑row sum‑4/5 pair at [0,4]‑[0,5].
💡 Hint 5: The complete solve from top to bottom
Place [0,4] with 0 at [0,0], 4 at [1,0]. [0,5] with 0 at [1,3], 5 at [0,3]. [0,1] with 0 at [2,2], 1 at [2,3]. [0,2] with 0 at [3,0], 2 at [2,0]. [0,3] with 0 at [2,5], 3 at [1,5]. [2,4] with 2 at [0,1], 4 at [1,1]. [1,3] with 1 at [0,2], 3 at [1,2]. [4,5] with 4 at [0,4], 5 at [0,5]. [1,4] with 1 at [3,4], 4 at [3,3]. [1,2] with 1 at [4,2], 2 at [3,2]. [2,3] with 2 at [2,1], 3 at [3,1]. [3,4] with 3 at [1,4], 4 at [2,4]. [2,5] with 2 at [3,5], 5 at [4,5]. [1,5] with 1 at [4,4], 5 at [4,3]. [3,5] fills the empty cells [4,0]=3, [4,1]=5.
🎨 Pips Solver
May 21, 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: The greater‑5 top cells demand sixes
Cells [0,0] and [0,5] each have a ‘greater than 5’ constraint, so both must be 6. The only dominos containing a 6 are [6,5] and [6,6]. Because both cells are in the top row, these two dominos must occupy the two pairs [0,0]‑[1,0] and [0,5]‑[1,5].
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Step 2: Column 0 sum‑8 fixes the five‑three split
Column 0 has a sum‑8 region covering [1,0] and [2,0]. If [6,5] goes into [0,0]‑[1,0], then [1,0] = 5. To reach a total of 8, [2,0] must be 3. The only remaining domino with a 3 is the [3,3] double, so it can be placed with 3 at [2,0] and 3 at [2,1].
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Step 3: Column 5 sum‑8 and the double two
With [6,6] covering [0,5] and [1,5] (both 6), the sum‑8 region below needs [2,5] = 2. The only domino left containing a 2 with a second 2 is the [2,2] double, so it fills [2,4]=2 and [2,5]=2.
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Step 4: The final sum‑6 and the four‑two domino
The remaining unfilled region is a sum‑6 covering [2,3] and [2,4]. [2,4] is already 2 from the previous step, so [2,3] must be 4. The [4,2] domino fits perfectly, placing 4 at [2,3] and 2 at [2,2]—and the grid is complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The less‑2 cell forces a one
Cell [0,2] is tagged ‘less than 2’, so its pip must be 1. The only domino that can easily supply a 1 at that position while also pairing with the sum‑2 region at [0,1] is [1,0]. Place it with 1 at [0,2] and 0 at [0,1].
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Step 2: Sum‑2 drives the next domino
The sum‑2 region covering [0,1] and [1,1] now has 0 at [0,1]; thus [1,1] needs a 2. The [0,2] domino can provide 2 at [1,1] and 0 at its other end, which fits into the sum‑6 column starting at [2,1]. Place [0,2] with 2 at [1,1] and 0 at [2,1].
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Step 3: Building the sum‑6 column
The vertical sum‑6 region in column 1 includes [2,1]=0, [3,1], and [4,1], so [3,1]+[4,1] must total 6. A domino with a 2 and a 6, [6,2], can place 2 at [3,1] and 6 at [3,2] to satisfy the equals region above. So [6,2] goes with 2 at [3,1], 6 at [3,2].
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Step 4: Equals pair at [4,2] and [3,2]
The equals region [3,2] and [4,2] demands both cells match. [3,2] is already 6, so [4,2] must also be 6. The [6,3] domino has a 6 and a 3, so it can place 6 at [4,2] and 3 at [5,2]. This also sets up the next equals pair.
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Step 5: Completing the bottom-right equals chains
The equals region [5,1]‑[6,1] and [5,2]‑[6,2] together with the sum‑3 at [6,0] and less‑2 at [6,3] lock in the final dominos. [4,5] gives 4 at [4,1] (to finish sum‑6 column: 0+2+4=6) and 5 at [5,1]; then [1,3] places 1 at [6,3] and 3 at [6,2] to match the equals demand. Finally, [3,5] places 3 at [6,0] and 5 at [6,1].
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Zero at top-left anchors the first domino
Cell [0,0] has a sum‑0 constraint, so it must be 0. Adjacent [1,0] has a sum‑4 constraint, so that cell must be 4. The only domino pairing 0 and 4 is [0,4]; place it vertically with 0 at [0,0] and 4 at [1,0].
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Step 2: A second zero fixes another pair
Cell [1,3] is also a sum‑0, forcing a 0 there. Its neighbor [0,3] is a sum‑5, requiring a 5. The [0,5] domino provides 0 and 5; place it with 0 at [1,3] and 5 at [0,3].
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Step 3: Zeros propagate to row 2 and row 3
Three more sum‑0 cells appear: [2,2], [3,0], and [2,5]. [2,2] is paired with [2,3] (sum 1), so [0,1] (0‑1) fits with 0 at [2,2] and 1 at [2,3]. [3,0] is paired with [2,0] (sum 2), so [0,2] fits with 0 at [3,0] and 2 at [2,0]. [2,5] pairs with the equals region [1,4]=[1,5], both of which must be 3; [0,3] provides 0 at [2,5] and 3 at [1,5].
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Step 4: Top row cells get their assigned values
Remaining top row cells [0,1] sum 2, [0,2] sum 1, [0,4] sum 4, [0,5] sum 5. [0,1] connects to [1,1] (sum 4), so the [2,4] domino gives 2 at [0,1] and 4 at [1,1]. [0,2] connects to [1,2] (sum 3), so [1,3] gives 1 at [0,2] and 3 at [1,2]. [0,4]‑[0,5] are the only adjacent pair left, needing 4 and 5; [4,5] fills them with 4 at [0,4] and 5 at [0,5].
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Step 5: Equals region and row 3 middle
The equals region [1,4]=[1,5]=3 is already 3 from Step 3; the domino covering [1,4] must also place the adjacent [2,4] (sum 4) value. [3,4] (3‑4) works: 3 at [1,4] and 4 at [2,4]. For [3,4] sum 1 paired with [3,3] sum 4, use [1,4] with 1 at [3,4] and 4 at [3,3].
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Step 6: Bottom rows and the empty cells
[4,2] sum 1 needs a 1, and [3,2] sum 2 needs a 2; the [1,2] domino gives 1 at [4,2] and 2 at [3,2]. [2,1] sum 2 and [3,1] sum 3 take [2,3] with 2 at [2,1] and 3 at [3,1]. [3,5] sum 2 and [4,5] sum 5 take [2,5] with 2 at [3,5] and 5 at [4,5]. [4,3] sum 5 and [4,4] sum 1 take [1,5] with 5 at [4,3] and 1 at [4,4]. Finally, the empty cells [4,0] and [4,1] greedily accept the remaining [3,5] domino: 3 at [4,0] and 5 at [4,1].
💡 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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