NYT Pips Hints & Answers for May 19, 2026

May 19, 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 easy, by Ian Livengood, is a confidence-builder — every domino is a double, and the grid has only five small regions. The puzzle solves linearly: the equals region locks in a 3-3, then a simple sum-5/less-4 interplay forces the rest. No guesswork, perfect for new solvers.

Medium, constructed by Rodolfo Kurchan, tightens the screws. The main bottleneck is a three-cell equals stretch along the bottom that demands two different dominoes sharing a 3. From there, two sum-11 regions branch out quickly. If you spot that initial equals demand, the entire right side cascades.

Hard, also by Kurchan, is a beast with two sprawling equals regions — a quartet of 1s at top and a quartet of 6s at bottom. The bottom-right 6s trigger forces three dominoes at once, then a sum-9 propagates upward into the 1s. The middle row sums close the loop. Expect a satisfying tussle with interlocking constraints.

💡 Progressive Hints

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

💡 Hint 1: Find the Forced Double
The puzzle is all doubles, but only one type of double can satisfy a region where all cells must be identical. Scan for that region; it's your starting domino.
💡 Hint 2: Trace the Sum and Less-Than Regions
After placing the equals region's domino in the bottom-center, the vertical sum-5 region above it narrows to exactly two possible doubles. Meanwhile, the less-than-4 region at top-left needs a small sum — pair the clues to assign the 1 and 2.
💡 Hint 3: Full Solution
Place [3,3] horizontally at (2,3)-(2,4). Then [1,1] vertically at (0,0)-(1,0), [4,4] vertically at (2,0)-(3,0), [2,2] horizontally at (0,1)-(0,2), and [6,6] horizontally at (3,1)-(3,2). This satisfies all constraints: less-4, empty, sum-5, equals, greater-9, and the final empty cell.
💡 Hint 1: The Triple Equals Trap
Look for a long region that requires all cells to match. It's the only way to force two different dominoes to cooperate today.
💡 Hint 2: Bottom-Row Cooperation
The equals region covers three consecutive cells in row 2, columns 3-5. To fill it, you need a [3,3] and a [3,6] with the 3 landing inside the region. This also plants a 6 next door, which triggers the adjacent sum-11.
💡 Hint 3: Full Solution
Place [3,3] horizontally at (2,4)-(2,5), [3,6] vertically at (2,3)-(2,2) (3 at (2,3)). The sum-11 at (2,1)-(2,2) forces [6,5] vertically (1,1)-(2,1) with 6 at (1,1). The top-left sum-11 uses [4,4] at (0,0)-(0,1) and [0,1] vertically (2,0)-(1,0) (1 at (1,0)). Greater-4 and -3 are satisfied by [6,4] vertically at (0,4)-(1,4) (6 at (0,4)). The sum-1 at (1,5) takes [1,2] horizontally (1,5)-(0,5) (1 at (1,5)).
💡 Hint 1: Twin Giants
Two large equals regions dominate the grid — one top, one bottom-right. Both demand four cells with the same pip. Start with the bottom-right cluster; it will lock in several high numbers immediately.
💡 Hint 2: Bottom-Right 6s First
The region at (5,6), (6,5), (6,6), (6,7) forces a value of 6. You'll need [6,6] on the rightmost pair, plus [4,6] and [5,6] to supply 6 to the other two cells. Their exact placements also set (6,4)=5 and (5,5)=4, which triggers the neighboring sum-9.
💡 Hint 3: Upward Cascade
With (5,5)=4, the sum-9 at (4,5)-(5,5) requires (4,5)=5, bringing in [1,5] with 1 at (4,4). That 1 feeds the sum-2 region (3,4)-(4,4), forcing (3,4)=1 and pulling [1,3] into (3,3)-(3,4) with 3 at (3,3). Now you have enough 1s to resolve the top equals.
💡 Hint 4: Conquer the Top Equals
The top equals region (0,4),(1,2),(1,3),(1,4) all become 1. Place [1,1] horizontally on (1,2)-(1,3), [0,1] horizontally (0,3)-(0,4) (1 at (0,4)), and [2,1] vertically (1,4)-(1,5) (1 at (1,4), 2 at (1,5)). The row-6 sum-8 then forces (6,3)=3 and (6,2)=5 via [5,3].
💡 Hint 5: Complete Layout
Full placements: [6,6] at (6,6)-(6,7); [4,6] at (5,5)-(5,6) (4,6); [5,6] at (6,4)-(6,5) (5,6); [1,5] at (4,4)-(4,5) (1,5); [1,3] at (3,3)-(3,4) (3,1); [1,1] at (1,2)-(1,3); [0,1] at (0,3)-(0,4) (0,1); [2,1] at (1,4)-(1,5) (1,2); [2,2] at (6,0)-(6,1); [5,3] at (6,2)-(6,3) (5,3); [3,3] at (4,2)-(4,3); [4,5] at (5,3)-(5,4) (5,4); [4,4] at (5,1)-(5,2).

