NYT Pips Hints & Answers for September 15, 2026

Sep 15, 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

Easy is a confidence-builder from Ian Livengood: two singleton less-than gates and a handful of two-cell sums make the opening nearly automatic. There are no branching deductions — once the top-left gate and the adjacent high sum are placed, the rest cascades cleanly.

Medium, by Rodolfo Kurchan, is a tighter knot. The bottleneck is the pair of same-value blocks that cross the grid; most solvers will stall until they see how the top-left two-cell sum feeds one value into the large equals region while another value escapes into a second equals block.

Hard, also by Rodolfo Kurchan, is a much bigger domino tiling. The difficulty is not obscure arithmetic but maintaining a chain of one-cell sums and four separate equals branches. Expect the bottom row and upper-left single cells to be the keys; if you anchor those early, the rest of today's NYT Pips hard unfolds without having to guess.

💡 Progressive Hints

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

💡 Start with the tiny gates
Look for the single-cell fewer-than regions and the high two-cell sums. They create forced values before you even think about the middle.
💡 Top-left is the key
The cell in row 0, column 1 has a less-than-4 target, and it sits right above a sum-12 pair. Once that first value lands, the neighboring domino is forced by the sum.
💡 Full easy chain
Place [3,6] vertical r0c1-r1c1 with r0c1=3 and r1c1=6; [0,6] vertical r2c0-r1c0 with r2c0=0 and r1c0=6; [6,6] horizontal r2c1-r2c2; [4,5] horizontal r1c3-r1c2 with r1c3=4 and r1c2=5; [5,5] horizontal r3c1-r3c2.
💡 Find the interlocking equals
This grid is dominated by same-value blocks, but the top-left two-cell sum is the spark. Expect one tile to feed two regions at once.
💡 Top row and zero block
The sum-6 pair at r0c0-r0c1 has to use two 3s, because no tile can put a 0/6 there without breaking the adjacent big equals block. That sends a 0 into the central equals zone.
💡 Full medium chain
Place [0,3] horizontal r0c1-r0c2 with r0c1=3 and r0c2=0; [3,5] vertical r0c0-r1c0 with r0c0=3 and r1c0=5; [0,5] horizontal r1c1-r1c2 with r1c1=5 and r1c2=0; [0,6] vertical r0c3-r1c3 with r0c3=0 and r1c3=6; [0,1] vertical r2c2-r3c2 with r2c2=0 and r3c2=1; [1,5] vertical r2c1-r3c1 with r2c1=5 and r3c1=1; [1,6] vertical r2c3-r3c3 with r2c3=6 and r3c3=1.
💡 Read the one-cell sums
The hard grid's tightest signals are the singleton sum cells along the bottom and upper middle. Let them lock values first; the equal clusters will follow.
💡 Bottom row is a chain
The three leftmost bottom cells have separate one-cell sum targets. The cell in r6c2 must be the smallest possible value, and it naturally pairs left with r6c1 to satisfy both their sums.
💡 Upper-left single-cell sum
The one-cell sum at r1c2 forces a 3 there. Its only viable neighbor is r1c1, which feeds the top-left equal cluster; a [3,2] vertical tile sets that cluster to 2.
💡 Work the cascades
With the bottom row and top-left clusters anchored, carry the 5-equals from r5c0 rightward into r5c1, the 4-equals from r3c1 into r4c0-r4c1, and the 6-equals from r3c6 through r4c6. The right-side 4/5 domino then closes the bottom edge.
💡 Full hard chain
Place [2,1] r6c1-r6c2 (2/1); [3,5] r6c0-r5c0 (3/5); [3,2] r1c2-r1c1 (3/2); [4,2] r1c0-r0c0 (4/2); [2,5] r0c1-r0c2 (2/5); [5,5] r0c3-r1c3 (5/5); [1,4] r3c2-r3c1 (1/4); [1,6] r2c2-r2c3 (1/6); [3,6] r3c5-r3c6 (3/6); [6,5] r5c2-r5c1 (6/5); [6,6] r5c3-r6c3 (6/6); [4,4] r4c0-r4c1 (4/4); [6,4] r4c6-r5c6 (6/4); [5,4] r6c7-r6c6 (5/4).

