🚨 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
Ian Livengood's easy is a five-tile board with two self-contained footholds and a centre region that can only be closed by subtraction. The teal sum-4 pair in the bottom-right is a single-tile fit; the bottom-left corner links the orange less-3 cell to the purple sum-9 vertical pair through one shared domino, so the low bound there propagates straight into the sum. The four-cell pink sum-9 block is the terminal node: it waits until both anchors have spent their pips, then resolves from what remains, with the uncoloured square travelling along as a free passenger.
The medium board shifts the weight from sums to equality. Livengood builds a less-3 run of three across the top, an equals triple in the bottom-left, an equals pair on the right edge, two sum-9 shapes and a lone greater-5 corner cell, so the deduction order runs from the extremes inward: the threshold regions collapse on the first pass, the doubles commit — 0/0 to the top run, 6/6 to the equals triple, 4/4 to the equals pair — and the two sum-9 regions absorb whatever pip values are left over.
Rodolfo Kurchan's hard is the structural centrepiece of today's NYT Pips set: a nine-row board, twenty regions, three separate sum-0 singletons, two equals shapes of different lengths, five greater-than thresholds and a scatter of uncoloured single squares along the bottom. The deduction graph fans out from the zeros — the tray can only supply three of them — then narrows through the purple sum-3 block in the upper right, which admits exactly one filling, and finally runs down the ordered equals column on the right edge before closing along the bottom row.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Start with the sums, not the single cells
Two of the five regions are two-cell sums, and a two-cell sum is the tightest thing on a small board — it is either satisfied by exactly one tile in the tray or by none at all. Before you place anything, lay the five tiles out beside the grid and ask which single domino could fill each small sum whole.
💡 The teal pair is your first lock
Look at the teal sum-4 pair sitting in the bottom-right corner. Because the pair is exactly one domino's worth of cells, the tile covering it must have pips adding to four — and only one tile in the tray does. With that in place, drop your eye along the bottom edge to the orange less-3 cell: it is a single square whose only neighbour is the lower half of the purple sum-9 vertical pair, so whatever low pip lands in it is precisely what forces that pair's high half.
💡 Complete answer — the easy layout
The 2/2 double lies flat across the bottom-right teal sum-4 pair. The 0/6 tile lies flat along the bottom-left, 0 on the left in the orange less-3 cell and 6 on the right in the lower half of the purple sum-9 vertical pair — it is the only tile in the tray with one pip below three and one well above it. That leaves the purple pair three short of nine at the top, so the 3/6 tile lies flat across the middle row, 3 on the left in the upper purple cell and 6 on the right in the bottom-left corner of the pink sum-9 block. The 2/1 tile lies flat at the top, 2 on the left in the uncoloured square and 1 on the right in the top-centre cell of the pink block, and the 1/1 double lies flat across the last two pink cells, bottom-centre and bottom-right of the block. Pink reads six plus three ones, which is nine.
💡 Hunt thresholds and equalities first
This board is not won by sums. Two regions are governed by a bound and three insist on identical pips, so the useful first move is to scan the tray for doubles. Whenever an equals shape sits on the board, it almost always wants a double plus one borrowed partner of the same value — count how many tiles can supply that borrowed pip before you commit.
💡 The top run sets the tone
The purple less-3 run is a three-cell horizontal strip along the very top and it must stay tiny. The only way the tray can stack three minimum-value pips on three adjacent cells is to spend both zeros here: the 0/0 double lies flat across the middle and right cells, while the 0/3 tile stands vertically at the run's left end so its 0 fills the third cell. Because that same 0/3 tile keeps its other half pointing down, the 3 drops into the blue sum-9 block below — and the pink equals triple in the bottom-left is immediately forced to be built from sixes.
