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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 builds today's NYT Pips easy as a study in mirrored pressure: six regions, five tiles, and two equals pairs set at right angles — one running vertically down the left side, one running horizontally across the top-right. The elegant part is that the pairs never touch, so each has to be completed by a domino arriving from a different direction, while a greater-than square on the top edge and a less-than square on the left edge act as the only two exits from that symmetry. It is a small board that reads like a chain reaction, every answer sitting next door to the next question.
Livengood's medium is the same designer in a more crowded mood. The board is a narrow four-wide grid, and three separate greater-than regions jostle for the same small pool of high pips while a sum-10 pair on the right quietly absorbs the best of them. Notice the deliberate placement of the two sum-3 squares — one green at the top-right corner, one purple at the bottom-left — as though the puzzle were pinned at both ends before the middle was allowed to move. There is no 6 anywhere in the tray, which narrows the whole architecture more than it first appears.
Rodolfo Kurchan's hard is a different animal: a sprawling 5×9 board, twenty-one regions, and a tray in which a striking share of the pips are zeros. Four separate sum-0 regions — three singletons and one three-cell block — all demand nothing but zero, and Kurchan has calibrated the tray to supply precisely enough of them with none to spare. That is the design in a line: the emptiest constraint on the board does all the early work, and the pink triple-equals block pressed up against the centre becomes the pressure valve once the zeros run out.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Read the constraint types first
Before touching a tile, catalogue what the board is asking for. Today you have two equals pairs (one vertical on the left, one horizontal in the top-right), two multi-cell sums at the bottom, and two lonely single cells — one that wants something greater than its target, one that wants something less. The single cells are where the chain begins: find them, then ask which half of the tray could possibly fill them.
💡 The orange square rules the pink pair
The orange greater-than-4 square on the top edge has exactly one neighbour — the left cell of the pink equals pair in the top-right corner — so it must be covered vertically, and pink inherits its partner. Because pink has to repeat that value, the partner must be something the tray can supply twice over. Look for the tile whose other half sets up that repeat, and the whole top-right folds.
💡 Full answer for the easy board
Place the 5/6 tile upright at the top: 5 in the orange greater-than-4 square, 6 in the pink pair's left cell. The 6/1 tile stands upright beside it, 6 in the pink pair's right cell and 1 dropping into the teal sum-5 pair's upper cell. The 5/4 tile lies flat along the bottom, 5 in the green sum-8 pair's right-hand cell and 4 in the teal pair's lower cell. The 1/3 tile stands vertically, 3 in green's left cell and 1 above it in the purple pair's lower cell. Last, the 0/1 tile lies flat on the left edge: 1 in the purple pair's upper cell, 0 in the blue less-than-3 square.
💡 Two kinds of constraint to chase
This grid leans on single-cell sums and on greater-than regions. There are two sum-3 squares (green at the top-right corner, purple at the bottom-left) and three regions that must beat a target rather than hit one. Start where a region has the fewest possible neighbours — a corner square has only one way out, and that is where the deduction lives.
💡 The green 3 corner opens everything
That green sum-3 square at the top-right can only be covered from the cell directly beneath it, so a 3 sits on top and its partner drops straight into the blue greater-than-6 pair. Now look at the purple sum-10 pair on the right: nothing in today's tray is worth more than 5, so ten can only be assembled one way. Work out what that forces and you have two of the four row-three cells pinned.
💡 Full answer for the medium board
The 3/4 tile stands upright at the top-right: 3 in the green sum-3 square, 4 below in the blue greater-than-6 pair. That pair's lower cell takes the 3 from the 3/5 tile, whose 5 drops to the right-hand cell of the purple sum-10 pair; the pair's left-hand cell takes 5 from the 5/1 tile, whose 1 drops to the right end of the pink greater-than-7 run. On the left, the teal greater-than-7 pair reads 5 then 3 — 5 from the 5/4 tile with 4 dropping into the pink run's left end, 3 from the 1/3 tile with 1 rising into the orange less-than-3 block. The orange block reads 1, 1, 0, with the 1/0 tile upright giving its top-right cell a 1 and its bottom-right a 0. The pink run finishes 4, 3, 1 for 8, its middle 3 coming from the 3/3 double whose other half fills the purple sum-3 square at the very bottom.
💡 One constraint type dominates
Three kinds of region share this board — equals blocks, sum cells, and a handful of greater/less signposts — but the one that decides the puzzle is the humblest: sum 0. Count how many cells across the grid are told to total nothing, then count the zeros in your tray. Those two numbers are effectively the entire solution.
💡 Two zero corners do the early talking
The green sum-0 square at the far top-right has a single neighbour, the blue sum-5 cell to its left, so the tile covering it resolves the instant you look at it — and it satisfies the blue cell exactly, leaving no room to argue. Down at the bottom, the other green sum-0 square sits flush against the bottom-left cell of the teal sum-0 block, so those two zeros lean on each other in the same row.
