NYT Pips Hints & Answers for August 20, 2026

Aug 20, 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.

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🎲 Today's Puzzle Overview

Ian Livengood's NYT Pips easy grid opens on two independent footholds: a single-cell less-than clamp at the corner and a pair of sum regions that act as algebraic locks. Once the clamp sets the low anchor, the domino's other pip is forced into an adjacent sum, and each subsequent domino cascades with almost no branching.

Ian Livengood's medium puzzle is more of an exclusion web. Two low-ceiling less-than cells sit at opposite ends of the board, while a top-row sum and two equality regions demand repeated values. The deduction graph is tighter because the top-row sum forces a double of the same pip, which then unblocks the left-side equality and the middle repeated-value block.

Rodolfo Kurchan's hard grid is a dense network of single-cell inequalities, equality triples, and low-target sums. The key structural insight is that a left-edge greater-than cell cannot be the lower value because it would strand the adjacent three-cell sum, so it forces a descending chain down the left column. After that, the equality regions at left, top-right, and center share dominoes in a way that creates coupled placements rather than isolated singles.

💡 Progressive Hints

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

💡 Spot the clamp and sum pairs
Look for a single-cell less-than region; it can only accept the smallest available pip. Then look for sum regions that already have one known partner value and another cell dangling.
💡 Corner clamp feeds the first sum
At [0,0], the less-than cell is forced to the smallest pip, and its domino's other side is 4 at [0,1]. That 4 feeds the sum region [0,1]/[1,1], forcing [1,1] to 6.
💡 Full easy chain
Place 4/0 horizontally [0,0]=0, [0,1]=4. Run 6/5 vertically [1,1]=6, [2,1]=5; place 5/1 vertically [2,2]=5, [3,2]=1 to complete the [2,1]/[2,2] sum. The remaining 3/3 goes horizontally [0,3]=3, [0,4]=3, and 4/1 goes vertically [1,3]=4, [2,3]=1.
💡 Follow the low ceilings
Look for the single-cell less-than regions and the top-row sum. One less-than cell anchors an edge domino, and the top-row sum then demands a doubled value.
💡 Right-edge less-than opens it
At [0,3], the less-than cell must be 1, forcing the 4/1 domino horizontally into [0,2]. The top-row sum then needs the two remaining cells to total 8, so the 4/4 double covers [0,0] and [0,1].
💡 Full medium chain
Place 4/1 [0,2]=4, [0,3]=1; 4/4 [0,0]=4, [0,1]=4. Place 0/2 vertical [1,0]=2, [2,0]=0. The left equality forces 2/5 [1,1]=2, [1,2]=5 and 1/2 [2,1]=2, [2,2]=1. Finish with 5/6 [1,3]=5, [1,4]=6 and 3/4 [2,3]=4, [2,4]=3.
💡 Find the left-edge pinch
Look for a single-cell greater-than that touches a three-cell sum on the left edge. The key is ruling out the value that would make the sum below impossible.
💡 Pin [4,0]
At [4,0], the greater-than cell cannot be 5 because the three-cell column sum would need an impossible split below. It is 6, so the 3/6 domino places 3 at [5,0].
💡 Column sum and left equality
With [5,0]=3, the sum-9 region needs [6,0]+[7,0]=6, so the 3/3 double runs vertically. After that, the left equality block [2,0],[3,0],[3,1] is forced to all 5s, with 5/5 covering [2,0]/[3,0] and 4/5 giving [3,1]=5, [3,2]=4.
💡 Top-right repeaters
The equality block [0,3],[1,3],[2,3] is all 6s. Use 6/6 vertically on [0,3]/[1,3], then 6/5 runs [2,3]=6 to [3,3]=5, which also satisfies the greater-than at [3,3].
💡 Full hard chain
Place 3/6 vertical [4,0]=6, [5,0]=3; 3/3 vertical [6,0]=3, [7,0]=3. Left block: 5/5 vertical [2,0]=5, [3,0]=5; 4/5 horizontal [3,1]=5, [3,2]=4. Top right: 6/6 vertical [0,3]=6, [1,3]=6; 6/5 vertical [2,3]=6, [3,3]=5. Top left: 0/1 horizontal [0,2]=0, [0,1]=1; 2/1 vertical [0,0]=2, [1,0]=1. Middle: 4/4 vertical [3,5]=4, [4,5]=4; 3/4 horizontal [3,6]=4, [3,7]=3. Finish: 0/0 vertical [4,7]=0, [5,7]=0; 2/5 horizontal [5,6]=2, [5,5]=5; 6/4 vertical [6,5]=6, [7,5]=4.

