🚨 SPOILER WARNING
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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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Easy is today's NYT Pips warm-up, and Ian Livengood keeps it deliberately small — a two-row strip with only five tiles in the tray. Two of the six regions are equals blocks, which do most of the thinking for you, and a single greater-than cell tucked into the bottom-right corner is effectively a gift. Expect clean, forced deductions with no branching; most solvers finish inside two minutes. Use it to learn the board's logic before the day's real work starts.
Medium is Rodolfo Kurchan's difficulty spike, and the bottleneck sits on the left edge: a three-cell column with a very high sum target that only one combination in the tray can satisfy. Spot that and the board collapses quickly — the orange equals block and the two little 'less than' cells along the bottom then fill in almost without thought. Miss it and you will churn through the tray twice. Budget five minutes, and start on the left.
Hard is Kurchan again, and it is a different species: thirty-two cells, every single region a one-cell sum, and all four corners clipped from the 6×6 grid. There are no equals blocks and no arithmetic to hide behind, so the opening comes from the extremes — the cells demanding the smallest and largest totals, and the one unconstrained square that no target touches. Expect ten to fifteen minutes: the first half is a counting exercise, the second half resolves in a rush once the low tiles are exhausted.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Start with the equals regions
Two regions today are marked 'equals' — every cell inside them carries the same pip value, and one of those regions is three cells big. Repeated values are the strongest anchors on a board this small, so hunt the equals regions down before you touch the tray.
💡 The greater-5 corner
Down in the bottom-right corner sits a blue one-cell region marked greater than 5. Only a single pip in the 0–6 range can satisfy it, so the tile reaching into that corner is decided for you — and it is joined by one domino to the orange equals pair in the row above, not sharing its value but handing it the next clue.
💡 Full easy solution
The 4/4 double stands vertically in the teal equals block down the centre: 4 in the block's top-right cell and 4 in its bottom-right cell, with the third 4 arriving on the 2/4 tile. The 6/5 tile stands vertically on the right edge — 6 in the blue greater-5 corner cell at the bottom, 5 up in the right cell of the orange equals pair; the 5/2 tile stands vertically immediately to its left, 5 up in the orange pair's left cell and 2 down in the uncoloured free square. On the left the 3/1 tile lies flat along the top row, 1 on the left in the upper purple sum-3 cell and 3 on the right in the pink sum-3 cell. The 2/4 tile lies flat along the bottom row, 2 on the left in the lower purple sum-3 cell (1+2=3) and 4 on the right in the teal block's last open cell.
💡 Hunt the tightest arithmetic
Three regions do the work here: a tall column that must hit a big total, a three-cell equals block, and two one-cell 'less than' regions that can only hold tiny pips. Start with whichever feels most restricted — one of them has essentially a single solution in the whole tray.
💡 The tall pink column on the left
The pink column down the left edge wants three cells to add up to 15. Pips cap at 6, so there is exactly one way to get there, and the tray's 5/5 double covers two of those three cells in a single move. That placement also tells you which tile must sit in the top-left purple greater-7 pair, because only one other tile in the tray carries the missing pip.
💡 Full medium solution
Begin at the top-left: the 5/6 tile stands vertically, 6 up in the left cell of the purple greater-7 pair and 5 down in the top cell of the pink sum-15 column. The 5/5 double then stands vertically over the two cells beneath it, finishing that column. The 2/2 double stands vertically between the purple pair's right cell — 2 up, 6+2 clearing the greater-7 bar — and the teal greater-1 cell, 2 down. The 4/4 double lies flat across the top of the orange equals block, 4 left and 4 right, and the 0/4 tile lies flat one row below it, 0 in the upper cell of the blue sum-4 pair and 4 in the block's third cell. The blue pair's lower cell takes the 4 of the 4/2 tile, with 2 dropping into the green less-3 cell below; the final 1/4 tile stands vertically at the bottom right, 4 up in the uncoloured free square and 1 down in the purple less-2 cell.
💡 One constraint type only
Every region on this board is a single cell carrying a sum target — not one equals block, not one inequality cell. That tells you the whole puzzle is about exact values rather than relationships between neighbours, so the extreme targets are where the logic lives.
💡 The one cell with no target
Look at the uncoloured square on the right edge, fourth row down — the only region with no constraint at all. Every other cell is capped by its own printed total, and the largest total anywhere on this grid is five, so that free square is the only place a six can possibly live. The 3/6 tile lies flat there: 6 in the free square, 3 to its left in the pink sum-3 cell.
