🚨 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 builds the easy grid around a single high-value sum constraint that quietly eliminates all but one possible opening move. That's a subtle but deliberate design choice — rather than layering multiple constraint types, Livengood lets arithmetic do the heavy lifting, creating a puzzle where one deduction cascades cleanly through the rest of the board with no branching. The result feels less like solving and more like watching dominoes fall.
Rodolfo Kurchan takes a sparser approach on the medium: a sum-12 region at the bottom functions as a mathematical lock, forcing both cells to their maximum value simultaneously. It's a confident construction — Kurchan trusts that solvers will recognize the ceiling immediately, and rewards that recognition with a clean chain that unravels upward through sum and less-than constraints.
The hard is where Kurchan's architectural ambition shows most clearly. Three separate sum-0 regions scattered across the grid create independent entry points that can be activated in any order — a design that keeps the puzzle feeling open even as constraints accumulate. The centerpiece is a four-cell equals region that threads together placements from different corners of the board, requiring all four members to converge on a single shared value before the less-than column at the center can finally seal everything shut.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 High sum, narrow options
One of the two-cell sum regions has a target so high that most domino combinations can't reach it. Start there — it rules out almost everything at once.
💡 Eleven means only one combination
The top-left region sums to 11. No single domino in the set adds up to 11, so two separate dominoes each contribute one cell. With 6 as the max pip, there's exactly one way to make 11 with two cells.
💡 6 and 5 unlock everything
[3-6] runs vertically in the top-left corner — its 6 sits at the very top, forcing the adjacent cell to be 5. Only [5-5] can provide that, laid horizontally across the top two cells. With those anchored, the sum-9 region in the middle falls quickly: [2-3] drops horizontal below [5-5], [0-2] runs vertically beneath that, and [3-0] closes the bottom row — confirming both equals regions at zero and three.
💡 One region solves itself
Scan the constraints. There's a sum target at the bottom so high that the grid barely gives you room to maneuver. Start there.
💡 Sum-12 forces both cells to maximum
The bottom sum region needs to hit 12. The highest pip in the game is 6, so the only way to reach 12 with two cells is 6+6. Both cells must be exactly 6 — find the two dominoes in your set that each carry a 6.
💡 Two 6s anchor the bottom row
[6-4] and [2-6] each carry a 6. [6-4] runs left in the bottom row (6 left, 4 right), [2-6] runs right (6 left, 2 right). From there a chain follows — the sum-6 region just above the right side resolves, which feeds the sum-8 region on the left, placing [4-2] vertically. [2-2] fills the center with sum-4, [6-0] handles the middle sum-6, and double-zero [0-0] closes the top column.
💡 Three regions that sum to zero
When a region sums to 0, every cell in it must be 0. Find those three regions in the grid — they're your unconditional starting points, no arithmetic required.
💡 Sum-0 means all zeros, no exceptions
There's a single-cell sum-0 region, a two-cell sum-0 in the middle-right area, and a two-cell sum-0 at the top. That's six cells you can fill immediately. The cell adjacent to the single-cell zero is in a greater-than-0 region, which tells you exactly which domino goes there.
💡 Lock the zeros, then find the equals region
[1-0] covers the single-cell zero and its greater-than-0 neighbor. [0-0] slots into the two-cell sum-0 in the middle — it's the only double-zero. [3-0] and [4-0] handle the top, confirmed by the adjacent sum-3 single cell. With those placed, look for the four-cell equals region in the center — all four cells must share one value, and [5-5] tells you what it is.
💡 The central equals region runs on 5s
The four-cell equals region needs a shared value. [5-5] spans two of those cells vertically — both show 5, locking the common value. [4-5] then places its 5-end into a third equals cell, and [5-2] fills the fourth. The 2-end of [5-2] feeds into a nearby equals constraint, which in turn anchors the outer column with [4-2].
💡 The less-than column closes everything out
A four-cell column running through the center must all be below 5. [1-1] covers the top two cells of that column (both 1s). [2-1] and [6-1] run horizontally at the lower levels with their 1-ends facing the column — their outward ends (2 and 6) satisfy the adjacent sum-8 region. [5-3] handles the bottom-right sum-8, and [4-4] closes the final equals region.
🎨 Pips Solver
Apr 18, 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: Sum-11 forces two separate dominoes
The two top-left cells are in a region targeting 11. Check the domino set — no single piece has pips that add up to 11, so two different dominoes must each contribute one cell here. With 6 as the max pip, the only way two cells sum to 11 is if one shows 6 and the other shows 5. That completely determines what values go in those spots.
2
Step 2: [3-6] anchors the top-left column
[3-6] is the only domino carrying a 6. Place it vertically in the top-left column: the 6-end faces up into the sum-11 region, and the 3-end drops to the cell below. That lower cell has an 'empty' label — no constraint — so either orientation is allowed, but the 6-at-top arrangement is the one that satisfies the sum.
3
Step 3: [5-5] fills the top row
With 6 in the top-left cell, the adjacent cell in the sum-11 region must be 5. [5-5] is the only remaining domino with a 5. Lay it horizontally across the top-center and top-right cells — both become 5. Now the sum-9 region comes into focus: it covers the top-right cell (5) plus two cells in the row below, so those two cells need to sum to 4.