🎨 Pips Solver

Sep 15, 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 15, 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 15, 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: Top-left low gate
The singleton less-than-4 region at r0c1 forces 3, and the only tile that can put a 3 there is [3,6]. Place it vertically at r0c1-r1c1 with r0c1=3 and r1c1=6.
2
Step 2: Sum-12 forces the first column
The sum-12 region r1c0+r1c1 now has r1c1=6, so r1c0 must also be 6. Domino [0,6] goes vertically at r2c0-r1c0, giving r2c0=0 and satisfying the singleton less-than-3 gate.
3
Step 3: Sum-11 pair and high domino
The sum-11 region r1c2+r2c2 needs 5+6. Domino [6,6] supplies the 6 horizontally at r2c1-r2c2, while [4,5] covers r1c3-r1c2 with r1c3=4 and r1c2=5.
4
Step 4: Finish the last five-sum
The final sum region r2c1+r3c1 must total 11. Since r2c1 is already 6 from the double, r3c1 must be 5. The remaining domino [5,5] fills r3c1-r3c2, and the r3c2 singleton sum-5 is satisfied.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Top-left sum sends a zero
The sum-6 pair at r0c0-r0c1 is forced to be two 3s. Domino [0,3] is the only tile that can put a 3 at r0c1 while sending its 0 into the big equals block at r0c2. Place it horizontally at r0c1-r0c2 with r0c1=3 and r0c2=0.
2
Step 2: Second 3 completes the sum
With r0c1=3, the top-left sum requires r0c0=3. Domino [3,5] covers r0c0-r1c0 vertically, giving r0c0=3 and r1c0=5, which starts the r1c0-r1c1-r2c1 equals block.
3
Step 3: Zero block and 6-equals
The big equals region now forces r0c3, r1c2, and r2c2 to 0. Place [0,5] horizontally at r1c1-r1c2 with r1c1=5 and r1c2=0; [0,6] vertically at r0c3-r1c3 with r0c3=0 and r1c3=6; and [0,1] vertically at r2c2-r3c2 with r2c2=0 and r3c2=1.
4
Step 4: The last 6 and the empty corner
The two-cell equals block r1c3-r2c3 now demands a 6 at r2c3. Domino [1,6] goes vertically at r2c3-r3c3, giving r2c3=6 and placing a harmless 1 in the empty r3c3 corner.
5
Step 5: Close the 5 and 1 equals
The r1c0-r1c1-r2c1 equals block still needs r2c1=5, and the r3c1-r3c2 equals block needs r3c1=1. Domino [1,5] covers r2c1-r3c1 vertically with r2c1=5 and r3c1=1, completing the grid.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Anchor the bottom 2/1
The one-cell sum at r6c2 forces a 1, and its neighbor r6c1 is a one-cell sum that needs a 2. Domino [2,1] is the only fit; place it horizontally across r6c1-r6c2 with r6c1=2 and r6c2=1.
2
Step 2: Force the left bottom 3
With r6c1 already covered, r6c0 cannot pair to the right. It must pair upward with r5c0. Place domino [3,5] vertically at r6c0-r5c0, giving r6c0=3 for its sum and r5c0=5 for the left-side equals pair.
3
Step 3: Unlock the upper-left 3
The single-cell sum at r1c2 forces 3. The adjacent r1c1 is its only neighbor that can absorb a 2 while joining the top-left equals cluster, so place [3,2] vertically at r1c2-r1c1 with r1c2=3 and r1c1=2.
4
Step 4: Top-left and top-right equals
The top-left equals block now makes r0c0 and r0c1 both 2. Domino [4,2] covers r1c0-r0c0 with r1c0=4 and r0c0=2; domino [2,5] covers r0c1-r0c2 with r0c1=2 and r0c2=5. The top-right equals block then takes [5,5] vertically at r0c3-r1c3 for two 5s.
5
Step 5: Middle sums and 6-equals
The two-cell sum at r2c2+r3c2 needs two 1s. Place [1,4] at r3c2-r3c1 with r3c2=1 and r3c1=4, beginning the 4-equals block. Place [1,6] at r2c2-r2c3 with r2c2=1 and r2c3=6. The right-side 6-equals is served by [3,6] at r3c5-r3c6 and [6,4] at r4c6-r5c6, with r3c5=3 and r5c6=4.
6
Step 6: Close all remaining clusters
Finish the 5-equals block r5c0-r5c1 with [6,5] at r5c2-r5c1, giving r5c2=6 and r5c1=5. Place [6,6] at r5c3-r6c3 for the 6-equals branch, and [4,4] at r4c0-r4c1 to join r3c1's 4-equals. Finally [5,4] closes the r6c7 sum-5 and the r6c6 4-equals via r6c7-r6c6.

💡 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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