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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, constructed by Ian Livengood, is a confidence-builder centered on a four‑cell sum‑1 region that instantly dictates the domino placements in that corner. With only five dominoes and a small equals region to chain off, the solve is direct and fork‑free — perfect for a smooth start.
Rodolfo Kurchan's medium puzzle expands the grid and introduces a large "unequal" region spanning seven cells. That single constraint is the tight bottleneck: each cell must hold a distinct pip, which forces you to use the entire 0‑6 set exactly once. Once you untangle which dominoes can fit that pattern, the rest of the grid — a pair of equals edges and a few small‑target sums — clicks into place with satisfying momentum.
Kurchan's hard puzzle is the day's heavy hitter. A six‑cell equals region in the lower half creates a cascade of interdependencies, as the only viable equal value (1) must be delivered by multiple carefully chosen dominoes. Overlapping sum‑10, sum‑5, and sum‑12 constraints then squeeze placement options further, making this a slow‑burn deduction that rewards systematic thinking. Expect to spend real time on the middle‑to‑late domino tangle.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Follow the tiny sum
Look for a constraint that asks for an impossibly small total. That region will lock down a cluster of cells right away and give you your first couple of domino placements.
💡 Hint 2: The four‑cell sum‑1 corner
Focus on the region covering [0,1], [0,2], [0,3], and [1,3]. It must sum to exactly 1. With only pip values 0‑6 available, the only possible combination is three zeros and a one. That tells you which domino must cover the two rightmost cells.
💡 Hint 3: Full resolution
The sum‑1 region forces three zeros and one one. Place the [0,0] domino vertically in [0,3]/[1,3] (both 0). The remaining pair [0,1]/[0,2] must take the [1,0] domino, giving 1 at [0,1] and 0 at [0,2]. The equals region on the left now forces [1,0] and [2,0] to match; place the [4,5] domino with the 4 at [1,0] (and 5 at [1,1]), making [2,0] also 4. To achieve that, the [6,4] domino spans [2,1] (6) and [2,0] (4). Finally, the sum‑7 region at [2,1]/[3,1] forces [3,1] to be 1, so the [3,1] domino goes to [3,1] (1) and [3,2] (3).
💡 Hint 1: Spot the equality ban
Look for a region where all cells must carry different pip values. That one constraint will dictate which dominoes can even enter that part of the grid, so it's your natural starting point.
💡 Hint 2: The seven-cell unequal strip
The unequal region covers cells [0,2], [0,3], [0,4], [1,3], [1,4], [2,3], and [2,4]. Since it demands seven different pips, it must contain the full set 0‑6 exactly once. Notice that the [3,3] domino can only survive if one of its 3s lands outside the region — and [0,1] is the only safe neighbor.
💡 Hint 3: Complete medium unlock
Place the [3,3] domino horizontally on [0,1]/[0,2], putting a 3 inside the unequal region at [0,2]. The [6,0] domino then must place its 0 at [1,4] and 6 at [2,4] (the only way to get 0 into the region without duplicating). The [2,4] domino covers [2,3] with 2 and [1,3] with 4; the [1,5] domino fills [0,3] (1) and [0,4] (5). The top‑right sum‑11 region [0,6]/[0,7] uses the [5,6] domino on [0,5]/[0,6] (5 and 6) and the 5 from [2,5] on [0,7]; [2,5]'s 2 goes to [1,7] to match the right equals edge with [2,7] via the [1,2] domino (1 at [3,7] for the sum‑1, 2 at [2,7]). The left equals edge takes [0,4] domino (0 at [0,0], 4 at [1,0]) and [5,4] (5 at [3,0], 4 at [2,0]). All dominoes are placed.
💡 Hint 1: Find the massive equal clan
Search for a region where a whole cluster of cells must all show the same pip value. That constraint will dominate the domino inventory and narrow the possibilities drastically.
💡 Hint 2: The six‑cell equals anchor
The region at [2,1], [3,1], [3,2], [4,1], [4,2], and [4,3] must all be identical. With the available dominoes, the only pip that can cover six cells is 1. This means multiple dominoes will need to supply a 1 into this zone — a major restriction.
