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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 is a confidence-builder—a single sum-1 cell at [1,1] forces the very first domino, and from there a linear chain of forced placements unwinds. The equals region in row 2 locks in doubles, and a greater-than on [0,4] slots the remaining pip values cleanly. You’ll finish in under a minute.
Medium ramps up with a large less-1 region at the top that compels zero placements, then an equals strip across row 2 that demands identical pip values. The bottleneck is that equals chain—once you realize every cell in that 4-cell region must be the same number, the available dominoes narrow drastically. Expect a brief mental pivot before the grid opens up.
Kurchan’s hard is a dense web of sum constraints, but the starting point is a brilliant sum-0 trio at the bottom right that forces three zeros and two dominoes immediately. From there, an equals chain along the bottom row locks in twos, while a sum-13 triplet at top left and a sum-22 quartet on the right demand careful pairing. The hardest part is untangling the region overlaps without guessing.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Identify the tightest constraint
Start by looking at the tightest numerical constraint—a single cell that must sum to a very low target. That forces one specific value there, and its domino partner will fall into place.
💡 Hint 2: Zoom in on the sum-1 cell
The sum-1 cell at [1,1] can only be 0 or 1, but the domino in play forces it to be 1. This immediately gives you the domino covering [0,1] and [1,1].
💡 Hint 3: Full solution walkthrough
The domino [4,1] goes on [0,1]-[1,1] with 4 on [0,1] and 1 on [1,1]. The greater-4 at [0,4] takes 5 from [1,5], which places its 1 on [1,4] to satisfy the equals region with [2,4]. Then the equals pair [2,0]-[2,1] takes [6,6]. The less-4 [2,5] and equals [2,4] force [2,1] domino covering [2,5]-[2,4] with 2 and 1. Finally, [3,2] less-1 takes 0 from [3,0], placing its 3 on the empty [3,3].
💡 Hint 1: Find the forced zeros
Search for a region where all cells must be less than 1—that means every cell there has only one possible pip value. This clarity will anchor the entire puzzle.
💡 Hint 2: The less-1 cluster
The less-1 region covers [0,3], [1,2], and [1,3], all forced to 0. The dominoes [0,0] and [1,0] fill these, with [0,0] spanning [1,2]-[1,3] and [1,0] covering [0,2]-[0,3] (where [0,2] must be less than 2, so it gets the 1 from that pair).
💡 Hint 3: Full solution walkthrough
The less-1 zeros go: [0,0] on [1,2]-[1,3]; [1,0] on [0,2]-[0,3] (1 at [0,2], 0 at [0,3]). That sets [2,0] greater-4 to 5 from [5,2] domino, which puts its 2 at [2,1] starting the equals region [2,1]-[2,4] all 2. So [2,1]-[2,4] gets [2,2] on [2,2]-[2,3] and [3,2] on [2,5]-[2,4] with 3 at [2,5] and 2 at [2,4]. The sum-4 region [3,2]-[3,3] uses [0,4] on [3,3]-[4,3] (0,4) and [6,4] on [4,2]-[3,2] (6,4). Greater-11 [4,2]-[5,2] takes 6 and 6, so [6,5] on [5,2]-[5,1] with 6,5; equals [4,3]-[5,3] both 4 via [0,4] and [2,4] domino on [5,4]-[5,3] (2 at sum-2 [5,4], 4 at [5,3]).
💡 Hint 1: Seek the zero-sum region
Identify the region that demands the smallest possible sum—zero. That region will force multiple cells to be 0, pulling in specific dominoes that must contain zeros.
💡 Hint 2: Locate the sum-0 trio
The sum-0 region sits at [4,5], [5,5], [5,6]. Since 0 is the only pip that can sum to zero in multiple cells, all three must be 0. The double-zero domino [0,0] fits perfectly on [5,5]-[5,6], while [4,5] needs a partner 0 that connects vertically to [3,5].
💡 Hint 3: Equals chain on bottom row
With [5,5] and [5,6] occupied by [0,0], and [4,5] getting 0 from the [0,5] domino (paired with 5 at [3,5]), turn to the equals region on row 5: [5,0], [5,1], [5,2] must all be identical. The double-2 [2,2] can cover [5,1]-[5,2], forcing [5,0] to also be 2.
💡 Hint 4: Completing the equals row and beyond
[5,0] must connect to [4,0] (since [5,1] is taken), so use [0,2] domino with 2 at [5,0] and 0 at [4,0]. Sum-3 region forces 3 at [3,0] from [3,5] domino placed on [3,0]-[2,0] (3,5). Sum-11 then demands 6 at [1,0] via [5,6] on [0,0]-[1,0] (5,6). Top-left sum-13 gets [4,4] on [0,1]-[0,2]; top-right sum-22 gets [5,5] on [0,5]-[0,6] and [6,6] on [0,7]-[1,7].
💡 Hint 5: Full solution walkthrough
Sum-0 region: [0,0] on [5,5]-[5,6]; [0,5] on [4,5]-[3,5] (0,5). Equals row 5: [2,2] on [5,1]-[5,2]; [0,2] on [4,0]-[5,0] (0,2) forcing all 2s. Sum-3: [3,5] on [3,0]-[2,0] (3,5). Sum-11: [5,6] on [0,0]-[1,0] (5,6). Sum-13: [4,4] on [0,1]-[0,2]. Sum-22: [5,5] on [0,5]-[0,6]; [6,6] on [0,7]-[1,7]; [4,6] on [2,5]-[1,5] (4,6). Sum-1: [0,1] on [1,3]-[2,3] (0,1); [0,6] on [3,3]-[4,3] (0,6). Sum-9s: [3,3] on [2,7]-[3,7]; [1,4] on [5,7]-[4,7] (1,4) completes sum-1 at [5,7] and sum-7 at [3,7]-[4,7].
