NYT Pips Hints & Answers for September 16, 2026

Sep 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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🎲 Today's Puzzle Overview

Ian Livengood's easy grid presents a compact deduction graph anchored by two classes of constraints: single-cell greater/less gates and two-cell equals locks. The equals blocks behave as pairwise equalizers, while the tightest less-than gate forces the lowest allowable pip and thereby orients an entire domino. The adjacent equals pair then propagates that value across the bottom row.

Livengood's medium expands the architecture into three interlocking equals regions plus one fixed-sum cell and one less-than outlier. The lower row's equals chain is the structural spine: three cells must share a pip, the adjacent two-cell equals region must share a different pip, and the boundary between them is where a mixed domino bridges the two. In the upper band, a greater-than single locks the high end of a mixed domino, which then feeds the upper-right equals cluster. This NYT Pips medium resolves less by arithmetic than by matching repeated values to available doubles.

Rodolfo Kurchan's hard grid is a sum-network with an unequal four-cell node woven through the left side. Single-cell sum constraints at opposite edges create early seeds; from a one-cell sum at the bottom, a two-cell sum pair and a three-cell 15-sum cascade upward. The upper-right cluster then resolves through a chain of two-cell sums feeding a three-cell sum, while a two-cell-wide strip at the middle forces a minimal sum split. The deduction graph is denser and more directional than in the easier grids.

💡 Progressive Hints

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

💡 Start with one-cell limits
Scan for the single-cell greater-than and less-than constraints. They cap or raise individual cells and can fix the orientation of any mixed domino that touches them.
💡 Let the equals pairs propagate
Focus on the two-cell equals regions at [3,1]-[3,2] and [5,1]-[5,2]. The lower one is tied to a very tight less-than cell at [5,0], so once that cell is fixed, the adjacent equal cell is forced, and the second equal cell must copy it.
💡 Full solve for easy
Place [3,0] on [5,1]-[5,0]; [5,3] on [5,3]-[5,2]; [4,4] on [3,1]-[3,2]; [6,1] on [1,0]-[0,0]; and [0,1] on [1,3]-[0,3]. The less-1 cell at [5,0] takes the 0, the two equals regions force 3s and 4s, and the greater-4 cell at [1,0] takes the 6.
💡 Find the arithmetic outliers
Look for the single-cell sum and less-than constraints, plus the single greater-than cell. They create narrow anchors, but the real work is coordinating the three equals regions that share values.
💡 Bridge the upper equals clusters
The greater-than cell at [0,1] fixes one end of a mixed domino into [1,1], and the sum cell at [2,0] fixes another into [2,1]. Together they force the three-cell equals region at [1,1]-[1,2]-[2,1] to all carry the same low pip, which in turn pushes the adjacent equals region at [1,3]-[2,2]-[2,3] toward a high double.
💡 Full solve for medium
Place [6,1] on [0,1]-[1,1]; [3,1] on [2,0]-[2,1]; [5,1] on [1,3]-[1,2]; [5,5] on [2,2]-[2,3]; [4,4] on [3,0]-[3,1]; [4,1] on [3,2]-[3,3]; and [1,0] on [3,4]-[2,4]. The greater-3 cell takes 6, the sum-3 cell takes 3, the upper equals regions become 1 and 5, and the lower row splits into 4s and 1s.
💡 Map the sum seeds
Start with the single-cell sum constraints and the tight sum pairs. These are the most restrictive nodes in the graph; solving one often pins an entire domino and sends a value into a neighboring sum region.
💡 Anchor the lower-left and middle
Check the single sum cell at [7,0], the two-cell sum region at [6,1]-[7,1], and the three-cell 15-sum at [4,1]-[4,2]-[5,1]. The first two lock a low pair, and the third then uses that shared value to force the middle of the grid.
💡 Chain the top-right sums
Once the lower-left is set, move to the top-right cluster. The single sum at [0,6] feeds the two-cell sum at [0,7]-[1,7], then the two-cell sum at [2,6]-[2,7], and finally the three-cell sum at [0,5]-[1,5]-[2,5]. This chain dictates a vertical stack of mixed and double dominos.
💡 Resolve the unequal block and left edge
The uneven region at [0,2]-[1,2]-[2,2]-[2,3] requires four different pip values. Use the remaining mixed dominos that can sit vertically and horizontally across that four-cell block, so the low and high values interlock without repeating.
💡 Full solve for hard
Place [2,4] on [7,1]-[7,0] (2,4); [5,2] on [5,1]-[6,1] (5,2); [5,5] on [4,1]-[4,2]; [5,4] on [0,7]-[0,6] (5,4); [5,6] on [1,7]-[2,7] (5,6); [5,3] on [2,6]-[2,5] (5,3); [2,2] on [0,5]-[1,5]; [3,2] on [0,2]-[1,2] (3,2); [5,1] on [2,3]-[2,2] (5,1); [4,1] on [4,4]-[4,5] (4,1); [1,6] on [4,6]-[5,6] (1,6); [0,0] on [4,7]-[4,8]; [4,4] on [6,6]-[7,6]. Sum 4 corners force the first dominos; sum 15 and the sum 2 strip complete the middle; sum 11 and sum 7 complete the top.