💡 Complete answer — the medium layout
The top run is all zeros: 0/3 stands vertically at its left end with 0 in the left cell and 3 dropping into the blue sum-9 block, and the 0/0 double lies flat across the middle and right cells. That commits the bottom-left equals triple to sixes — the 6/6 double lies flat across the block's bottom-left and bottom-right cells, and the 1/6 tile stands vertically so its 6 fills the block's top cell while its 1 rises into the blue sum-9 block. In the bottom-right corner the green greater-5 cell can only hold a 6, and it takes the 6 from the 5/6 tile, which lies flat with 5 on the left in the lower half of the orange sum-9 pair; that pair then needs 4 above, supplied by the 5/4 tile lying flat across the middle row with 5 on the left in the blue block and 4 on the right in the upper orange cell. The 4/4 double stands vertically in the teal equals pair on the right edge. Blue reads one plus five plus three — nine; orange reads four plus five — nine.
💡 Chase the sum regions, especially the single-cell ones
Before anything else, catalogue every sum region on this board. Three of them are single cells, and a single-cell sum has no arithmetic to do — one pip either satisfies it completely or the puzzle breaks. Scan the full tray and count exactly how many tiles contain the value those three cells demand; if the count is small, the whole opening falls out of that number.
💡 Three cells, three zero-bearing tiles
The three single-cell sums to spot are the orange sum-0 in the top-right corner, the green sum-0 near the centre of the board, and the teal sum-0 down in the bottom-left. Each needs a zero, and a sweep of all sixteen tiles shows only three of them contain a zero at all — so those three tiles are spoken for immediately, one per cell, and nothing else on the board can be built around a zero.
💡 Which zero-tile goes where
The orange sum-0 corner has only one neighbour with a constraint: the teal less-4 cell to its left. Whatever half the covering tile carries into that neighbour must stay under four, so it is the 3 from the 0/3 tile — the tile lies flat on the top row, 3 left in the teal cell, 0 right in the corner. The green sum-0 cell in the centre touches the pink greater-8 pair above and the teal less-5 cell below; a 4 fits comfortably under five while a 5 fits neither, so the 4/0 tile stands vertically with 0 up and 4 down into the teal less-5 cell. That leaves the 0/5 tile for the bottom-left: it stands vertically, 0 down in the teal sum-0 cell and 5 up in the uncoloured square above it.
💡 The sum-3 block can only be three ones
With every zero now spent, the purple sum-3 block in the upper right has to reach exactly three from three ordinary pips — the only shape that does it is 1+1+1. That needs the 1/1 double and the 1 half of the 1/5 tile. The double stands vertically, 1 up in the block's top-left cell and 1 down in its bottom cell; the 1/5 tile lies flat, 1 left in the block's third cell and 5 right in the upper cell of the top-right pink greater-7 pair. That 5 alone already drives the next two placements: the pair must beat seven, so its lower cell takes the 3 from the 2/3 tile, whose 2 drops into the right cell of the blue sum-6 pair below; blue then needs 4 in its left cell, and the 2/4 tile lies flat with 2 left at the top of the orange equals column and 4 right in the blue pair.
💡 Complete answer — the hard layout
Top row: 3/4 lies flat in the purple greater-6 pair at the top-left, 3 left and 4 right; 5/6 stands vertically in the pink greater-8 pair, 5 up and 6 down; 0/3 lies flat at the top-right, 3 left in the teal less-4 cell and 0 right in the orange sum-0 corner. Upper right: the purple sum-3 block is three 1s — the 1/1 double stands vertically with 1 up in its top-left cell and 1 down in its bottom cell, and the 1/5 tile lies flat with 1 left in the block's third cell and 5 right in the upper cell of the top-right pink greater-7 pair; under that 5 the 2/3 tile stands vertically, 3 up in the pair's lower cell and 2 down in the right cell of the blue sum-6 pair. Right edge: the 2/4 tile lies flat, 2 left at the top of the orange equals column and 4 right in the blue sum-6 pair, and the 2/2 double stands vertically below it with 2 in the middle cell and 2 in the lower cell. Bottom-right: the 3/5 tile stands vertically in the bottom-right pink greater-7 pair, 3 up and 5 down. Centre: the 4/0 tile stands vertically, 0 up in the green sum-0 cell and 4 down in the teal less-5 cell. Bottom-left: the 0/5 tile stands vertically, 5 up in the uncoloured square and 0 down in the teal sum-0 cell; the 5/5 double stands vertically in the green equals pair; the 1/6 tile lies flat on the bottom row, 6 left in the orange sum-6 cell and 1 right in the uncoloured square. Left side: the 1/2 tile stands vertically in the blue less-4 pair, 1 up and 2 down; the 3/1 tile stands vertically in the purple less-5 pair at the bottom, 1 up and 3 down; and the 4/1 tile stands vertically near the bottom-centre, 1 up in the uncoloured square and 4 down in the blue greater-1 cell.
