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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 constructs today's easy NYT Pips with an elegant double lock: a sum-9 pair anchoring the top and a triple-equals column of sixes along the left edge. The domino set is small but precise, forcing a clean dependency between the equals regions and a tiny sum-2 region. It’s a puzzle that rewards scanning for duplicate pips early.
Rodolfo Kurchan's medium puzzle weaves a web of sum constraints, most notably a single-cell sum-6 that acts as a crucible. The region layout, with cascading sums linking rows 0 through 3, demands careful accounting of pip totals. Kurchan’s signature lies in creating interlocking arithmetic that feels organic yet tightly controlled.
Kurchan's hard NYT Pips is a masterclass in expansive equals regions: a sprawling six-cell block of sixes and a four-cell line of zeros carve out the board, while less-than and unequal constraints add texture. The puzzle opens with a clever domino placement that cascades into these equals chains, showcasing his ability to build a cathedral of logic from a few simple planks.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Spot the duplicates
Look for regions that demand identical numbers—equal-constraint zones are often the tightest starting points, especially when they span multiple cells.
💡 Zero in on the left column
The vertical three-cell equals region covering [2,0], [3,0], and [3,1] can only be one pip value. Consider which dominoes can supply three copies of the same number. Also note the tiny sum-2 region at bottom right.
💡 Full solve path
The three-cell equals must be 6s: only [6,6] supplies a pair, and [1,6] gives the third. Place [6,6] vertically at [2,0]/[3,0], then [1,6] with 6 at [3,1] and 1 at [3,2]. The equals region [3,2]/[3,3] forces both 1s, so [3,3]=1. The sum-2 region at [2,2]/[2,3] needs 2+0; place [1,0] vertically with 0 at [2,3] and 1 at [3,3]. Finally, sum-9 at [1,1]/[1,2] takes 5+4: [5,5] puts 5 at [1,1] and [0,1]; [2,4] puts 2 at [2,2] and 4 at [1,2].
💡 Single-cell standouts
Medium's many sum regions create a domino-placement puzzle; start by identifying single-cell sum constraints—they dictate exact values and quickly lock a high-number domino.
💡 Top-row tens and the lonely six
The single-cell sum-6 at [1,1] forces a 6, which pulls in the [6,4] domino. On the top row, the sum-10 region at [0,2]/[0,3] can't use 6+4 (that 6 is taken), so it must be 5+5. Scan your 5-dominoes—[5,1], [5,2], and [5,3] will each play a role.
💡 Complete chain
Place [6,4] at [1,1]=6, [1,0]=4. Now sum-6 region [0,0]/[1,0] gets 2+4: use [0,2] at [0,0]=2, [0,1]=0. Top sum-10 needs two 5s; set [5,1] at [0,3]=5, [1,3]=1 and [5,2] at [0,2]=5, [1,2]=2, satisfying sum-3 at [1,2]/[1,3] neatly. Row 2's sum-6 zones: [2,0]/[2,1] needs 3+3 using [1,3] (3 at [2,0]) and [5,3] (3 at [2,1]); [2,2]/[2,3] sums to 6 with [5,3]'s 5 at [2,2] and [1,0]'s 1 at [2,3], putting 0 at [3,3]. Finally, bottom sum-6 at [3,0]/[3,1]/[3,2] has [3,0]=1 from earlier; add [1,4] as [3,1]=1, [3,2]=4 for total 1+1+4=6.
💡 Backbone equals
The hard puzzle is defined by massive equals regions. Before touching any domino, identify the large zones that must share a single pip value—these will dictate your high-number placements.
💡 The six-cell colossus
A six-cell equals region spans columns 2–3 from rows 1 through 4. This many cells of the same number demand multiple copies of that pip; look at the available 6-dominoes ([6,1], [6,0], [6,3], [6,6], [2,6]) and you'll see which number is forced.
💡 Sum-4 ignition
The top-right sum-4 region at [0,3]/[0,4] requires a 1+3 or 2+2. Since 2s are needed elsewhere, 1+3 is the key; the 1 comes from the [6,1] domino, which simultaneously places its 6 into the big equals zone at [1,3].
💡 Cascading zeroes and twos
With the 6-engine running, the zero-equals region (rows 1–4, column 4) emerges. The [6,0] domino places 0 at [2,4]; then [0,3] and [4,0] and [1,0] fill the remaining zero slots. Up top, the four-cell equals block ([0,0]–[1,0] except [0,2]?) must become all 2s, anchored by [2,2] and [1,2].
💡 Full solution map
Start with [6,1] at [1,3]=6, [0,3]=1. Place [6,6] vertically at [2,2]/[3,2]=6,6. Set [2,6] at [0,2]=2, [1,2]=6. Place [6,0] at [2,3]=6, [2,4]=0. Use [0,3] at [1,4]=0, [0,4]=3 to complete sum-4. Set [6,3] at [4,2]=6, [4,3]=3. Put [4,0] at [3,3]=4, [3,4]=0. Place [1,0] at [5,4]=1, [4,4]=0. Top-left equals demands 2s: [2,2] at [0,0]/[0,1]=2,2; [1,2] at [2,0]=1, [1,0]=2. Bottom left: [3,3] at [3,0]/[4,0]=3,3; [4,2] at [5,0]=4, [6,0]=2 (unequal). Finally [1,5] at [6,4]=1, [6,3]=5; [2,5] at [6,1]=2, [6,2]=5 (equals). All regions satisfied.
