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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 gentle warm-up. A chunky equals region in the bottom-left corner seizes control early, forcing a double-number domino and setting off a cascade that reaches every other constraint. The top-row paired equals regions and a greater-than cell offer clean, fork-free deductions. Expect to finish quickly with a satisfying click.
Livengood's medium puzzle revolves around a razor-thin bottleneck: a single-cell sum-1 region. That tiny cell dictates exactly which domino can cover it, and the surrounding equals chains then resolve in rapid succession. The puzzle feels taut but never unfair—find that one critical cell and the rest opens up smoothly.
Rodolfo Kurchan's hard grid is a densely packed challenge. A sprawling six-cell unequal region dominates the center, while tiny sum targets (3, 2, 6) lock in extreme values right out of the gate. Cross-referencing the sum-3 and sum-6 cells with the unequal region creates a tight weave that demands careful ordering. It's a slow burn, but the payoffs are deeply satisfying.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Spot the anchor
Look for the largest equals region on the board—one that spans multiple cells and will demand a very specific type of domino with matching pips.
💡 Hint 2: Bottom-left cluster
The equals region covering [2,0], [3,0], and [3,1] must have three identical values. Pair that requirement with the neighboring less-3 region at [3,2]-[3,3] to narrow down which dominoes can land there.
💡 Hint 3: Full solve
The triple-equals region forces the 0-0 domino onto [2,0] and [3,0]. The 0-1 domino then places its 0 at [3,1] and its 1 at [3,2] (satisfying less-3). The greater-3 cell at [1,4] takes the 5 from the 5-3 domino, with 3 at [0,4]. The top-row equals pair [0,3]/[0,4] both become 3 via the 3-4 domino (3 at [0,3], 4 at [0,2]). The 2-4 domino finishes the top with 4 at [0,1] and 2 at [1,1]. Finally, the 3-1 domino covers [2,3] with 3 and [3,3] with 1.
💡 Hint 1: The tiniest target
A single cell with a sum constraint acts as the puzzle's narrowest bottleneck. Identify that region first—it will dictate the very first domino you place.
💡 Hint 2: The 1-cell and the zero block
The sum-1 cell at [3,2] can only hold the value 1. Meanwhile, the large equals region at the bottom right ([3,3],[3,4],[3,5]) needs three identical pips, forcing a double-number domino. Those two facts together will unlock the bottom half of the grid.
💡 Hint 3: Full solve
The sum-1 cell [3,2] forces the 1-3 domino: 1 at [3,2], 3 at [3,1]. That makes the equals pair [3,0]/[3,1] both 3, with the 3 coming from the 2-3 domino (3 at [3,0], 2 at [2,0]). The bottom-right equals block gets the 0-0 domino covering [3,3] and [3,4]; the 6-0 domino then supplies 0 at [3,5] and 6 at [2,5]. The top-row equals [0,2]/[0,3] becomes 5s from the 1-5 domino (5 at [0,2], 1 at [0,1]), while [0,4]/[0,5] are 4s from the 4-5 domino (4 at [0,4], 5 at [0,3]). The 6-4 domino puts 6 at [1,5] and 4 at [0,5]; the 2-2 domino gives 2s to [0,0] and [1,0], completing the left equals chain with [2,0]'s existing 2.
💡 Hint 1: Extreme values
Focus on the smallest sum regions and the lone single-cell sum. They will force extreme values early, while a large unequal region will tightly restrict which digits can appear where.
💡 Hint 2: Sum-3 and sum-6 cells
The sum-3 region at [2,0], [2,1], [3,0] must be three 1s. The single-cell sum-6 at [6,1] demands exactly 6. The sum-2 cell at [3,3] forces a 2. These three hot spots will dictate your opening domino placements.
💡 Hint 3: Building the bottom
Place the 1-1 domino on [2,0] and [2,1]. The 4-6 domino then puts 6 on [6,1] and 4 on [5,1] (entering the unequal region). The 3-2 domino gives 2 to [3,3] and 3 to [3,2]. Now consider how the sum-8 region [4,0]/[5,0] and the sum-6 region [6,0],[7,0],[7,1] will interact.
