NYT Pips Hints & Answers for April 12, 2026

Apr 12, 2026

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 12th opens at the far right of the top row. A single-cell sum=4 constraint at the row's edge names the only domino that can place a 4-pip face there, and its partner carries a 5 into the adjacent top-row cell. The four-cell sum=22 spanning the rest of the top row then closes around the [6|6] double — the only tile that can supply the required bulk — while a 5-pip face from the last available domino completes the total. Below the midpoint, the [2|2] double drops cleanly into the two-cell sum=4 region, and two non-double dominoes with matching 4-pip faces resolve the equals constraint and the lone greater-than check to finish the board.

Ian Livengood's medium puzzle for April 12th is anchored by a three-cell equals region in the lower-center of the board. Only the [5|5] double can cover the horizontal pair of cells in that region with two equal values, forcing the third cell to follow. That single placement cascades immediately into the sum=10 region to the right — two dominoes close it in sequence — and a sum=1 constraint at the corner, the tightest two-cell constraint on today's medium board, names its domino without ambiguity. The two single-cell inequality constraints at opposite ends of the grid are each answered by the one remaining domino that satisfies them once the anchor chain is complete.

Rodolfo Kurchan's hard puzzle for April 12th is defined by three separate sum=0 regions. A single-cell sum=0 names its domino immediately — only one tile can place a 0-pip face at that isolated cell — and the partner face plants a known value in the adjacent cell above it. That value propagates through a three-cell sum=6 at the top-left corner, where the [2|2] double fills the two remaining cells and closes the region in one move. Two double-zero constraints elsewhere on the board are each satisfied by a pair of distinct dominoes contributing a 0-pip face apiece, and a four-cell equals region in the lower-left column — the board's longest equals run — is anchored entirely by the [4|4] double, which pins three of the four cells to 4 while the fourth arrives through the chain above.

