🚨 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!
Click here to play today's official NYT Pips game first.
Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Today's NYT Pips puzzles, edited by Ian Livengood, showcase a range of elegant constructions from Livengood himself and Rodolfo Kurchan. Each difficulty level demonstrates a distinct design philosophy, from the minimalist twin anchors of the easy grid to the tightly interlocking equals chains in Kurchan's hard puzzle. Livengood's easy puzzle makes clever use of two isolated sum-1 cells, each demanding a pip value of 1. This forces immediate placement of the [1,1] double domino against the grid's edge, while a sum-2 region works in tandem with an equals region to lock down the remaining dominos with satisfying precision. Kurchan's medium leans heavily on equals regions, creating a structural rhythm where dominoes must match across multiple cells. The single less-4 cell provides a gentle entry point, but the unequal trio quickly tightens the possibilities. His hard puzzle ups the stakes with a sprawling arrangement of sum and equals constraints. A four-cell equals chain on the right side of the board forces the value 5, setting off a cascade that resolves the entire lower half. It's a masterclass in constraint interdependence.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1
Look for single-cell sum regions that force a specific pip value. These tiny constraints are the key to unlocking the grid.
💡 Hint 2
The two sum-1 cells at the top corners each demand a 1. Since only one domino carries a pair of 1s, one of those corners must use the double-1. The other corner will pair with a 2 from the only other domino containing a 1.
💡 Hint 3
Place the [1,1] domino vertically covering the top-right sum-1 cell and the cell below it. Place the [2,1] domino vertically in the top-left corner, with the 1 on the sum-1 cell and the 2 below. The sum-6 region below then forces [4,6] placed vertically with 4 above and 6 at [2,1], satisfying the equals region that requires two 6s. The sum-2 region takes [0,3] horizontally, and the sum-4 region finishes with the 3 from that domino next to the placed 1.
💡 Hint 1
Start by scanning for inequality and equals constraints. The grid's few single-cell less-than and greater-than regions will point you to the first dominoes.
💡 Hint 2
The lone less-4 cell at [0,3] must be a small pip; the only domino that can pair it vertically with the equals region next to it forces a specific 6-1 domino there. The greater-3 cell at [2,0] then demands a high pip, which in turn dictates the neighboring equals chains.
💡 Hint 3
Place the [6,1] domino vertically at [0,3]-[1,3] with 1 at the top and 6 at [1,3], making the equals pair at [1,3]-[1,4] both 6. The greater-3 cell forces [5,2] horizontally at [2,0]-[2,1] with 5 and 2. The equals region spanning [2,1],[3,1],[3,2] then requires all 2s, so place the [2,2] double horizontally at [3,1]-[3,2]. The unequal trio at top-middle takes [2,3] horizontally on [1,1]-[1,2] for distinct values. The remaining equals pairs fill in with [4,4] vertically and [4,2] horizontally, and finally [2,6] ties off the right side.
💡 Hint 1
Identify the dual-cell greater-than region and the overlapping sum-10 region that share a column. The pip ceiling of 6 makes the solution inevitable.
💡 Hint 2
The greater-10 region in the top-left corner and the sum-10 region immediately below it force the placement of the [6,6] double domino vertically spanning rows 1 and 2. This, in turn, forces the value in the sum-10 region's other cell to be 4.
💡 Hint 3
With [3,0] set to 4, the single sum-3 cell at [4,0] ensures the vertical domino connecting them must be [4,3], giving [4,0]=3. Meanwhile, the sum-4 cell at [0,3] forces a 4 there; placing the [4,5] domino horizontally with the 5 on the left connects to the sum-8 region on the top edge.
💡 Hint 4
The sum-8 region at [0,1]-[0,2] now needs a 3, which comes from the [6,3] domino bridging [0,0]-[0,1] to also satisfy the greater-10 region. The sum-11 region at rows 2-3 mid-grid forces the [4,4] double horizontally, and the connected cell [3,3] gets a 3 from the [1,3] domino, which also sets up the equals region in column 3.
💡 Hint 5
Complete the top: [6,6] at [1,0]-[2,0] (6,6); [4,3] at [3,0]-[4,0] (4,3); [4,5] at [0,2]-[0,3] (5,4); [6,3] at [0,0]-[0,1] (6,3). Mid-grid: [4,4] at [2,2]-[2,3] (4,4); [1,3] at [3,3]-[4,3] (3,1) giving [4,3]=1 and setting equals [5,3]=1 via [6,1] at [5,2]-[5,3] (6,1). That forces [5,1]=2 from sum-8 → place [2,2] at [5,0]-[5,1] (2,2). Right side: equals chain of four 5s driven by [5,5] vertical at [4,5]-[5,5] (5,5); [1,5] at [3,5]-[3,6] (5,1) giving [3,6]=1, sum-3 requires [3,7]=2 via [0,2] at [3,7]-[4,7] (2,0); [5,0] at [5,6]-[5,7] (5,0) completes the chain. Lower right finishes with [0,1] at [6,7]-[7,7] (0,1), [0,0] at [7,5]-[8,5] (0,0), and [1,1] at [8,6]-[8,7] (1,1).
🎨 Pips Solver
Jul 10, 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: Single-cell sum-1 anchors
The two isolated sum-1 cells at [0,0] and [0,5] each need exactly 1. Among available dominos, only [1,1] and [2,1] contain a 1. The [1,1] double can satisfy one of them, but since it covers two adjacent cells, it must be placed vertically on the right, covering [0,5] and [1,5] both as 1. The left sum-1 [0,0] then must use the [2,1] domino, placed vertically with the 1 at [0,0] and the 2 at [1,0].
