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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 20, 2026 brings three puzzles with distinct personalities. Ian Livengood handles easy and medium with the clean, methodical style he's known for — every constraint feels intentional, and the solution tends to emerge in one smooth chain once you find the right starting point.
The easy grid splits into two mirrored four-cell equals regions, with a less-than and a greater-than anchor sitting at opposite ends of the top row. Those two polarity constraints nearly solve themselves, and the equals regions fill in right after.
Rodolfo Kurchan takes the medium and hard slots. His medium packs a sum-11 region that demands a specific pair of large pips, alongside a high-threshold greater-than that eliminates most options immediately. The hard puzzle is a different beast — dominoes are clustered at all four corners and in the center of an 8×8 grid, with each cluster holding its own tight set of constraints. You can crack each cluster nearly independently and then admire the symmetry at the end.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Two constraints on opposite ends of the top row
There's a less-than constraint on one end of the top row and a greater-than on the other. They each rule out most of your dominoes — figure out which end is most restricted and start there.
💡 The right end of the top row is nearly forced
The less-than constraint at the far right is quite strict. Only two pip values can satisfy it, and once you find the right domino, the entire right half of the board starts to lock in through the equals region.
💡 Full answer
The less-than-2 cell at [0,5] needs a 0 or 1. Place [0–6] horizontally with 0 at [0,5] and 6 at [0,4]. That 6 anchors the right-side equals region — [0,4], [1,3], [1,4], [2,3] must all be 6. Place [6–6] across [1,3]–[1,4], then orient [6–4] with 6 at [2,3] and 4 going left to [2,2]. That 4 anchors the left-side equals region: [0,1], [1,1], [2,1], [2,2] all equal 4. Place [4–4] vertically down column 1 rows 1–2, then [4–3] horizontally at row 0 with 4 at [0,1] and 3 at [0,0]. The 3 satisfies the greater-than-2 constraint — puzzle complete.
💡 One cell has a strict minimum
Find the cell with the highest minimum threshold in the grid. That single constraint eliminates most of your dominoes and gives you a strong starting point.
💡 Sum-11 demands two large pips
The two-cell sum-11 region needs a total very close to the maximum possible. Think about which domino can provide two pips that high — there aren't many options.
💡 Full answer
The greater-than-4 cell at [1,2] needs a 5 or 6. Place [6–6] vertically at [0,2] and [1,2], both showing 6. The equals region at [1,3]–[2,3] wants matching values — [4–4] fits cleanly. Sum-11 at [2,1]+[3,1] needs two large pips: [6–5] goes vertically with 6 at [1,1] and 5 at [2,1]. Then [6–4] places 6 at [3,1] and 4 at [3,0] — less-than-6 confirmed. The bottom-row sum-10 gets [3–0] (3 at [3,2], 0 at [2,2]) and [4–3] (4 at [3,3], 3 at [3,4]): 3+4+3=10.
💡 One cell is nearly impossible to satisfy
There's a less-than constraint somewhere with an extremely low threshold — only one pip value can fit. That cell is your first guaranteed placement.
💡 Two constrained cells sit side by side and trigger each other
The nearly-impossible cell from hint 1 shares a neighborhood with another tightly constrained cell. Resolving the first one gives you the second almost instantly — and together they determine which two dominoes belong in that area.
💡 Each corner of the grid is its own mini-puzzle
The hard puzzle scatters dominoes across four corners and a center cluster. Once you see that the clusters are mostly independent, you can solve each one separately without worrying about how they connect.
💡 The top-left chain flows into a sum-11 region
After pinning the first two cells, the domino ends that land in the top row directly feed the sum-11 region. The values cascade naturally — no guessing needed.
💡 Full answer
Less-than-1 at [1,1] forces 0 — place [0–5] with 0 at [1,1] and 5 at [0,1]. Sum=1 at [1,0] forces 1 — place [6–1] with 6 at [0,0] and 1 at [1,0]. Sum-11 confirms: 6+5=11 ✓. Top-right cluster: [4–1] places 4 at [0,7] (greater-than-2 ✓) and 1 at [0,6] (less-than-2 ✓); [2–2] fills both sum=2 cells at [1,6] and [1,7]. Center cluster: [3–6] places 3 at [4,3] (sum=3 ✓) and 6 at [4,4] (greater-than-3 ✓); [0–3] places 0 at [3,4] and 3 at [3,3] — sum=3 pair: 3+0=3 ✓. Bottom-right: [0–6] places 0 at [6,6] (less-than-5 ✓) and 6 at [6,7] (greater-than-5 ✓); [1–1] fills the equals pair at [7,6]=[7,7]=1. Bottom-left: [1–2] vertical with 1 at [6,1] and 2 at [7,1]; [2–3] vertical with 2 at [7,0] and 3 at [6,0] — sum=4: 3+1=4 ✓; equals: 2=2 ✓.
🎨 Pips Solver
Feb 20, 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: The less-than constraint decides the right edge
The cell at [0,5] sits inside a less-than-2 region — its value must be 0 or 1. Of your five dominoes, [0–6] is the only one with a 0. Orient it horizontally along the top row: 0 lands at [0,5] and 6 slides left to [0,4].
2
Step 2: The 6 anchors the right-side equals region
The right-side equals region covers four cells: [0,4], [1,3], [1,4], and [2,3]. You've just established [0,4]=6, so all four must be 6. Place [6–6] horizontally across row 1 at columns 3 and 4 — both show 6.