🎨 Pips Solver
Nov 12, 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
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Step 1: One tile fits the teal pair
The teal sum-4 pair in the bottom-right corner is exactly two cells wide, which means a single domino has to satisfy it alone. Run through the tray looking for a tile whose pips add to four and there is only one candidate: the 2/2 double. Lay it flat across the pair.
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Step 2: The orange cell dictates the bottom-left
Now look at the orange less-3 cell on the bottom-left edge. It is a single square whose only constrained neighbour is the lower half of the purple sum-9 vertical pair, so the tile covering it must carry one pip under three into the orange cell and a much larger pip into the purple pair. Scanning what is left, only the 0/6 tile offers a low pip and a high pip on the same domino — place it flat, 0 on the left in the orange cell and 6 on the right in the lower purple cell.
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Step 3: Purple only needs three more
The purple sum-9 vertical pair now holds a 6 at its base, so its upper cell needs exactly three. The 3/6 tile is the only remaining source of a 3, and its 6 has to go somewhere adjacent — the natural home is the bottom-left corner of the pink sum-9 block immediately to its right. Lay the tile flat across the middle row: 3 on the left in the upper purple cell, 6 on the right in the pink block.
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Step 4: Pink closes on the leftovers
The pink block is a four-cell sum-9 and it already carries a 6, so its three remaining cells must total three — three 1s, and nothing else fits. The 1/1 double lies flat across the two lower cells, bottom-centre and bottom-right, and the 2/1 tile lies flat along the top so its 1 drops into the block's top-centre cell while its 2 sits in the uncoloured square to the left. Every tray tile is now used and all five constraints check out.
🔧 Step-by-Step Answer Walkthrough For Medium Level
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Step 1: The greater-5 corner pins a six
The green greater-5 cell in the bottom-right corner is a single square, so it needs the biggest pip available — a 6. Which 6, though? The 6/6 double is needed intact for the bottom-left equals triple, and the 1/6 tile would strand its 1 beside the orange sum-9 pair, which needs a partner that can reach nine. That leaves the 5/6 tile: it lies flat, 5 on the left in the lower half of the orange pair and 6 on the right in the green cell.
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Step 2: Orange asks for four
The orange sum-9 vertical pair at the bottom now has a 5 in its lower cell, so the cell above it must be 4. The 5/4 tile supplies exactly that — lie it flat across the middle row with 5 on the left in the blue sum-9 block and 4 on the right in the upper orange cell.
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Step 3: The top run is all zeros
The purple less-3 run of three along the very top has to stay as small as the tray allows. The only tiles that can put repeated tiny pips on adjacent cells are the 0/0 double and the 0 half of the 0/3 tile, so those two fill the run. The 0/3 tile stands vertically at the left end with its 0 in the run's left cell and its 3 dropping into the blue sum-9 block below; the 0/0 double lies flat across the run's middle and right cells.