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Sep 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 Sept 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 Sept 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: Tight less-than seed
The single-cell less-than constraint at [5,0] forces the lowest possible pip. The only available domino that can place that value at [5,0] and still reach [5,1] is [3,0], so orient it with 0 at [5,0] and 3 at [5,1].
2
Step 2: Equal bottom pair propagates
With [5,1] now pinned to 3, the equals region at [5,1]-[5,2] requires [5,2] to be 3 as well. That 3 must come from the [5,3] domino, placed across [5,3]-[5,2]; the other end puts 5 in the less-than-6 cell at [5,3], which fits.
3
Step 3: The double locks the middle
The equals region at [3,1]-[3,2] demands two identical values. The only remaining double is [4,4], so it covers both cells with 4s.
4
Step 4: Top cells finish the grid
The greater-than-4 cell at [1,0] needs the largest remaining value, so [6,1] runs from [1,0] to [0,0] with 6 and 1. The final domino [0,1] covers [1,3]-[0,3], placing the 0 end in the less-than-4 cell and the 1 end in the empty cell.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Greater-than anchor
The greater-than-3 cell at [0,1] must be a high pip and is paired with [1,1]. The [6,1] domino supplies that high value at [0,1] and puts a 1 at [1,1], which feeds the three-cell equals region below.
2
Step 2: Sum cell locks the left
The single-cell sum-3 at [2,0] fixes that cell to 3. The [3,1] domino covers [2,0]-[2,1] with 3 at [2,0] and 1 at [2,1]. Now the equals region [1,1]-[1,2]-[2,1] is confirmed as three 1s.
3
Step 3: Upper-right equals becomes high
Because [1,2] must be 1, the domino covering [1,2] and [1,3] is [5,1], placing 5 at [1,3]. Then the adjacent equals region [1,3]-[2,2]-[2,3] forces all three to be 5, so [5,5] covers [2,2]-[2,3].
4
Step 4: Lower row splits by equals
The lower left equals region [3,0]-[3,1]-[3,2] must all carry the same value. [4,4] covers [3,0]-[3,1] with 4s; then [3,2] also needs to be 4, so [4,1] reaches across [3,2]-[3,3] with 4 at [3,2] and 1 at [3,3].
5
Step 5: Less-than tail closes the grid
The lower right equals region [3,3]-[3,4] already has 1 at [3,3], so [3,4] must be 1. The [1,0] domino covers [3,4]-[2,4], giving 1 at [3,4] and 0 at [2,4], which satisfies the less-than-3 cell.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Bottom-left sum seed
The single-cell sum-4 constraint at [7,0] pins that cell to 4. The [2,4] domino is the only way to put 4 at [7,0] and still cover the adjacent vertical cell [7,1], so it places 2 at [7,1].
2
Step 2: Pair sum and 15-sum cascade
The two-cell sum-4 at [6,1]-[7,1] now sees [7,1]=2, so [6,1] must be 2. That forces the [5,2] domino to cover [5,1]-[6,1], with 5 at [5,1] and 2 at [6,1]. The 5 at [5,1] feeds the three-cell 15-sum [4,1]-[4,2]-[5,1], leaving 10 for the two upper cells; only the [5,5] domino can fill [4,1]-[4,2] with 5 and 5.
3
Step 3: Top-right sum chain
The single-cell sum-4 at [0,6] forces a 4 there. The [5,4] domino covers [0,7]-[0,6] with 5 at [0,7] and 4 at [0,6], so the two-cell sum-10 at [0,7]-[1,7] makes [1,7]=5. The [5,6] domino then covers [1,7]-[2,7] with 5 at [1,7] and 6 at [2,7], which satisfies the sum-11 pair with [2,6]=5. That 5 is supplied by [5,3] at [2,6]-[2,5], putting 3 at [2,5]. Finally, the sum-7 region [0,5]-[1,5]-[2,5] leaves 4 total for [0,5]+[1,5]; the [2,2] domino supplies two 2s.
4
Step 4: Unequal block interlocks
The four-cell unequal region at [0,2]-[1,2]-[2,2]-[2,3] demands four different values. Cover the left vertical pair with [3,2], placing 3 at [0,2] and 2 at [1,2]. Then cover the lower horizontal pair with [5,1], placing 5 at [2,3] and 1 at [2,2]. This keeps all four values distinct.
5
Step 5: Sum-2 strip and sum-14 column
The less-than-5 cell at [4,4] must take 4. The [4,1] domino covers [4,4]-[4,5] with 4 and 1, which contributes the first 1 to the sum-2 strip [4,5]-[4,8]. To total 2, [4,6] must also be 1 and [4,7]-[4,8] must both be 0. Thus [1,6] covers [4,6]-[5,6] with 1 and 6, and [0,0] covers [4,7]-[4,8] with two 0s. The 6 at [5,6] then leaves 8 for the sum-14 region [6,6]-[7,6], so [4,4] covers those two cells with 4s.
6
Step 6: Verify all remaining constraints
With the thirteen dominos placed, the sum targets are satisfied: [7,0]=4; [6,1]-[7,1]=4; [4,1]-[4,2]-[5,1]=15; [0,6]=4; [0,7]-[1,7]=10; [2,6]-[2,7]=11; [0,5]-[1,5]-[2,5]=7; [4,5]-[4,8]=2; [5,6]-[6,6]-[7,6]=14; and the unequal four-cell block remains all different.

💡 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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