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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
February 23, 2026 brings three puzzles in ascending complexity. Ian Livengood's easy grid is built around sum constraints that mostly share the same target — a self-reinforcing structure where resolving one region immediately feeds the next.
Rodolfo Kurchan takes both medium and hard. The medium is a compact grid with three sum-7 regions chained together: the value you place in one flows directly into the next constraint, giving the puzzle a satisfying domino-chain feel. The hard scales up to a much larger grid with high-stakes regions scattered across it — a sum-17, a sum-12, and a sum-11 among them. Those extreme targets narrow the domino choices dramatically and give you strong footholds even in an otherwise open board.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Multiple constraints share the same target
Look at the sum regions in this grid — several of them have identical targets. That narrows down the pip values dramatically. Find the single-cell sum regions first; they give you free information with no arithmetic required.
💡 Two single-cell sums sit in the same column
There are two solo sum regions stacked in the leftmost column, both with the same target. Each one locks in a specific pip value — and once you place the right domino in each spot, the bottom pair constraint in that column falls into place automatically.
💡 Full answer
The single-cell sum=6 at [1,0] forces a 6 there — place [6–5] vertically with 6 at [1,0] and 5 at [1,1]. The single-cell sum=6 at [2,0] forces another 6 — place [6–4] vertically with 6 at [2,0] and 4 at [3,0]. The pair sum=6 at [3,0]+[3,1] now has [3,0]=4, so [3,1] must be 2. Place [0–2] horizontally with 2 at [3,1] and 0 at [3,2]. The equals region at [2,2]–[3,2] needs [2,2]=0 — place [4–0] horizontally with 4 at [2,3] (sum=4 ✓) and 0 at [2,2]. One domino left: [1–1] fills the pair sum=6 at [0,1]+[1,1] — [1,1]=5 already, so [0,1] must be 1. Place [1–1] with 1 at [0,1] and 1 at [0,2] (empty).
💡 Three sum regions share the same target
Every sum constraint in this puzzle has an identical target value. Start by identifying which region has the fewest possible pip combinations that could reach it — that's your sharpest entry point.
💡 The top-row sum and the vertical sum are connected
The two-cell sum in the top row and the vertical two-cell sum in column 3 share a dependency through [1,3]. The value you establish in the top pair filters directly into the vertical pair below it.
💡 Full answer
The top-row sum=7 at [0,2]+[0,3] gets [6–1] horizontally: 6 at [0,2] and 1 at [0,3]. The greater-than-2 constraint at [2,2] needs at least 3 — place [2–3] vertically with 3 at [2,2] (✓) and 2 at [3,2]. The vertical sum=7 at [1,3]+[2,3] needs to total 7 — place [5–4] vertically with 5 at [2,3] and 4 at [3,3]. Then [0–2] runs vertically: 0 at [1,2] (empty) and 2 at [1,3]. Check: 2+5=7 ✓. Bottom-row sum=7 at [3,2]+[3,3]+[3,4]: 2+4+1=7, so [3,4]=1 — [1–0] fills [3,4]=1 and [3,5]=0 (empty). The last domino [0–0] fills both empty cells at [3,0] and [3,1].
💡 Aim for the extreme sum regions first
Several regions in this puzzle have unusually high targets. When a sum is close to the maximum possible, your pip options shrink fast. Find the highest-target region and ask: how few ways can dominoes reach that total?
💡 The sum-17 region tells you almost exactly what goes there
A three-cell sum=17 is one of the highest you'll encounter in Pips. The only way to hit it is with two 6s and a 5. Figure out which dominoes carry those values and which orientation fits the region.
💡 Sum-12 in two cells also forces large pips
Two cells summing to 12 means both must be 6. That's a hard constraint — only one combination works. Find the two-cell sum=12 region and place the pair of sixes there.
💡 A single-cell sum region gives a free pip value somewhere in the grid
There's a lone sum constraint targeting a moderate value. It hands you one pip count directly, no calculation needed. Use that as an anchor point alongside the high-value regions you've already filled.
