NYT Pips Hints & Answers for April 10, 2026

Apr 10, 2026

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 10th is bookended by two greater-than-5 constraints — one at the top-left corner, one at the bottom-right. Both cells must show 6 pips, and there are exactly two dominoes in today's set with a 6-pip face. Once each 6 is placed, their partners drop into position and a neat three-cell sum=6 region fills from three identical 2-pip faces. The center of the board is held together by an equals pair and a sum=5 that each resolve from what the flanking placements have already established.

Ian Livengood's medium puzzle is controlled by two tight inequality constraints that sit at opposite corners. The less-than-2 cell at the lower-left is the entry point: only pip 0 or pip 1 qualifies, and of those, only pip 0 can supply a useful face for the domino that fits that slot. That zero placement feeds a greater-than-4 constraint one column over, which in turn locks the three-cell equals region running through the center of the board to a single shared value of 6. A second three-cell equals region below it fills entirely with 4s, anchored by the only double in today's set. The top row settles last, closed by a two-cell equals pair and a single-cell sum.

Rodolfo Kurchan's hard puzzle for April 10th contains four doubles — [0|0], [1|1], [5|5], and [6|6] — and each one anchors a distinct region. The four-cell equals region in the upper-left corner is the natural starting point: only [5|5] can lock two adjacent cells there and force the remaining two cells to match. From that anchor, sum=4 constraints along the top row resolve one after the other, triggering a cascade of equals regions that flows diagonally down the right side of the board: 4s, then 0s. The [6|6] double surfaces at the bottom to close a sum=6 cell and feed the seven-cell unequal region — which demands all seven pip values from 0 through 6, and receives them in order once the surrounding constraints are satisfied.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Two corners, two sixes — find both 6-pip dominoes first
The top-left corner demands a pip value strictly greater than 5 — only 6 qualifies. The bottom-right corner has the same constraint. Scan today's easy set: exactly two dominoes carry a 6-pip face. Each one belongs at one of the two corners. Place them, and the rest of the board resolves from their partners.
💡 Three identical partners fill the middle sum
Once both 6-pip faces are seated in their corners, look at the three-cell sum=6 region in the middle row. Each of the three remaining dominoes contributes its non-6 face to that row — and all three of those faces show the same pip count. Three identical values totaling 6 close the middle in one sweep.
💡 Full solution
Greater-than-5 at (0,0): [0|6] places 6 at (0,0) ✓ and 0 at (1,0). Greater-than-5 at (3,4): [2|6] places 6 at (3,4) ✓ and 2 at (2,4). Sum=6 at (2,2)+(2,3)+(2,4): (2,4)=2. [2|3] places 2 at (2,2) and 3 at (2,1) (empty ✓). [2|0] places 2 at (2,3) and 0 at (1,3). Sum check: 2+2+2=6 ✓. Equals at (1,0),(1,1): (1,0)=0, so (1,1)=0. [0|5] covers (1,1) and (1,2): 0 at (1,1) ✓, 5 at (1,2). Sum=5 at (1,2)+(1,3): (1,2)=5, (1,3)=0 → 5+0=5 ✓. All six constraints satisfied. Puzzle complete.
💡 Less-than-2 is the tightest constraint — start at the bottom-left
