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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
In this NYT Pips trio, Ian Livengood’s easy is a confidence-builder — a four-cell equals region anchors the grid and eliminates guesswork, letting the solve flow cleanly from a single forced double-domino. It’s a swift, satisfying warm-up.
Rodolfo Kurchan’s medium introduces more intricate interplay: multiple sum-3 pockets and a central equals cell create a narrow bottleneck where a specific low-pip domino must bridge two sums. Once that lock picks, the surrounding cells resolve in a tidy chain. Expect a middle-of-the-road difficulty that rewards careful pairing.
Kurchan’s hard is the main event. Dense sum-6 targets and cascading equals regions demand systematic logic. The grid forces an early breakthrough from a single-cell sum constraint that locks a high-value double, triggering a domino effect across every remaining region. This one will feel like a tightly woven puzzle where each deduction builds inevitably toward the finish.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Type of anchor
Start by identifying the region where every cell must share the same pip value. A region that large can only be satisfied by a domino with identical pips.
💡 Hint 2: Focus on the lower right
The four-cell equals region at [2,3], [3,3], [4,2], [4,3] forms the backbone. Check the available double-number dominos. The less-2 cell at [4,4] sits just outside this block and will force the orientation of the domino that completes it.
💡 Hint 3: Full solution
Place the double [2,2] domino vertically at [2,3] and [3,3] (both pip 2). The [1,2] domino goes to [4,4] (pip 1, satisfying less-2) and [4,3] (pip 2). The [2,3] domino then fills [4,2] (2) and [4,1] (3). The [5,1] domino covers [1,1] (5) and [2,1] (1) to satisfy greater-13 and greater-0. Top row: [0,5] fills [0,0] (0) and [0,1] (5), leaving [3,6] for [0,3] (3) and [0,2] (6), completing the sum-3 cell.
💡 Hint 1: Sum-3 overtones
Multiple regions require pairs that sum exactly to three. The low numbers will be dictated by these tiny sums. Look for a double-domino that can lock an equals region silently.
💡 Hint 2: The central row bottleneck
The row with two adjacent sum-3 regions at [2,2]–[3,2] and [2,3]–[3,3] interacts with an equals region in the top row and a greater-2 cell at [2,1]. The double [0,0] must cover the equals cells [1,2] and [1,3], forcing the [1,0]–[1,1] sum-3 pair.
💡 Hint 3: Full solution
Place [0,0] at [1,2] and [1,3] (both 0). For the sum-3 at [1,0]–[1,1], use [1,4] vertically at [1,0] (1) and [2,0] (4), and [0,2] at [0,1] (0) and [1,1] (2). The equals [2,0]–[3,0] forces [3,0]=4, filled by [4,5] horizontally at [3,0] (4) and [3,1] (5). The greater-2 cell [2,1] takes [6,2] vertically at [2,1] (6) and [2,2] (2). The sum-3 at [2,2]–[3,2] needs [3,2]=1, so place [1,1] at [3,2] and [3,3] (both 1). Finally [2,2] goes to [2,3] (2) and [2,4] (2) to satisfy the last sum-3 and greater-0.
💡 Hint 1: Equals signposts
There are two key equals regions where cells must match. One of them, in the middle-left, offers a powerful pivot because it sits next to a single-cell sum-6 region that can only accept one pip value.
💡 Hint 2: The sum-6 singles out a double
The equals region at [2,1]–[2,2] forces a double-domino. The isolated sum-6 cell at [3,2] demands a specific high pip. Among the available doubles, only one can coexist with the required domino that covers [3,2] and [4,2]. Check the double [6,6] and the [6,2] domino.
💡 Hint 3: Cascade downward
Once [6,6] anchors [2,1]–[2,2], the [6,2] domino must run vertically from [3,2] to [4,2] (6 and 2). The sum-6 pair at [4,1]–[4,2] then forces [4,1]=4, which couples with the bottom-left sum-6 region [3,0]–[4,0]. The [4,2] domino will settle that corner.
💡 Hint 4: Top-row choreography
With the left side locking in, the [5,4] domino fills [2,0]–[3,0] (5 and 4), forcing [1,0]=1 via the upper sum-6. The [1,1] double then sets [0,0]=1 and [0,1]=5, which feeds the [5,3] domino into the top-left equals region. The [3,3] double will complete the equals chain on row 0.
💡 Hint 5: Full solution
Place [6,6] horizontally at [2,1]–[2,2] (both 6). The sum-6 single [3,2] becomes 6, using [6,2] vertically at [3,2] (6) and [4,2] (2). The sum-6 [4,1]–[4,2] forces [4,1]=4, so [4,2] runs horizontally at [4,1] (4) and [4,0] (2). That yields [3,0]=4 via sum-6 [3,0]–[4,0], so [5,4] goes vertically at [2,0] (5) and [3,0] (4). Then [1,0]=1 from sum-6 [1,0]–[2,0], and [1,1] fills [0,0] (1) and [1,0] (1). The sum-6 [0,0]–[0,1] makes [0,1]=5, and [5,3] goes horizontally at [0,1] (5) and [0,2] (3). The equals [0,2]–[0,3] sets [0,3]=3, so [3,3] covers [0,3]–[0,4] (both 3). The sum-7 [0,4]–[0,5] needs [0,5]=4, provided by [3,4] at [0,5] (4) and [0,6] (3). The sum-7 [0,6]–[1,6] demands [1,6]=4, so [6,4] runs vertically at [1,6] (4) and [2,6] (6). Equals [2,6]–[3,6] forces [3,6]=6, and [6,0] fills [3,6] (6) and [4,6] (0).