💡 Hint 4: Sum-8 and sum-6 interplay
The sum-8 pair [4,0]/[5,0] must sum to 8. With [5,1] already 4, the 2-4 domino can place 4 at [5,0] and 2 at [6,0]. That gives the sum-6 region (now 2 at [6,0]) a need for two more 2s—the 2-2 domino fills [7,0] and [7,1]. Then the sum-8 demands 4 at [4,0], so the 4-1 domino places 4 there and 1 at [3,0], completing the sum-3 region.
💡 Hint 5: Full solve
The 1-1 domino covers [2,0] and [2,1]. The 4-1 domino puts 1 at [3,0] and 4 at [4,0]. The 2-4 domino sets 4 at [5,0] and 2 at [6,0]. The 4-6 domino places 6 at [6,1] and 4 at [5,1]. The 3-2 domino gives 3 at [3,2] and 2 at [3,3]. The 2-0 domino covers [3,1] with 2 and [4,1] with 0. The 1-5 domino fills [2,2] with 1 and [2,3] with 5. The 6-5 domino places 6 at [0,3] and 5 at [0,4] (sum 11). The 4-4 domino sets 4s at [0,5] and [1,5]; the 4-0 domino puts 4 at [1,4] and 0 at [1,3] (sum 12). The 0-0 domino finishes with 0s at [0,2] and [1,2]. The 2-2 domino gives 2s to [7,0] and [7,1] (sum 6). All unequal region cells are distinct: 1,5,2,3,0,4.
🎨 Pips Solver
May 3, 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: Triple-equals anchor
The equals region covering [2,0], [3,0], and [3,1] requires three identical pips. The only way to achieve this is with the double-0 domino covering two of these cells (say [2,0] and [3,0]), forcing both to 0. The third cell, [3,1], must also be 0, which will come from a domino that also contains a 0.
2
Step 2: Less-3 region and the 0-1 domino
The less-3 region at [3,2] and [3,3] can only accept values 0, 1, or 2. Since [3,1] still needs a 0 and is adjacent to [3,2], the 0-1 domino is the perfect fit: place its 0 at [3,1] and its 1 at [3,2]. This satisfies both the equals (0) and less-3 (1) constraints for those cells.
3
Step 3: Greater-3 and top-row equals pairs
Cell [1,4] is subject to a greater-3 constraint (must be 4,5,6). The only available domino with a pip above 3 that can reach this cell is the 5-3. Place 5 at [1,4] and 3 at [0,4]. Now the top-row equals pair [0,3] and [0,4] must match—[0,4] is 3, so [0,3] must also be 3. That forces the 3-4 domino: 3 at [0,3], 4 at [0,2].
4
Step 4: Remaining top and final pieces
The other top-row equals pair [0,1] and [0,2] must now match. Since [0,2] is already 4, [0,1] must be 4. The last domino containing a 4 is the 2-4, so place 4 at [0,1] and 2 at [1,1] (the empty cell below it). Finally, the remaining 3-1 domino covers the leftover cells [2,3] (empty) and [3,3] (less-3) with 3 and 1 respectively, completing the puzzle.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The sum-1 bottleneck
The single-cell sum-1 region at [3,2] must equal 1. Only the 1-3 and 1-5 dominoes contain a 1. Whichever is chosen, its 1-pip will go in that cell, and the other pip will land in the adjacent cell [3,1], which is part of the equals region with [3,0].
2
Step 2: Forcing the 1-3 domino
If the 1-5 domino were used, [3,1] would become 5, forcing [3,0] to also be 5. No remaining domino can supply a 5 to [3,0] while also satisfying the equals chain at the left. Therefore, the 1-3 domino must be placed: 1 at [3,2], 3 at [3,1]. This means [3,0] must also be 3, accomplished by the 2-3 domino (3 at [3,0], 2 at [2,0]).
