🚨 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!
Click here to play today's official NYT Pips game first.
Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Ian Livengood's easy grid for April 6th is anchored at both ends of the same column. A single-cell sum constraint at the top of the grid names an exact pip value, and a second single-cell constraint at the far end of a diagonal run demands a matching total. The only double in today's easy set — a [3|3] — turns out to be the last piece placed, not the first. The opening move comes from reading which domino can simultaneously satisfy two constraints at once by orientating a 6-pip face into the right spot, and everything fans out from there through a clean chain of sums.
Rodolfo Kurchan's medium for today is built around a five-cell equals region that snakes down the right side of the board. Two doubles sit in today's set — and the larger of the two is the one that cracks the puzzle open. Placing the [6|6] double inside that long equals region forces every cell in the chain to the same value, resolving three further dominoes in one fell swoop. From there, a pair of tight less-than constraints at the foot of the grid each name their domino outright, and a small two-cell equals region in the center closes the board.
Rodolfo Kurchan's hard puzzle for April 6th also features three doubles — and there are exactly three equals regions waiting to receive them. The largest equals region spans four cells and anchors in the middle of the board at a pip value of 4. A separate three-cell column of zeros locks down in a single step once the [0|0] double is placed. And the [6|6] double — the only tile that can satisfy a sum-7 constraint when paired with a 1 — drops into an empty cell beside the constrained region and sets off a two-step chain of sums down the left and across the diagonal.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Two sum constraints, one pip value in common
The grid carries a two-cell sum constraint along the left column and a single-cell sum constraint just to its right — both targeting the same total. Scan today's easy set and ask: which pip value could satisfy both a shared column sum and a standalone cell at once? That value points directly to the domino you need and its orientation.
💡 Place that domino, then follow the chain of sums
Once the opening domino is placed, two more sum regions come into focus in sequence. The first uses the pip value you just placed to name the other half of its pair. That second value feeds the next sum region below it — and a three-cell equals region near the bottom can only resolve after you know what value threads through it.
💡 Full solution
Place [6|0] horizontally in the top row: 6 at (0,1) (satisfying the single-cell sum=6), 0 at (0,0). Sum=6 at (0,0)+(1,0)=6: (0,0)=0, so (1,0)=6. The [3|6] domino orients with 6 at (1,0) and 3 at (2,0), placed vertically. Sum=5 at (1,2)+(2,2)=5: (1,2) is from the [3|0] domino — 3 at (1,2) and 0 at (1,3). So (2,2)=2. The [2|6] domino orients with 2 at (2,2) and 6 at (3,2), placed vertically. Sum=6 at (3,2)=6 ✓. Equals region at (2,0),(3,0),(3,1): (2,0)=3, so all three must be 3. The [3|3] double fills (3,0)=3 and (3,1)=3 horizontally ✓. Puzzle complete.
💡 The five-cell equals region has a double inside it
A single equals region snakes from the top of the grid down through five cells. Today's set contains two doubles — and one of them can be placed inside that region such that both of its identical faces land in constrained cells. Once that double is placed, all five cells lock to the same value instantly.
💡 Two less-than constraints at the bottom each identify their domino outright
The bottom two cells each carry less-than constraints with tight thresholds. Check the remaining dominoes after the equals region is resolved — for each threshold, only one pip face from the available tiles can fit below the limit. Both cells resolve independently and then cascade upward into the two-cell equals region in the center.
💡 Full solution
The five-cell equals region spans (0,1),(1,1),(1,2),(2,2),(3,2). The [6|6] double placed vertically at (1,2) and (2,2) forces both to 6 — all five cells must therefore equal 6. The [0|6] domino covers (0,0) and (0,1) horizontally: 0 at (0,0), 6 at (0,1) ✓. The [5|6] domino covers (1,0) and (1,1) horizontally: 5 at (1,0), 6 at (1,1) ✓. The [3|3] double fills the two empty cells (0,2) and (0,3) horizontally. Less-than-2 at (4,1): only pip 0 or 1 qualifies; [0|4] placed vertically with 0 at (4,1) and 4 at (3,1). Less-than-4 at (4,2): [2|6] placed vertically with 2 at (4,2) and 6 at (3,2) — but (3,2) must equal 6 ✓. Equals region (2,1),(3,1): (3,1)=4, so (2,1)=4. The [4|2] domino covers (2,1) and (2,0): 4 at (2,1), 2 at (2,0). Unequal check at (0,0),(1,0),(2,0): {0, 5, 2} — all distinct ✓. Puzzle complete.
💡 Count the doubles, count the equals regions
Three doubles in today's hard set, three equals regions on the board. Before placing anything, figure out which double belongs to which region. The sizes of the regions and the pip values of the doubles will narrow this down quickly — one assignment is forced by the sum constraints nearby.
💡 The [6|6] double and the sum-7 region work together
One of the three equals regions isn't the obvious home for the [6|6] double — a sum constraint next to it tells you why. A sum-7 region adjacent to an empty cell can only be satisfied if the [6|6] double occupies the empty cell and the constrained pair. Place the double there and the sum-7 forces one pip value, which then cascades into two more sum constraints below it.
