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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 on July 22 is a masterclass in constrained minimalism. With only four dominos and a 3×3 grid, every cell is yoked by forceful sum targets: a top-row duo demands 10, a vertical pair 11, and a column 4. The less-than-4 region in the southeast corner then neatly partitions the available pips between the remaining dominos, ensuring a tight, satisfying solve.
Rodolfo Kurchan brings a midweek step up with a medium grid dominated by equals pairs and a few stray sums. Two stacked equals regions create a vertical strip of identical pips, while a sum-6 duo and a greater-than-1 cell constrain the lower edge. The puzzle’s elegance lies in how the equals regions propagate values downward, leading to a harmonious placement of the double-6 and double-1 dominos.
Kurchan’s hard puzzle is a labyrinth of interlocking constraints across eight rows. Equals regions return as anchoring threads, but the true core is a trio of zero-sum cells that force a cascade of zeros and paired equals. The designer weaves in a greater-than-4 outlier and a sum-12 target that isolate the double-6, creating a layered deduction that feels like peeling an onion. This NYT Pips hard rewards methodical unwinding of the zero-sum corridor before the rest falls into place.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Where the big sums lead
Start by looking at the highest sum constraint—it must be satisfied by a domino that contains a specific high number. The available pip list holds only one pair that can make it work.
💡 The left-column anchor
The sum-11 region occupies [1,0] and [2,0]. That sum forces a 5 and a 6, so the only domino with both pips must cover those two cells. But watch out: the sum-10 region at [0,0] and [0,1] also needs a 6, which will steer the orientation.
💡 Full eastern breakthrough
Place domino [5,6] with the 6 on [0,0] and the 5 on [1,0]. That satisfies sum-11 and gives sum-10 its 6. Then domino [0,4] goes to [0,1]=4 and [0,2]=0 (completing sum-10 and handing the 0 to the sum-4 region). Domino [6,0] fills [2,0]=6 and [2,1]=0, and finally domino [3,4] delivers 4 to [1,2] and 3 to [2,2], satisfying sum-4 and the less-4 region.
💡 Follow the equals
Several regions insist on equal values. The most restrictive are two vertical pairs that share a column; they demand the same pip in four cells, which can only come from a domino that carries a repeated number.
💡 The double-6 lock
The equals regions at [1,2]/[2,2] and [3,1]/[3,2] intersect indirectly. The top pair must be fully equal, so the only way to fill both is with the [6,6] domino. Its placement then forces the lower equals pair to also receive 6s from two different dominos.
💡 Unlocking the grid
Take the [6,6] domino onto [1,2] and [2,2]. Next, the [4,6] domino slides into [2,1]=4 and [3,1]=6, satisfying the lower left equals and giving the [1,1]/[2,1] pair a 4. To finish the lower right equals, place [6,5] at [3,2]=6 and [3,3]=5, which also satisfies the sum-6 region when [2,3] gets a 1 from the [1,1] domino. The greater-1 cell at [4,0] takes the 5 from [3,5] on [4,0]=5 and [4,1]=3, and the final sum-6 region [4,1]/[4,2] is completed by the [3,0] domino’s 3 on [4,2] and 0 on [4,3].
💡 Zero in on the nothing
A sum-0 region demands that three cells all contain a zero. That constraint seems impossible at first, but the domino set carries several zeros. Locate which zeros must serve this region and how they can tile the three cells.
💡 The zero-sum cluster
The zero-sum cells sit at [2,3], [3,3], and [3,4]. Because each must be 0, you need to place dominoes that supply those zeros. The [3,0] domino is your first candidate—it can anchor one zero and a 3 next to it.
💡 Placing the first zero
Place the [3,0] domino so that its 0 covers [3,3] and its 3 occupies [3,2]. That satisfies the equals pair [2,2]/[3,2] later. Now you still need zeros at [2,3] and [3,4]—look for dominoes [0,2] and [0,4] to claim those spots.
💡 Equals cascade from zero
With [3,0] set, the equals region [2,0]/[3,0] pushes you to supply identical pips there. The [2,4] domino (2 and 4) can go to [2,1]=2 and [2,0]=4, covering the sum-2 cell. Then the [5,4] domino places a 4 at [3,0] and a 5 at [3,1], completing the equals pair. The [0,4] domino finishes the zero corridor by putting 0 at [3,4] and 4 at [2,4].
💡 Full solution unraveled
Step through: [2,6] onto [4,2]=2,[4,1]=6. [3,0] onto [3,2]=3,[3,3]=0. [2,1] onto [4,3]=2,[4,4]=1. [0,5] onto [5,3]=0,[6,3]=5. [2,4] onto [2,1]=2,[2,0]=4. [3,5] onto [0,5]=3,[0,6]=5. [0,2] onto [2,3]=0,[1,3]=2. [3,3] onto [7,3]=3,[7,4]=3. [5,4] onto [3,1]=5,[3,0]=4. [2,3] onto [2,6]=2,[3,6]=3. [4,1] onto [2,5]=4,[1,5]=1. [6,6] onto [1,6]=6,[1,7]=6. [0,4] onto [3,4]=0,[2,4]=4. [3,1] onto [2,2]=3,[1,2]=1. [5,5] onto [3,5]=5,[4,5]=5. [3,4] onto [0,4]=3,[1,4]=4. This sequences the zero anchors, then the equals web, leaving the double-6 and high sums to lock into place.
🎨 Pips Solver
Jul 22, 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 sum-11 imperative
The region in column 0, rows 1 and 2, must sum to 11. The only way to do this with the provided pips is 5+6. Glance at the domino list: [5,6] is the sole domino with both a 5 and a 6, so it must occupy those two cells. Because [1,0] and [2,0] are vertically adjacent, place the domino there—but the trick is which direction gives 6 to the top. The sum-10 region at [0,0] and [0,1] also needs a 6, so the 6 from [5,6] should sit at [0,0] with the 5 at [1,0], leaving [2,0] open for a later 6.
