NYT Pips Hints & Answers for April 27, 2026

Apr 27, 2026

🚨 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!

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Want hints instead? Scroll down for progressive clues that won't spoil the fun.

🎲 Today's Puzzle Overview

Today’s NYT Pips lineup for April 27, 2026, offers a satisfying gradient. The easy puzzle from Ian Livengood is a quick win — two greater‑5 cells immediately demand sixes, and a compact sum‑5 region alongside an equals pair sweeps up the rest in a handful of placements. No guesswork needed.

The medium, also by Livengood, turns the dial up a notch. A top‑left equals region gifts you a zero anchor, but the key bottleneck is a high‑target greater constraint in the bottom rows that forces a pair of maximum pips, which then unlocks a tidy sum‑8 cluster. It’s a neat chain of forced moves.

Rodolfo Kurchan’s hard is where the real tangle lies. A triple‑equals region spanning three cells in the center mandates all zeros — the only way to satisfy it with the given domino set. From that central lock, single‑cell sum regions on the upper edge and a web of sum‑and‑greater constraints propagate outward, making this the most involved grid of the day.

💡 Progressive Hints

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

💡 Hint 1: Spot the High‑Value Bottleneck
The grid contains two ‘greater 5’ regions. Only one pip value fits a ‘greater 5’ constraint, and that immediately locks in two cells.
💡 Hint 2: Anchor the Right Side
The greater‑5 cell in the top‑right corner forces a 6 there and a 5 in the cell directly below it. That pairing also dictates which domino covers the top‑left greater‑5 cell, thanks to a sum‑5 region beneath.
💡 Hint 3: Full Chain
Place [6,5] vertically at [0,4]–[1,4]. Then [1,6] vertically at [0,0]–[1,0] to satisfy both the left greater‑5 and the sum‑5 (1+4). The sum‑5 needs a 4 at [1,1], so [4,4] goes vertically to [1,1]–[2,1], making [2,1] a 4 that pairs with a second 4 from [1,4] placed horizontally at [2,2]–[2,3] (giving 4–1). Finally, the equals region [1,2]–[1,3] takes the [3,3] domino horizontally.
💡 Hint 1: Start with Equality
Look for a region that forces two cells to be identical — it’s your zero‑cost entry point.
💡 Hint 2: Top‑Left Lock
The equals region at [0,0] and [1,0] can only be fulfilled by a double‑zero domino. Placing it gives you two zeros and opens up the adjacent sum‑3 region at [0,1]/[1,1], which will require the numbers 2 and 1.
💡 Hint 3: Full Walkthrough
1) Place [0,0] vertically for [0,0]=0, [1,0]=0 (equals). 2) The sum‑3 pair [0,1]/[1,1] needs 2+1; use [4,2] to put 2 at [0,1] and 4 at [0,2] (creating an equals pair with [0,3]=4 from [4,6]). 3) [4,6] vertically at [0,3]=4, [1,3]=6; then [5,1] gives [1,2]=5 and [1,1]=1, completing sum‑11 (5+6) and sum‑3. 4) The greater‑4 cell [2,0] takes the 5 from [5,6], which also places 6 at [3,0] for the high‑sum greater region. That region needs a second 6, so [6,1] goes horizontally at [3,1]=6, [3,2]=1. 5) Finally, the sum‑8 cluster [2,2],[2,3],[2,4],[3,2] is resolved with [3,3] for 3‑3 and [0,1] vertical giving [2,4]=1, [3,4]=0 (satisfying the less‑1 cell).
💡 Hint 1: The Central Lock
Concentrate on the multi‑cell equals region — it’s the only way to get a firm grip on this grid.
💡 Hint 2: Zero Out the Equals
The equals region at [1,2], [2,1], and [2,2] spans three cells. With the available dominoes, only pip 0 can occupy all three; you’ll need the [0,0] and [0,3] dominoes to make it work.
💡 Hint 3: Neighboring Sums
Once zeros sit in that cluster, the single‑cell sum‑3 at [0,0], the sum‑4 at [1,0], and the sum‑3 at [2,0] become forced. The sum‑4 at [0,6] also becomes a lone cell that must be a 4, pulling in the [0,4] domino.
💡 Hint 4: Long‑Range Dominoes
The sum‑10 region [2,6]/[3,6] demands two 5s; [6,5] and [1,5] will cover it together with the greater‑4 cell at [4,6]. Meanwhile, the upper‑right sum‑10 pair and the sum‑4 pair at [0,8]/[1,8] will use the [5,5] and [3,1] dominoes, with [2,3] domino tying off the sum‑3 at [2,7]/[2,8].
💡 Hint 5: Complete Solution
Place dominoes in this order: [0,0] → [1,2]=0, [2,2]=0; [0,3] → [2,1]=0, [2,0]=3; [4,3] → [1,0]=4, [0,0]=3; [0,4] → [1,6]=0, [0,6]=4; [6,5] → [4,6]=6, [3,6]=5; [1,5] → [2,7]=1, [2,6]=5; [2,3] → [2,8]=2, [1,8]=3; [3,1] → [0,7]=3, [0,8]=1; [5,5] → [0,1]=5, [0,2]=5; [5,4] → [1,4]=5, [2,4]=4; [2,2] → [3,0]=2, [4,0]=2; [6,6] → [3,4]=6, [4,4]=6. This resolves all constraints.

