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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 designs today's NYT Pips easy around a rare oversized equality: a five-cell block that essentially spends all the high-value tiles in one statement. A tiny greater-than singleton then becomes the release valve, and a plain two-cell equality on the left is the puzzle's quiet second anchor. It's a compact, elegant construction that reads like a domino haiku.
Livengood's medium keeps the grid small but treats the arithmetic more like interlocking gears. A dangling top cell forces a vertical domino, two one-cell sums at the bottom reach upward, and two- and three-cell sum regions close the chain. The design's pleasure is in how each region dovetails into its neighbor without feeling cluttered.
Rodolfo Kurchan's hard is a different animal: a sparse grid with many one-cell sum constraints that act as hair-trigger locks. The constructor hides a long left-edge cascade behind a lone low-sum singleton, while the top and middle are bound by parallel equals runs and a top-right sum showcase. It's structurally demanding, with a clear emphasis on forced vertical chains.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Spot the oversized equals
Start by scanning the equals regions. One is much larger than the others, so it will dominate which matching-pip tiles have to fit around it.
💡 Let the high pips flow
The five-cell equals block at [0,1] through [1,3] uses the highest available matching value, so every tile carrying that value is called into the block. The one tile without it is then forced toward the smaller [0,0]/[1,0] equals pair.
💡 Easy full chain
Place [6,4] horizontally at [0,1]/[0,0] with 6 at [0,1] and 4 at [0,0]. Set [4,5] vertically at [1,0]/[2,0] so 4 matches [0,0] and 5 lands at [2,0]. Double [6,6] fills [1,1]/[1,2]; [6,1] goes vertical at [1,3]/[2,3] with 1 in the empty cell; [6,2] covers [0,2]/[0,3] with 2 satisfying greater-1.
💡 Follow the sum spine
The medium's logic lives in its sum regions. Look for a dangling cell at the top of a three-cell column; it gives away the first domino shape.
💡 Bottom singletons reach up
The one-cell sum at [4,2] only has one useful neighbor, so it forces a tile upward into the three-cell sum at [1,2]/[2,2]/[3,2]. The top-column sum at [0,3]/[1,3]/[2,3] then locks the [0,0] tile.
💡 Medium complete chain
Place [0,0] vertically at [0,3]/[1,3], then [3,6] horizontally at [2,2]/[2,3] with 3 at [2,2] and 6 at [2,3]. Put [4,2] vertically at [3,2]/[4,2] with 4 above and 2 below; [0,4] goes horizontally at [1,2]/[1,1] with 0 at [1,2] and 4 at [1,1]. Then [4,1] runs vertically at [3,1]/[2,1] with 4 at [3,1] and 1 at [2,1]; [4,4] fills [4,1]/[5,1] vertically as two 4s; and [2,1] finishes vertically at [3,0]/[2,0] with 2 over 1.
💡 Find the lonely sums
In the hard, don't start with the big regions—look for the one-cell sum constraints that are almost completely walled off. A very low target on the left edge is the key that starts a long vertical cascade.
💡 Start at [2,0]
The sum-1 cell at [2,0] has only one neighbor, so its domino is forced. That placement flips the [3,0]/[4,0] equals pair and feeds the [5,0]/[6,0] sum, which then resolves [6,1].
💡 Use the equality runs
After the left edge, tackle the double-zero opportunity in the top equality at [0,3]/[0,4]/[1,4]. Then let [0,2]'s sum force a matching column through [1,2]/[2,2]/[3,2].
💡 Chain the 4s and the right side
The 4-equality at [2,3]/[2,4]/[3,4] connects to the sum-3 [4,4]. On the right, the sum-15 trio at [1,8]/[2,7]/[2,8] and the bottom sum-5 triple at [5,6]/[5,7]/[5,8] snap into place once their neighboring singletons are known.
💡 Hard complete chain
Work the left spine first: [3,1] vertical at [3,0]/[2,0] (3 at [3,0], 1 at [2,0]); [5,3] vertical at [5,0]/[4,0] (5 at [5,0], 3 at [4,0]); [4,5] horizontal at [6,0]/[6,1] (5 at [6,0], 4 at [6,1]). Top equality: [0,0] horizontal at [0,3]/[0,4] (both 0); [4,0] vertical at [2,4]/[1,4] (4 at [2,4], 0 at [1,4]). Middle: [3,2] vertical at [0,2]/[1,2] (3 at [0,2], 2 at [1,2]); [2,4] horizontal at [2,2]/[2,3] (2 at [2,2], 4 at [2,3]); [5,2] vertical at [4,2]/[3,2] (5 at [4,2], 2 at [3,2]); [3,4] vertical at [4,4]/[3,4] (3 at [4,4], 4 at [3,4]). Right side: [6,4] vertical at [1,8]/[2,8] (6 at [1,8], 4 at [2,8]); [0,5] vertical at [3,7]/[2,7] (0 at [3,7], 5 at [2,7]); [3,0] vertical at [5,7]/[4,7] (3 at [5,7], 0 at [4,7]); [2,1] vertical at [2,6]/[1,6] (2 at [2,6], 1 at [1,6]); [6,1] vertical at [6,6]/[5,6] (6 at [6,6], 1 at [5,6]); [4,1] vertical at [6,8]/[5,8] (4 at [6,8], 1 at [5,8]).
🎨 Pips Solver
Sep 14, 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
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
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: The big equals block
The region covering [0,1], [0,2], [1,1], [1,2], and [1,3] is an equals constraint. Since five cells must match, the only way to supply that many identical pips is to use every domino carrying a 6. So all five cells are locked as 6.