2
Step 2: Feeding the sum-10 pair
With [0,0] now a 6, the top row pair [0,0]+[0,1]=10 demands a 4 in [0,1]. The domino [0,4] contains a 0 and a 4. Place it horizontally so that [0,1] gets the 4 and [0,2] gets the 0. This satisfies both sum-10 and gives the sum-4 region at [0,2],[1,2] its needed 0.
3
Step 3: Completing the left column
The sum-11 region still needs a 6 at [2,0]. The only remaining domino with a 6 is [6,0], so it must drop its 6 on [2,0] and its 0 on the adjacent [2,1]. That fulfills sum-11 and also places a 0 into the less-4 region at [2,1],[2,2].
4
Step 4: Finishing the southeast corner
The last domino [3,4] must fill [2,2] and [1,2]. Since the less-4 region [2,1],[2,2] can only accept pips below 4, and [2,1] already holds a 0, the 3 from [3,4] goes to [2,2] (pips 0 and 3, both less than 4). The 4 then lands on [1,2], exactly what the sum-4 region needed alongside the 0 at [0,2]. The grid is complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Spotting the equals anchors
Two equals regions dominate the center-left: [1,2] equals [2,2], and [3,1] equals [3,2]. The upper pair sits entirely in column 2, so those two cells must contain the same pip. The only available domino with a repeated number is [6,6] (the double-1 would conflict with nearby constraints later). Therefore, [6,6] must go to [1,2] and [2,2], both becoming 6.
2
Step 2: Extending the 6s downward
With column 2 locked in, the lower equals region [3,1]/[3,2] also needs identical pips. To maintain the propagated 6-values, [3,2] must be a 6. The domino [6,5] can supply that 6 and a 5. Place it so that [3,2] gets the 6 and [3,3] receives the 5. This simultaneously fills the sum-6 region [2,3]+[3,3] because [2,3] will later get a 1 from the double-1 domino.
3
Step 3: Solving the left-side equals column
The equals region [1,1]/[2,1] must now match. Because [2,1] sits adjacent to [3,1] (also part of the lower equals), and [3,1] still needs a value, you can use the [4,6] domino. Place it with 4 on [2,1] and 6 on [3,1]. That gives [2,1] a 4, forcing [1,1] to also be 4, and supplies the required 6 for [3,1]/[3,2] equality.
4
Step 4: Filling the top and the 1s
Now the less-3 cell at [0,1] must be under 3. The domino [4,0] is perfect: place it so [1,1] gets the 4 (already needed) and [0,1] gets the 0, which is less than 3. Then the double-1 domino [1,1] slots into [2,3] and [2,4], giving [2,3] the expected 1 (finishing sum 6: 1+5) and leaving [2,4] as an empty-region 1.
5
Step 5: The lower sum-6 and greater-1
The greater-1 region at [4,0] needs a pip above 1. Use the [3,5] domino with 5 on [4,0] and 3 on [4,1]. Next, the sum-6 region at [4,1],[4,2] demands 3+3=6. The [3,0] domino supplies the second 3 at [4,2] and a harmless 0 at [4,3] (empty region). All constraints are satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Zero-sum forces the first three
The sum-0 region at [2,3], [3,3], and [3,4] requires every cell to be 0. To begin, place the [3,0] domino so that its 0 covers [3,3] and its 3 occupies [3,2]. Next, use [0,2] to put its 0 on [2,3] and its 2 on [1,3] (satisfying part of the sum-3 region [1,2]+[1,3]=3 later). Then place [0,4] with its 0 on [3,4] and its 4 on [2,4]. The zero corridor is now complete.
2
Step 2: Equals pair [2,0]/[3,0] harnesses a 4
The equals region at the very left forces [2,0] and [3,0] to match. Since [2,1] has a sum-2 constraint, use the [2,4] domino (2 and 4) to place 2 on [2,1] and 4 on [2,0]. That gives [2,0] a 4, so [3,0] must also become 4. The [5,4] domino accommodates this: put its 4 on [3,0] and its 5 on [3,1].
3
Step 3: Equals [2,2]/[3,2] and sum-3 fall into place
Now [2,2] and [3,2] must equal each other. [3,2] is already 3 from Step 1, so [2,2] must be 3 as well. The [3,1] domino (3 and 1) is the tool: place 3 on [2,2] and 1 on [1,2]. That 1 combines with the 2 at [1,3] (from [0,2]) to make the sum-3 region [1,2]+[1,3]=3.
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Step 4: The sum-13 triple and sum-1 pockets
The sum-13 region at [3,5], [3,6], and [4,5] needs 5+3+5=13. Deploy the [5,5] domino to put 5 on both [3,5] and [4,5]. Then the [2,3] domino (2 and 3) fills [2,6] with 2 (sum-2 cell) and [3,6] with 3, completing the sum-13. The sum-1 cell at [4,4] is satisfied by the 1 from the [2,1] domino on [4,3]=2 and [4,4]=1.
5
Step 5: High-value isolations and the equals finale
The sum-12 region [1,6],[1,7] can only be made by [6,6], placing 6s there. The equals [0,4]/[0,5] pair must match; the [3,4] domino supplies a 3 to [0,4] and a 4 to [1,4], so [0,5] gets 3 from the [3,5] domino’s 3 (and its 5 occupies the sum-5 cell [0,6]). Meanwhile, the [4,1] domino puts 4 on [2,5] and 1 on [1,5] (sum-1 region), while the remaining equals [7,3]/[7,4] takes the double-3 domino [3,3] at [7,3]=3, [7,4]=3. The grid locks shut.
💡 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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