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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's easy grid for April 4th opens at a corner. A single-cell sum constraint sits at the top-left, naming the exact pip value that belongs there — and only one domino in today's set can fill that role with its 6-pip face pointing outward. From that locked starting point, two "equals" regions running through the center of the grid come into focus. The available dominoes make both easier to satisfy than they look: the same high pip value threads through both pairs, and the remaining sums fall in line without any guessing.
The medium puzzle, also from Livengood, anchors from the bottom. A "sum=0" constraint at the foot of the grid is the most restrictive you can have — both cells must be completely blank, which immediately identifies the only two dominoes that can stand there. Once those two are locked in vertically, a chain of matching sum-5 constraints climbs upward through the board. The final pair at the top has a "less than" limit that rules out one of the two remaining orientations cleanly.
Rodolfo Kurchan's hard puzzle for April 4th is a vertical-only board: every one of the sixteen active domino slots is a standing column pair, spanning two adjacent rows. The grid splits into four horizontal bands, each with its own cluster of constraints. Two single-cell anchors at the bottom corners pin the first two dominoes immediately — a blank face on the left, a single pip on the right — and a "greater than 9" region in the upper-middle narrows the candidate list to just one domino. From those starting points, the remaining slots yield through a combination of sum targets and inequality limits.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 One corner, one pip value, one domino
The top-left corner of the grid carries an exact sum constraint that names a specific pip value. Only one domino in today's easy set has that face available and can orient it into that corner. Find it and you have your starting point — everything else connects from there.
💡 Two equals regions share the same key value
Once the corner domino is placed, look at the two "equals" regions in the center of the grid. Both turn out to need the same pip value — and you have two separate dominoes that each bring it to the table. Slot them in horizontally and the center column lines up cleanly.
💡 Full solution
Place [6|1] horizontally in the top row: 6 at the corner (satisfying sum=6) and 1 alongside it. Then [2|6] horizontally in the next row: 6 in the left part of the equals column, 2 to the right. [4|6] horizontally below it: 6 in the same equals column (matching [2|6] above it), 4 to the right. Sum=7 at the bottom of that column: 4+3=7, so [5|3] slides in with 3 below the 4 and 5 in the corner. Sum=5 in the remaining pair: 0+5=5, so [0|2] stands vertically with 0 below and 2 above — and that 2 also satisfies the second equals region where [2|6] already placed a 2 on the other side. All six constraints check out.
💡 The bottom row does all the heavy lifting
Scan the constraints and find the most locked-down one first. The bottom row has a "sum=0" — which means both cells there must show zero pips. That constraint doesn't just hint at two specific dominoes, it demands them.
💡 Sum=0 forces two dominoes vertical, and then the fives appear
Once the two zero-carrying dominoes stand in the bottom row, their upper halves are fixed. The pair of sum=5 constraints directly above now have one side already filled in — and from those, a sum=10 column and the final two dominoes in the top block resolve in order.
💡 Full solution
Sum=0 at the bottom forces [3|0] vertically: 3 above, 0 at the foot of the left slot. And [1|0] vertically: 1 above, 0 at the foot of the right slot. Sum=5 in the lower-middle pair: 1 (from [1|0]) + 4 = 5, so [3|4] stands vertically with 4 at the bottom and 3 above. Sum=5 in the upper-middle pair: 3 (from [3|4]) + 2 = 5, so [2|3] lies horizontally with 2 left and 3 right — which also feeds the sum=6 pair below (3+3=6 ✓). Sum=10 in the left column: only [5|5] delivers 10, so it stands vertically. Two dominoes remain for the top block: [6|4] and [2|5]. Less-than-6 on the single top-left cell rules out placing a 6 there, so [2|5] goes vertically with 5 at the top (5<6 ✓) and 2 below. [6|4] fills the right side vertically. Less-than-7 check: 2+4=6, just under the limit. Puzzle complete.
💡 Every domino stands upright
Before solving a single constraint, notice the board structure: all active cells sit in vertical pairs. Every domino in today's hard set will be placed standing upright, connecting two cells in the same column. That means each of the sixteen slots is independent — your job is just to match the right domino to the right column.
💡 Pin the four exact-value corners first
Several single-cell regions name a specific pip count exactly. "Sum=0" means that cell must be blank. "Sum=1" means exactly one pip. "Sum=2" (appearing twice in the middle band) and "sum=6" in the top-right corner each point to a specific domino face. These four anchors give you a head start before you touch the inequality constraints.