3
Step 3: The bottom-right zero block
The equals region [3,3], [3,4], [3,5] requires three identical pips. The only way is to use a double-number domino to provide two of them. The 0-0 domino is ideal: place it covering [3,3] and [3,4] (both 0). The cell [3,5] still needs a 0; that will come from the 6-0 domino placed later at [2,5] and [3,5] with 6 above and 0 below.
4
Step 4: Completing the left equals chain
The equals region [0,0], [1,0], [2,0] requires all three cells to match. We already have [2,0]=2 from the 2-3 domino, so [0,0] and [1,0] must be 2. The 2-2 domino is the only one that can provide two 2s; place it at [0,0] and [1,0].
5
Step 5: Top-row and right-side fills
The remaining dominoes slot into the top: the 1-5 domino places 5 at [0,2] and 1 at [0,1] (empty), establishing the [0,2]/[0,3] equals pair as both 5 (the 5 for [0,3] comes from the 4-5 domino, with 4 at [0,4]). The 4-5 domino thus covers [0,4] and [0,3] with 4 and 5. The equals pair [0,4]/[0,5] both become 4, with the 6-4 domino supplying 6 at [1,5] and 4 at [0,5]. Finally, the 6-0 domino finishes the grid: 6 at [2,5] (matching [1,5]=6) and 0 at [3,5], completing the triple-zero block.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum-3 region forces 1-1
The sum-3 region at [2,0], [2,1], [3,0] must add to 3 with three cells—the only feasible split is 1,1,1. Place the 1-1 domino on [2,0] and [2,1] to establish two 1s. The third cell, [3,0], will later receive a 1 from the 4-1 domino.
2
Step 2: Lone sum-6 cell dictates 4-6
The single-cell sum-6 region at [6,1] can only hold the value 6. The 4-6 domino is the only one containing a 6. Place it with 6 at [6,1] and 4 at [5,1]; this introduces 4 into the large unequal region.
3
Step 3: Sum-2 cell demands 3-2
The sum-2 cell at [3,3] must be 2. The 3-2 domino is the only remaining option that can supply a 2 and also fit the adjacent cell [3,2] (which will become 3). Place 2 at [3,3] and 3 at [3,2], adding two more distinct values to the unequal region.
4
Step 4: Sum-8 and sum-6 interplay
The sum-8 pair [4,0]/[5,0] must sum to 8. The sum-6 region [6,0],[7,0],[7,1] must sum to 6. Notice [5,0] is adjacent to [6,0] and [5,1] (already 4). To balance both, place the 2-4 domino with 4 at [5,0] and 2 at [6,0]. This sets [6,0]=2, leaving the sum-6 region needing two more 2s to make 6. Place the 2-2 domino on [7,0] and [7,1] (both 2). Now the sum-8 region has 4 from [5,0], so [4,0] must be 4. The 4-1 domino provides that 4 at [4,0] and finally delivers the missing 1 to [3,0], completing the sum-3 region with three 1s.
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Step 5: Unequal region midsection
The unequal region still needs distinct values at [2,2], [2,3], [3,1], and [4,1]. The 2-0 domino places 2 at [3,1] and 0 at [4,1]. The 1-5 domino fills [2,2] with 1 and [2,3] with 5. Now the six cells of the unequal region hold 1,5,2,3,0,4—all distinct.
6
Step 6: Top sums and empty cells
The sum-11 pair [0,3]/[0,4] must be 6 and 5 (in either order). The 6-5 domino settles this: 6 at [0,3], 5 at [0,4]. The sum-12 region ([0,5],[1,4],[1,5]) needs a total of 12; the only possible arrangement with remaining dominoes is 4,4,4. The 4-4 domino covers [0,5] and [1,5] (both 4), and the 4-0 domino provides the third 4 at [1,4] and 0 at [1,3]. Finally, the 0-0 domino fills the empty cells [0,2] and [1,2] with 0s, 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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