🚨 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!
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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Today's NYT Pips easy puzzle by Ian Livengood hands you a cluster of equals regions right away — you’ll quickly spot the trio across the top and the vertical pair down the left that all but fill themselves. A single-cell sum and a less-than-1 cell in the bottom row shrink the remaining possibilities to zero. The solve feels like tidying up a set of dominoes that already know where they belong.
The medium grid from Rodolfo Kurchan changes the tempo. You start by staring at a sum-10 region that demands a pair of fives, and when you realize the 5s must come from two different dominoes, the interlocking equals and less constraints start to cascade. One cell at a time, the bottom-row greater-4 and the top-row sum-5 pull the last pieces into place, leaving a neat trail of forced placements.
The hard puzzle — also by Kurchan — spreads ten columns wide, but don’t let the size intimidate you. A slim sum-4 column on the right and a lone sum-4 cell at top-left are your early handholds. Once you plant those, a vast equals region of sixes ripples across the middle, and a matching zero trio settles at the bottom. The real delight is watching how the less-than and greater-than singletons steer the remaining dominoes home.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Follow the equals signs
Start with the regions that demand identical pips. When several cells must match, you can often lock in a domino that forces the whole group.
💡 Hint 2: Zoom in on the top row and the bottom right
The three cells along the top are all equal. Meanwhile, the cell at row 3, column 2 has a ‘less than 1’ rule — meaning only one pip value fits there.
💡 Hint 3: The final layout
Place [3,3] horizontally across the top-row equals cells (0,1)-(0,2), turning all three into 3. The sum-6 cell at (2,1) gets 6 from domino [4,6] placed horizontally with 4 at (2,0). That forces the column-0 equals pair to be 4, so place [4,3] vertically with 4 at (1,0) and 3 at (0,0). The middle-left equals trio at (1,1)-(1,2)-(2,2) all become 1: use [1,1] horizontally across (1,1)-(1,2). Finally, the less-1 cell at (3,2) demands a 0 — use [0,1] vertically with 0 at (3,2) and the 1 extending into the already-1 cell (2,2).
💡 Hint 1: Look for sum regions that demand specific pairs
A sum-10 region that spans two cells can only be satisfied by a limited set of pip pairs. Combine that with adjacent equals and greater-than rules to eliminate possibilities.
💡 Hint 2: Attack the top-right and the middle row
The cells (1,2) and (1,3) must add to 10. Since no single domino carries a 10, you’ll need two different dominoes to supply the fives. Right below them, the pair (2,1)-(2,2) must be equal.
💡 Hint 3: The forced placement chain
The sum-10 force makes (1,2)=5 and (1,3)=5. Place [4,5] vertically with 5 at (1,2) and 4 at (2,2); then place [5,0] vertically with 5 at (1,3) and 0 at (2,3). The equals pair (2,1)-(2,2) forces (2,1)=4, so use [0,4] horizontally with 0 at (3,1) and 4 at (2,1). That sets (3,2)=0 via the equals region (3,1)-(3,2). The greater-4 at (3,3) needs >4, so place [0,6] horizontally with 0 at (3,2) and 6 at (3,3). The less-5 pair (1,0)-(1,1) takes [2,1] (2 at (1,0), 1 at (1,1)). Finally, the sum-5 region (0,3)-(0,4) gets [3,2] horizontally (3 at (0,4), 2 at (0,3)).
💡 Hint 1: Anchor on sum and less regions that involve single cells
A sum-4 cell and a less-4 cell sit adjacent in the top-left corner. Their values are practically forced, and the domino covering them will be the first key.
💡 Hint 2: The narrow sum-4 column on the right
The four cells in column 7 must add to exactly 4. With four cells to fill, only a few small-pip combinations can work. The top two cells there will take the only domino that fits that tight ceiling.
💡 Hint 3: The six-equals core
Once the sum-4 column is settled, look at the equals region that ties together (1,2), (2,0), (2,1), and (2,2). A domino with a 6 placed vertically will set the entire group to 6, which then interacts with the less-4 cell at (0,2).
💡 Hint 4: Zero trio and the bottom row
A three-cell equals region in rows 5-6 will all become 0. The domino that covers two of them forces the third, which then feeds into the less-4 cell at (5,4) — a 2 will fit perfectly.
💡 Hint 5: The completing sequence
Place [4,0] vertically at (0,0)-(1,0): (0,0)=4, (1,0)=0. Column 7 sum 4 starts with [1,0] vertically at (0,7)-(1,7): (0,7)=1, (1,7)=0, then [3,0] vertically at (2,7)-(3,7): (2,7)=0, (3,7)=3. The six-equals block: place [4,6] vertically at (3,2)-(2,2): (3,2)=4, (2,2)=6; then [6,6] horizontally at (2,0)-(2,1) for 6s; and [0,6] vertically at (0,2)-(1,2): (0,2)=0, (1,2)=6. Zero trio: [0,0] vertically at (5,6)-(6,6) gives 0s; then [0,2] horizontally at (5,5)-(5,4): (5,5)=0, (5,4)=2. Row 8 sum-4: [2,2] horizontally at (8,1)-(8,2). The equals-1 trio (7,5)-(7,6)-(8,6): [1,1] horizontally at (7,5)-(7,6) gives 1s, and [3,1] vertically at (9,6)-(8,6) with 1 at (8,6). Last, sum-11 in row 9: [6,2] horizontally at (9,4)-(9,5) gives 6+2, plus 3 from [3,1] at (9,6) for total 6+2+3=11. Remaining dominoes: [5,6] at (6,2)-(6,1): (6,2)=5, (6,1)=6; and [4,4] at (2,4)-(2,5).