🎨 Pips Solver

Aug 20, 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 August 20, 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 August 20, 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: Anchor the less-than corner
The single-cell region at [0,0] can only satisfy the less-than-1 constraint by taking the 0 side of the 4/0 domino. That domino therefore runs horizontally, putting its 4 at [0,1].
2
Step 2: Complete the first sum
The sum region [0,1]/[1,1] must total 10. Since [0,1] is already 4, [1,1] is forced to 6. The only available 6 is paired with 5, so that domino runs vertically from [1,1] to [2,1], putting 5 below.
3
Step 3: Resolve the second sum pair
Now the sum region [2,1]/[2,2] must total 10. With [2,1] already 5, [2,2] is forced to 5. That is the 5 side of the 5/1 domino, so [3,2] gets the 1.
4
Step 4: Finish the three-cell sum
The only unfinished sum is [0,3]/[0,4]/[1,3]. The remaining 3/3 double must lie horizontally at [0,3] and [0,4], and the 4/1 domino must run vertically at [1,3]/[2,3], giving 4 above and 1 in the empty cell below.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Right-edge less-than settles an edge domino
At [0,3], a single-cell less-than-2 constraint forces the 1 side of the 4/1 domino. That domino must stretch left into [0,2], so [0,2] becomes 4.
2
Step 2: Top-row sum forces a double
The top-row sum across [0,0],[0,1],[0,2] has [0,2]=4 already, so the remaining two cells must sum to 8. The only available way is the 4/4 double, placed horizontally at [0,0] and [0,1].
3
Step 3: Left less-than feeds an equality block
The single-cell less-than-2 at [2,0] forces 0, so the 0/2 domino runs upward to [1,0] with 2. Then the equality block [1,0],[1,1],[2,1] all must be 2; the 1/2 domino later covers [2,1]=2 and [2,2]=1.
4
Step 4: Middle equality creates the 5 repeat
With [1,1]=2, the 2/5 domino runs [1,1]=2 to [1,2]=5. The equality region [1,2],[1,3] forces [1,3]=5 as well, so the 5/6 domino runs right to [1,4]=6.
5
Step 5: Close the lower sums
The sum-9 region [1,4],[2,4] has [1,4]=6, so [2,4]=3. The 3/4 domino runs left to [2,3]=4, and the sum-5 region [2,2],[2,3] then forces [2,2]=1, matching the 1/2 domino already placed at [2,1].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Pin the left-edge greater-than
At [4,0], the greater-than-4 cell cannot be 5: if it were, the three-cell sum below would need an impossible 4 split after [5,0] becomes 5. Therefore [4,0]=6, and the 3/6 domino places [5,0]=3.
2
Step 2: Resolve the column sum
The sum-9 region at [5,0],[6,0],[7,0] now has 3 at the top. The bottom two must sum to 6, which forces the 3/3 double to run vertically from [6,0] to [7,0].
3
Step 3: Set the left equality block to 5s
The equality block [2,0],[3,0],[3,1] must repeat one value. With the 3/3 double used and the 4/4 double reserved for the middle equality, the only viable repeat is 5. The 5/5 domino covers [2,0]/[3,0], and the 4/5 domino gives [3,1]=5, [3,2]=4 to satisfy the greater-than-3 cell.
4
Step 4: Top-right equality and [3,3] greater-than
The equality region [0,3],[1,3],[2,3] is forced to 6. The 6/6 double runs vertically on [0,3]/[1,3], while the 6/5 domino puts 6 at [2,3] and 5 at [3,3]; that 5 satisfies the greater-than-4 cell at [3,3].
5
Step 5: Unlock the top-left sum and empty cell
The empty cell at [0,2] takes the 0 side of the 0/1 domino, so [0,1]=1. Since [0,0]+[0,1] must sum to 3, [0,0]=2. The 2/1 domino then runs vertically from [0,0] to [1,0], putting 1 in the greater-than-0 cell.
6
Step 6: Finish the middle and bottom-right interlocks
The middle equality [3,5],[3,6],[4,5] is all 4s: 4/4 runs [3,5]/[4,5], and 3/4 runs [3,6]=4 to [3,7]=3, satisfying the sum-3 pair [3,7]/[4,7] with 0/0 at [4,7]/[5,7]. Finally, 2/5 runs [5,5]=5 to [5,6]=2, and 6/4 runs [6,5]=6 to [7,5]=4, satisfying the greater-than-10 and less-than-3 regions.

💡 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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