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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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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Today's Pips for March 31, 2026, edited by Ian Livengood, closes out the month with three puzzles that each reward a different kind of entry point. Livengood's Easy grid is a flat two-row strip: four dominoes laid across eight cells with no gaps or cutouts. The shape is deliberately plain, which means every deduction has to come from reading the constraints rather than from the layout. One of the sum regions is so tight it names its domino outright — and from there the rest of the board falls in order.
The Medium puzzle comes from constructor Rodolfo Kurchan and uses a 4×4 frame with two cells quietly left empty in the lower-left corner. Seven dominoes fill the rest. The key tension is a pair of sum-of-2 regions sitting at opposite ends of the board — both too small to be reached by most of today's dominoes — while a sum-of-10 pair stretches across adjacent cells and asks for maximum coverage from the limited set.
Kurchan also built today's Hard puzzle, which uses all fifteen non-double dominoes from the 0–5 set and fills a 5×6 grid completely. What makes this one unusual is that two separate zero-sum regions — one spanning two cells, one spanning three — together demand a zero pip in exactly five cells. There are exactly five zero-pip dominoes in the set. That isn't a coincidence; it's where the puzzle begins.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 One cell has almost no room
A single cell in the top row carries a sum constraint that leaves almost no options. Look at it before you consider anything else — it narrows the opening move to exactly one domino.
💡 That sum is zero
The constrained cell must show a blank pip — zero. Only one of today's four dominoes contains a zero at all, so it has to go there. Figure out which way it has to face, and the bottom row starts to fill itself in.
💡 Full solution
The sum=0 cell in the top row forces [2|0] there vertically: blank at the top, 2 below it. That 2 is now in the bottom-left pair (sum=7), so its partner to the left must be 5. Only [5|1] carries a 5 — place it vertically with 5 at the bottom and 1 above. Moving right: the top row's sum=5 cell needs a 5-pip. Of the remaining two dominoes, only [3|5] has one. Place it horizontally with 5 in that cell and 3 to its right. The last domino [2|2] drops into the bottom-right pair: 2+2 = 4 exactly. Done.
💡 Two sum regions share a very small target
Two separate regions in today's grid both require a sum that's unusually low — almost nothing across two adjacent cells. Start with the one that's most constrained by the available dominoes.
💡 Only one domino reaches that sum
The left-column pair needs a total of just 2. Among today's seven dominoes, only one has both pips low enough to hit that target. Find it, place it, and the board starts opening up from there.
💡 Full solution
The sum=2 left column forces [1|1] vertically: 1 at each cell. With the left column fixed, the sum=10 constraint on the adjacent top pair needs both cells to reach ten. [5|4] placed horizontally at row 1 gives a 5 at column 1 and 4 at column 2; [1|5] placed horizontally at row 0 puts 5 at column 1 and 1 at column 2 — 5+5=10 ✓. The vertical pair below the 4 must also sum to 10: (1,2)=4, so (2,2)=6. Only [3|6] has a 6 — place it horizontally at row 2, giving 3 at column 1 and 6 at column 2. The equals constraint at the top right requires both cells to match: (0,2)=1 from [1|5], so (0,3)=1 as well. [2|1] placed vertically at column 3 gives 2 at (1,3) and 1 at (0,3) ✓. The sum=2 pair below then has (1,3)=2, requiring (2,3)=0. [0|5] placed vertically: 0 at (2,3) and 5 at (3,3). Last domino [3|5] fills the bottom row horizontally: 3 at (3,1) and 5 at (3,2). Sum check: 5+5=10 ✓. Puzzle complete.
💡 Count the zeros before you place anything
Two zero-sum regions appear in today's grid. Before touching the board, count how many cells must be zero and compare that to how many dominoes in the set carry a zero pip at all.
💡 Every zero-pip domino is spoken for
Five cells across the two zero regions must each show a blank. The set has exactly five zero-pip dominoes — one per zero cell, no extras. That tells you where five dominoes must go before you've reasoned about anything else.
💡 A single-cell constraint pins the first zero placement
One cell adjacent to the two-cell zero region carries its own sum constraint that accepts only one value. That value matches exactly one zero-pip domino's non-zero end. Use it to anchor your first placement.
💡 The top row and row-1 constraints form a tight chain
Once the zero-region anchor is placed, the single-cell constraints in rows 0 and 1 resolve in sequence. Several cells must each show a specific value, and the available dominoes leave only one valid assignment for each. Follow the chain left to right across the top before working downward.