🎨 Pips Solver
Jul 23, 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: Sum-1 forces the first domino
The sum-1 cell at [1,1] forces that cell to be 1 (the only value summing to 1 on its own) and its domino partner [0,1] must take the other pip from the same domino. Looking at the domino list, only [4,1] can supply a 1 and a 4, so it must go covering [0,1] and [1,1] with 4 on the empty cell.
2
Step 2: Greater-4 and equals lock the next pair
The greater-4 region at [0,4] requires a value >4, so 5 or 6. The only remaining domino with a 5 is [1,5]. Placing it horizontally means its 1 falls on [1,4], which is part of an equals region with [2,4], so [2,4] must also be 1.
3
Step 3: Double-six satisfies the equals-pair
The equals region at [2,0]-[2,1] demands identical values. The only double remaining is [6,6], so place it there. Now the less-4 cell [2,5] must be <4, and it's adjacent to the already-placed 1 at [2,4], so a domino covering [2,5]-[2,4] can use the [2,1] domino with 2 on the less-4 and 1 matching the equals region.
4
Step 4: Final zero and empty cell
Finally, the less-1 cell [3,2] must be 0. The last domino [3,0] covers it, placing 0 on [3,2] and 3 on the empty [3,3].
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Zeros from the less-1 region
Find all cells that must be 0: the less-1 region on [0,3], [1,2], [1,3] can only hold 0. The [0,0] domino fits perfectly on [1,2]-[1,3], placing zeros. The [1,0] domino then covers [0,2]-[0,3] with 1 on [0,2] (which is less-2) and 0 on [0,3].
2
Step 2: Greater-4 kickstarts the equals strip
The greater-4 at [2,0] needs >4, so 5 or 6. The [5,2] domino has a 5; placing it with 5 at [2,0] forces its 2 into [2,1], which is part of the large equals region [2,1]-[2,4]—all those cells must now be 2.
3
Step 3: Completing the equals chain
Fill the equals strip: use [2,2] domino on [2,2]-[2,3] (both 2). The cell [2,4] still needs 2; it pairs with [2,5], whose greater-2 constraint allows 3 from the [3,2] domino. So [3,2] goes on [2,5]-[2,4] giving 3 and 2.
4
Step 4: Sum-4 region pairs with [0,4] and [6,4]
The sum-4 region [3,2]-[3,3] requires a sum of 4. Place [0,4] vertically: 0 at [3,3], 4 at [4,3]. Then [3,2] still empty—use [6,4] domino on [4,2]-[3,2] placing 6 at [4,2] and 4 at [3,2], so sum becomes 4+0=4.
5
Step 5: Greater-11 and final placements
Greater-11 [4,2]-[5,2] demands total >11, so they must be 6 and 6. [6,4] already gave 6 to [4,2]; [5,2] gets 6 from the [6,5] domino, which places 5 at [5,1] (empty). The equals [4,3]-[5,3] takes the 4s: [4,3] already 4, so [5,3] gets 4 from [2,4] domino on [5,4]-[5,3] with 2 at [5,4] to satisfy sum-2.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum-0 forces zeros
Start with the sum-0 region at [4,5], [5,5], [5,6]. All three must be 0. Place [0,0] domino on [5,5]-[5,6] (both 0). Then [4,5] must be 0, so it pairs with [3,5] using [0,5] domino, placing 0 at [4,5] and 5 at [3,5].
2
Step 2: Equals row builds a 2-chain
Turn to the equals region on row 5: [5,0], [5,1], [5,2] must all share the same value. The [2,2] domino can cover [5,1]-[5,2] with 2s. That forces [5,0] to also be 2, so it pairs with [4,0] using [0,2] domino (0 at [4,0], 2 at [5,0]).
3
Step 3: Sum-3 and sum-11 resolve the left column
Now the sum-3 region [3,0]-[4,0] has 0 at [4,0], so [3,0] must be 3. The [3,5] domino supplies a 3 and a 5; place it on [3,0]-[2,0] with 3 at [3,0] and 5 at [2,0]. The sum-11 region [1,0]-[2,0] now has 5 at [2,0], demanding 6 at [1,0]. Use the [5,6] domino on [0,0]-[1,0] with 5 at [0,0] and 6 at [1,0].
4
Step 4: Top-left sum-13 and top-right sum-22
The sum-13 top-left region [0,0]-[0,2] has 5, so needs 8 more; place [4,4] on [0,1]-[0,2] giving 4+4. For sum-22 on the right: [0,5],[0,6] get 5,5 from [5,5]; [0,7],[1,7] get 6,6 from [6,6]. That leaves [1,5] needing 6 to reach 5+5+6+6=22; use [4,6] on [2,5]-[1,5] (6 at [1,5], 4 at [2,5]) which also satisfies the sum-9 region [2,5]-[3,5] (4+5=9).
5
Step 5: Remaining sum-1 and sum-9 regions
Sum-1 region [1,3],[2,3],[3,3] requires 0,1,0. Place [0,1] on [1,3]-[2,3] (0,1). Then [3,3] gets 0 from [0,6] domino, placing 6 at [4,3] (greater-0). Sum-9 [1,7]-[2,7] has 6 at [1,7], so [2,7] needs 3—use [3,3] on [2,7]-[3,7]. Sum-7 [3,7]-[4,7] now has 3 at [3,7], so [4,7] gets 4 from [1,4] domino on [5,7]-[4,7] (1 at sum-1 [5,7], 4 at [4,7]). All regions satisfied.
💡 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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