NYT Pips Hints & Answers for April 15, 2026

Apr 15, 2026

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🎲 Today's Puzzle Overview

Ian Livengood's easy grid for April 15th is anchored by the greater-than-4 cell at (1,6). The only pip values that satisfy the constraint are 5 and 6, and of the six dominoes in today's easy set only [0|5] carries either — its 5-pip face drops straight into (1,6) and deposits a 0 at (1,5). That 0 immediately seeds the three-cell equals region at (0,4),(0,5),(1,5): all three must be 0, and the [0|0] double fills two of them outright. The three-cell equals column at (1,3),(2,3),(3,3) follows next — [3|3] covers two cells and forces the third — and a short chain through two equals pairs and a less-than-2 cell closes the board.

Ian Livengood's medium puzzle for April 15th opens on the bottom-right. The greater-than-5 cell at (3,4) admits only pip 6, and [1|6] is the only tile that can deliver it, depositing a 1 at (3,3). A perpendicular entry at the single-cell sum=2 constraint at (3,1) identifies [3|2] without ambiguity, giving (3,2)=3. With (3,2)=3 and (3,3)=1 already placed in the four-cell unequal region, [5|2] is the only remaining domino whose two faces are both different from 3, 1, and each other. From there the less-than-1 constraint forces both (0,3) and (1,3) to pip 0, a cascade of equals constraints flows across row 1, and two single-cell sum=3 clues close the board.

Rodolfo Kurchan's hard puzzle for April 15th is built around four interlocking equals regions and a four-cell sum column whose total of 23 nearly saturates the maximum. The entry point is the five-cell equals region stretching across the upper-left corner — five cells that must all carry the same pip. Only the [4|4] double can seed two of them from a single tile, and once placed, three more dominoes complete the region and deposit a 1 at (0,2). That value feeds a sum=4 constraint that reads off (0,3)=3 and (0,4)=5, which in turn forces the entire sum=23 column to commit to three 6s. From there, the [6|6] double places itself at (1,4) and (2,4), and the cascade continues through two more equals regions, two sum constraints at the bottom, and a less-than-5 cell before the final domino closes the board.

