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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 is a linear chain: one sum constraint anchors the top-left corner, and every subsequent domino falls from that single entry point in sequence. No branching, no backtracking — just a clean propagation from corner to corner.
Rodolfo Kurchan steps in for medium and hard. The medium organizes four sum-7 regions around a sparse grid of dominos; a strict less-than constraint at one edge cell cuts the search space immediately, triggering a cascade through the interior. The hard expands to a 5×9 grid with 16 dominos, and Kurchan deploys a distinctive structural signature: three consecutive single-cell sum constraints in row 0 force their values without any cross-checking, giving solvers a full scaffold before the real deductions begin. The greater-than-5 constraints in row 1 lock in two more cells, and from there the puzzle resolves through a series of equal-value propagations and tight less-than boundaries.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Check the top-left corner
There's a one-cell sum region in the top-left corner. Because it contains exactly one cell, you know the pip value there without any cross-referencing. That single anchor is all you need to get started.
💡 Follow the sum region below
After placing the corner domino, look at the two-cell region directly beneath it — the target is 10. With one value now fixed from the corner, the other cell is completely determined. That second cell feeds directly into a three-cell equals region running across the middle of the grid.
💡 Full solution
The sum-3 corner forces [0,0]=3, placing domino [5,3] vertically: 5 at [1,0], 3 at [0,0]. The sum-10 region [1,0]+[1,1]=10 → [1,1]=5, so domino [0,5] goes horizontal in row 1 with 5 at [1,1] and 0 at [1,2]. The three-cell equals region [0,2],[0,3],[1,2] all equal 0: domino [2,0] places [0,1]=2, [0,2]=0; domino [0,3] places [0,3]=0, [0,4]=3. The equals pair [0,4]=[1,4]=3 places domino [3,1] at [1,3]-[1,4] with [1,4]=3 and [1,3]=1 — satisfying the less-than-2 constraint.
💡 Find the tightest constraint
One cell near the right side of the grid carries a less-than constraint that leaves very little wiggle room. Start there — that cell's domino placement immediately cascades into a two-cell sum region right beside it.
💡 The right edge starts a chain
The less-than-2 cell can only hold 0 or 1. Once you determine which, its domino partner lands in a two-cell sum-7 region — fixing the adjacent cell precisely. Continue to the lower-right: a less-than-3 cell there connects to another sum-7 region in the row above.
💡 Full solution
[1,5]<2 → [1,5]=0, domino [0,1] vertical, [2,5]=1. Sum [2,5]+[2,6]=7 → [2,6]=6, domino [6,3] horizontal, [2,7]=3. [4,6]<3 and sum [3,5]+[3,6]=7: domino [2,5] vertical gives [4,6]=2, [3,6]=5; then [3,5]=2, domino [3,2] horizontal with [3,4]=3. Sum [0,1]+[1,1]+[2,1]=7: domino [3,3] gives [2,0]=3 (<5 ✓), [2,1]=3; domino [3,1] gives [0,1]=3, [1,1]=1. Sum [1,2]+[2,2]+[3,2]=7: domino [1,1] gives [2,2]=[3,2]=1; [1,2]=5, domino [5,5] horizontal at [1,2]-[1,3].
💡 Count your free anchors
Several cells in this grid are completely determined by their constraints before you place a single domino — no crossing deductions required. Look for sum constraints on isolated cells, and greater-than constraints tight enough to admit only one value. Count how many such anchors you can identify before you start placing anything.
💡 Three in a row at the top
The first three cells of row 0 each sit inside their own single-cell sum constraint, fixing their pip values precisely. Two cells in row 1 carry greater-than constraints that leave only one possible value. Those five cells are certain — work out which dominos they require before tackling the rest.
💡 Top scaffold placements
With [0,0], [0,1], [0,2] pinned by sum constraints and [1,2],[1,3] forced to 6 by greater-than-5, the first four dominos are: [2,5] horizontal at [0,0]-[0,1]; [0,6] vertical at [0,2]-[1,2]; [1,6] vertical at [0,3]-[1,3] with 1 on top; [3,0] reversed at [0,5]-[0,4], satisfying [0,5]>2 and both cells in [0,3],[0,4]<2.