🎨 Pips Solver
Jun 7, 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: Anchor the equals block
The four-cell equals region at [2,3], [3,3], [4,2], [4,3] must all show the same pip. Only a double-domino can cover two cells with the same value, and we’ll need one double to occupy two of these cells while another domino completes the rest with the same pip. The only available double is [2,2], so place it vertically at [2,3] and [3,3] — both become 2.
2
Step 2: The less-2 cell dictates the edge
Cell [4,4] is under a less-2 constraint, so it must be 0 or 1. The adjacent cell [4,3] must match the equals region value of 2. The only domino that provides a 2 and a number less than 2 is [1,2]. Place [1,2] horizontally: [4,4] gets 1 (satisfying less-2) and [4,3] gets 2, completing the equals block’s bottom edge.
3
Step 3: Fill the remaining equals cells and empty neighbor
With [4,3]=2, the equals region demands [4,2]=2. The empty cell [4,1] sits next to it. The [2,3] domino can deliver a 2 and a 3. Place [2,3] horizontally at [4,2] (2) and [4,1] (3) — this fills the last equals cell and satisfies the empty region with no conflict.
4
Step 4: Top-row and left-over constraints
The greater-13 region includes [0,1], [0,2], [1,1] and needs high pips. The [5,1] domino fits beautifully: place it vertically at [1,1] (5) and [2,1] (1) — the 5 helps the greater-13 total, while [2,1] satisfies its greater-0 constraint. Now [0,5] goes to [0,0] (0) and [0,1] (5), completing the top-left. Finally [3,6] fills the remaining top cells at [0,3] (3, satisfying the sum-3 cell) and [0,2] (6) — the greater-13 total becomes 5+6+5=16, well above 13.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Lock the equals region on row 1
The equals region [1,2]–[1,3] both must have the same pip. Among available dominoes, only the [0,0] double can sit naturally here. Place it horizontally: [1,2]=0, [1,3]=0.
2
Step 2: Solve the left-side sum-3 pair
The sum-3 region [1,0]–[1,1] needs a total of 3. With [1,1] not yet filled, we check dominoes that can reach it. The [1,4] domino can place 1 at [1,0] and 4 at [2,0] (vertical), which would force [1,1]=2 to reach sum 3. The [0,2] domino then fits horizontally at [0,1] (0) and [1,1] (2) — completing the sum and filling the empty cell [0,1].
3
Step 3: Propagate the vertical equals constraint
The equals region [2,0]–[3,0] forces [3,0] to match [2,0] (which is 4). The [4,5] domino contains a 4; place it horizontally at [3,0] (4) and [3,1] (5), filling both the equals region cell and the empty [3,1].
4
Step 4: Greater-2 and row 2 sum-3 chain
Cell [2,1] requires a value greater than 2. The [6,2] domino fits perfectly: vertical placement at [2,1] (6) and [2,2] (2). Now the sum-3 region [2,2]–[3,2] has [2,2]=2, so [3,2] must be 1. The double [1,1] can cover that and its neighbor: place [1,1] horizontally at [3,2] (1) and [3,3] (1).
5
Step 5: Close out the right side
The sum-3 region [2,3]–[3,3] now has [3,3]=1, so [2,3] must be 2. The [2,2] domino (the only remaining domino) goes horizontally at [2,3] (2) and [2,4] (2). This satisfies sum-3 and the greater-0 cell [2,4] with a safe 2, completing the grid.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Single-cell sum-6 forces the first double
Cell [3,2] is a standalone sum-6 region, so it must contain exactly 6. The only way to place a 6 there is to use a domino with a 6 that can reach it. The equals region [2,1]–[2,2] requires a double, and the double [6,6] fits nicely: place [6,6] horizontally at [2,1] and [2,2] (both 6). Then the [6,2] domino runs vertically at [3,2] (6) and [4,2] (2).
2
Step 2: Bottom-left sum-6 domino placement
The sum-6 pair [4,1]–[4,2] now has [4,2]=2, so [4,1] must be 4. The only remaining domino with a 4 and another number that fits the adjacent [4,0] is [4,2] (pips 4,2). Place it horizontally at [4,1] (4) and [4,0] (2). This satisfies that sum-6 region and sets [4,0]=2.
3
Step 3: Cascade upward through two sum-6 regions
The sum-6 region [3,0]–[4,0] now demands [3,0]=4 (since [4,0]=2). The sum-6 region [1,0]–[2,0] above needs a total of 6. The [5,4] domino (pips 5,4) can be placed vertically at [2,0] (5) and [3,0] (4) — that supplies 4 to [3,0] and gives [2,0]=5. Then [1,0] must be 1 to complete sum 6, so the double [1,1] goes vertically at [0,0] (1) and [1,0] (1).
4
Step 4: Top row’s sum-6 and equals dominoes
The sum-6 region [0,0]–[0,1] with [0,0]=1 forces [0,1]=5. The [5,3] domino fits horizontally at [0,1] (5) and [0,2] (3). The equals region [0,2]–[0,3] then forces [0,3]=3. To cover that and the sum-7 region [0,4]–[0,5], use the double [3,3] horizontally at [0,3] (3) and [0,4] (3). Now the sum-7 pair [0,4]–[0,5] has 3+[0,5]=7, so [0,5]=4. The [3,4] domino goes horizontally at [0,5] (4) and [0,6] (3), feeding the next sum-7.
5
Step 5: Right-side equals and sum-7 wrap-up
The sum-7 region [0,6]–[1,6] has [0,6]=3, so [1,6] must be 4. The [6,4] domino can be placed vertically at [1,6] (4) and [2,6] (6). The equals region [2,6]–[3,6] then enforces [3,6]=6. Finally, the [6,0] domino runs vertically at [3,6] (6) and [4,6] (0), satisfying the empty cell [4,6] and completing the puzzle.
💡 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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