🎨 Pips Solver
Apr 26, 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 forces sixes
The three-cell equals region at [2,0], [3,0], [3,1] must hold the same pip value. Only the [6,6] domino provides a pair of identical numbers, and you need a third 6 from another domino. The [1,6] domino is the only other source of a 6, so these two must fill the region.
2
Step 2: Anchor the left column
Place [6,6] vertically to cover [2,0] and [3,0], giving both cells 6. The remaining cell [3,1] gets its 6 from [1,6], which forces the domino to sit horizontally with 6 at [3,1] and 1 at [3,2].
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Step 3: Sum-2 and the bottom-right equals
The equals region [3,2]/[3,3] now forces a 1 at [3,3]. The sum-2 region [2,2]/[2,3] must sum to 2; 1+1 is impossible, so it needs 2+0. Only [1,0] can provide the 0 and the other 1. Place [1,0] vertically with 0 at [2,3] and 1 at [3,3], satisfying both constraints.
4
Step 4: Final sum-9 falls into place
The last region is sum-9 at [1,1]/[1,2]. Remaining dominoes are [5,5] and [2,4]. [5,5] must put a 5 at [1,1] and a 5 at the empty cell [0,1]. [2,4] then places its 2 at [2,2] (now vacated by the earlier deduction) and 4 at [1,2], giving 5+4=9.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Lonely six locks a domino
The single-cell sum-6 region at [1,1] forces that cell to be exactly 6. The only domino carrying a 6 is [6,4]. Place it so that [1,1]=6 and [1,0]=4. This instantly puts a 4 into the sum-6 region [0,0]/[1,0].
2
Step 2: Complete the top-left sum-6
With [1,0]=4, the partner cell [0,0] in the sum-6 region must be 2. The domino [0,2] supplies 0 and 2; place it as [0,0]=2 and [0,1]=0. The empty cell [0,1] happily accepts the 0.
3
Step 3: Double five for sum-10
The top sum-10 region at [0,2]/[0,3] needs two numbers adding to 10. Since the 6 is used, only 5+5 works. Available 5-dominoes are [5,1], [5,2], and [5,3]. Place [5,1] at [0,3]=5, [1,3]=1 and [5,2] at [0,2]=5, [1,2]=2. This serendipitously satisfies the adjacent sum-3 region [1,2]/[1,3] with 2+1.
4
Step 4: Row 2 sum-6 pairings
The region [2,0]/[2,1] must sum to 6. Remaining pip values include 3,3,1,4,0. The only way to make 6 is 3+3. Use [1,3] to give 3 at [2,0] (and 1 at [3,0]), and [5,3] to give 3 at [2,1] (and 5 at [2,2]). Next, [2,2]/[2,3] also sums to 6; [2,2] already has 5, so [2,3] needs a 1. Place [1,0] with 1 at [2,3] and 0 at [3,3].
5
Step 5: Bottom row sum-6 cleanup
The region [3,0]/[3,1]/[3,2] must total 6. [3,0] is already 1. The remaining cells need to sum to 5. The [1,4] domino provides 1 and 4; place it as [3,1]=1 and [3,2]=4, giving 1+1+4=6. All constraints are now satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The six-cell equals dictates sixes
The enormous equals region covering [1,2], [1,3], [2,2], [2,3], [3,2], [4,2] forces all six cells to hold the same pip. Among all dominoes, only 6 appears frequently enough (via [6,6], [6,1], [6,0], [6,3], [2,6]) to supply six cells with copies, so the value must be 6.
2
Step 2: Sum-4 triggers the six cascade
The sum-4 region at [0,3]/[0,4] needs 1+3 or 2+2. Since 2s are reserved for the top-left equals, 1+3 is the key. The [6,1] domino can place its 6 into the equals region at [1,3] and its 1 at [0,3]. Place it as [1,3]=6, [0,3]=1. This seeds the entire 6 network.
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Step 3: Pillars of the equals region
With the 6 commitment, deploy the remaining 6-bearing dominoes into the equals area. [6,6] goes vertically at [2,2]/[3,2]. [2,6] places 6 at [1,2] and 2 at [0,2] (linking to the top-left equals). [6,0] sets 6 at [2,3] and 0 at [2,4], starting the zero-equals region.
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Step 4: The zero-equals line
The equals region in column 4 (rows 1–4) must all be 0. Already [2,4]=0 from [6,0]. Next, [0,3] gives 0 at [1,4] (and 3 at [0,4], completing sum-4). [4,0] puts 0 at [3,4] (with 4 at [3,3] for sum-7). Finally, [1,0] places 0 at [4,4] (and 1 at [5,4] for the bottom-right equals).
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Step 5: Top-left two equals
The top-left equals region ([0,0],[0,1],[0,2],[1,0]) requires all cells to have the same value. From [2,6] we have [0,2]=2, so the value is 2. Place [2,2] at [0,0]/[0,1]=2,2. Then [1,0] needs a 2: use [1,2] as [2,0]=1, [1,0]=2. Now the less-4 region [2,0] gets a valid 1.
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Step 6: Remaining gaps and final checks
Fill the bottom-left empty and unequal: [3,3] places at [3,0]=3, [4,0]=3 (both empty). [4,2] goes to [5,0]=4, [6,0]=2, satisfying the unequal constraint. The sum-7 region [3,3]/[4,3] already has 4 at [3,3] and 3 at [4,3] from [6,3] (which also gave 6 at [4,2]). Finally, [1,5] sets [6,4]=1, [6,3]=5 and [2,5] sets [6,1]=2, [6,2]=5 to satisfy the bottom equals pairs. The board is complete.
💡 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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