💡 Symmetric structure at the bottom
The bottom strip mirrors the top: single-cell sum constraints fix [4,5]=2, [4,6]=5, [4,7]=0. The right side of row 1 has its own sum anchors at [1,6]=2, [1,7]=5, [1,8]=0. These pin four more dominos. Then the equals and greater-than-5 constraints in rows 2–4 close off the interior.
💡 Full solution
Top: [2,5]→[0,0]-[0,1]; [0,6]→[0,2]-[1,2]; [1,6]→[0,3]-[1,3] (1 at top, 6 below); [3,0]→[0,5]-[0,4] (3 at col5, 0 at col4). Row 1 right: [0,5]→[1,8]-[1,7] (0,5). Interior: [2,6]→[2,0]-[3,0]; [2,4]→[2,1]-[2,2]; [0,0]→[2,3]-[2,4]; [1,3]→[2,5]-[3,5]; [2,1]→[1,6]-[2,6]; [1,1]→[2,8]-[3,8]. Bottom: [4,6]→[4,1]-[4,0]; [4,1]→[4,2]-[4,3]; [0,2]→[4,4]-[4,5]; [5,1]→[4,6]-[3,6]; [0,1]→[4,7]-[4,8].
🎨 Pips Solver
Apr 24, 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 locked corner
The sum-3 constraint governs the single cell at [0,0]. Because the region contains exactly one cell, the pip there must equal 3 — no arithmetic needed, just reading the constraint. The only domino that places a 3 at [0,0] while connecting to an adjacent cell is [5,3], running vertically: it places 3 at [0,0] and 5 at [1,0].
2
Step 2: Sum of 10 determines the neighbor
The region covering [1,0] and [1,1] must sum to 10. [1,0] is now fixed at 5, so [1,1] must be exactly 5 as well. Among the remaining dominos, only [0,5] carries a 5. Placed horizontally across [1,1] and [1,2], it puts 5 at [1,1] and 0 at [1,2].
3
Step 3: The equals region snaps into place
The equals region covering [0,2], [0,3], and [1,2] requires all three cells to share the same pip value. [1,2] is already 0, so both [0,2] and [0,3] must also be 0. Domino [2,0] fills [0,1]-[0,2] with 2 at [0,1] and 0 at [0,2]. Domino [0,3] fills [0,3]-[0,4]: placing 0 at [0,3] satisfies the equals constraint, leaving 3 at [0,4].
4
Step 4: Equals pair closes the grid
The two-cell equals region at [0,4] and [1,4] requires matching values. [0,4]=3 forces [1,4]=3. Domino [3,1] spans [1,4] and [1,3] horizontally — 3 at [1,4] and 1 at [1,3]. The single-cell less-than-2 constraint at [1,3] is satisfied: 1 < 2.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The right-edge less-than forces the first domino
The single-cell region at [1,5] carries a less-than-2 constraint, meaning the pip there is 0 or 1. Among the available dominos, [0,1] can sit vertically at [1,5]-[2,5] with 0 at [1,5] — satisfying the constraint. No other domino can reach [1,5] with a valid value given the grid geometry. This placement is forced: [1,5]=0, [2,5]=1.
2
Step 2: Sum-7 cascades right
The two-cell sum region at [2,5] and [2,6] must total 7. [2,5] is now 1, so [2,6] must be exactly 6. The only unused domino carrying a 6 is [6,3]. Placed horizontally at [2,6]-[2,7], it gives [2,6]=6 and [2,7]=3.
3
Step 3: Bottom-right corner resolves through two constraints
The less-than-3 constraint on [4,6] limits it to 0, 1, or 2. The two-cell region [3,5]+[3,6] must sum to 7. Domino [2,5] running vertically at [4,6]-[3,6] with [4,6]=2 satisfies the less-than-3 constraint, and places [3,6]=5. That makes [3,5]+5=7, so [3,5]=2. Domino [3,2] then sits horizontally at [3,4]-[3,5]: [3,4]=3, [3,5]=2.