💡 Spend the zero budget
Six cells are told to total zero and the tray holds exactly six zeros across five tiles: the 0/0 double plus four single-zero tiles. Every zero is therefore committed to a sum-0 region before anything else moves. The 0/5 tile lies flat in the top-right corner, 0 on the right in green and 5 on the left in blue; the 0/0 double lies flat at the bottom, covering the green sum-0 square and the teal block's bottom-left cell.
💡 Reaching the block's far corners
Two cells of the teal block still lack a zero. Its top-right cell can only be reached from above, from the teal sum-4 pair, so the 0/1 tile stands upright there — 0 in the block, 1 rising into the sum-4 pair. Its bottom-right cell can only be reached from its right, from the purple sum-4 pair, so the 0/2 tile lies flat, 0 in the block and 2 heading right into purple. That fixes both remaining cells of the purple sum-4 pair at 2.
💡 Follow the knock-ons down the right edge
Teal sum-4 now holds a 1 and needs a 3, which can only arrive from the orange sum-4 square above — the 3/4 tile, 3 below in teal and 4 above in orange. Down at the bottom-right the 1/2 tile puts the second 2 into the purple sum-4 pair and drops a 1 into the last cell of the orange less-than-4 column. That column's top two cells are left holding nothing but the 1/1 double.
💡 Full answer for the hard board
Top row, left to right: the 2/5 tile stands upright with 5 in the purple sum-5 corner and 2 below in the purple equals pair; the 1/3 tile puts 3 above the pink equals pair (which reads 1 and 1); the 1/4 tile puts 4 in the pink sum-4 square; the 2/2 double lies flat in the teal equals pair; the 3/4 tile puts 4 in the orange sum-4; the 0/5 tile lies flat, 5 in the blue sum-5 and 0 in the green sum-0 corner. Down the right edge the 1/1 double fills the top two cells of the orange less-than-4 column, and the 1/2 tile puts 1 in its third cell with 2 below in the purple sum-4 pair's right cell. Bottom row: the 0/0 double covers the green sum-0 square and the teal block's bottom-left cell; the 0/2 tile puts 0 in the block's bottom-right cell and 2 in the purple sum-4 pair's left cell; the 4/5 tile lies flat with 4 in the bottom-left orange sum-4 and 5 in the bottom blue sum-5. In the middle, the 0/1 tile stands upright, 0 in the teal block's top-right cell and 1 in the teal sum-4 pair, whose upper cell takes 3 from the same 3/4 tile; the 1/5 tile stands upright with 5 in the green sum-5 square and 1 in the uncoloured free square; the 0/3 tile lies flat, 0 in the purple sum-0 centre cell and 3 in the pink equals block, which completes 3-3-3 with the 3/3 double; and the 2/4 tile lies flat with 2 in the purple equals pair's lower cell and 4 in the blue sum-4 square.
🎨 Pips Solver
Nov 10, 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
1
Step 1: The extreme tile goes up top
The orange greater-than-4 square on the top edge accepts only a 5 or a 6, and it has exactly one neighbour: the left cell of the pink equals pair in the top-right corner. So it must be covered vertically, and pink inherits that tile's partner — a value pink will then have to repeat. The 5/6 tile is the one that makes the repeat work: 5 stays in orange, 6 drops into pink.
2
Step 2: Pink completes itself from below
Pink's right-hand cell now needs a second 6. The only other 6 in the tray sits on the 6/1 tile, and that tile can only reach the pink cell from underneath. It stands upright: 6 in pink, 1 falling into the upper cell of the teal sum-5 pair tucked into the bottom-right corner.
3
Step 3: Teal needs a 4 beside its 1
Teal holds a 1 and needs four more to reach five. The 5/4 tile supplies exactly that, lying flat along the bottom row — 5 in the right-hand cell of the green sum-8 pair, 4 in the teal cell to its right. Nothing else in the tray offers a 4, so this placement is a single option rather than a choice.
4
Step 4: Green and purple close the board
Green now holds 5 and wants 3. The 1/3 tile stands vertically at the bottom-left: 3 into green's left-hand cell, 1 above it in the lower cell of the purple equals pair on the left. Purple has to match, and the last remaining tile is the 0/1 — it lies flat on the left edge, putting 1 in purple's upper cell and 0 in the blue less-than-3 square beside it.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The corner square has one way out
The green sum-3 square at the top-right corner is the most constrained region on the board: its only usable neighbour is the cell directly beneath it in the blue greater-than-6 pair. So the tile covering it must stand vertically, and the green square demands a 3. The 3/4 tile obliges — 3 on top in green, 4 falling into the blue pair.
