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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
August 18, 2025 is the day NYT Pips launched — puzzles #1, #2, and #3. Rodolfo Kurchan opens with the easy, and Heidi Erwin handles both medium and hard. Together they set the template for what Pips would become: precise constraint networks that reward logical deduction over guessing.
The easy intro is a tight 3×4 grid with four dominoes. A single-cell sum and a pair sum at opposite ends of the board serve as the entry points, and the four-cell equals region in the middle fills in right after.
The medium steps up with a sum=18 three-cell region — a target so high that all three cells must hold 6 — alongside a strict less-than and a low single-cell sum that each eliminate most options immediately. The hard widens the grid and introduces a four-cell sum=0 region, where every placed cell must be zero, plus another sum=18 cluster that demands three 6s.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Two sum regions anchor opposite ends of the board
One corner holds a pair sum with a very high target, and another corner holds a single-cell sum with a moderate target. The high-target pair is so demanding that only one domino in your set can satisfy it — start there.
💡 A double domino locks in the bottom, then the single cell follows
Once the pair sum is filled with its double, look at the single-cell sum on the other end. It tells you exactly which pip value goes there — and its domino partner lands inside the equals region, giving you the value all four cells must share.
💡 Full answer
The pair sum=12 at [2,1]+[2,2] needs two cells totaling 12 — only [6–6] can do that. Place it horizontally with 6 at each cell. The single-cell sum=5 at [0,3] must hold 5 — place [5–2] horizontally with 5 at [0,3] and 2 at [0,2]. That 2 sits inside the four-cell equals region covering [0,1],[0,2],[1,1],[1,2] — all must be 2. Place [2–2] horizontally with 2 at [1,1] and 2 at [1,2]. One domino left: [2–3] completes the equals region with 2 at [0,1] ✓ and places 3 at [0,0] (empty).
💡 A single-cell sum on the bottom row is a free move
One cell in the bottom row carries a solo sum constraint. Its target directly tells you the pip value — no calculation needed. Place the matching domino end there and note where the other half lands.
💡 Sum-18 in three cells demands near-maximum pips across the board
A three-cell region targeting 18 means every single cell must be 6 — there's no other way to reach it. Once you figure out which two cells are covered by a single domino, the third cell's domino becomes clear.
💡 Full answer
Sum=2 at [2,1] forces a 2 there — place [2–6] horizontally with 2 at [2,1] and 6 at [2,2]. Greater>4 at [2,4] needs 5 or 6 — place [5–6] horizontally with 5 at [2,4] and 6 at [2,3]. Sum=18 at [1,3]+[2,2]+[2,3]: [2,2]=6 and [2,3]=6 already; [1,3] must be 6. Less<3 at [1,4] forces 0 or 1 — place [1–6] horizontally with 1 at [1,4] ✓ and 6 at [1,3] ✓. Sum-18 confirmed: 6+6+6=18 ✓. Place [4–4] horizontally with 4 at [1,0] and 4 at [1,1]. Less<13 region at [0,0]+[1,0]+[1,1]: need [0,0]<5 — place [6–3] horizontally with 6 at [0,1] (empty ✓) and 3 at [0,0]. Check: 3+4+4=11<13 ✓.
💡 A four-cell sum region targets zero — every cell must be 0
Somewhere in the grid a region covers four cells and targets a sum of zero. That leaves no room for any pip value other than 0. Think about which dominoes in your set can place four zeros across those positions.
💡 Sum-18 in three cells means three 6s — find the double-six first
Just like in the medium, a three-cell sum=18 region requires all three cells to be 6. The [6–6] double domino handles two of them in one placement. The third 6 comes from a different domino that also feeds an adjacent equals constraint.
💡 Equals constraints chain together across the middle row
Several equals regions sit along row 1. Once you know one cell's value from a sum constraint, the paired cell inherits it — and that value may cascade into another equals pair nearby.
💡 The sum-1 cell and the less-than-3 pair each point to small-pip domino ends
A single-cell sum=1 in the middle of row 1 is nearly as forced as a sum=0. Combine that with the two-cell less-than-3 region above it — both constraints demand small values, and the dominoes that satisfy them overlap in a useful way.
💡 Full answer
Sum=0 across [2,0]+[2,1]+[2,2]+[2,3]: all four must be 0. Place [0–0] horizontally at [2,1]=0 and [2,2]=0. Place [0–6] vertically with 0 at [2,0] and 6 at [1,0]. Place [2–0] horizontally with 2 at [2,4] and 0 at [2,3]. Row 2 sum: 0+0+0+0=0 ✓. Sum=18 at [1,5]+[1,6]+[2,6]: place [6–6] vertically with 6 at [1,6] and 6 at [2,6]. Place [2–6] horizontally with 2 at [1,4] and 6 at [1,5]. 6+6+6=18 ✓. Equals [1,4]=[2,4]: both are 2 ✓. Sum=1 at [1,2]: place [1–6] horizontally with 1 at [1,2] and 6 at [1,1]. Equals [1,0]=[1,1]: [1,1]=6, [1,0]=6 ✓. Less<3 at [0,1]+[0,2]: sum must be <3 — place [1–1] horizontally with 1 at [0,1] and 1 at [0,2]. 1+1=2<3 ✓. Equals [0,7]=[1,7]: place [3–1] horizontally with 3 at [0,7] and 1 at [0,6] (empty ✓). Place [3–5] horizontally with 3 at [1,7] and 5 at [1,8]. 3=3 ✓. Sum=10 at [1,8]+[2,8]: [1,8]=5, so [2,8]=5. Sum=4 at [2,7]: must be 4. Place [4–5] horizontally with 4 at [2,7] and 5 at [2,8]. ✓
🎨 Pips Solver
Aug 18, 2025
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 high-target pair locks in the bottom row
The pair sum=12 at [2,1] and [2,2] is demanding — two cells must total 12. The only way to get there is with two 6s. Place [6–6] horizontally: 6 at [2,1] and 6 at [2,2].