4
Step 4: The column sum-7 narrows the left side
The three-cell column region at [0,1], [1,1], [2,1] must sum to 7. The less-than-5 constraint at [2,0] allows 0–4. Domino [3,3] spans [2,0]-[2,1] horizontally: [2,0]=3 (satisfies <5), [2,1]=3. The remaining sum is [0,1]+[1,1]=4. Domino [3,1] placed vertically at [0,1]-[1,1] gives 3+1=4, with [0,1]=3 and [1,1]=1.
5
Step 5: The final column sum places the double-five
The three-cell column region at [1,2], [2,2], [3,2] sums to 7. Domino [1,1] spans [2,2]-[3,2] vertically: [2,2]=1, [3,2]=1. So [1,2]+1+1=7 → [1,2]=5. Only domino [5,5] (double-five) fits at [1,2]-[1,3]: [1,2]=5, [1,3]=5.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Five anchors before the first placement
Three consecutive single-cell sum constraints open row 0: [0,0]=2, [0,1]=5, [0,2]=0 — no cross-referencing needed, each cell is pinned by its own region. Two greater-than-5 constraints in row 1 at [1,2] and [1,3] must each hold 6 (the maximum pip value). These five anchors immediately force two dominos: [2,5] sits horizontally at [0,0]-[0,1], and [0,6] sits vertically at [0,2]-[1,2] with 0 on top and 6 below.
2
Step 2: Vertical dominos complete the top cluster
[1,3]=6 requires a domino half worth 6 there. The only remaining 6-carrying domino is [1,6]; placed vertically at [0,3]-[1,3], it gives [0,3]=1 and [1,3]=6. The two-cell region [0,3],[0,4] carries a less-than-2 constraint — both cells must be 0 or 1. [0,3]=1 is fine; [0,4] must be 0 or 1. The cell [0,5] requires a value greater than 2. Domino [3,0] placed reversed at [0,5]-[0,4] gives [0,5]=3 (>2 ✓) and [0,4]=0 (<2 ✓).
3
Step 3: Bottom strip and row-1 right side mirror the top
Single-cell sum constraints fix the bottom strip: [4,5]=2, [4,6]=5, [4,7]=0. Domino [0,2] sits horizontally at [4,4]-[4,5]: [4,4]=0, [4,5]=2. Domino [0,1] sits horizontally at [4,7]-[4,8]: [4,7]=0, [4,8]=1. Domino [5,1] covers [4,6]-[3,6] vertically: [4,6]=5, [3,6]=1. On the right side of row 1, sum constraints pin [1,6]=2, [1,7]=5, [1,8]=0; domino [0,5] runs reversed at [1,8]-[1,7]: [1,8]=0, [1,7]=5.
4
Step 4: Greater-than chains resolve the left column
Greater-than-5 constraints at [3,0] and [4,0] both demand 6. Domino [2,6] spans [2,0]-[3,0] vertically: [2,0]=2, [3,0]=6 (>5 ✓). The equals region [2,0],[2,1] forces [2,1]=2. Domino [2,4] spans [2,1]-[2,2]: [2,1]=2, [2,2]=4. Domino [4,6] spans [4,1]-[4,0]: [4,1]=4, [4,0]=6 (>5 ✓). Equals region [4,1],[4,2] forces [4,2]=4. Domino [4,1] spans [4,2]-[4,3]: [4,2]=4, [4,3]=1. Both cells in [4,3],[4,4]<2: [4,3]=1 ✓, [4,4]=0 ✓ (already placed).
5
Step 5: Center rows close on less-than and equals constraints
The three-cell less-than-2 region [2,3],[2,4],[2,5] forces all values to 0 or 1. Domino [0,0] (double-zero) fills [2,3]-[2,4]: both cells = 0. [2,5] must then be 1; domino [1,3] spans [2,5]-[3,5] vertically: [2,5]=1, [3,5]=3 (<5 ✓). Sum region [2,6]+[3,6]=2: [3,6]=1 (placed), so [2,6]=1. Domino [2,1] spans [1,6]-[2,6] vertically: [1,6]=2 (sum=2 ✓), [2,6]=1. Equals region [2,8],[3,8],[4,8]: [4,8]=1 forces all three to 1; domino [1,1] spans [2,8]-[3,8]: [2,8]=1, [3,8]=1 ✓.
💡 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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