💡 Full solution
Start with (2,2): it must be 1, and it's adjacent to the zero region at (1,2). [0|1] vertical: (1,2)=0, (2,2)=1 ✓. Top-right: (0,5)=1 and the pair (0,4)+(1,4)=8. [1|4] horizontal: (0,5)=1, (0,4)=4. [0|4] horizontal: (1,3)=0, (1,4)=4. Sum check: 4+4=8 ✓. Top row left side: (0,0)=1 and (0,3)=3. [1|5] horizontal: (0,0)=1, (0,1)=5. [3|5] horizontal: (0,3)=3, (0,2)=5. Sum check: 5+5=10 ✓. Row 1: (1,0)=4 and (1,1)=3. [3|4] horizontal: (1,1)=3, (1,0)=4 ✓. (2,0)=1 and (2,4)=3. [1|2] horizontal: (2,0)=1, (2,1)=2. [2|3] horizontal: (2,3)=2, (2,4)=3. Region sum=10 with five cells: (2,1)=2 and (2,3)=2 placed — the remaining three cells (3,1), (3,2), (3,3) must each be 2. [0|2] horizontal: (3,4)=0, (3,3)=2 ✓. [2|5] horizontal: (3,1)=2, (3,0)=5. [2|4] vertical: (3,2)=2, (4,2)=4. Sum=8 at (3,0)+(4,0): 5+3=8 — [1|3] horizontal: (4,1)=1 ✓, (4,0)=3 ✓. [0|3] vertical: (2,5)=0, (1,5)=3 ✓. [0|5] vertical: (3,5)=0, (4,5)=5. [4|5] horizontal: (4,3)=4, (4,4)=5. Sum checks: (4,2)+(4,3)=4+4=8 ✓, (4,4)+(4,5)=5+5=10 ✓. Puzzle complete.
🎨 Pips Solver
Mar 31, 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 sum=0 cell names its domino
The second cell of the top row must show exactly zero — its sum constraint allows no other value. Check all four dominoes in today's set: only [2|0] has a blank pip. For that blank to land in the top row, the domino must be placed vertically, with 0 pointing up and 2 pointing down. That places (0,1)=0 and (1,1)=2.
2
Step 2: The bottom-left pair follows from what's now known
The two-cell region in the bottom row's left half must sum to 7. Cell (1,1) is already set to 2. Its partner (1,0) therefore needs 5. Scan the remaining dominoes — [5|1], [2|2], [3|5] — and [5|1] is the only one with a 5-pip. Place it vertically in the first column: 5 at the bottom, 1 at the top. That sets (1,0)=5 and (0,0)=1.
3
Step 3: The sum=5 cell forces the next placement
The third cell of the top row must show exactly 5. Two dominoes remain: [2|2] and [3|5]. Only [3|5] carries a 5. Place it horizontally across the top-right pair: 5 at (0,2) and 3 at (0,3). The two empty-type cells at (0,0) and (0,3) require no constraint check.
4
Step 4: The last domino closes the puzzle
One domino remains — [2|2] — and one region remains: the bottom-right pair at (1,2) and (1,3), which must sum to 4. Place [2|2] horizontally: 2+2=4. Every constraint is satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Sum=2 on the left column rules out everything but one domino
Cells (0,0) and (1,0) together must sum to 2. That's a tight target — only a pip-pair totaling 2 will work. Go through today's seven dominoes and ask which one adds up to exactly 2 across its two sides: only [1|1] does. Place it vertically in the left column with 1 at the top and 1 below.
2
Step 2: The top-row sum=10 pair resolves with two horizontally placed dominoes
Cells (0,1) and (1,1) must sum to 10. No single domino straddles both cells — each belongs to a different domino placed along its own row. [5|4] placed horizontally at row 1, columns 1–2, gives (1,1)=5 and (1,2)=4. With (1,1)=5 locked in, (0,1) needs 5 to reach the target. [1|5] placed horizontally at row 0, columns 2–1, gives (0,2)=1 and (0,1)=5. Sum check: 5+5=10 ✓.
3
Step 3: Sum=10 on the inner vertical pair forces [3|6]
Cells (1,2) and (2,2) share a sum constraint of 10. Cell (1,2)=4 from the previous step, so (2,2) must be 6. Among the remaining dominoes, only [3|6] has a 6. Place it horizontally at row 2: (2,1)=3 and (2,2)=6. The cell at (2,1) falls in an empty region — no constraint to check.
4
Step 4: The equals constraint and sum=2 resolve together
Cells (0,2) and (0,3) must be equal. Cell (0,2)=1 from [1|5] already placed, so (0,3) must also be 1. [2|1] placed vertically at column 3 delivers exactly that: (1,3)=2 at the bottom, (0,3)=1 at the top. The sum=2 constraint on (1,3)+(2,3) then requires (2,3)=0. [0|5] placed vertically at column 3, rows 2–3: (2,3)=0 and (3,3)=5.