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Apr 27, 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 27, 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 27, 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: Spot the Greater‑5 Cells
Both [0,0] and [0,4] must be greater than 5, and the only pip that qualifies is 6. So those two corners are forced to 6.
2
Step 2: Fill the Right Corner
Domino [6,5] can supply a 6 and a 5. Placing it vertically with 6 in [0,4] and 5 in [1,4] satisfies the right greater‑5 and fills the empty cell below.
3
Step 3: Satisfy the Left and the Sum
The left corner [0,0] still needs its 6. Domino [1,6] placed vertically puts 6 at [0,0] and 1 at [1,0]. That 1 contributes to the sum‑5 region [1,0]–[1,1], which still requires a 4. Domino [4,4] fits perfectly by placing 4 at [1,1] and another 4 at [2,1], which also initiates the equals region with [2,2].
4
Step 4: Clean Up the Equals
Now [2,1]=4 and [2,2] need to be equal; domino [1,4] placed horizontally covers [2,2]=4 and [2,3]=1. Finally, the remaining equals region [1,2]–[1,3] takes the [3,3] domino for a matching pair of 3s.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Lock the Top‑Left Equals
The region covering [0,0] and [1,0] forces both cells to share the same pip. The only domino capable of providing two identical low numbers is [0,0], so place it vertically: [0,0]=0, [1,0]=0.
2
Step 2: Tackle the Adjacent Sum‑3
Next to those zeros, cells [0,1] and [1,1] must sum to 3. Domino [4,2] can place a 2 at [0,1] and a 4 at [0,2], providing the 2 without violating the adjacent equals‑region that will require two 4s at [0,2] and [0,3].
3
Step 3: Finish the Top Row Equals and Sum‑11
To make [0,3] a 4, use [4,6] vertically (4 at top, 6 at [1,3]). The sum‑11 region [1,2]–[1,3] now needs a 5 to pair with the 6, so [5,1] domino places 5 at [1,2] and 1 at [1,1], completing the sum‑3 as well.
4
Step 4: Handle the High‑Stakes Greater Regions
The single cell [2,0] must be greater than 4, so it takes the 5 from [5,6] while the 6 goes to [3,0]. The greater (high‑target) region [3,0]–[3,1] needs a second 6, so [6,1] domino supplies 6 at [3,1] and 1 at [3,2].
5
Step 5: Resolve the Sum‑8 Cluster
The four‑cell sum‑8 region at [2,2],[2,3],[2,4],[3,2] now has 1 from [3,2] and needs 3+3+1. Domino [3,3] goes horizontally at [2,2]–[2,3] for the two 3s, and [0,1] placed vertically gives [2,4]=1 and [3,4]=0, satisfying the less‑1 requirement.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Zero Out the Triple‑Equals
The three‑cell equals region at [1,2], [2,1], [2,2] must hold identical pips. The only pip that can appear in three separate cells with the given dominoes is 0. Place [0,0] to set [1,2]=0 and [2,2]=0; then place [0,3] to set [2,1]=0 and, simultaneously, satisfy the sum‑3 at [2,0] with the accompanying 3.
2
Step 2: Solve the Left‑Edge Sums
The isolated sum‑3 at [0,0] and sum‑4 at [1,0] are adjacent. Domino [4,3] placed vertically gives [1,0]=4 and [0,0]=3, completing both. Meanwhile, the equals region [3,0]–[4,0] is satisfied by [2,2] domino placing a 2 in each.
3
Step 3: Nail the Right‑Side Sum‑4 and Sum‑10
The single‑cell sum‑4 at [0,6] forces a 4 there. Use [0,4] vertically: [1,6]=0, [0,6]=4. The sum‑10 pair [2,6]–[3,6] must total 10; place [6,5] to give [4,6]=6 (greater‑4) and [3,6]=5, and [1,5] to give [2,6]=5 and [2,7]=1, finishing that cluster while feeding the sum‑3 region at [2,7]–[2,8].
4
Step 4: Fill the Upper‑Right Sum‑10 and Sum‑4
The sum‑10 region at [0,1]–[0,2] requires 5+5 — the [5,5] domino fits perfectly horizontally. For the sum‑4 pair [0,8]–[1,8], place [3,1] to get [0,7]=3 and [0,8]=1, then [2,3] to get [1,8]=3 and [2,8]=2, which also satisfies the sum‑3 at [2,7]–[2,8] (1+2 with the 1 already placed).
5
Step 5: Resolve the Remaining Greater Regions
The greater‑4 cell [1,4] gets its 5 from [5,4] (placing 5 there and 4 at [2,4] for greater‑3). Finally, the greater‑10 pair [3,4]–[4,4] must both be 6 — only one domino remains: [6,6], placed vertically.

💡 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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