💡 Progressive Hints

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

💡 Single-cell sum=4 on the right edge names the first domino
Cell (0,4) at the far right of the top row must show exactly 4 pips. Scan today's easy set for a domino with a 4-pip face that can reach that position — there is only one valid placement. Its partner face carries a 5 into the adjacent top-row cell (0,3), giving you two fixed values before anything else.
💡 Sum=22 across the top row: the [6|6] double and a 5-pip face
With (0,3)=5 already known, the four-cell sum=22 needs 17 more pips from three cells. The [6|6] double is the only tile that supplies enough bulk — place it at (0,0) and (0,1) for 12 pips combined, leaving (0,2) needing exactly 5. One domino in today's set has a 5-pip face pointing into (0,2), and its partner lands in the empty cell below at (1,2). The entire top row is now closed: 6+6+5+5=22.
💡 Full solution
Sum=4 at (0,4): [4,5] covers (0,4) and (0,3) — 4 at (0,4) ✓, 5 at (0,3). Sum=22 at (0,0)+(0,1)+(0,2)+(0,3): [6,6] double covers (0,0) and (0,1) — 6 at each. [5,0] covers (0,2) and (1,2) — 5 at (0,2), 0 at (1,2) (empty ✓). Sum: 6+6+5+5=22 ✓. Sum=4 at (2,0)+(2,1): [2,2] double covers (2,0) and (2,1) — 2+2=4 ✓. Equals at (2,2)+(2,3): [1,4] covers (3,2) and (2,2) — 1 at (3,2) (empty ✓), 4 at (2,2). [2,4] covers (2,4) and (2,3) — 2 at (2,4), 4 at (2,3). Equals: 4=4 ✓. Greater-than-1 at (2,4)=2 ✓. All seven constraints satisfied. Puzzle complete.
💡 The [5|5] double locks the three-cell equals region
The equals region at (2,1), (2,2), and (3,1) requires all three cells to share one pip value. Of today's medium dominoes, only the [5|5] double can cover the horizontal pair (2,1) and (2,2) with two equal values — no non-double fits cleanly here under the surrounding constraints. That placement forces (3,1) to also equal 5, resolving three cells at once and opening the rest of the board.
💡 Sum=10 and sum=1 cascade from the anchor
With (3,1)=5 fixed, a domino placing 5 at (3,1) also reveals the value at (3,0), satisfying the less-than-3 constraint there. From (3,1), the sum=10 region to the right closes in two steps: one domino covers the empty cell and supplies a 5 into (2,5), and another places 5 at (3,5) and 0 at (3,4). That 0 at (3,4) immediately solves the sum=1 constraint at (3,3)+(3,4) — the tightest two-cell constraint on today's medium board — naming the domino that occupies the corner.
💡 Full solution
Equals at (2,1),(2,2),(3,1): [5,5] double covers (2,1) and (2,2) — 5+5. (3,1) must also be 5. [2,5] covers (3,0) and (3,1) — 2 at (3,0) < 3 ✓, 5 at (3,1) ✓. Sum=10 at (2,5)+(3,5): [4,5] covers (2,4) and (2,5) — 4 at (2,4) (empty ✓), 5 at (2,5). [5,0] covers (3,5) and (3,4) — 5 at (3,5), 0 at (3,4). Sum: 5+5=10 ✓. Sum=1 at (3,3)+(3,4): (3,4)=0, so (3,3)=1. [1,3] covers (3,3) and (4,3) — 1 at (3,3) ✓, 3 at (4,3) (empty ✓). Greater-than-5 at (1,0): must be 6. [6,4] covers (1,0) and (1,1) — 6 at (1,0) ✓, 4 at (1,1). Equals at (1,1),(1,2): (1,1)=4, so (1,2)=4. [4,2] covers (1,2) and (0,2) — 4 at (1,2) ✓, 2 at (0,2). Less-than-3 at (0,2)=2 ✓. Equals at (1,3),(2,3): [6,6] double covers (1,3) and (2,3) — 6=6 ✓. All ten constraints satisfied. Puzzle complete.
💡 Sum=0 at a single cell names its domino and starts the chain
The single-cell region at (2,0) must hold exactly 0 pips. Only one domino in today's hard set can place a 0-pip face at that cell — its partner face lands in the adjacent cell above, fixing that value as well. That known value propagates immediately into the three-cell sum=6 at the top-left corner, where the [2|2] double is the only tile that can fill the remaining two cells and reach the required total.
💡 Three sum=0 regions, each solved by 0-pip contributions from distinct dominoes
Today's hard board has two additional double-zero constraints — two-cell regions that must sum to 0 — meaning both cells in each must hold 0 pips. Neither can be solved by a [0|0] double (none exists in today's set); instead, each is satisfied by two separate dominoes each contributing one 0-pip face. Identifying which dominoes supply these four 0-pip placements unlocks a chain that runs through the sum=6 and less-than-6 constraints across the top of the board and down into the four-cell equals region in the lower-left column.
💡 Full solution
Sum=0 at (2,0): (2,0)=0. [2,0] covers (1,0) and (2,0) — 2 at (1,0), 0 at (2,0) ✓. Sum=6 at (0,0),(0,1),(1,0): (1,0)=2, so (0,0)+(0,1)=4. [2,2] double covers (0,0) and (0,1) — 2+2=4 ✓. Sum=0 at (0,2)+(1,2): both cells must be 0. [3,0] covers (0,3) and (0,2) — 3 at (0,3), 0 at (0,2). [6,0] covers (2,2) and (1,2) — 6 at (2,2) (empty ✓), 0 at (1,2). Sum: 0+0=0 ✓. Less-than-6 at (0,3)+(0,4): (0,3)=3. [2,1] covers (0,4) and (1,4) — 2 at (0,4), 1 at (1,4). Sum: 3+2=5 < 6 ✓. Sum=6 at (1,4)+(2,4): (1,4)=1, so (2,4)=5. Sum=0 at (3,4)+(4,4): both cells must be 0. [0,5] covers (3,4) and (2,4) — 0 at (3,4), 5 at (2,4) ✓. [1,0] covers (5,4) and (4,4) — 1 at (5,4), 0 at (4,4). Sum: 0+0=0 ✓. Sum=6 at (5,4)+(6,4): (5,4)=1, so (6,4)=5. [6,5] covers (6,3) and (6,4) — 6 at (6,3), 5 at (6,4) ✓. Equals at (6,2),(6,3): (6,3)=6, so (6,2)=6. [4,6] covers (6,1) and (6,2) — 4 at (6,1), 6 at (6,2) ✓. Equals at (4,0),(5,0),(6,0),(6,1): (6,1)=4, so all four must be 4. [4,4] double covers (5,0) and (6,0) — 4+4. [2,4] covers (3,0) and (4,0) — 2 at (3,0) (empty ✓), 4 at (4,0) ✓. All four cells = 4 ✓. Sum=6 at (4,2)+(5,2): [3,3] double covers (4,2) and (5,2) — 3+3=6 ✓. All twelve constraints satisfied. Puzzle complete.