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May 19, 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 May 19, 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 May 19, 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: Place the Equals Domino
The equals region at (2,3)-(2,4) must hold identical numbers. Since every domino today is a double, the only way to fill it is with a double that matches. [3,3] fits perfectly, placed horizontally.
2
Step 2: Solve the Sum-5 Region
Region sum-5 at (1,0)-(2,0) needs two numbers that add to 5. With only doubles left, the sum formula 2x=5 has no integer solution, so the two cells must come from different doubles. The only remaining pair that sums to 5 is 1 (from [1,1]) and 4 (from [4,4]).
3
Step 3: Tie in the Less-4 Region
The less-4 region at (0,0)-(0,1) must have a total under 4. The [1,1] provides a 1, and [2,2] provides a 2; together they sum to 3. To satisfy both the sum-5 and less-4, [1,1] must cover (0,0)-(1,0) vertically, putting 1s in both cells. Then [4,4] goes vertically at (2,0)-(3,0) to give the 4 needed for sum-5.
4
Step 4: Clean Up with [2,2] and [6,6]
Now the greater-9 region at (3,0)-(3,1) has 4 at (3,0). To exceed 9, (3,1) must be 6; place [6,6] horizontally at (3,1)-(3,2). Finally, [2,2] fills the remaining empty cell (0,1) and the empty region (0,2) horizontally, both as 2.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Fill the Three-Cell Equals
The equals region (2,3)-(2,4)-(2,5) needs three identical numbers. The dominoes [3,3] and [3,6] each carry a 3. Place [3,3] horizontally at (2,4)-(2,5) and [3,6] vertically such that its 3 falls on (2,3) and the 6 lands on (2,2).
2
Step 2: Trigger the Adjacent Sum-11
With (2,2)=6, the sum-11 region (2,1)-(2,2) forces (2,1)=5. The only domino with a 5 is [6,5]; place it vertically (1,1)-(2,1) with 6 at (1,1).
3
Step 3: Resolve the Top-Left Sum-11
The region (0,0)-(1,0)-(1,1) now has 6 at (1,1). The remaining 5 must come from (0,0) and (1,0). The pair 4 and 1 fits; use [4,4] horizontally (0,0)-(0,1) for the 4, and [0,1] vertically (2,0)-(1,0) with 1 at (1,0). This also satisfies less-2 at (2,0) with 0.
4
Step 4: Place the Greater Constraints
The greater-4 cell (0,4) needs a sum >4, and greater-3 cell (1,4) needs sum >3. [6,4] delivers 6 and 4; place it vertically (0,4)-(1,4) with 6 above.
5
Step 5: Finish with the Sum-1
The sum-1 region at (1,5) requires exactly 1. The [1,2] domino can place a 1 there, with 2 going to the empty (0,5). Place [1,2] horizontally (1,5)-(0,5).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the Bottom-Right Equals
The equals region spanning (5,6), (6,5), (6,6), (6,7) needs four identical pips. Use [6,6] for (6,6)-(6,7). The two other cells must also be 6; place [4,6] vertically (5,5)-(5,6) with 6 at (5,6) and 4 at (5,5), and [5,6] horizontally (6,4)-(6,5) with 6 at (6,5) and 5 at (6,4).
2
Step 2: Propagate into the Sum-9 Region
The sum-9 region (4,5)-(5,5) has 4 at (5,5), so (4,5)=5. That forces [1,5] vertically (4,4)-(4,5) with 1 at (4,4) and 5 at (4,5).
3
Step 3: Use the Sum-2 to Reach the 1s
The sum-2 region (3,4)-(4,4) now has 1 at (4,4), so (3,4)=1. Place [1,3] vertically (3,3)-(3,4) with 3 at (3,3) and 1 at (3,4).
4
Step 4: Resolve the Top Equals with 1s
The top equals region requires (0,4),(1,2),(1,3),(1,4) to be 1. Place [1,1] horizontally at (1,2)-(1,3). Then [0,1] horizontally (0,3)-(0,4) with 1 at (0,4) and 0 at (0,3). Finally, [2,1] vertically (1,4)-(1,5) puts 1 at (1,4) and 2 at (1,5), satisfying greater-0.
5
Step 5: Close the Bottom Row Sums
The sum-8 region (6,3)-(6,4) already has 5 at (6,4), so (6,3)=3. Place [5,3] horizontally (6,2)-(6,3) with 5 at (6,2). The row-6 sum-9 (6,0)-(6,2) now has (6,2)=5, so (6,0)+(6,1)=4; place [2,2] there.
6
Step 6: Fill the Last Middle Regions
The sum-9 region (3,3)-(4,2)-(4,3) has 3 at (3,3); the remaining 6 must be 3+3 from [3,3] placed horizontally at (4,2)-(4,3). The sum-9 at (5,3)-(5,4) is satisfied by [4,5] placed horizontally with 5 at (5,3) and 4 at (5,4). Finally, the sum-8 at (5,1)-(5,2) takes [4,4] horizontally.

💡 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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