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Step 4: Pink becomes the sixes
With the zeros spent at the top, the bottom-left equals triple has no choice but to be three sixes. The 6/6 double lies flat across the block's bottom-left and bottom-right cells, and the 1/6 tile stands vertically so its 6 fills the block's top cell while its 1 rises into the blue sum-9 block — the 1 that blue was waiting for.
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Step 5: The last pair is the 4/4
Blue already holds a 1 and a 5, so its third cell takes the 3 that the 0/3 tile dropped in — nine exactly. Orange holds 4 and 5 — nine exactly. The only region still bare is the teal equals vertical pair on the right edge, and the only double still unused is 4/4, so it stands vertically, 4 up and 4 down. All seven tiles placed, all six constraints satisfied.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Three zeros, three zero-bearing tiles
Three separate regions on this board sum to zero, and a single pip of zero is the only thing that satisfies them: the orange sum-0 in the top-right corner, the green sum-0 near the centre, and the teal sum-0 down in the bottom-left. Sweep all sixteen tiles and exactly three contain a zero — the 0/3, the 4/0 and the 0/5. Each sum-0 cell claims one of those three tiles, and every other placement on the board has to work around that.
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Step 2: The top-right corner settles the first zero
The orange sum-0 corner cell has a single constrained neighbour, the teal less-4 cell immediately to its left. The tile covering the corner must push its other half into that neighbour, and that half has to stay under four. Only the 3 of the 0/3 tile qualifies, so the tile lies flat on the top row: 3 on the left in the teal less-4 cell, 0 on the right in the orange corner.
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Step 3: The centre and the bottom-left zeros
The green sum-0 cell near the centre touches the pink greater-8 pair above it and the teal less-5 cell below it. A 4 slides comfortably under five; the 5 from the other zero-tile would satisfy neither neighbour. So the 4/0 tile stands vertically, 0 up in the green cell and 4 down in the teal less-5 cell. The 0/5 tile is therefore left for the bottom-left teal sum-0 cell, and it stands vertically too — 0 down in the sum-0 cell, 5 up in the uncoloured square above it.
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Step 4: The sum-3 block can only be three ones
The purple sum-3 block in the upper right must reach exactly three across three cells, and every zero in the tray is now committed elsewhere. The only shape that lands on three is 1+1+1, so the block takes the 1/1 double and the 1 half of the 1/5 tile. The double stands vertically, 1 up in the block's top-left cell and 1 down in its bottom cell; the 1/5 tile lies flat, 1 left filling the block's third cell and 5 right in the upper cell of the top-right pink greater-7 pair.
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Step 5: The 2/3 tile and the equals column
That 5 in the top-right pink greater-7 pair already leaves the pair needing to beat seven, so its lower cell takes the 3 from the 2/3 tile, which stands vertically and drops its 2 into the right cell of the blue sum-6 pair below. Blue now needs 4 in its left cell, and the 2/4 tile supplies it while lying flat — 2 on the left at the top of the orange equals column, 4 on the right in the blue pair. The equals column repeats that 2 down the right edge, so the 2/2 double stands vertically beneath it, filling the middle and lower cells.
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Step 6: The bottom band and the remaining pairs
The green equals pair in the bottom-left must be a double, and the only one left is 5/5, standing vertically, 5 up and 5 down. The blue less-4 pair takes the 1/2 tile vertically (1 up, 2 down, totalling three), the purple less-5 pair takes the 3/1 tile vertically (1 up, 3 down, totalling four), and the top-left purple greater-6 pair takes the 3/4 tile lying flat with 3 left and 4 right. The pink greater-8 pair at the top grabs the highest remaining tile, 5/6, standing vertically with 5 up and 6 down, and the bottom-right pink greater-7 pair takes 3/5, standing 3 up and 5 down. The orange sum-6 single on the bottom row can only be the 6 half of the 1/6 tile, lying flat with 1 to its right in the uncoloured square; and the blue greater-1 cell at the bottom takes the 4 from the 4/1 tile, standing vertically with the 1 above it in its uncoloured square.
💡 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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