The cell at the bottom-left must be strictly less than 2: only pip 0 or pip 1. Check which dominoes in today's set can deliver either of those faces into that position. Only one arrangement works cleanly once you account for where the other half of the domino lands — it deposits a value directly into the greater-than-4 constraint one cell up.
💡 Greater-than-4 locks a three-cell equals region to 6
Once the bottom-left domino is placed, its partner cell feeds the greater-than-4 constraint at (1,0). Only pip 5 or 6 qualifies — and from the remaining dominoes, only one tile supplies that face. Its other half seeds the three-cell equals region running diagonally, locking all three cells to a value of 6. The [4|4] double then closes the second three-cell equals region below.
💡 Full solution
Less-than-2 at (3,0)=0: [0|2] covers (3,0) and (4,0): 0 at (3,0) ✓, 2 at (4,0) (empty ✓). Greater-than-4 at (1,0): [5|6] covers (1,0) and (2,0): 5 at (1,0) — 5>4 ✓ — and 6 at (2,0). Three-cell equals (2,0),(2,1),(3,1): (2,0)=6, so all three must equal 6. [3|6] covers (1,1) and (2,1): 3 at (1,1), 6 at (2,1) ✓. [4|6] covers (4,1) and (3,1): 4 at (4,1), 6 at (3,1) ✓. Three-cell equals (3,2),(4,1),(4,2): (4,1)=4, so all three must equal 4. [4|4] double covers (3,2) and (4,2): 4 at (3,2) ✓, 4 at (4,2) ✓. Equals (0,1),(1,1): (1,1)=3, so (0,1)=3. [4|3] covers (0,2) and (0,1): 4 at (0,2) (empty ✓), 3 at (0,1) ✓. Sum=2 at (1,3)=2: [6|2] covers (0,3) and (1,3): 6 at (0,3) (empty ✓), 2 at (1,3) ✓. All nine constraints satisfied. Puzzle complete.
💡 Four doubles, four regions — the four-cell equals corner is first
Today's hard set contains four doubles: [0|0], [1|1], [5|5], and [6|6]. The board has four regions that naturally host a double. The four-cell equals region in the upper-left corner demands the most attention: only [5|5] can anchor two adjacent cells there with a value high enough that all four cells share it. Place it and the cascade begins.
💡 Sum=4 twice — across the top row, then across the middle
After the four-cell equals region is placed, two separate sum=4 regions come into focus: one in the top row and one two rows down. Each one resolves from a value already fixed by earlier steps, and each one hands off a pip to an equals pair below it. The equals pairs form a diagonal chain: 4, then 0, with the [0|0] double anchoring the last one.
💡 Full solution
Four-cell equals (0,0),(0,1),(1,0),(2,0): [5|5] double covers (0,0) and (1,0) — both 5. (0,1)=5 and (2,0)=5. [1|5] covers (0,2) and (0,1): 1 at (0,2), 5 at (0,1) ✓. Sum=4 at (0,2)+(0,3): (0,2)=1, so (0,3)=3. [3|4] covers (0,3) and (0,4): 3 at (0,3) ✓, 4 at (0,4). Equals (0,4),(1,4): (0,4)=4, so (1,4)=4. [0|4] covers (2,4) and (1,4): 0 at (2,4), 4 at (1,4) ✓. Equals (2,4),(3,4): (2,4)=0, so (3,4)=0. [0|0] double covers (3,4) and (4,4): 0 at (3,4) ✓, 0 at (4,4). Sum=4 at (2,2)+(2,3): [1|3] covers (2,2) and (2,3): 1 at (2,2), 3 at (2,3) → 1+3=4 ✓. Greater-than-4 sum at (3,0)+(4,0): [3|5] covers (3,0) and (2,0): 3 at (3,0), 5 at (2,0) ✓. [2|6] covers (4,0) and (5,0): 2 at (4,0), 6 at (5,0). Sum: 3+2=5>4 ✓. Equals (5,0),(6,0): (5,0)=6, so (6,0)=6. [3|6] covers (6,1) and (6,0): 3 at (6,1), 6 at (6,0) ✓. [6|6] double covers (6,3) and (6,4): 6 at (6,3), 6 at (6,4). Sum=6 at (6,4)=6 ✓. Sum=4 at (6,1)+(6,2)=4: (6,1)=3, so (6,2)=1. [1|1] double covers (5,2) and (6,2): 1 at (5,2), 1 at (6,2) ✓. Unequal region (4,2),(4,3),(4,4),(5,2),(5,3),(5,4),(6,3): [2|3] covers (4,3) and (4,2): 2 at (4,3), 3 at (4,2). [4|5] covers (5,3) and (5,4): 4 at (5,3), 5 at (5,4). Values: {3,2,0,1,4,5,6} — all seven distinct ✓. All ten constraints satisfied. Puzzle complete.