💡 Progressive Hints

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

💡 Greater>4 at (1,6) names one domino outright
The only pip values strictly greater than 4 are 5 and 6. Scan today's easy set for a domino that carries a 5 or a 6: only [0|5] qualifies. Place it with 5 at (1,6) — greater>4 ✓ — and its other face, 0, sitting at (1,5). The three-cell equals region at (0,4),(0,5),(1,5) now has its anchor pip.
💡 (1,5)=0 seeds two equals regions; [0|0] double and [3|3] double close them both
All three cells in the equals region at (0,4),(0,5),(1,5) must match (1,5)=0. The [0|0] double covers (0,4) and (0,5): 0 at both ✓. Across the board, the three-cell equals at (1,3),(2,3),(3,3) is waiting for the [3|3] double — it covers (2,3) and (3,3), forcing (1,3)=3 as well, which locks the domino above it.
💡 Full solution
Greater>4 at (1,6)=5: [0|5] covers (1,5) and (1,6) — 0 at (1,5), 5 at (1,6) ✓. Equals (0,4),(0,5),(1,5): all must be 0. [0|0] covers (0,4) and (0,5): 0 at both ✓. Equals (1,3),(2,3),(3,3): [3|3] covers (2,3) and (3,3): 3 at both ✓. So (1,3)=3. [2|3] covers (0,3) and (1,3): 3 at (1,3) ✓, 2 at (0,3). Equals (0,2),(0,3): (0,3)=2, so (0,2)=2. [2|1] covers (0,2) and (0,1): 2 at (0,2) ✓, 1 at (0,1). Equals (0,1),(1,1): (0,1)=1, so (1,1)=1. [1|1] covers (1,0) and (1,1): 1 at both ✓. Less<2 at (1,0)=1 ✓. All six constraints satisfied. Puzzle complete.
💡 Greater>5 at (3,4) is the sharpest entry — only pip 6 qualifies, only [1|6] delivers it
Cell (3,4) must be strictly greater than 5 — so pip 6, the only possibility. Scan today's medium domino set: [1|6] is the sole tile with a 6-pip face. Place it with 6 at (3,4) — greater>5 ✓ — and 1 at (3,3). That 1 immediately enters the four-cell unequal region below.
💡 Sum=2 at (3,1) anchors [3|2]; unequal region fills; less<1 forces (0,3) and (1,3) both to pip 0
Sum=2 at (3,1) means (3,1)=2 exactly. Only [3|2] can cover (3,1) and (3,2) with 2 at (3,1): that gives (3,2)=3. The unequal region now has (3,2)=3 and (3,3)=1 — the remaining two cells (4,2) and (4,3) must each differ from 3, 1, and each other: [5|2] provides pips 5 and 2, all four values distinct ✓. Separately, less<1 at (0,3) and (1,3) forces both to pip 0; [0|1] and [4|0] deliver them.
💡 Full solution
Greater>5 at (3,4)=6: [1|6] covers (3,3) and (3,4) — 1 at (3,3), 6 at (3,4) ✓. Sum=2 at (3,1)=2: [3|2] covers (3,2) and (3,1) — 3 at (3,2), 2 at (3,1) ✓. Unequal (3,2),(3,3),(4,2),(4,3): values 3 and 1 are taken; [5|2] covers (4,2) and (4,3) — 5 and 2, all four distinct ✓. Less<1 at (0,3),(1,3): both must be 0. [0|1] covers (0,3) and (0,2) — 0 at (0,3) ✓, 1 at (0,2) — empty ✓. [4|0] covers (1,4) and (1,3) — 4 at (1,4), 0 at (1,3) ✓. Equals (1,4),(1,5): (1,4)=4, so (1,5)=4. [4|3] covers (1,5) and (2,5) — 4 at (1,5) ✓, 3 at (2,5) — sum=3 ✓. Equals (1,0),(1,1): [5|3] covers (1,0) and (2,0) — 5 at (1,0), 3 at (2,0) — sum=3 ✓. [5|1] covers (1,1) and (1,2) — 5 at (1,1) — equals ✓, 1 at (1,2) — empty ✓. All nine constraints satisfied. Puzzle complete.
💡 The five-cell equals region dominates the upper left — [4|4] double is the only anchor
Five cells — (0,0),(0,1),(1,0),(2,0),(2,1) — must all carry the same pip. A double is the only way to guarantee two identical faces from one tile. Today's hard set contains [4|4], [5|5], [6|6], and [2|2] as doubles; only [4|4] can fit inside this region. Place it at (0,0) and (1,0): pip 4 at both. The remaining three cells of the region must also equal 4, and three dominoes — [1|4], [4|0], and [4|5] — each contribute one 4-face to complete it.
💡 Sum=23 in the four-cell column nearly saturates the maximum; [6|6] is forced, then a cascade of equals regions follows
Cells (0,4),(1,4),(2,4),(3,4) must sum to 23. The maximum four-cell sum is 24 (four 6s), so the column must hold three 6s and one 5. [1|4] placed in step 1 deposits 1 at (0,2), which feeds sum=4 at (0,2)+(0,3)=4: (0,3)=3. [3|5] places 3 at (0,3) and 5 at (0,4) — that 5 is the lone non-6 in the column, so [6|6] goes at (1,4) and (2,4), and (3,4)=6. From (3,4)=6, [5|6] places 5 at (4,4), seeding the three-cell equals region at (3,3),(4,3),(4,4) to pip 5.
💡 Full solution
[4|4] covers (0,0) and (1,0): 4 at both ✓ — five-cell equals seeds at pip 4. [1|4] covers (0,2) and (0,1): 4 at (0,1) ✓, 1 at (0,2). [4|0] covers (2,0) and (3,0): 4 at (2,0) ✓, 0 at (3,0) — empty ✓. [4|5] covers (2,1) and (3,1): 4 at (2,1) ✓, 5 at (3,1). Sum=4 at (0,2),(0,3): (0,2)=1, so (0,3)=3. [3|5] covers (0,3) and (0,4): 3 at (0,3) ✓, 5 at (0,4). Sum=23 at (0,4)–(3,4): (0,4)=5; remaining three must sum to 18 — all must be 6. [6|6] covers (1,4) and (2,4): 6 at both ✓. (3,4)=6. [5|6] covers (4,4) and (3,4): 6 at (3,4) ✓, 5 at (4,4). Equals (3,3),(4,3),(4,4)=5. [0|5] covers (2,3) and (3,3): 5 at (3,3) ✓, 0 at (2,3) — less<5 ✓. [5|1] covers (4,3) and (5,3): 5 at (4,3) ✓, 1 at (5,3). Equals (3,1),(4,1),(5,1): (3,1)=5. [5|5] covers (4,1) and (5,1): 5 at both ✓. [0|3] covers (4,0) and (5,0): 0 at (4,0) — empty ✓, 3 at (5,0). Sum=4 at (5,0),(6,0): (5,0)=3, so (6,0)=1. [1|3] covers (6,0) and (6,1): 1 at (6,0) ✓, 3 at (6,1). Sum=7 at (5,3),(5,4),(6,3),(6,4): (5,3)=1. [2|2] covers (5,4) and (6,4): 2 at both. 1+2+(6,3)+2=7 → (6,3)=2. [2|3] covers (6,3) and (6,2): 2 at (6,3) ✓, 3 at (6,2). Equals (6,1),(6,2): (6,1)=3, (6,2)=3 ✓. All eleven constraints satisfied. Puzzle complete.