💡 Hint 3: Top rows fall into line
Now turn to the top‑left sum‑10 region ([0,0]/[1,0]) and the adjacent sum‑5 region ([0,1]/[1,1]). Sum‑10 requires two 5s, so [0,0] and [1,0] are both 5. This forces the [5,3] domino to occupy [0,0]/[0,1] (5 and 3) and the [5,6] domino to place its 5 at [1,0] with 6 at [2,0]. The sum‑5 region then gets a 2 from the [2,1] domino at [1,1] (with the 1 going to [2,1], feeding the equals zone).
💡 Hint 4: Bottom‑left equals trio and the center
The bottom‑left region [5,0]/[5,1]/[5,2] is also an equals cluster. The only way to produce three matching values there is to use the [4,4] and [4,6] dominoes, setting all three cells to 4. That places 4 at [5,0] (from [4,6]) and 4s at [5,1] and [5,2] (from [4,4]), with the [4,6] domino also giving 6 at [4,0]. The sum‑12 left edge then forces [3,0] to be 0, satisfied by the [1,0] domino (0 at [3,0], 1 at [3,1]). For the middle, the sum‑10 pair [2,2]/[2,3] needs 5+5; the [5,1] domino supplies 5 at [2,2] (1 at [3,2]) and the [4,5] domino gives 5 at [2,3] (4 at [3,3]).
💡 Hint 5: Final hard placements
The single‑cell sum‑6 at [4,4] gets 6 from the [6,6] domino, which also puts 6 at [4,5]. The sum‑9 region [4,5]/[5,5] then needs a 3 at [5,5], so the [3,3] domino goes to [5,4]/[5,5] (both 3). The sum‑6 region [5,3]/[5,4] gets its 3 at [5,3] from the [1,3] domino (1 at [4,3] into the equals clan, 3 at [5,3]). All placements: [5,6] at [1,0]/[2,0]; [1,1] at [4,1]/[4,2]; [4,4] at [5,1]/[5,2]; [5,3] at [0,0]/[0,1]; [2,1] at [1,1]/[2,1]; [4,5] at [3,3]/[2,3]; [1,3] at [4,3]/[5,3]; [4,6] at [5,0]/[4,0]; [3,3] at [5,4]/[5,5]; [1,0] at [3,1]/[3,0]; [6,6] at [4,4]/[4,5]; [5,1] at [2,2]/[3,2].
🎨 Pips Solver
May 15, 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: Crack the sum‑1 region
The four‑cell region at [0,1], [0,2], [0,3], [1,3] must sum to 1. The only way with pip values 0‑6 is to have three 0s and one 1. The domino [0,0] is the only double‑zero available, so it must cover two of those zeros. The only adjacent pair that can both be zero is [0,3] and [1,3] (vertical) — place the [0,0] domino there.
2
Step 2: Fill the rest of the sum‑1 region
With the two rightmost cells now zero, the remaining cells [0,1] and [0,2] must be one and zero. The [1,0] domino fits horizontally, giving 1 at [0,1] and 0 at [0,2], completing the sum‑1 constraint.
3
Step 3: Satisfy the equals column
The region at [1,0] and [2,0] requires equal values. [1,0] is currently empty. The only unused domino that can pair with a matching neighbor on [2,0] is [4,5] and [6,4] working together: place the [4,5] domino with the 4 at [1,0] (5 goes to [1,1]), setting [1,0]=4. To make [2,0] also 4, the [6,4] domino must go horizontally on [2,0] (4) and [2,1] (6).
4
Step 4: Resolve the bottom row
The sum‑7 region covers [2,1] (already 6) and [3,1]. So [3,1] must be 1. The remaining domino [3,1] completes the grid: place it vertically with 1 at [3,1] and 3 at [3,2].
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Understand the unequal region
The seven‑cell region at [0,2], [0,3], [0,4], [1,3], [1,4], [2,3], [2,4] demands all different pip values. With seven cells and pips 0‑6, it must use each pip exactly once. This means no double domino can have both ends inside the region unless one value can appear outside — and 3 is the only candidate because [3,3] will place one 3 inside ([0,2]) and one 3 outside ([0,1]).
2
Step 2: Place the [3,3] and start the unequal chain
Put the [3,3] domino horizontally at [0,1]/[0,2], giving a 3 to [0,2] inside the unequal zone. Now look at the remaining six cells in that zone: they must contain 0,1,2,4,5,6. The only domino with a 0 that can fit is [6,0], and 0 cannot repeat, so place it vertically with 0 at [1,4] and 6 at [2,4].