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Apr 10, 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

Pips hint for April 10, 2026 – hard level puzzle grid with critical first placements and strategy

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

NYT Pips April 10, 2026 hard puzzle full solution grid showing final answer with hints

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: Greater-than-5 at the top-left corner — only 6 qualifies
Cell (0,0) must show a pip value strictly greater than 5. The only pip count that qualifies is 6. Today's set has two dominoes with a 6-pip face: [0|6] and [2|6]. The [0|6] domino covers (1,0) and (0,0): 0 at (1,0), 6 at (0,0) — greater-than-5 ✓.
2
Step 2: Greater-than-5 at the bottom-right corner — the second 6
Cell (3,4) must also show 6. The remaining domino with a 6-pip face is [2|6]. It covers (2,4) and (3,4): 2 at (2,4), 6 at (3,4) — greater-than-5 ✓.
3
Step 3: Sum=6 across the middle row — three 2-pip faces
The region at (2,2),(2,3),(2,4) must total 6. Cell (2,4)=2 is already fixed. The remaining dominoes [2|3] and [2|0] each carry a 2-pip face. [2|3] covers (2,2) and (2,1): 2 at (2,2), 3 at (2,1) — (2,1) is empty ✓. [2|0] covers (2,3) and (1,3): 2 at (2,3), 0 at (1,3). Sum: 2+2+2=6 ✓.
4
Step 4: Equals pair and sum=5 close the middle row
Equals region at (1,0),(1,1): (1,0)=0 is fixed. So (1,1) must also be 0. The [0|5] domino covers (1,1) and (1,2): 0 at (1,1) ✓, 5 at (1,2). Sum=5 at (1,2)+(1,3): (1,2)=5, (1,3)=0 — 5+0=5 ✓. All six constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Less-than-2 at (3,0) forces pip 0
Cell (3,0) must be 0 or 1. The only domino in today's set with a 0-pip face that can place it at (3,0) is [0|2]. Place [0|2] vertically: 0 at (3,0) ✓, 2 at (4,0). Cell (4,0) carries an empty constraint — any value is valid ✓.
2
Step 2: Greater-than-4 at (1,0) names the domino
Cell (1,0) must exceed 4 — so pip 5 or 6. Scanning the remaining dominoes for faces above 4: [5|6] qualifies. Place [5|6] vertically: 5 at (1,0) — 5>4 ✓ — and 6 at (2,0).
3
Step 3: Three-cell equals region locks to 6
The equals region spans (2,0),(2,1),(3,1). Cell (2,0)=6, so all three must equal 6. [3|6] covers (1,1) and (2,1): 3 at (1,1), 6 at (2,1) ✓. [4|6] covers (4,1) and (3,1): 4 at (4,1), 6 at (3,1) ✓.
4
Step 4: Three-cell equals region below locks to 4
The equals region spans (3,2),(4,1),(4,2). Cell (4,1)=4 is now fixed, so all three must equal 4. The [4|4] double covers (3,2) and (4,2): 4 at (3,2) ✓, 4 at (4,2) ✓.
5
Step 5: Equals pair at the top and sum=2 close the puzzle
Equals at (0,1),(1,1): (1,1)=3 from Step 3, so (0,1) must also be 3. [4|3] covers (0,2) and (0,1): 4 at (0,2) — empty ✓ — and 3 at (0,1) ✓. Sum=2 at (1,3)=2: [6|2] covers (0,3) and (1,3): 6 at (0,3) — empty ✓ — and 2 at (1,3) ✓. All nine constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Four-cell equals region in the upper-left — [5|5] is the only fit
The region (0,0),(0,1),(1,0),(2,0) requires all four cells to share one value. The [5|5] double placed vertically at (0,0) and (1,0) sets both to 5, forcing (0,1)=5 and (2,0)=5. No other double in today's set can anchor two adjacent cells in this region with a consistent shared value.
2
Step 2: [1|5] feeds sum=4 and hands off across the top row
Cell (0,1)=5 is fixed. The [1|5] domino covers (0,2) and (0,1): 1 at (0,2), 5 at (0,1) ✓. Sum=4 at (0,2)+(0,3)=4: (0,2)=1, so (0,3)=3. [3|4] covers (0,3) and (0,4): 3 at (0,3) ✓, 4 at (0,4).
3
Step 3: Equals pair at (0,4),(1,4) and the zero cascade
Equals at (0,4),(1,4): (0,4)=4, so (1,4)=4. [0|4] covers (2,4) and (1,4): 0 at (2,4), 4 at (1,4) ✓. Equals at (2,4),(3,4): (2,4)=0, so (3,4)=0. [0|0] double covers (3,4) and (4,4): 0 at (3,4) ✓, 0 at (4,4).
4
Step 4: Sum=4 in the middle row
The region at (2,2) and (2,3) must total 4. [1|3] covers (2,2) and (2,3): 1 at (2,2), 3 at (2,3) → 1+3=4 ✓.
5
Step 5: Greater-than-4 sum at the left edge and the equals pair below
The region (3,0)+(4,0) must exceed 4 as a combined total. [3|5] covers (3,0) and (2,0): 3 at (3,0), 5 at (2,0) ✓. [2|6] covers (4,0) and (5,0): 2 at (4,0), 6 at (5,0). Sum: 3+2=5>4 ✓. Equals (5,0),(6,0): (5,0)=6, so (6,0)=6. [3|6] covers (6,1) and (6,0): 3 at (6,1), 6 at (6,0) ✓.
6
Step 6: [6|6] double closes sum=6 and feeds the bottom sum=4
Sum=6 at (6,4)=6: the [6|6] double covers (6,3) and (6,4): 6 at (6,3), 6 at (6,4) ✓. Sum=4 at (6,1)+(6,2)=4: (6,1)=3, so (6,2)=1. [1|1] double covers (5,2) and (6,2): 1 at (5,2), 1 at (6,2) ✓.
7
Step 7: The seven-cell unequal region fills with all pip values 0–6
The unequal region (4,2),(4,3),(4,4),(5,2),(5,3),(5,4),(6,3) requires all seven cells to show distinct pip values. Four cells are already fixed: (4,4)=0, (5,2)=1, (6,3)=6. [2|3] covers (4,3) and (4,2): 2 at (4,3), 3 at (4,2). [4|5] covers (5,3) and (5,4): 4 at (5,3), 5 at (5,4). Final set: {3,2,0,1,4,5,6} — all seven pip values represented ✓. All ten 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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