🎨 Pips Solver

Apr 15, 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 15, 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 15, 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>4 at (1,6) — only pip 5 or 6 qualifies, and only [0|5] carries one
Cell (1,6) must be strictly greater than 4. Pip values 5 and 6 are the only candidates. Scan today's easy set: [0|5] is the only domino that has a 5-pip or 6-pip face. Place it with 5 at (1,6) — greater>4 ✓ — and 0 at (1,5).
2
Step 2: (1,5)=0 forces the three-cell equals at (0,4),(0,5),(1,5) to pip 0; [0|0] fills it
The equals region at (0,4),(0,5),(1,5) requires all three cells to match. (1,5)=0 from Step 1, so all must be 0. The [0|0] double covers (0,4) and (0,5): 0 at both ✓. Three-cell equals satisfied.
3
Step 3: Three-cell equals at (1,3),(2,3),(3,3) — [3|3] double covers two cells; [2|3] forces the third
The equals region at (1,3),(2,3),(3,3) requires all three to match. [3|3] double covers (2,3) and (3,3): 3 at both ✓. So (1,3) must also equal 3. [2|3] is the only remaining domino that can place a 3 at (1,3): orient it with 3 at (1,3) ✓ and 2 at (0,3).
4
Step 4: Equals (0,2),(0,3) and equals (0,1),(1,1) with less<2 close the puzzle
(0,3)=2 from Step 3. The equals region at (0,2) and (0,3) forces (0,2)=2. [2|1] covers (0,2) and (0,1): 2 at (0,2) ✓, 1 at (0,1). Equals at (0,1) and (1,1): (0,1)=1, so (1,1)=1. [1|1] double covers (1,0) and (1,1): 1 at both ✓. Less<2 at (1,0)=1 — 1<2 ✓. All six constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Greater>5 at (3,4) identifies [1|6]
Cell (3,4) must be strictly greater than 5, so (3,4)=6. Scan today's medium dominoes: [1|6] is the only tile with a 6-pip face. Orient it with 6 at (3,4) — greater>5 ✓ — and 1 at (3,3).
2
Step 2: Sum=2 at (3,1) identifies [3|2] and fixes (3,2)=3
Cell (3,1) must sum to 2 on its own, so (3,1)=2 exactly. Only [3|2] can cover (3,1) and (3,2) with 2 at (3,1). Place it with 2 at (3,1) — sum=2 ✓ — and 3 at (3,2).
3
Step 3: Unequal region {(3,2),(3,3),(4,2),(4,3)}: (3,2)=3 and (3,3)=1; [5|2] is the only fit
The unequal region requires all four cells to carry distinct pips. Two are already determined: (3,2)=3 and (3,3)=1. The remaining tile must place two pips both different from 3, 1, and from each other. [5|2] gives pips 5 and 2 — {3,1,5,2} are all distinct ✓. Place [5|2] at (4,2) and (4,3).
4
Step 4: Less<1 at (0,3),(1,3) forces pip 0 at both; [0|1] and [4|0] deliver
Both cells in the less-than-1 region must be strictly less than 1, so (0,3)=0 and (1,3)=0. [0|1] covers (0,3) and (0,2): 0 at (0,3) ✓, 1 at (0,2) — empty ✓. [4|0] covers (1,4) and (1,3): 0 at (1,3) ✓, 4 at (1,4).
5
Step 5: Equals (1,4),(1,5): (1,4)=4 forces (1,5)=4; [4|3] closes sum=3 at (2,5)
The equals region at (1,4) and (1,5) requires both to match. (1,4)=4 from Step 4, so (1,5)=4. [4|3] covers (1,5) and (2,5): 4 at (1,5) ✓, 3 at (2,5) — sum=3 ✓.
6
Step 6: Equals (1,0),(1,1) and sum=3 at (2,0) close with [5|3] and [5|1]