2
Step 2: Ten can only be five plus five
The purple sum-10 pair on the right cannot reach ten with a 6, because today's tray tops out at 5. That leaves exactly one combination: two 5s. Both purple cells must hold a 5, and the tray offers three tiles carrying a 5 — the 5/1, the 5/4 and the 3/5 — so one of them will be spent somewhere else entirely.
3
Step 3: The blue pair picks the odd 5 out
The blue greater-than-6 pair already holds a 4, so its second cell must be at least 3 for the total to clear six. Of the three 5-bearing tiles, only the 3/5 pairs a 5 with a value that large. It stands upright: 3 up into the blue pair, 5 down into the right-hand cell of the purple sum-10 pair. The purple pair's left-hand cell is left waiting for a 5 from either the 5/1 or the 5/4.
4
Step 4: The orange block forces the left side
The orange less-than-3 block is the tightest region here: three cells totalling under three can only be 1, 1 and 0. The tray holds a single 0, on the 1/0 tile, so it must sit inside the block — the tile stands upright in the block's right-hand column, 1 in the top-right cell and 0 in the bottom-right. The block's third cell takes the 1 from the 1/3 tile, whose 3 drops into the right-hand cell of the teal greater-than-7 pair. With 3 already banked, teal needs a 5 alongside, and the 5/4 tile fits: 5 on top in teal, 4 falling into the pink run's left end.
5
Step 5: The pink run finishes the tray
The pink greater-than-7 run now reads 4, 3 and 1 from left to right — eight in total. The 4 came from the 5/4 tile and the 1 from the 5/1 tile, which simultaneously gave the purple sum-10 pair its left-hand 5. The middle 3 arrives from the 3/3 double, and that tile's lower half fills the purple sum-3 square at the very bottom of the board. Every tile in the tray is now accounted for.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Count the zeros before anything else
Four regions on this board are told to total zero: the green sum-0 square at the far top-right, the purple sum-0 cell in the centre, the green sum-0 square near the bottom, and the teal three-cell sum-0 block in the bottom-right corner. That is six cells demanding nothing. Now scan the tray: the zeros sit on the 0/0 double, the 0/1, the 0/2, the 0/3 and the 0/5 — six zeros, five tiles. The supply exactly matches the demand, so every zero is committed from the start and no sum-0 region can hold anything else.
2
Step 2: The top-right corner resolves itself
The green sum-0 square at the far top-right has a single neighbour — the blue sum-5 cell immediately to its left — so the tile covering it has to lie flat and carry a zero into green. That means the 0/5: 0 on the right in the green square, 5 on the left in blue. The blue cell wants exactly five and gets it in one move, so no second tile can reach it.
3
Step 3: The bottom zero pair locks together
The other green sum-0 square sits in the bottom row, flush against the bottom-left cell of the teal sum-0 block. Both cells need a zero and both are free, so they take the 0/0 double lying flat across them — the green square on the left, the teal block's bottom-left cell on the right. The 0/0 is now spent, which matters later: it can never be used to rescue another zero cell.
4
Step 4: Reaching the block's two far corners
Two cells of the teal block are still without a zero. Its top-right cell can only be approached from above, from the teal sum-4 pair, so the 0/1 tile stands upright — 0 in the block, 1 rising into the sum-4 pair. Its bottom-right cell can only be approached from its right, from the purple sum-4 pair, so the 0/2 tile lies flat — 0 in the block, 2 heading right into purple. Purple now holds one 2 and must match it, so its second cell is also a 2.
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Step 5: The right edge cascades
Teal sum-4 holds a 1, so its upper cell needs a 3 — and the only square that cell can reach is the orange sum-4 directly above it. The 3/4 tile stands upright, 3 below in teal and 4 above in orange. At the bottom-right, the purple pair's second 2 comes from the 1/2 tile, which leaves a 1 in the last cell of the orange less-than-4 column — a perfect fit, since everything in that column must stay under four. The column's top two cells are then left with only the 1/1 double to fill them.
6
Step 6: The left side and the centre fall in order
The purple sum-5 corner square touches only the cell below it, so the 2/5 tile stands upright: 5 in the corner, 2 down in the purple equals pair, whose lower cell then needs a 2 as well. That cell's only free neighbour is the blue sum-4 square, so the 2/4 tile lies flat, 2 in purple and 4 in blue. Moving right, the pink less-than-4 cell above the pink equals pair takes a 3 from the 1/3 tile, while the pink sum-4 square beside it takes the 4 from the 1/4 tile — both leaving 1s in the equals pair below. The green sum-5 square pairs with the uncoloured free square through the 1/5 tile, 5 above and 1 below. The purple sum-0 centre cell can only escape into the pink equals block, so the 0/3 tile lies flat with 0 on the left and 3 on the right, and the block completes 3-3-3 with the 3/3 double. The last two cells, the orange sum-4 at bottom-left and the blue sum-5 beside it, are filled by the 4/5 tile lying flat.
💡 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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