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Apr 12, 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 12, 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 12, 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: Sum=4 at (0,4) identifies the first placement
Cell (0,4) must show exactly 4 pips. Today's easy set includes [2,4], [1,4], [4,5], and [2,2] as dominoes with a 4-pip face. Of these, only [4,5] can orient its 4-pip face into (0,4) from the adjacent cell (0,3). Place [4,5] horizontally: 4 at (0,4) ✓, 5 at (0,3).
2
Step 2: Sum=22 across the top row — [6|6] and the 5-pip domino
The region at (0,0),(0,1),(0,2),(0,3) must total 22. With (0,3)=5 fixed, the remaining three cells must sum to 17. The [6,6] double at (0,0) and (0,1) contributes 12, leaving (0,2) needing exactly 5. The [5,0] domino covers (0,2) and (1,2): 5 at (0,2) gives the required pip, and 0 at (1,2) satisfies the empty constraint. Sum: 6+6+5+5=22 ✓.
3
Step 3: Sum=4 at (2,0)+(2,1) — the [2|2] double
The two-cell region at (2,0) and (2,1) must total 4. The [2,2] double covers both cells: 2+2=4 ✓. No other remaining domino reaches this total across these two positions.
4
Step 4: Equals at (2,2)+(2,3) and greater-than-1 at (2,4) close the board
The equals region requires (2,2) and (2,3) to match. Remaining dominoes are [2,4] and [1,4]. [1,4] covers (3,2) and (2,2): 1 at (3,2) (empty ✓), 4 at (2,2). [2,4] covers (2,4) and (2,3): 2 at (2,4), 4 at (2,3). Equals: 4=4 ✓. Greater-than-1 at (2,4)=2 ✓. All seven constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: [5|5] double anchors the three-cell equals region
The equals region at (2,1),(2,2),(3,1) requires all three cells to share one value. The [5,5] double placed horizontally at (2,1) and (2,2) pins both to 5, forcing (3,1) to also equal 5. Three cells resolved in one placement.
2
Step 2: (3,1)=5 names the domino for the left-bottom corner
Cell (3,1)=5 is now fixed. The [2,5] domino covers (3,0) and (3,1): 2 at (3,0), 5 at (3,1) ✓. Less-than-3 at (3,0)=2 ✓.
3
Step 3: Sum=10 at (2,5)+(3,5) — two dominoes in sequence
The sum=10 region needs total 10 from two cells. [4,5] covers (2,4) and (2,5): 4 at (2,4) (empty ✓), 5 at (2,5). That forces (3,5)=5 for the sum to work. [5,0] covers (3,5) and (3,4): 5 at (3,5) ✓, 0 at (3,4). Sum: 5+5=10 ✓.
4
Step 4: Sum=1 at (3,3)+(3,4) — the tightest constraint on the board
Cell (3,4)=0 is now fixed. The two-cell sum=1 requires (3,3)+(3,4)=1, so (3,3)=1. The [1,3] domino covers (3,3) and (4,3): 1 at (3,3) ✓, 3 at (4,3) (empty ✓).
5
Step 5: Greater-than-5, equals chains, and the [6|6] double close the board
Greater-than-5 at (1,0) requires 6. [6,4] covers (1,0) and (1,1): 6 at (1,0) ✓, 4 at (1,1). Equals at (1,1),(1,2): (1,1)=4, so (1,2) must also be 4. [4,2] covers (1,2) and (0,2): 4 at (1,2) ✓, 2 at (0,2). Less-than-3 at (0,2)=2 ✓. Equals at (1,3),(2,3): the remaining [6,6] double covers both — 6=6 ✓. All ten constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum=0 at (2,0) — the board's entry point
A single cell forced to 0 pips means the domino covering it must contribute a 0-pip face there. Of today's hard set, the [2,0] domino covers (1,0) and (2,0): 2 at (1,0), 0 at (2,0) ✓. Cell (1,0)=2 is now fixed.
2
Step 2: Sum=6 at (0,0),(0,1),(1,0) — the [2|2] double
With (1,0)=2, the three-cell sum=6 needs (0,0)+(0,1)=4. The [2,2] double placed at (0,0) and (0,1) gives 2+2=4 ✓. Sum: 2+2+2=6 ✓.
3
Step 3: Sum=0 at (0,2)+(1,2) — two separate 0-pip contributions
Both cells must hold 0 pips. There is no [0|0] double in today's set, so two separate dominoes each supply a 0. [3,0] covers (0,3) and (0,2): 3 at (0,3), 0 at (0,2). [6,0] covers (2,2) and (1,2): 6 at (2,2) (empty ✓), 0 at (1,2). Sum: 0+0=0 ✓.
4
Step 4: Less-than-6 at (0,3)+(0,4) and sum=6 at (1,4)+(2,4)
(0,3)=3 is now fixed. The less-than-6 region needs 3+(0,4)<6, so (0,4)<3. [2,1] covers (0,4) and (1,4): 2 at (0,4), 1 at (1,4). Sum: 3+2=5<6 ✓. Sum=6 at (1,4)+(2,4): (1,4)=1, so (2,4)=5.
5
Step 5: Sum=0 at (3,4)+(4,4) — another double-zero constraint
Both cells must be 0. [0,5] covers (3,4) and (2,4): 0 at (3,4), 5 at (2,4) ✓ (matching the required 5 from step 4). [1,0] covers (5,4) and (4,4): 1 at (5,4), 0 at (4,4). Sum: 0+0=0 ✓.
6
Step 6: Sum=6 at (5,4)+(6,4) and equals at (6,2),(6,3)
(5,4)=1, so (6,4)=5. [6,5] covers (6,3) and (6,4): 6 at (6,3), 5 at (6,4) ✓. Equals at (6,2),(6,3): (6,3)=6, so (6,2) must also be 6. [4,6] covers (6,1) and (6,2): 4 at (6,1), 6 at (6,2) ✓.
7
Step 7: Four-cell equals at (4,0),(5,0),(6,0),(6,1) — the [4|4] double
(6,1)=4, so all four cells must be 4. [4,4] double covers (5,0) and (6,0): 4+4. [2,4] covers (3,0) and (4,0): 2 at (3,0) (empty ✓), 4 at (4,0) ✓. All four cells confirmed equal to 4 ✓.
8
Step 8: Sum=6 at (4,2)+(5,2) — the last domino
The only remaining domino is [3,3]. It covers (4,2) and (5,2): 3+3=6 ✓. All twelve 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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