💡 Full solution
Bottom-left: sum=0 at (6,0) + less=4 at (7,0) → [3|0] with 0 above and 3 below. Bottom-right-of-left: sum=1 at (6,2) → [6|1] with 1 above and 6 below. Bottom-right: sum=5 at (6,6) + less=3 at (7,6) → [2|5] with 5 above and 2 below (2<3 ✓). Bottom-center: sum=7 at (6,4)+(7,4) → [3|4] (3+4=7). Top-right: sum=6 at (0,7) + less=4 at (1,7) → [6|3] with 6 above and 3 below. Left second band: sum=2 at (2,0) + less=3 at (3,0) → [2|2] with 2 above and 2 below (2<3 ✓). Right second band: greater=4 at (2,6) + less=4 at (3,6) → [5|3] with 5 above (5>4 ✓) and 3 below (3<4 ✓). Center second band: less=4 at (2,4)+(3,4) sum<4 → [1|2] (1+2=3<4 ✓). Center left: sum=2 at (2,2) + less=4 at (3,2) → [2|3] with 2 above and 3 below. Middle equals at (4,1)+(5,1) → [5|5] (the only double with both values equal). Sum=8 at (4,3)+(5,3) → [4|4] (4+4=8). Sum=5 at (4,5)+(5,5) → [0|5] (0+5=5). Both greater=4 at (4,7) and (5,7) → [5|6]: 6 above (6>4 ✓), 5 below (5>4 ✓). Greater=9 at (0,5)+(1,5) → [4|6] (4+6=10>9 ✓). Top-left sum=9 at (0,1)+(1,1) → [5|4] (5+4=9). Last slot: sum=4 at (0,3) → [2|4] with 4 above and 2 below. All 25 constraints satisfied.
🎨 Pips Solver
Apr 4, 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 corner constraint names your first domino
The top-left cell carries a "sum=6" constraint — the pip value there must be exactly 6. Three dominoes in today's easy set have a 6-pip face: [2|6], [4|6], and [6|1]. Only [6|1] can be oriented to put 6 at that specific corner position, since the other two would require placing a non-6 face there first. So [6|1] goes in the top row horizontally: 6 at the corner, 1 alongside it.
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Step 2: Two equals regions need the same value — and you have two matching dominoes
With [6|1] placed, two 6-faced dominoes remain: [2|6] and [4|6]. The "equals" region at (1,1) and (2,1) requires both cells to show the same pip value. Neither cell has its own constraint, so you need to find which value they naturally share. Both [2|6] and [4|6] can contribute a 6 to that column — place [2|6] horizontally with 6 at (1,1) and 2 at (1,2), and [4|6] horizontally below it with 6 at (2,1) and 4 at (2,2). Equals: 6=6 ✓.
3
Step 3: Sum=7 forces [5|3] into the lower block
The region at (2,2) and (3,2) must total 7. Cell (2,2)=4 is now fixed from Step 2. So (3,2) must be 3. The only remaining domino with a 3-pip face is [5|3] — orient it horizontally with 3 at (3,2) and 5 at (3,3).
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Step 4: Sum=5 points to the last domino's orientation
The region at (2,3) and (3,3) must total 5. Cell (3,3)=5 is just placed in Step 3. So (2,3) must be 0. The last domino [0|2] stands vertically: 0 at (2,3), 2 at (1,3). Both constraints check out: sum=5 gives 0+5=5 ✓.
5
Step 5: The second equals region confirms the solution
The "equals" region at (1,2) and (1,3) is the final check. Cell (1,2)=2 came from [2|6] in Step 2, and (1,3)=2 just placed by [0|2] in Step 4. Both show 2 — equals ✓. Every constraint on the board is satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
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Step 1: Sum=0 at the bottom forces two specific dominoes
The "sum=0" region spans (4,1) and (4,2) at the bottom of the grid. A sum of zero is only possible if both cells show 0 pips. Scan the domino set: the only dominoes carrying a 0-pip face are [3|0] and [1|0]. Both must be placed vertically here with their 0 face pointing down. [3|0] goes in the left slot: 3 at (3,1), 0 at (4,1). [1|0] goes in the right slot: 1 at (3,2), 0 at (4,2).
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Step 2: Sum=5 in the lower row — right side leads
The region at (3,2) and (3,3) must total 5. Cell (3,2)=1 is fixed from Step 1. So (3,3) must be 4. The domino [3|4] fits perfectly standing vertically: 3 at (2,3) and 4 at (3,3). That also fills (2,3)=3.
3
Step 3: Sum=5 again — left side of the same row
The region at (2,2) and (2,3) must total 5. Cell (2,3)=3 is now known. So (2,2) must be 2. The domino [2|3] lies horizontally: 2 at (2,2) and 3 at (2,1). Check the sum=6 region below it: (2,1)+(3,1)=3+3=6 ✓. Two constraints resolved in one step.