5
Step 5: The last domino finishes the bottom row
One domino remains — [3|5] — and one region remains: (3,2) and (3,3) must sum to 10. Cell (3,3)=5 from the previous step, so (3,2) must be 5 as well. Place [3|5] horizontally at row 3: (3,1)=3 and (3,2)=5. Sum check: 5+5=10 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Map the zero cells to their dominoes
Two regions require every pip to be zero: [1,2]+[1,3] (sum=0) and [2,5]+[3,4]+[3,5] (sum=0). That's five zero-constrained cells. The set of 15 non-double dominoes contains exactly five zero-pip tiles: [0|1], [0|2], [0|3], [0|4], and [0|5]. Each one must send its zero end into one of these five cells — no domino goes to waste and none is available for other placements. Establishing this mapping is the structural first move.
2
Step 2: A standalone constraint pins [0|1]
Cell (2,2) must equal exactly 1 — it carries a standalone sum=1 constraint. It sits directly below (1,2), which is a zero-region cell. A domino bridging (1,2) and (2,2) would carry pips 0 and 1 — that's [0|1] and nothing else. Place [0|1] vertically: (1,2)=0, (2,2)=1 ✓. This locks the anchor for the two-cell zero region and confirms [0|1]'s placement in the left zero cluster.
3
Step 3: The top-right corner resolves as a connected trio
Cell (0,5) must be 1 (standalone sum=1). Cell (1,3) is in the zero region and must be 0. Cells (0,4) and (1,4) must sum to 8. [1|4] placed horizontally at the top right fills (0,5)=1 and (0,4)=4. [0|4] placed horizontally at row 1 fills (1,3)=0 and (1,4)=4. Sum check: (0,4)+(1,4)=4+4=8 ✓. [0|4] also accounts for the zero-region cell at (1,3).
4
Step 4: The top row resolves left to right with four dominoes
Cell (0,0) must be 1 and cell (0,3) must be 3. [1|5] placed horizontally: (0,0)=1, (0,1)=5. [3|5] placed horizontally: (0,3)=3, (0,2)=5. Sum check: (0,1)+(0,2)=5+5=10 ✓. Constraints (0,0)=1 ✓ and (0,3)=3 ✓.
5
Step 5: Two standalone cells in row 1 force [3|4]
Cells (1,0) and (1,1) are both standalone: (1,0) must be 4, and (1,1) must be 3. [3|4] placed horizontally bridges them: (1,1)=3, (1,0)=4. Both constraints satisfied simultaneously.
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Step 6: Four interior cells resolve with two dominoes
Cell (2,0) must be 1 and cell (2,4) must be 3. [1|2] placed horizontally: (2,0)=1, (2,1)=2. [2|3] placed horizontally: (2,3)=2, (2,4)=3 ✓. These two placements also seed the large non-contiguous region in the next step.
7
Step 7: The five-cell interior region forces all remaining values to 2
Cells [2,1]+[2,3]+[3,1]+[3,2]+[3,3] must sum to 10. Two of those cells are now fixed: (2,1)=2 and (2,3)=2. The remaining three — (3,1), (3,2), and (3,3) — must sum to 6. The only way to partition 6 across three non-negative pip values using the available dominoes is 2+2+2. Each of those three cells must equal 2.
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Step 8: The three-cell zero region distributes its dominoes
Cells (3,4), (3,5), and (2,5) must all be zero. [0|2] placed horizontally at row 3: (3,4)=0 and (3,3)=2 ✓. [0|3] placed vertically at column 5, rows 1–2: (1,5)=3 ✓ (standalone sum=3) and (2,5)=0 ✓. [0|5] placed vertically at column 5, rows 3–4: (3,5)=0 and (4,5)=5.
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Step 9: The interior-2 cells resolve with their column and row partners
Cells (3,1) and (3,2) still need domino assignments. [2|5] placed horizontally: (3,1)=2 ✓ and (3,0)=5. [2|4] placed vertically: (3,2)=2 ✓ and (4,2)=4. The sum=8 constraint on (3,0)+(4,0): (3,0)=5, so (4,0) must be 3. [1|3] placed horizontally: (4,1)=1 ✓ (standalone sum=1) and (4,0)=3 ✓.
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Step 10: The bottom row closes the puzzle
One domino remains — [4|5]. The sum=8 pair at (4,2)+(4,3) has (4,2)=4 already; (4,3) must be 4 as well. The sum=10 pair at (4,4)+(4,5) has (4,5)=5 already; (4,4) must be 5. Place [4|5] horizontally: (4,3)=4 and (4,4)=5. Both constraints verified: 4+4=8 ✓, 5+5=10 ✓. Puzzle complete.
💡 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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