💡 Full solution
Three doubles: [6,6], [4,4], [0,0]. Three equals regions: three-cell (0,5),(0,6),(1,5); three-cell (1,4),(2,4),(3,4); four-cell (2,5),(2,6),(3,5),(3,6). [6,6] goes to the empty cell + sum-7: place [6,6] horizontally at (1,0) and (1,1); (1,1)=6, so (1,2)=1 via sum-7 at (1,1)+(1,2)=7. The [4,1] domino covers (2,2) and (1,2): 4 at (2,2), 1 at (1,2) ✓. Sum-7 at (2,2)+(3,2)=7: (2,2)=4, so (3,2)=3. The [3,0] domino covers (3,2) and (3,1): 3 at (3,2), 0 at (3,1). Less-than-6 at (3,1)=0 ✓. [0,0] double → three-cell equals (1,4),(2,4),(3,4): place [0,0] vertically at (2,4),(3,4); (1,4)=0 from [2,0] domino (pip0 at (1,4), pip2 at (1,5)). Equals check: all three cells = 0 ✓. Three-cell equals (0,5),(0,6),(1,5): (1,5)=2, so all three = 2. The [2,3] domino covers (0,5),(0,4): 2 at (0,5) ✓, 3 at (0,4). Sum-3 at (0,4)=3 ✓. The [2,5] domino covers (0,6),(0,7): 2 at (0,6) ✓, 5 at (0,7). Sum-6 at (0,7)+(1,7)=6: (0,7)=5, so (1,7)=1. The [1,0] domino covers (1,7),(2,7): 1 at (1,7) ✓, 0 at (2,7). Equals at (2,7),(3,7): (2,7)=0, so (3,7)=0. The [0,4] domino covers (3,7),(3,6): 0 at (3,7) ✓, 4 at (3,6). [4,4] double → four-cell equals (2,5),(2,6),(3,5),(3,6): place [4,4] at (2,5),(3,5) — both = 4. (3,6)=4 ✓. The [4,6] domino covers (2,6),(1,6): 4 at (2,6) ✓, 6 at (1,6). Sum-6 at (1,6)=6 ✓. Sum-11 at (2,0)+(3,0)=11: the [6,5] domino covers (3,0),(2,0): 6 at (3,0), 5 at (2,0). 5+6=11 ✓. Puzzle complete.
🎨 Pips Solver
Apr 6, 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: Single-cell sum=6 at (0,1) names your first domino
Cell (0,1) must show exactly 6 pips. Today's easy set has three dominoes with a 6-pip face: [3|6], [2|6], and [6|0]. Only [6|0] can orient its 6-pip face into (0,1) while placing the other half in the adjacent cell (0,0). Place [6|0] horizontally: 6 at (0,1) ✓, 0 at (0,0).
2
Step 2: Sum=6 in the left column forces the second domino
The region at (0,0) and (1,0) must total 6. Cell (0,0)=0 is now fixed, so (1,0) must be exactly 6. The only remaining domino with a 6-pip face that can reach (1,0) is [3|6]. Place it vertically: 6 at (1,0), 3 at (2,0).
3
Step 3: Sum=5 across the middle row places two more dominoes
The region at (1,2) and (2,2) must total 5. No value is fixed for either cell yet, but the remaining dominoes constrain your choices. The [3|0] domino covers (1,2) and (1,3) horizontally: 3 at (1,2), 0 at (1,3) — (1,3) carries an empty constraint, any value is valid. With (1,2)=3, cell (2,2) must be 2. The [2|6] domino covers (2,2) and (3,2) vertically: 2 at (2,2), 6 at (3,2).
4
Step 4: Sum=6 at (3,2) checks out, and the double closes the equals region
Cell (3,2)=6 satisfies the single-cell sum=6 constraint ✓. The final region — equals at (2,0),(3,0),(3,1) — requires all three cells to share the same value. Cell (2,0)=3 is already fixed from Step 2. So (3,0) and (3,1) must both be 3. The [3|3] double placed horizontally at (3,0) and (3,1) delivers exactly that. Equals check: {3, 3, 3} ✓. All constraints satisfied. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The [6|6] double anchors the five-cell equals region
The equals region spans five cells: (0,1),(1,1),(1,2),(2,2),(3,2). Any double placed inside it immediately pins two adjacent cells to the same value, forcing every other cell in the region to match. The [6|6] double is the natural anchor: placed vertically at (1,2) and (2,2), it sets both to 6. All five cells must therefore equal 6.
2
Step 2: Two more dominoes fill the equals region from its ends
With (1,1) and (0,1) both needing to equal 6, and (3,2) also requiring 6, the remaining dominoes with 6-pip faces slot in immediately. The [0|6] domino covers (0,0) and (0,1) horizontally: 0 at (0,0), 6 at (0,1) ✓. The [5|6] domino covers (1,0) and (1,1) horizontally: 5 at (1,0), 6 at (1,1) ✓. The [2|6] domino covers (4,2) and (3,2) vertically: 2 at (4,2), 6 at (3,2) ✓.