💡 Count the zeros before anything else
Five regions want a total of zero: the purple sum-0 at the top-left, the orange sum-0 at the top-right, the orange sum-0 on the right edge, the blue sum-0 on the left edge, and the orange sum-0 down at the bottom-left. The tray holds exactly five zero-carrying tiles — 0/1, 0/2, 0/3, 0/4 and 0/5 — so all five are locked into those cells, each leaving its partner pip in a neighbouring cell.
💡 The zero that picks itself
Take the orange sum-0 on the right edge first: both of its open neighbours read 1, so only the 0/1 tile fits, standing vertically with 0 down in the zero cell and 1 up in the orange sum-1. The top-right orange zero can then only pair with a 1 or a 3, and the 0/1 is committed — so 0/3 goes in the corner, 3 dropping into the teal sum-3 below. The top-left purple zero takes 0/4, lying flat with 0 on the left and 4 on the right in the pink sum-4. That leaves 0/2 and 0/5 for the bottom: the left-edge blue zero can only touch a 3, 4 or 5, so it must be 0/5 with 5 dropping into the orange sum-5, and 0/2 stands at the bottom-left with 2 rising into the blue sum-2.
💡 Full hard solution
The 3/6 lies flat with 6 in the uncoloured free square on the right edge, fourth row down, and 3 in the pink sum-3 cell to its left. The zeros: 0/1 stands vertically with 0 down in the orange sum-0 on the right edge and 1 up in the orange sum-1; 0/3 stands vertically at the top-right corner, 0 up in the orange sum-0 and 3 down in the teal sum-3; 0/4 lies flat along the top, 0 left in the purple sum-0 and 4 right in the pink sum-4; 0/2 stands vertically at the bottom-left, 0 down in the orange sum-0 and 2 up in the blue sum-2; 0/5 stands vertically on the left edge, 0 up in the blue sum-0 and 5 down in the orange sum-5. The rest: 1/2 stands vertically in the third column, 1 up in the teal sum-1 on the top row and 2 down in the pink sum-2; 2/3 stands vertically in the left column, 2 up in the blue sum-2 and 3 down in the blue sum-3; 1/3 lies flat on the second row, 1 left in the green sum-1 and 3 right in the purple sum-3; 3/4 stands vertically below that, 3 up in the green sum-3 and 4 down in the green sum-4; 2/4 stands vertically down the centre, 2 up in the purple sum-2 and 4 down in the purple sum-4; 1/4 lies flat with 4 left in the pink sum-4 near the centre and 1 right in the teal sum-1 on the third row; 1/5 stands vertically on the right edge, 1 up in the teal sum-1 at the far right of the fourth row and 5 down in the teal sum-5; 3/5 stands vertically just left of it, 5 up in the pink sum-5 and 3 down in the purple sum-3; 4/5 lies flat one row above the bottom, 5 left in the green sum-5 and 4 right in the purple sum-4; 2/5 lies flat along the bottom row, 5 left in the blue sum-5 and 2 right in the green sum-2.
🎨 Pips Solver
Oct 23, 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
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Step 1: The corner that only takes a six
The blue one-cell region at the bottom-right is marked greater than 5. Pips run from 0 to 6, so exactly one value qualifies, and of the five tiles in the tray only the 6/5 carries it. That tile must stand vertically in the right-hand column with the 6 sitting in the blue cell.
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Step 2: The equals pair fills itself
The orange equals pair runs along the top-right and shares a domino with the blue cell, so the 5 from the 6/5 already occupies its right-hand cell. Both orange cells must match, so the left one is a 5 as well — and the only other 5 in the tray belongs to the 5/2 tile. Lay it vertically one column to the left, its 2 dropping into the uncoloured free square below.
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Step 3: Three equal cells in the centre
The teal block is the only three-cell equals region, so three identical pips have to come out of just two tiles. The 4/4 double supplies two of them: stand it vertically in the block's right-hand column, 4 on top and 4 on the bottom, and the third 4 rides on the remaining tile that carries a 4 — the 2/4.
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Step 4: The left column closes out
Two tiles are left and the purple sum-3 pair still needs its total. The 2/4's 4 is already claimed by the teal block, so its 2 drops into the lower purple cell; that leaves 1 for the upper purple cell, since 1+2=3. The 3/1 tile provides it, lying flat along the top row — 1 on the left in purple, 3 on the right in the pink single cell, which is precisely the 3 that cell wants.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Three cells, one total
The pink column on the left edge sums to 15 across three cells, and no pip can exceed 6. The only combination that fits is the same value three times over, so every cell in that column is identical — which commits both of the tray's pips of that size to the column at once.