4
Step 4: [2-3] and [0-2] fill the middle
From the three remaining dominoes — [0-2], [3-0], [2-3] — find the pair whose values sum to 4. [2-3] contributes a 2, and [0-2] also contributes a 2: place [2-3] horizontally in the middle row (2 on the left, 3 on the right), then [0-2] vertically below the left cell of [5-5] (2 at top feeding the sum-9 region, 0 at bottom).
5
Step 5: [3-0] completes the bottom row
The last piece, [3-0], fills the bottom-right corner. Place it with 3 on the right and 0 in the center-bottom. Both bottom cells are now 0 — the equals region there confirms ✓. The right-side equals region also checks out: the middle-row rightmost cell (3 from [2-3]) matches the bottom-right cell (3 from [3-0]) ✓.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Sum-12 is the mathematical lock
Scan all constraints. The bottom row has a sum region targeting 12. Since the maximum pip is 6, the only way two cells reach 12 is 6+6. Both cells must be exactly 6 — this is forced with no alternatives to consider.
2
Step 2: [6-4] and [2-6] anchor the bottom row
[6-4] and [2-6] are the only dominoes each carrying a 6. [6-4] runs left in the bottom row: 6 on the left side of the sum-12 region, 4 extending further left. [2-6] runs right: 6 on the left side of the sum-12 region, 2 extending further right. The bottom row now reads 4, 6, 6, 2 from left to right.
3
Step 3: Right column resolves upward
The rightmost bottom cell is 2 (from [2-6]). The region spanning it and the cell above targets 6: 2 + ? = 6, so the upper cell must be 4. [3-4] placed horizontally covers that cell with its 4-end pointing left into the region and 3 extending outward to the right.
4
Step 4: Left column resolves upward
The leftmost bottom cell is 4 (from [6-4]). The region spanning it and the cell above targets 8: 4 + ? = 8, so the upper cell is also 4. [4-2] placed vertically does exactly that — 4 at the bottom, 2 above. The 2 satisfies the less-than-3 constraint on that cell: 2 < 3 ✓.
5
Step 5: Three dominoes close the grid
The center sum-4 region takes [2-2]: both cells equal 2, and 2+2=4 ✓. The middle sum-6 region takes [6-0]: 6+0=6 ✓. The double [0-0] finishes the top, placed vertically: both the single top cell and the middle-row cell beside it show 0, each comfortably below 7 ✓.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Find the three sum-0 regions
A region summing to zero forces every pip in it to be zero — no math needed, just placement. There are three such regions in this puzzle: one single-cell, one two-cell spanning the middle-right area, and one two-cell at the top. Locate all three before making any moves. These six cells are your unconditional starting positions.
2
Step 2: [1-0] handles the single-cell zero
The single-cell sum-0 region forces that pip to 0. The cell immediately beside it is in a greater-than-0 region, meaning it must be at least 1. [1-0] placed horizontally covers both at once: 1 on the greater-than-0 side, 0 on the sum-0 side. No other domino with a 0 can satisfy both constraints in the same move.
3
Step 3: [0-0] locks into the middle sum-0
The two-cell sum-0 region in the middle-right area needs both cells to be 0. [0-0] is the only double-zero domino in the set. It slots in vertically here — completely forced, no orientation choice.
4
Step 4: [3-0] and [4-0] resolve the top
The two-cell sum-0 at the top forces both cells to 0. The lone cell immediately to their left is in a sum-3 region — it must be 3. [3-0] covers both: 3 on the left, 0 on the right ✓. The cell directly below the rightmost top-zero is in an equals region. [4-0] placed vertically fills those two cells: 0 at top (matching the sum-0 constraint above it) and 4 below, entering the equals region.
5
Step 5: [5-5] reveals the central equals value
Four cells scattered through the center belong to the same equals region — all four must share one pip value. [5-5] is a double and can span two of those cells vertically, giving both a value of 5. That immediately locks the common value for all four cells at 5. The remaining two cells in the equals region must also show 5.
6
Step 6: [4-5] and [5-2] extend the equals chain
[4-5] placed horizontally puts a 5 into one of the remaining equals cells, with 4 extending outward — and that 4 aligns with the cell above from [4-0], satisfying a side equals region ✓. [5-2] placed vertically fills the last equals cell with 5 at the top and 2 below.
7
Step 7: [4-2] closes the equals chain and outer column
The 2 from [5-2] falls into a two-cell equals region — its neighbor must also be 2. [4-2] placed horizontally satisfies this: 2 on the equals side, 4 extending to the outer column. That outer column now has its top cell pinned at 4. The sum-3 single-cell region in the adjacent area confirms the 3-end of a nearby domino ✓, and the empty regions place no additional burden.
8
Step 8: The less-than column and bottom rows
A four-cell column through the center must all be below 5. [1-1] fills the top two cells of that column (both 1 < 5 ✓). [2-1] runs horizontally at the next level with its 1-end facing the column (1 < 5 ✓), its 2-end extending outward. [6-1] runs horizontally at the bottom level with its 1-end in the column (1 < 5 ✓), its 6-end extending outward. The 2 and 6 together satisfy the adjacent sum-8 region: 2+6=8 ✓. [5-3] closes the bottom-right sum-8: 5+3=8 ✓. [4-4] fills the final equals region: both cells equal 4 ✓.
💡 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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