NYT Pips Hints & Answers for August 16, 2026

Aug 16, 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 easy grid is a compact deduction graph. Two tight two-cell sum regions and a single-cell sum provide independent footholds: one two-cell sum locks a matched double, while the single-cell sum pulls a mixed domino across its row and cascades through adjacent sum regions to the bottom. The remaining dominoes are consumed as a chain, not a search.

Livengood's medium starts from a large equals region, which is sharply constrained by the overall distribution of matching pips in the domino set. Once that equal block is placed, the neighboring second equals region and a vertical three-cell sum tighten the rest of the grid into two parallel deduction tracks.

Rodolfo Kurchan's hard is a more fragmented deduction graph, built from single-cell less/greater hooks and small sums around the perimeter. The lower-left sum-plus-sum pair forms an early lock, while a less-than-2 pair in the upper-right and a two-cell equals region in the top-center split the board into independent chains that reconnect through the central three-cell sum. This NYT Pips hard rewards tracking which dominoes carry the scarce low and high pips.

💡 Progressive Hints

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

💡 Start with the tight sums
Scan for compact sum regions first; one two-cell sum has only one possible matched placement, and a single-cell sum will force a mixed domino into the row below.
💡 Use the top row and [1,0]
The two-cell sum at [0,1]–[0,2] is your matched-pair anchor. After that, the standalone sum at [1,0] forces a specific mixed domino to extend to [1,1], which locks the neighboring row-1 sum.
💡 Complete the chain
Place [3,3] horizontally at [0,1]-[0,2]. Place [5,2] at [1,1]-[1,0] with the 2 in [1,0]. The row-1 sum then makes [1,2] a 5, so [4,5] runs vertically to [2,2] (4). The [2,1]-[2,2] sum forces [2,1] to be 4, so [6,4] runs upward to [3,1]; finally [0,6] fills [3,2]-[3,3] with 6 and 0.
💡 Count the equals block
Locate the four-cell equals region; because it needs the same pip in every cell, the number of available matching pips in the domino set is the key constraint.
💡 The [0,2] block is the linchpin
The equals block at [0,2], [0,3], [1,2], and [2,2] consumes every domino carrying the only pip value that appears enough times. Its placement also forces the value at [1,1] and the zero-side cell at [2,3].
💡 Full medium answer
Place [5,5] horizontally at [0,2]-[0,3]; [5,1] at [1,2]-[1,1] with 1 in [1,1]; [5,0] horizontally at [2,2]-[2,3] with 5 in [2,2], 0 in [2,3]. The column sum at [1,1],[2,1],[3,1] then forces [2,1]=3 and [3,1]=3, so place [4,3] at [2,0]-[2,1] (4,3) and [3,0] at [3,1]-[3,2] (3,0). The zero-equals region then forces [1,3]=0 via [2,0] at [1,4]-[1,3] (2,0) and [3,3]=0 via [0,4] at [3,3]-[3,4] (0,4). The remaining [3,3] fills [2,4]-[2,5] horizontally.
💡 Follow the one-cell chokepoints
The most forcing spots are single-cell less/greater and sum constraints. They act as chokepoints that force adjacent cells to take very specific dominoes, so work the perimeter before the center.
💡 Lock the lower-right and lower-left
Start at [4,6]/[3,6] and [3,8]/[4,8] on the right; then move to the single-cell sum at [3,0]. The sum-9 region at [4,0]-[4,1] and the column reaching up to [3,1] will respond immediately.
💡 Track the central row-2 sum
Once [3,1] is fixed, the three-cell sum at [2,1],[2,2],[3,1] consumes the remaining high pip and a 1. That splits into two dominoes: one places a 5 at the single-cell sum [2,3], the other covers [2,0]-[2,1] with a matched 1.
💡 Top-left and top-right locks
The less-than-5 cell at [1,1] forces a double-four vertical with [0,1], and the sum-9 forces [0,0] and [1,0]. In the upper-right, the less-than-2 pair takes two zero-carrying dominoes, one vertical with [0,6] and one horizontal with [2,7]; the [0,8]/[1,8] column is also forced.
💡 Full hard answer
Place [1,2] vertically at [4,6]-[3,6] (1 below, 2 above); [3,4] vertically at [3,8]-[4,8] (3 above, 4 below); [3,3] vertically at [3,0]-[4,0]. Then [2,6] at [4,1]-[3,1] (6 below, 2 above). Central: [5,6] at [2,2]-[2,3] (6 left, 5 right) and [1,1] at [2,0]-[2,1]. Top-left: [4,4] at [0,1]-[1,1], [1,5] at [0,0]-[1,0] (5 top, 1 bottom). Top-center/right: [5,5] at [0,4]-[1,4], [0,2] at [0,6]-[1,6] (2 top, 0 bottom), [0,4] at [2,6]-[2,7] (0 left, 4 right), [0,5] at [0,8]-[1,8] (0 top, 5 bottom). Remaining [1,4] completes [3,4]-[4,4] (1 above, 4 below).