💡 Full answer
Single-cell sum=4 at [5,2] forces 4 — place [4–0] vertically with 4 at [5,2] and 0 at [5,1] (empty ✓). Sum=17 needs two 6s and a 5: place [6–6] horizontally at [3,5]=6 and [3,6]=6; place [2–5] vertically with 5 at [4,5] and 2 at [5,5]. Sum=12 needs two 6s: place [0–6] vertically with 0 at [4,3] (empty ✓) and 6 at [5,3]; place [6–3] vertically with 6 at [6,3] and 3 at [7,3]. 6+6=12 ✓. Sum=9 at [7,3]+[7,4]+[7,5]: [7,3]=3, so [7,4]+[7,5]=6 — place [3–3] horizontally with 3 at [7,4] and 3 at [7,5]. Sum=11 at [3,1]+[4,1]: place [5–0] vertically with 5 at [3,1] and 0 at [3,0] (empty ✓); place [2–6] horizontally with 2 at [4,2] and 6 at [4,1]. 5+6=11 ✓. Sum=4 at [3,2]+[4,2]: [4,2]=2, so [3,2]=2 — place [0–2] vertically with 0 at [2,2] (empty ✓) and 2 at [3,2]. Sum=4 at [0,3]+[1,3]: place [2–2] vertically with 2 at [0,3] and 2 at [1,3]. The four-cell sum=4 at [2,3]+[3,3]+[3,4]+[4,4]: place [1–1] vertically with 1 at [2,3] and 1 at [3,3]; place [0–1] vertically with 0 at [2,4] (empty ✓) and 1 at [3,4]; place [1–2] vertically with 1 at [4,4] and 2 at [5,4]. 1+1+1+1=4 ✓. Sum=4 at [5,4]+[5,5]: 2+2=4 ✓. Last domino [0–0] fills empty cells [7,1] and [7,2].
🎨 Pips Solver
Feb 23, 2026
Click a domino to place it on the board. You can also click the board, and the correct domino will appear.
✅ 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: Two single-cell sums unlock the left column
Cell [1,0] is a lone sum=6 region — it must hold exactly 6. The [6–5] domino delivers: place it vertically with 6 at [1,0] and 5 dropping down to [1,1].
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Step 2: The second single-cell sum mirrors the first
Cell [2,0] is also a lone sum=6 region — another 6. The [6–4] domino fits neatly: place it vertically with 6 at [2,0] and 4 below at [3,0].
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Step 3: The bottom-row pair clicks into place
The sum=6 region covering [3,0] and [3,1] now has [3,0]=4 locked in. Subtract: [3,1] must be 2. The [0–2] domino threads both remaining cells — place it horizontally with 2 at [3,1] and 0 going right to [3,2].
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Step 4: Equals region forces the domino above
The equals constraint at [2,2] and [3,2] demands both cells match. [3,2]=0, so [2,2] must also be 0. The [4–0] domino handles this row: place it horizontally with 4 at [2,3] and 0 at [2,2]. The single-cell sum=4 at [2,3] confirms: 4=4 ✓.
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Step 5: The top pair closes out the puzzle
One domino and one region remain. The pair sum=6 at [0,1]+[1,1] has [1,1]=5 already placed. Subtract: [0,1] must be 1. The [1–1] domino fits: 1 at [0,1] and 1 at [0,2] (the unconstrained empty cell). Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Start at the top — sum-7 in the first row
The sum=7 region spans [0,2] and [0,3] at the top of the board. Your dominoes include [6–1], which gives exactly 6+1=7. Place it horizontally: 6 at [0,2] and 1 at [0,3].
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Step 2: Handle the greater-than constraint
Cell [2,2] has a greater-than-2 constraint — it needs a pip count of 3 or more. The [2–3] domino can put 3 there. Place it vertically: 3 at [2,2] (constraint satisfied) and 2 going down to [3,2].