2
Step 2: The single-cell sum fixes the top-right corner
The single-cell sum=5 at [0,3] must hold exactly 5. The [5–2] domino fits: place it horizontally with 5 at [0,3] and 2 going left to [0,2].
3
Step 3: The equals region fills itself in
The four-cell equals region at [0,1],[0,2],[1,1],[1,2] now has [0,2]=2 anchored. All four cells must be equal — so all four are 2. Place [2–2] horizontally at [1,1] and [1,2], both showing 2.
4
Step 4: Last domino completes the board
One domino remains: [2–3]. Cell [0,1] must be 2 (equals region), so the 2-end goes there. The 3-end drops into [0,0], the empty cell. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Single-cell sum=2 gives away the first domino end
Cell [2,1] is a lone sum=2 region — it holds exactly 2. The [2–6] domino places 2 there and 6 going right to [2,2].
2
Step 2: Greater-than fills the right end of the bottom row
Cell [2,4] needs a value greater than 4 — so 5 or 6. The [5–6] domino handles it: place it horizontally with 5 at [2,4] and 6 at [2,3].
3
Step 3: Sum-18 forces the last six into row 1
The sum=18 region at [1,3]+[2,2]+[2,3] already has [2,2]=6 and [2,3]=6. That means [1,3] must also be 6 to reach 18. The less-than-3 constraint at [1,4] forces 0 or 1 there. Place [1–6] horizontally: 1 at [1,4] ✓ and 6 at [1,3]. Sum check: 6+6+6=18 ✓.
4
Step 4: The double-four fills the middle of row 1
Two cells in the middle of row 1 remain open and unconstrained on their own. Place [4–4] horizontally with 4 at [1,0] and 4 at [1,1].
5
Step 5: Less-than-13 confirms the last placement
One domino remains: [6–3]. The less-than-13 region covering [0,0]+[1,0]+[1,1] already has [1,0]+[1,1]=8. So [0,0] must be less than 5. Place [6–3] horizontally with 6 at [0,1] (empty ✓) and 3 at [0,0]. Final check: 3+4+4=11<13 ✓. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Wipe the entire bottom row to zero
The sum=0 region at [2,0],[2,1],[2,2],[2,3] demands all four cells be 0. Three dominoes share the load: place [0–0] horizontally at [2,1]=0 and [2,2]=0. Place [0–6] vertically with 0 at [2,0] and 6 at [1,0]. Place [2–0] horizontally with 2 at [2,4] and 0 at [2,3]. All four bottom cells: 0+0+0+0=0 ✓.
2
Step 2: Fill the sum-18 cluster with three sixes
The sum=18 region at [1,5],[1,6],[2,6] needs every cell to be 6. Place [6–6] vertically with 6 at [1,6] and 6 at [2,6]. Place [2–6] horizontally with 2 at [1,4] and 6 at [1,5]. All three cells confirmed as 6. ✓
3
Step 3: The equals pair in column 4 confirms itself
The equals region at [1,4] and [2,4] demands both cells match. [1,4]=2 from step 2 and [2,4]=2 from step 1 — they already agree. ✓ No additional placement needed.
4
Step 4: Sum-1 and the adjacent equals lock row 1's center
Cell [1,2] is a single-cell sum=1 — it must be exactly 1. Place [1–6] horizontally: 1 at [1,2] and 6 at [1,1]. The equals region at [1,0] and [1,1] requires both to match — [1,1]=6 is confirmed, and [1,0]=6 from step 1. ✓
5
Step 5: Less-than-3 pins the top-middle pair
The sum of [0,1] and [0,2] must be less than 3 — so the total is 0, 1, or 2. The [1–1] domino places 1 at [0,1] and 1 at [0,2]. Sum: 1+1=2<3 ✓.
6
Step 6: Resolve the top-right equals pair
The equals region at [0,7] and [1,7] needs both cells to match. Place [3–1] horizontally with 3 at [0,7] and 1 at [0,6] (empty ✓). Place [3–5] horizontally with 3 at [1,7] and 5 at [1,8]. Equals confirmed: [0,7]=[1,7]=3 ✓.
7
Step 7: Close the bottom-right with the last domino
The sum=10 region at [1,8]+[2,8] has [1,8]=5; so [2,8]=5. The single-cell sum=4 at [2,7] must hold 4. Place [4–5] horizontally: 4 at [2,7] ✓ and 5 at [2,8] ✓. Every constraint satisfied — 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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