3
Step 3: Fill the remaining unequal cells
The [2,4] domino now covers [2,3] and [1,3]. To avoid repeating pips already placed, it must give 2 at [2,3] and 4 at [1,3]. The [1,5] domino then naturally fills [0,3] and [0,4] with 1 and 5. The unequal region is fully assigned.
4
Step 4: Top‑right sum‑11 and right equals edge
The sum‑11 region [0,6]/[0,7] needs two numbers adding to 11. The only unused domino with a high enough pair is [5,6]. Place it on [0,5] and [0,6] (5 and 6). Then [0,7] must be 5 from the [2,5] domino, whose 2 lands at [1,7]. The right equals region forces [2,7] to also be 2; the [1,2] domino satisfies this with the 2 at [2,7] and its 1 at [3,7] to meet the sum‑1 target.
5
Step 5: Left side equals and bottom sums
The left equals region [1,0]/[2,0] forces matching values. Place the [0,4] domino with 0 at [0,0] and 4 at [1,0]. To match, [2,0] must be 4, provided by the [5,4] domino (5 at [3,0] to satisfy the sum‑5 there and 4 at [2,0]). All dominoes are placed.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Lock the equals region to 1
The six‑cell region at [2,1], [3,1], [3,2], [4,1], [4,2], [4,3] must all share the same pip. Checking the domino list, only pip 1 appears enough times across different dominoes — specifically in [2,1] (one 1), [1,1] (two 1s), [1,3] (one 1), [1,0] (one 1), and [5,1] (one 1). No other pip can cover six cells. So this entire region must be 1.
2
Step 2: Top‑left sums force 5s
The sum‑10 region [0,0]/[1,0] needs 5+5. The only dominoes with 5 that can be placed here are [5,3] (5,3) and [5,6] (5,6). Place [5,3] horizontally on [0,0]/[0,1], setting [0,0]=5 and [0,1]=3. Then place [5,6] vertically with 5 at [1,0] and 6 at [2,0], completing sum‑10. The sum‑5 region [0,1]/[1,1] now has [0,1]=3, so [1,1] must be 2. The [2,1] domino provides this: 2 at [1,1] and 1 at [2,1] (adding a 1 to the equals zone).
3
Step 3: Bottom‑left equals trio
The region [5,0]/[5,1]/[5,2] must be equal. The only way to get three matching cells from the remaining dominoes is to use [4,4] and [4,6]. Place [4,6] with 4 at [5,0] and 6 at [4,0]; place [4,4] with 4 at [5,1] and 4 at [5,2]. This forces [4,0]=6. The sum‑12 left column ([2,0]=6, [3,0], [4,0]=6) then requires [3,0]=0. The [1,0] domino fits perfectly: 0 at [3,0] and 1 at [3,1] (another 1 for the equals region).
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Step 4: Middle sum‑10 pair
The region [2,2]/[2,3] needs a total of 10, so both cells must be 5. The [5,1] domino places 5 at [2,2] and 1 at [3,2] (equals zone), while the [4,5] domino gives 5 at [2,3] and 4 at [3,3], satisfying the single‑cell sum‑4 at [3,3].
5
Step 5: Top‑right sum‑6 and the equals leftovers
The sum‑6 cell at [4,4] is filled by the [6,6] domino, which also places 6 at [4,5] (giving the first value for the sum‑9 region [4,5]/[5,5]). The sum‑9 now needs 3 at [5,5], so the [3,3] domino goes to [5,4]/[5,5] (both 3). The remaining sum‑6 region [5,3]/[5,4] already has a 3 at [5,4]; place the [1,3] domino with 3 at [5,3] and 1 at [4,3] (the final cell of the equals region).
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Step 6: Verify the full layout
All dominoes are now placed: [5,6] (index 0) at [1,0]/[2,0]; [1,1] (1) at [4,1]/[4,2]; [4,4] (2) at [5,1]/[5,2]; [5,3] (3) at [0,0]/[0,1]; [2,1] (4) at [1,1]/[2,1]; [4,5] (5) at [3,3]/[2,3]; [1,3] (6) at [4,3]/[5,3]; [4,6] (7) at [5,0]/[4,0]; [3,3] (8) at [5,4]/[5,5]; [1,0] (9) at [3,1]/[3,0]; [6,6] (10) at [4,4]/[4,5]; [5,1] (11) at [2,2]/[3,2]. Every constraint is met.
💡 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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