3
Step 3: Complete the right-side equals with the last 6 needed
Cell [2,3] still needs a 6. The [6–4] domino delivers: orient it horizontally with 6 at [2,3] and 4 going left to [2,2]. The right-side equals region is now fully satisfied.
4
Step 4: The 4 at [2,2] anchors the left-side equals region
The left-side equals region covers [0,1], [1,1], [2,1], and [2,2]. With [2,2]=4 now placed, every cell in this region must also be 4. Place [4–4] vertically down column 1, spanning rows 1 and 2.
5
Step 5: Finish the top row
Cell [0,1] still needs a 4. One domino remains: [4–3]. Place it horizontally with 4 at [0,1] and 3 at [0,0]. The greater-than-2 constraint at [0,0] checks out — 3 is comfortably above 2. Every cell is placed; puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The greater-than threshold eliminates most options
Cell [1,2] has a greater-than-4 constraint — it needs a pip value of 5 or 6. Scanning your dominoes for one that can reach [1,2] with a 5 or higher, [6–6] is the clearest fit. Place it vertically: 6 at [0,2] and 6 at [1,2]. Both empty cells filled, threshold satisfied.
2
Step 2: Lock the equals pair in column 3
The equals region at [1,3] and [2,3] needs both cells to match. The [4–4] double domino is tailor-made for this — place it vertically with 4 at both rows. Clean and certain.
3
Step 3: Target the sum-11 region
The two-cell sum-11 at [2,1] and [3,1] demands a total very close to the maximum. The [6–5] domino gives exactly that. Place it vertically in column 1: 6 at row 1 (the unconstrained empty cell [1,1]) and 5 at row 2. [2,1]=5 is now confirmed.
4
Step 4: Fill the left column and satisfy less-than-6
The sum-11 region needs [3,1]=6. The [6–4] domino provides that: place it vertically with 6 at [3,1] and 4 at [3,0]. The less-than-6 constraint at [3,0] is satisfied — 4 is safely below 6.
5
Step 5: Complete the bottom row
Two dominoes remain — [3–0] and [4–3] — filling three cells in the bottom row's sum-10 region plus one cell above. Place [3–0] vertically with 3 at [3,2] and 0 going up to [2,2]. Then place [4–3] horizontally with 4 at [3,3] and 3 at [3,4]. Bottom row check: 3+4+3=10. ✓ Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The tightest constraint on the board
Cell [1,1] carries a less-than-1 constraint — the only valid pip value is 0. Look through your dominoes for one with a 0 that can reach this cell. The [0–5] domino fits: place it vertically with 0 at [1,1] and 5 going up to [0,1].
2
Step 2: Sum-1 resolves immediately next door
Cell [1,0] has a sum=1 constraint — it must hold exactly 1. Adjacent to it, [0,0] is part of a sum-11 region with [0,1]. You now know [0,1]=5, so [0,0] must be 6 to reach 11. The [6–1] domino delivers: place it vertically with 6 at [0,0] and 1 at [1,0]. Sum=11 confirmed: 6+5=11. Sum=1 confirmed: 1. Both cells nailed in one placement.
3
Step 3: Decode the top-right cluster
The top-right pair [0,6] and [0,7] each carry polarity constraints: less-than-2 and greater-than-2. Cell [0,6] needs 0 or 1; [0,7] needs 3 or more. The [4–1] domino satisfies both at once — place it horizontally with 4 at [0,7] and 1 at [0,6]. 4>2 ✓, 1<2 ✓.
4
Step 4: Fill the top-right sum cells
Cells [1,6] and [1,7] each need a pip count of exactly 2. The [2–2] domino is the only double-two available — place it horizontally across both cells. Both sum=2 regions confirmed in one move.
5
Step 5: Solve the center cluster
Cell [4,3] is a single-cell sum=3 region — it must hold a 3. Cell [4,4] has a greater-than-3 constraint — it needs 4 or more. The [3–6] domino handles both: place it horizontally with 3 at [4,3] and 6 at [4,4]. Sum=3 ✓, greater-than-3 ✓.
6
Step 6: Complete the center with the sum-3 pair
The two-cell sum=3 region at [3,3] and [3,4] needs to total 3. The [0–3] domino fits: place it horizontally with 3 at [3,3] and 0 at [3,4]. 3+0=3 ✓. The center cluster is fully solved.
7
Step 7: Crack the bottom-right corner
Cell [6,7] needs a value greater than 5 — only 6 works. Cell [6,6] needs less than 5 — so 4 or below. The [0–6] domino threads both constraints: place it horizontally with 0 at [6,6] and 6 at [6,7]. The equals region at [7,6]–[7,7] takes the [1–1] double — both cells become 1.
8
Step 8: Seal the bottom-left corner
Two dominoes remain: [1–2] and [2–3]. The sum=4 region at [6,0]+[6,1] and the equals region at [7,0]=[7,1] need to be satisfied together. Place [1–2] vertically with 1 at [6,1] and 2 at [7,1]. Place [2–3] vertically with 2 at [7,0] and 3 at [6,0]. Sum check: 3+1=4 ✓. Equals check: 2=2 ✓. Every cluster solved — 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.
💬 Community Discussion
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