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Aug 16, 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 16, 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 16, 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: Lock the matched double
The top two-cell sum at [0,1]-[0,2] must be made from identical pips. Scanning the domino list, the only matched pair available is [3,3], so it must sit horizontally in those two cells.
2
Step 2: Pull in the single-cell sum
The lone cell at [1,0] has a single-cell sum that requires a 2. The only domino containing a 2 is [5,2], so it must extend from [1,0] to [1,1], placing the 5 in [1,1].
3
Step 3: Cascade through the adjacent sums
With [1,1]=5, the row-1 sum at [1,1]-[1,2] forces [1,2] to be 5. That consumes [4,5] vertically at [1,2]-[2,2], giving [2,2]=4. The sum at [2,1]-[2,2] then forces [2,1]=4, so [6,4] fills [2,1]-[3,1] vertically with 6 in [3,1].
4
Step 4: Close the bottom with the last mix
The bottom sum at [3,1]-[3,2] already has 6 at [3,1], so [3,2] must be 6. The last unused domino [0,6] therefore completes the grid as [3,2]=6 and [3,3]=0.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Resolve the all-equals block
The four-cell equals region at [0,2], [0,3], [1,2], and [2,2] requires the same pip in all four cells. Only one pip value has enough copies across the domino set: the four 5s. Thus [5,5] must begin the block at [0,2]-[0,3].
2
Step 2: Place the mixed 5s around the block
[5,1] must cover [1,2], and its 1 cannot go to [1,3] because that cell is in the zero-equals region, so it lands in [1,1]. [5,0] must cover [2,2], and its 0 must avoid the already-filled [1,2], so it lands in [2,3].
3
Step 3: Satisfy the column sum
The vertical sum at [1,1],[2,1],[3,1] already has 1 at [1,1]. The remaining total must be split as 3+3. That places [4,3] at [2,0]-[2,1] with 4 above and 3 below, and [3,0] at [3,1]-[3,2] with 3 in [3,1] and 0 in [3,2].
4
Step 4: Fill the zero-equals region
The remaining cells in the all-zero equals region are [1,3] and [3,3]. Use [2,0] horizontally at [1,3]-[1,4] with 0 in [1,3] and 2 in [1,4]; then use [0,4] vertically at [3,3]-[3,4] with 0 in [3,3] and 4 in [3,4].
5
Step 5: Finish with the double 3
The bottom-right sum at [2,4]-[3,4] has 4 at [3,4], so [2,4] must be 3. The single-cell sum at [2,5] also needs 3. The remaining [3,3] domino completes both, placed horizontally at [2,4]-[2,5].

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the lower-right single cells
The single-cell sum at [4,6] forces a 1. The cell directly above at [3,6] must be greater than 1, so the mixed domino [1,2] fits vertically with 1 below and 2 above. Nearby, the single-cell sum at [3,8] forces a 3, and the empty cell at [4,8] takes its partner 4 from [3,4].
2
Step 2: Unlock the lower-left sum pair
The single-cell sum at [3,0] forces the [3,3] double to run vertically into [4,0]. That gives [4,0]=3, so the sum-9 region at [4,0]-[4,1] forces [4,1]=6. The [2,6] domino then runs vertically at [3,1]-[4,1], placing 2 above and 6 below.
3
Step 3: Resolve the central row-2 sum
With [3,1]=2, the central three-cell sum at [2,1],[2,2],[3,1] still needs 7 total from [2,1] and [2,2]. The only remaining high pip available is 6, so [2,2]=6 and [2,1]=1. This forces [5,6] horizontally at [2,2]-[2,3] with 6 left and 5 right, and [1,1] horizontally at [2,0]-[2,1].
4
Step 4: Close the top-left sum chain
The less-than-5 cell at [1,1] must be 4, and the only remaining double 4 forces [4,4] vertically at [0,1]-[1,1]. The sum-9 at [0,0]-[0,1] then makes [0,0]=5. Since [2,0] is already 1 from the central step, the sum-2 at [1,0]-[2,0] forces [1,0]=1, so [1,5] runs vertically at [0,0]-[1,0] with 5 top and 1 bottom.
5
Step 5: Complete the top-center and upper-right
The two-cell equals region at [0,4]-[1,4] takes [5,5] vertically. The less-than-2 pair at [1,6]-[2,6] forces both cells to 0: [0,2] runs vertically at [0,6]-[1,6] with 2 above and 0 below, while [0,4] runs horizontally at [2,6]-[2,7] with 0 left and 4 right. The less-than-4 cell at [0,8] and greater-than-4 cell at [1,8] take [0,5] vertically with 0 top and 5 bottom.
6
Step 6: Place the last middle-right domino
The final unused domino is [1,4]. It completes the sum-5 region at [3,4]-[4,4] vertically, placing 1 in [3,4] and 4 in [4,4].

💡 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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