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Step 3: Fill the vertical sum-7 in column 3
The sum=7 region at [1,3] and [2,3] needs to total 7. The [5–4] domino provides the right pair — place it vertically with 5 at [2,3] and 4 at [3,3]. Now [1,3] needs to make [1,3]+5=7, so [1,3]=2. The [0–2] domino drops in vertically: 0 at [1,2] (empty) and 2 at [1,3]. Sum check: 2+5=7 ✓.
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Step 4: Resolve the bottom-row sum-7
The three-cell sum=7 at [3,2]+[3,3]+[3,4] now has two values known: [3,2]=2 and [3,3]=4. The third cell [3,4] must be 1 to reach 7. The [1–0] domino places 1 at [3,4] and 0 at [3,5] (empty). Sum check: 2+4+1=7 ✓.
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Step 5: Last domino fills the remaining empties
Cells [3,0] and [3,1] are both empty — no constraints. The [0–0] double domino fills them both with zero. Every region satisfied; puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The lone sum cell is the easiest free value on the board
Cell [5,2] is a single-cell sum=4 region — it holds exactly 4. The [4–0] domino delivers: place it vertically with 4 at [5,2] and 0 at [5,1] (empty constraint satisfied).
2
Step 2: Attack the sum-17 region with near-maximum pips
The sum=17 region spans [3,5], [3,6], and [4,5]. Three cells must total 17 — the only viable breakdown is 6+6+5. Place [6–6] horizontally across [3,5] and [3,6], both showing 6. Then place [2–5] vertically with 5 at [4,5] and 2 at [5,5]. Sum check: 6+6+5=17 ✓.
3
Step 3: Sum-12 demands two sixes in a vertical pair
The sum=12 region at [5,3] and [6,3] requires both cells to be 6. Place [0–6] vertically with 0 at [4,3] (empty ✓) and 6 at [5,3]. Place [6–3] vertically with 6 at [6,3] and 3 at [7,3]. Sum check: 6+6=12 ✓.
4
Step 4: Sum-9 across the bottom row follows directly
The sum=9 region at [7,3], [7,4], and [7,5] already has [7,3]=3 from step 3. The remaining two cells need 3+3=6. Place [3–3] horizontally: 3 at [7,4] and 3 at [7,5]. Sum check: 3+3+3=9 ✓.
5
Step 5: Crack the sum-11 pair in column 1
The sum=11 region at [3,1] and [4,1] needs two large values. Place [5–0] vertically with 5 at [3,1] and 0 at [3,0] (empty ✓). Place [2–6] horizontally with 2 at [4,2] and 6 at [4,1]. Sum check: 5+6=11 ✓.
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Step 6: Sum-4 in the adjacent vertical pair
The sum=4 region at [3,2] and [4,2] already has [4,2]=2 from step 5. So [3,2] must be 2. Place [0–2] vertically with 0 at [2,2] (empty ✓) and 2 at [3,2]. Sum check: 2+2=4 ✓.
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Step 7: Resolve the top sum pair and the four-cell region
Sum=4 at [0,3]+[1,3]: place [2–2] vertically with 2 at [0,3] and 2 at [1,3]. 2+2=4 ✓. The four-cell sum=4 at [2,3]+[3,3]+[3,4]+[4,4]: place [1–1] vertically with 1 at [2,3] and 1 at [3,3]; place [0–1] vertically with 0 at [2,4] (empty ✓) and 1 at [3,4]; place [1–2] vertically with 1 at [4,4] and 2 at [5,4]. Sum check: 1+1+1+1=4 ✓. Sum=4 at [5,4]+[5,5]: 2+2=4 ✓ — both already placed.
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Step 8: Last domino wraps the bottom empties
Cells [7,1] and [7,2] carry no constraints. The [0–0] double domino fills both with zero. Every region verified; puzzle complete.
💡 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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