2
Step 2: Sum-6 region in row 1
The region comprising [1,0] and [1,1] must sum to 6. With [1,0]=2 already from the previous step, [1,1] must be 4. The only remaining domino with a 4 is [4,6], so place it vertically covering [1,1] (4) and [2,1] (6).
3
Step 3: Equals region forces a pair
The equals region at [2,1] and [2,2] requires identical values. We now have [2,1]=6, so [2,2] must also be 6. The [6,2] domino can supply a 6 and a 2; place it horizontally on [2,2]-[2,3] with the 6 in [2,2] and 2 in [2,3].
4
Step 4: Final sums lock it in
The sum-2 region at [2,3]-[2,4] needs a total of 2. With [2,3]=2, [2,4] must be 0. The [0,3] domino fits perfectly placed horizontally with 0 at [2,4] and 3 at [1,4]. Then the sum-4 region at [1,4]-[1,5] with [1,4]=3 and [1,5]=1 (already from [1,1]) completes the grid.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Single less-4 cell
The cell [0,3] must be less than 4. It only connects downward to [1,3]. Trying the [2,3] domino would place a 2 or 3 at [0,3] but would force [1,3] to match the equals region at [1,4] with no available 3. Thus the only viable option is the [6,1] domino placed vertically: [0,3]=1, [1,3]=6, which then forces [1,4]=6 via the equals constraint.
2
Step 2: Greater-3 cell points right
The single cell [2,0] needs a pip greater than 3. Only the [5,2] domino can supply a 5 here, placed horizontally covering [2,0]=5 and [2,1]=2. This immediately sets the equals chain below: [2,1],[3,1],[3,2] must all be 2, so the [2,2] double domino is placed horizontally at [3,1]-[3,2].
3
Step 3: Unequal trio tightens the top
The region [1,1],[1,2],[2,2] demands all three values be distinct. With the [2,3] domino placed horizontally at [1,1]-[1,2], we supply 2 and 3. The remaining cell [2,2] will later receive a 4 from the [4,4] double, making the trio 2,3,4 — all distinct.
4
Step 4: Closing the equals loops
Place the [4,4] double vertically covering [2,2] and [2,3] with 4s, fulfilling the unequal requirement and satisfying the equals region at [2,3]-[3,3]. Then place the [4,2] domino horizontally at [3,3]-[3,4] with 4 and 2, meeting both the equals region for [3,3] and the equals region for [3,4].
5
Step 5: Final horizontal fill
The only remaining gap is at [2,4]. Place the [2,6] domino horizontally covering [2,4]-[1,4] with 2 and 6, locking in [2,4]=2 (matching its equals partner [3,4]) and [1,4]=6 (already set). The puzzle is complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Greater-10 and Sum-10 lock
The region [0,0],[1,0] requires sum >10, so both pips must be high (at least one 6). The sum-10 region [2,0],[3,0] shares the column. The only way to give [1,0] a 6 while also providing [2,0] a value that pairs with [3,0] to sum 10 is to use the [6,6] double. Placing it vertically at [1,0]-[2,0] sets both to 6, satisfying the greater-10 region's need for a high pip in [1,0] and giving [2,0]=6. Consequently, [3,0] must be 4 to reach sum 10.
2
Step 2: Sum-3 cell triggers column fill
Cell [4,0] is a sum-3 single, so it must be 3. To connect [3,0]=4 and [4,0]=3, the only domino is [4,3], placed vertically at that spot. This locks in the left column values.
3
Step 3: Top-right sum-4 starts a chain
The isolated sum-4 cell at [0,3] forces a 4 there. The only domino that can supply a 4 to [0,3] while also feeding the neighboring sum-8 region [0,1]-[0,2] is [4,5], placed horizontally with the 5 at [0,2] and 4 at [0,3]. This pushes the sum-8 region: with [0,2]=5, then [0,1] must be 3.
4
Step 4: Top-left completion
With [0,1]=3 required, and the greater-10 region still needing a high pip in [0,0], the [6,3] domino is the perfect fit horizontally across [0,0]-[0,1], giving [0,0]=6 and [0,1]=3. Now the top row is complete, and the greater-10 region finds its full 12 sum.
5
Step 5: Sum-11 cluster and the 5 chain
The three-cell sum-11 region at [2,2],[2,3],[3,3] demands total 11. The [4,4] double placed horizontally at [2,2]-[2,3] gives two 4s, forcing [3,3]=3. The [1,3] domino then drops vertically at [3,3]-[4,3] with 3 and 1, setting [4,3]=1 and aligning with the equals region [4,3],[5,3] (so [5,3] will be 1). The four-cell equals group [3,5],[4,5],[5,5],[5,6] all must be identical; the [5,5] double placed vertically at [4,5]-[5,5] establishes the value 5, and the [1,5] domino at [3,5]-[3,6] adds another 5, while [5,0] placed at [5,6]-[5,7] provides the fourth 5. This chain forces surrounding sums.
6
Step 6: Finishing the lower half
With [5,3]=1 secured by the equals region, the sum-8 region at [5,1]-[5,2] needs an 8; placing [6,1] horizontally at [5,2]-[5,3] gives [5,2]=6 and [5,3]=1, so [5,1] must be 2. The [2,2] domino then fits horizontally at [5,0]-[5,1] with 2,2. The right-side sum-3 region [3,6]-[3,7] with [3,6]=1 forces [3,7]=2 via the [0,2] domino placed vertically at [3,7]-[4,7] (2,0), which in turn sets off the equals column in 7 making [4,7],[5,7],[6,7] all 0. Lower right finishes with [0,1] at [6,7]-[7,7] (0,1), [0,0] at [7,5]-[8,5] (0,0), and [1,1] at [8,6]-[8,7] (1,1), exactly satisfying the final sum-3 region.
💡 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
Leave your comment