2
Step 2: Leftover [4,5] finds its place
The only domino without a 6 is [4,5]. It cannot enter the big equals region, and the adjacent pair [0,0]/[1,0] must be equal, so [4,5] cannot lie across that pair. It is forced vertical at [1,0]/[2,0], with 4 at [1,0] to match [0,0] and 5 at the empty [2,0].
3
Step 3: Top row falls in
With [1,0] already filled, [0,0] must be paired with [0,1]. Tile [6,4] goes horizontally at [0,1]/[0,0], putting 6 in the big equals and 4 in [0,0]. Then [0,3] needs a value greater than 1; it is adjacent to [0,2], so [6,2] fills [0,2]/[0,3] with 6 and 2.
4
Step 4: Finish the interior
The double [6,6] must cover the two interior cells [1,1]/[1,2]. What remains is [6,1] vertical at [1,3]/[2,3], placing 6 in the equals region and leaving 1 in the empty [2,3] cell.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Top column locks a double
The cell [0,3] has only one neighbor, [1,3], and sits in the sum-6 region [0,3]/[1,3]/[2,3]. Forcing the [0,0] tile vertically at [0,3]/[1,3] gives two zeros, leaving [2,3] to make the missing sum. [2,3] then has only [2,2] free, so the [3,6] tile fills [2,2]/[2,3] with 3 and 6.
2
Step 2: Bottom singleton reaches upward
The single-cell sum-2 region at [4,2] must be exactly 2. Its only useful neighbor is [3,2], so the [4,2] tile goes vertical with 2 at [4,2] and 4 at [3,2]. This also satisfies part of the sum-7 region at [1,2]/[2,2]/[3,2].
3
Step 3: The sum-7 middle completes
With [2,2]=3 and [3,2]=4, the sum-7 region needs [1,2]=0. Since [1,2] is adjacent to [1,1], place [0,4] horizontally at [1,2]/[1,1] with 0 at [1,2] and 4 at [1,1].
4
Step 4: Sum-5 and sum-8 interlock
The sum-5 region [1,1]/[2,1] now has 4 at [1,1], so [2,1]=1. Pair [2,1] with [3,1] via [4,1], giving [3,1]=4. That leaves the sum-8 pair [3,1]/[4,1] needing 4 at [4,1], so [4,4] goes vertical at [4,1]/[5,1] with two 4s, satisfying the empty [5,1].
5
Step 5: Final less-2 and sum-2 stack
The remaining cells [2,0] and [3,0] form a vertical stack. [2,0] must be less than 2, and [3,0] has a sum-2 singleton, so the [2,1] tile fits with 1 at [2,0] and 2 at [3,0].
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The left-edge sum-1 trigger
The single-cell sum-1 region at [2,0] gives [2,0]=1. Its only neighbor is [3,0], so domino [3,1] is forced vertically: [2,0]=1 and [3,0]=3.
2
Step 2: Equals and sum-10 down the left spine
The equals pair [3,0]/[4,0] now requires [4,0]=3. Since [4,0] must connect to [5,0], domino [5,3] goes vertical: [4,0]=3 and [5,0]=5. The sum-10 pair [5,0]/[6,0] then makes [6,0]=5, and the only free neighbor is [6,1]; domino [4,5] fills horizontally at [6,0]/[6,1] with 5 and 4, which also satisfies the sum-4 singleton at [6,1].
3
Step 3: Top equality's zero lock
The equals triple [0,3]/[0,4]/[1,4] can only be zeros because the sole matching pair available is double-zero. Domino [0,0] lies horizontally at [0,3]/[0,4]. Then [1,4] must also be 0 and its only free neighbor is [2,4], so domino [4,0] goes vertical at [2,4]/[1,4] with 4 below and 0 above.
4
Step 4: Sum-3 seed and the 2-equality chain
The sum-3 singleton at [0,2] forces 3. Its neighbor [0,3] is already occupied, so [0,2] pairs with [1,2] via domino [3,2] vertically, giving [1,2]=2. The equals triple [1,2]/[2,2]/[3,2] now forces 2s at [2,2] and [3,2]. Domino [2,4] covers [2,2]/[2,3] horizontally with 2 and 4; domino [5,2] covers [3,2]/[4,2] vertically with 2 and 5, satisfying the sum-5 singleton at [4,2].
5
Step 5: The 4-equality center
The equals triple [2,3]/[2,4]/[3,4] already has 4s at [2,3] and [2,4], so [3,4] must be 4. It pairs with [4,4], whose sum-3 singleton forces a 3; domino [3,4] goes vertical at [3,4]/[4,4] with 4 above and 3 below.
6
Step 6: Right-side sums and equalities resolve
Sum-15 at [1,8]/[2,7]/[2,8] is 6+5+4: place [6,4] vertical at [1,8]/[2,8] (6 above, 4 below) and [0,5] vertical at [3,7]/[2,7] (0 at [3,7], 5 at [2,7]). The equals pair [3,7]/[4,7] then forces [4,7]=0 and [5,7]=3 with [3,0] vertical. Sum-3 at [1,6]/[2,6] uses [2,1] vertical with 2 at [2,6] and 1 at [1,6]. Finally, bottom sum-5 at [5,6]/[5,7]/[5,8] and adjacent singletons force [6,1] vertical at [6,6]/[5,6] with 6 at [6,6] and 1 at [5,6], plus [4,1] vertical at [6,8]/[5,8] with 4 at [6,8] and 1 at [5,8].
💡 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.
💬 Community Discussion
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