3
Step 3: The [3|3] double fills the two empty cells
The two empty-type cells (0,2) and (0,3) accept any value — and the [3|3] double is the only remaining domino that fits that slot. Place it horizontally: 3 at (0,2), 3 at (0,3). Both empty constraints satisfied ✓.
4
Step 4: Less-than constraints at the bottom each lock a domino
Cell (4,1) must be strictly less than 2 — so it holds 0 or 1. Scanning the remaining dominoes, [0|4] is the only tile with a face below 2. Place [0|4] vertically: 0 at (4,1) — less-than-2 ✓ — and 4 at (3,1). Cell (4,2) must be strictly less than 4 — so 0, 1, 2, or 3. The remaining domino is [4|2]. Place it vertically: 2 at (4,2) — less-than-4 ✓ — and 4 at (3,2). But wait — (3,2) is already 6 from Step 2. Check [2|6] again: (4,2)=2 and (3,2)=6 ✓. The final domino [4|2] covers (2,1) and (2,0): 4 at (2,1), 2 at (2,0).
5
Step 5: The center equals region and the unequal check close the puzzle
Equals region at (2,1) and (3,1): (3,1)=4 from Step 4, so (2,1) must also be 4. Cell (2,1)=4 is confirmed from [4|2] ✓. Unequal region at (0,0),(1,0),(2,0): values are {0, 5, 2} — all distinct ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Identify which double belongs to which equals region
Three doubles in today's set: [6|6], [4|4], [0|0]. Four equals-type regions on the board: three-cell (0,5),(0,6),(1,5); three-cell (1,4),(2,4),(3,4); four-cell (2,5),(2,6),(3,5),(3,6); two-cell (2,7),(3,7). The two-cell region doesn't require a double. Assign the three doubles to the three multi-cell regions by checking which pip values are consistent with surrounding sum constraints.
2
Step 2: [6|6] goes beside the empty cell — and sets off sum-7
The empty cell at (1,0) is adjacent to the sum-7 region at (1,1),(1,2). The [6|6] double placed horizontally at (1,0) and (1,1) puts 6 into (1,1). The sum-7 constraint then forces (1,2)=1. The [4|1] domino covers (2,2) and (1,2) vertically: 4 at (2,2), 1 at (1,2) ✓.
3
Step 3: Sum-7 at (2,2)+(3,2) and the less-than-6 cell
With (2,2)=4, the sum-7 region at (2,2)+(3,2) requires (3,2)=3. The [3|0] domino covers (3,2) and (3,1): 3 at (3,2), 0 at (3,1). Less-than-6 at (3,1)=0 — well under the limit ✓.
4
Step 4: [0|0] double anchors the three-cell equals column of zeros
The equals region (1,4),(2,4),(3,4) must all share one value. The [0|0] double placed vertically at (2,4) and (3,4) pins both to 0, forcing (1,4)=0 as well. The [2|0] domino covers (1,5) and (1,4): 2 at (1,5), 0 at (1,4) ✓.
5
Step 5: Three-cell equals at (0,5),(0,6),(1,5) — all must be 2
Cell (1,5)=2 is now fixed from Step 4. The equals region requires (0,5) and (0,6) to both equal 2. The [2|3] domino covers (0,5) and (0,4): 2 at (0,5) ✓, 3 at (0,4). Single-cell sum-3 at (0,4)=3 ✓. The [2|5] domino covers (0,6) and (0,7): 2 at (0,6) ✓, 5 at (0,7).
6
Step 6: Sum-6 at (0,7)+(1,7) and the two-cell equals region
Cell (0,7)=5 is fixed from Step 5. Sum-6 requires (1,7)=1. The [1|0] domino covers (1,7) and (2,7): 1 at (1,7) ✓, 0 at (2,7). Equals at (2,7),(3,7): (2,7)=0, so (3,7)=0. The [0|4] domino covers (3,7) and (3,6): 0 at (3,7) ✓, 4 at (3,6).
7
Step 7: [4|4] double locks the four-cell equals region to 4
The equals region (2,5),(2,6),(3,5),(3,6) requires one shared value. Cell (3,6)=4 is now fixed from Step 6. So all four cells must be 4. The [4|4] double placed vertically at (2,5) and (3,5) delivers 4 to both cells ✓. The [4|6] domino covers (2,6) and (1,6): 4 at (2,6) ✓, 6 at (1,6). Single-cell sum-6 at (1,6)=6 ✓.
8
Step 8: Sum-11 at the left edge closes the puzzle
The final region at (2,0) and (3,0) must total 11. The only remaining domino is [6|5]. Place it vertically: 6 at (3,0), 5 at (2,0) — 5+6=11 ✓. All twelve constraints across the board are 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.
💬 Community Discussion
Leave your comment