The equals region at (1,0) and (1,1) requires both to match. [5|3] covers (1,0) and (2,0): 5 at (1,0), 3 at (2,0) — sum=3 ✓. [5|1] covers (1,1) and (1,2): 5 at (1,1) — equals (1,0)=5 ✓, 1 at (1,2) — empty ✓. All nine constraints satisfied. Puzzle complete.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: [4|4] double is the only anchor for the five-cell equals region
The equals region at (0,0),(0,1),(1,0),(2,0),(2,1) requires all five cells to carry the same pip. The only way to guarantee two identical faces from a single tile is a double. Among today's hard doubles — [2|2],[4|4],[5|5],[6|6] — only [4|4] fits the region. Place it at (0,0) and (1,0): pip 4 at both. All remaining cells in the region must also be 4.
2
Step 2: Three dominoes complete the five-cell equals region and deposit 1 at (0,2)
Three cells remain in the region: (0,1),(2,0),(2,1) — each must be 4. [1|4] covers (0,2) and (0,1): 4 at (0,1) ✓, 1 at (0,2). [4|0] covers (2,0) and (3,0): 4 at (2,0) ✓, 0 at (3,0) — empty ✓. [4|5] covers (2,1) and (3,1): 4 at (2,1) ✓, 5 at (3,1).
3
Step 3: Sum=4 at (0,2),(0,3): (0,2)=1, so (0,3)=3; [3|5] deposits 5 at (0,4)
The sum region at (0,2) and (0,3) must total 4. (0,2)=1 from Step 2, so (0,3)=3. [3|5] covers (0,3) and (0,4): 3 at (0,3) — sum=1+3=4 ✓ — 5 at (0,4).
4
Step 4: Sum=23 at (0,4)–(3,4): (0,4)=5 forces three 6s; [6|6] double places itself, (3,4)=6
The sum region at (0,4),(1,4),(2,4),(3,4) must total 23. (0,4)=5 from Step 3, so the remaining three cells sum to 18 — the maximum possible, meaning all three must be 6. [6|6] double covers (1,4) and (2,4): 6 at both ✓. (3,4) must also be 6.
5
Step 5: [5|6] at (4,4),(3,4) feeds equals {(3,3),(4,3),(4,4)}=5; [0|5] closes less<5; [5|1] places (5,3)=1
[5|6] covers (4,4) and (3,4): 6 at (3,4) ✓, 5 at (4,4). The equals region at (3,3),(4,3),(4,4) requires all three to match: (4,4)=5, so all must be 5. [0|5] covers (2,3) and (3,3): 5 at (3,3) ✓, 0 at (2,3) — less<5 ✓. [5|1] covers (4,3) and (5,3): 5 at (4,3) ✓, 1 at (5,3).
6
Step 6: Equals {(3,1),(4,1),(5,1)}=5: [5|5] double fills the last two cells
The equals region at (3,1),(4,1),(5,1) requires all three to match. (3,1)=5 from Step 2. [5|5] double covers (4,1) and (5,1): 5 at both ✓.
7
Step 7: [0|3] fills the two empty cells; sum=4 at (5,0),(6,0) places [1|3]
[0|3] covers (4,0) and (5,0): 0 at (4,0) — empty ✓, 3 at (5,0). The sum region at (5,0) and (6,0) must total 4: (5,0)=3, so (6,0)=1. [1|3] covers (6,0) and (6,1): 1 at (6,0) — sum=3+1=4 ✓, 3 at (6,1).
8
Step 8: Sum=7 at (5,3)–(6,4) and equals (6,1),(6,2) close the puzzle
The sum region at (5,3),(5,4),(6,3),(6,4) must total 7. (5,3)=1 from Step 5. [2|2] double covers (5,4) and (6,4): 2 at both. So 1+2+(6,3)+2=7 → (6,3)=2. [2|3] covers (6,3) and (6,2): 2 at (6,3) ✓, 3 at (6,2). Equals (6,1),(6,2): (6,1)=3, (6,2)=3 ✓. All eleven 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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