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Step 2: The double goes vertical
Stand the 5/5 double vertically over the middle and lower cells of the pink column. The third matching pip has to come from the 5/6 tile, and the only cell adjacent to that column is the left cell of the purple greater-7 pair in the top-left corner — so the 6 lands up there and the 5 drops into the column's top cell.
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Step 3: Clearing the greater-7 bar
With 6 already in the purple pair, the right-hand cell only needs to push the pair's total past 7. The 2/2 double does it tidily: 2 up into the purple pair for a total of 8, and 2 down into the teal greater-1 cell directly below, which only has to beat 1.
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Step 4: The orange equals block
Three orange cells must match. The 4/4 double lies flat across the block's top row — 4 left, 4 right — and the third 4 comes from the 0/4 tile lying flat one row beneath, its 0 sliding into the upper cell of the blue sum-4 pair.
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Step 5: The last three tiles
The blue pair's lower cell still needs a 4, and the 4/2 tile supplies it while dropping its 2 into the green less-3 cell below, comfortably under that cap. The final 1/4 tile stands vertically at the bottom right: 4 up in the uncoloured free square, 1 down in the purple less-2 cell — the only pip under 2 still left in the tray.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Only one home for the six
Every region on this grid is a single cell with a sum target, and the largest target anywhere is 5. Pips go up to 6, so the 6 can only sit in the one unconstrained square, on the right edge of the fourth row. Its neighbours read 3, 1, 1 and 5, and only the pink sum-3 cell to its left matches the 3 on the 3/6 tile — so that tile lies flat, 6 in the free square and 3 in the pink cell.
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Step 2: Count the zeros
Five cells want a total of zero: the purple sum-0 at the top-left, the orange sum-0 at the top-right, the orange sum-0 on the right edge, the blue sum-0 on the left edge, and the orange sum-0 down at the bottom-left. The tray has exactly five zero-carrying tiles — 0/1, 0/2, 0/3, 0/4 and 0/5 — so every one of them is committed, and each drops its partner into an adjacent cell.
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Step 3: The right-edge zero picks the 0/1
The orange sum-0 on the right edge has just two open neighbours and both read 1, so only the 0/1 tile fits: it stands vertically, 0 down in the zero cell and 1 up in the orange sum-1 above it. That kills the top-right orange zero's 1 option — its neighbours read 1 and 3 — so 0/3 goes into the corner, 0 up and 3 down into the teal sum-3. The top-left purple zero is then left with only the pink sum-4 on its right, so 0/4 lies flat, 0 on the left and 4 on the right.
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Step 4: The last two zeros split cleanly
The bottom-left orange zero sits between a 2 and a 5; the left-edge blue zero sits between a 3, a 4 and a 5. With 0/3 and 0/4 already spent, the blue zero has only one option — 0/5 stands vertically, 0 up in the blue sum-0 and 5 down in the orange sum-5. That leaves 0/2 for the bottom-left corner: 0 down in the orange zero, 2 up in the blue sum-2.
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Step 5: The top of the board chains
The teal sum-1 on the top row still needs a 1, and its only open side now points down into the pink sum-2 — so the 1/2 stands vertically there and fills both cells in one go. That seals the two cells directly beneath, and the blue sum-2 in the left column can only pair downwards with the blue sum-3: the 2/3 stands vertically, 2 up and 3 down.
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Step 6: The rest falls in one sweep
The green sum-1 on the second row has a 3 to its right, so the 1/3 lies flat there, and the green sum-3 directly beneath it takes the 3 of the 3/4 with 4 dropping into the green sum-4. Down the centre the purple sum-2 takes the 2 of the 2/4, its 4 falling into the purple sum-4 below; the pink sum-4 near the centre takes the 4 of the 1/4, whose 1 heads right into the teal sum-1. On the right, the 1/5 stands vertically — 1 up in the teal sum-1 at the far right of the fourth row, 5 down in the teal sum-5 — and the 3/5 stands vertically just left of it, 5 up in the pink sum-5 and 3 down in the purple sum-3. The last two lie flat: 4/5 one row above the bottom, 5 left in the green sum-5 and 4 right in the purple sum-4; 2/5 along the bottom row, 5 left in the blue sum-5 and 2 right in the green sum-2.
💡 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.
💬 Community Discussion
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