4
Step 4: Sum=10 in the left column — only one domino adds to 10
The region at (2,0) and (3,0) must total 10. Look at the remaining dominoes: only [5|5] delivers a sum of 10 (5+5). Place it vertically: 5 at (2,0), 5 at (3,0). The "sum=2" constraint on (2,0) — wait, that's sum on a single cell in the hard puzzle, not here. Both cells simply need to total 10, and [5|5] is the only match.
5
Step 5: Two dominoes left, one constraint rules out one arrangement
The remaining dominoes are [6|4] and [2|5]. They fill the top block: (0,1), (0,2), (1,1), (1,2). The "less than 6" constraint on (0,1) means the value there must be 5 or less. [2|5] placed vertically in that slot gives 5 at (0,1) — 5<6 ✓ — and 2 at (1,1). [6|4] fills the adjacent slot vertically: 6 at (0,2) (unconstrained "empty" region), 4 at (1,2). Final check: less-than-7 at (1,1)+(1,2)=2+4=6 — safely under 7 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The board is all vertical pairs — map the four bands
Every active cell on today's hard board is part of a vertical domino slot: rows 0–1 (odd columns), rows 2–3 (even columns), rows 4–5 (odd columns), rows 6–7 (even columns). That gives 16 standing slots for 16 dominoes. Start with the bottom band (rows 6–7), where three single-cell exact constraints immediately identify three dominoes.
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Step 2: Three bottom-band slots resolve from exact-value constraints
"Sum=0" at (6,0) means that cell holds 0 pips. With "less than 4" at (7,0), the domino must be [3|0]: 0 above (satisfying sum=0), 3 below (3<4 ✓). "Sum=1" at (6,2) means that cell holds 1 pip. No constraint on (7,2), so [6|1] goes there flipped: 1 at (6,2), 6 at (7,2). "Sum=5" at (6,6) with "less than 3" at (7,6): value 5 above, and 0 or 1 or 2 below. [2|5] delivers this: 5 at (6,6), 2 at (7,6) — 2<3 ✓.
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Step 3: Bottom-center and top-right fill from sum targets
"Sum=7" at (6,4)+(7,4): need two values totaling 7. [3|4] gives 3+4=7. Place it with 3 at (6,4), 4 at (7,4). Top-right: "sum=6" at (0,7) fixes that cell as 6. With "less than 4" at (1,7), [6|3] fits perfectly: 6 at (0,7), 3 at (1,7) — 3<4 ✓.
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Step 4: The second band resolves through individual cell constraints
"Sum=2" at (2,0) and "less than 3" at (3,0): value 2 above, and under 3 below. [2|2] places 2 at both cells — 2<3 ✓. "Sum=2" at (2,2) and "less than 4" at (3,2): [2|3] fits with 2 above and 3 below — 3<4 ✓. "Greater than 4" at (2,6) and "less than 4" at (3,6): upper cell must be 5 or 6, lower must be 0, 1, 2, or 3. [5|3] delivers 5 above (5>4 ✓) and 3 below (3<4 ✓). "Less than 4" sum at (2,4)+(3,4): total must be under 4 — so 0, 1, 2, or 3. [1|2] gives 1+2=3<4 ✓.
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Step 5: Middle band — equals, sum=8, sum=5, and the two greater-than singles
"Equals" at (4,1)+(5,1): both cells must match. [5|5] is the only double available that fits — 5=5 ✓. "Sum=8" at (4,3)+(5,3): [4|4] gives 4+4=8 ✓. "Sum=5" at (4,5)+(5,5): [0|5] gives 0+5=5 ✓. "Greater than 4" at both (4,7) and (5,7): each cell must exceed 4, so both values must be 5 or 6. [5|6] covers this: 6 at (4,7) (6>4 ✓), 5 at (5,7) (5>4 ✓).
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Step 6: Top band — the greater-than-9 region and the last two slots
"Greater than 9" at (0,5)+(1,5): the two-cell sum must exceed 9, meaning 10 or more. Check the remaining dominoes: [4|6] gives 4+6=10>9 ✓. Place it with 4 at (0,5) and 6 at (1,5). "Sum=9" at (0,1)+(1,1): [5|4] gives 5+4=9 ✓. Last slot: "sum=4" at (0,3) means that cell holds 4. [2|4] placed with 4 at (0,3) and 2 at (1,3) — (1,3) carries an "empty" constraint, so any value is valid. All 25 constraints satisfied. 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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