🎨 Pips Solver
Apr 29, 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: The top-row equals trio
Cells [0,0], [0,1], [0,2] must all be equal. The only domino with two identical pips that can cover two of them is [3,3]. Placing it horizontally from [0,1] to [0,2] forces all three cells to 3, because the third cell must match. So [0,1]=3, [0,2]=3, [0,0]=3.
2
Step 2: The single-cell sum-6
Cell [2,1] needs a sum of 6 alone. With pip values 0–6, the only way to reach 6 is to put a 6 there. Domino [4,6] is the only one containing a 6. It must cover [2,1], so place it horizontally with 6 at [2,1] and 4 at the adjacent [2,0]. That satisfies the sum.
3
Step 3: The column-0 equals pair
Cells [1,0] and [2,0] must be equal, and [2,0] is already 4. So [1,0] must be 4. Domino [4,3] has a 4 and a 3. To get the 4 into [1,0] and the 3 into the still-free [0,0] (which we earlier deduced must be 3), place it vertically: 4 at [1,0], 3 at [0,0]. Now the top-left equals region is consistent.
4
Step 4: The middle-left equals and the less-1
Cells [1,1], [1,2], [2,2] are all equal. The remaining domino with identical pips is [1,1]. Place it horizontally from [1,1] to [1,2], giving both 1, so [2,2] also becomes 1. The less-1 region at [3,2] demands a pip less than 1, so only 0 works. Domino [0,1] can place its 0 there and its 1 in the cell above, [2,2], which needs to be 1. So place it vertically: 0 at [3,2], 1 at [2,2], completing the puzzle.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Unlocking the sum-10 region
The cells [1,2] and [1,3] must sum to 10. The only possible pair with pips 0–6 is 5+5 (since 6+4 would need a 4 that cannot simultaneously satisfy other constraints). Therefore, both cells must be 5. No single domino carries two 5s, so the 5s will come from two different dominoes.
2
Step 2: Placing the two fives
Domino [4,5] holds a 5; place it vertically with the 5 in [1,2] and the 4 in [2,2]. Domino [5,0] also has a 5; place it vertically with the 5 in [1,3] and the 0 in [2,3]. This leaves the ‘empty’ cell [2,3] without constraint, and the equals pair [2,1]-[2,2] now has [2,2]=4.
3
Step 3: The equals pair and the bottom row
Because [2,1] must equal [2,2], [2,1] is forced to 4. The domino that can supply a 4 to [2,1] while also covering [3,1] is [0,4]. Place it horizontally with 0 at [3,1] and 4 at [2,1]. Now the equals region [3,1]-[3,2] forces [3,2]=0. Cell [3,3] must be greater than 4, so it must be 5 or 6.
4
Step 4: The greater-4 and the remaining bottom cell
Domino [0,6] can cover [3,2] and [3,3]: place it horizontally with 0 at [3,2] (already 0) and 6 at [3,3]. 6 > 4, satisfying the greater constraint. The cell [3,2]’s value is confirmed.
5
Step 5: The less-5 and sum-5 finish
Cells [1,0] and [1,1] must both be less than 5. The remaining dominoes are [2,1] and [3,2]. Place [2,1] horizontally at [1,0]-[1,1] with 2 and 1 (both <5). The sum-5 region [0,3]-[0,4] needs a total of 5; the only remaining domino with a 3 and a 2 is [3,2]. Place it horizontally with 3 at [0,4] and 2 at [0,3] (2+3=5). Everything fits.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Fixing the sum-4 right column
Cells [0,7], [1,7], [2,7], [3,7] sum to 4. With four cells, the only feasible small-pip combination is 1, 0, 0, 3. The top two cells are vertically adjacent and can be covered by domino [1,0]: place it vertically with 1 at [0,7] and 0 at [1,7]. The bottom two cells then must sum to 3 (since 4−1−0=3). Domino [3,0] provides exactly a 3 and a 0; place it vertically with 0 at [2,7] and 3 at [3,7]. The column is satisfied.
2
Step 2: Top-left sum and less
Cell [0,0] has a single sum target of 4, so it must be 4. Cell [1,0] has a less-than-4 constraint, so it must be 0, 1, 2, or 3. Domino [4,0] can place a 4 at [0,0] and a 0 at [1,0]. Place it vertically: [0,0]=4, [1,0]=0. The less is satisfied.
3
Step 3: The equals-6 block across the middle
The region [1,2], [2,0], [2,1], [2,2] must all be equal. The cell [3,2] is a greater-than-3, so it must be 4, 5, or 6. Domino [4,6] can cover [3,2] and [2,2] vertically: place it with 4 at [3,2] (satisfying >3) and 6 at [2,2]. Now the entire equals block becomes 6. So [2,0] and [2,1] need 6s — use [6,6] horizontally there. [1,2] must be 6, and the only remaining domino with a 6 is [0,6]; place it vertically with 0 at [0,2] (which happily satisfies the less-4 there) and 6 at [1,2].
4
Step 4: The zero trio at the bottom
Cells [5,5], [5,6], [6,6] are all equal. Domino [0,0] can set two of these to 0; place it vertically with 0 at [5,6] and 0 at [6,6]. Then [5,5] must also be 0. To cover [5,5] and the less-4 cell [5,4], use domino [0,2] horizontally: 0 at [5,5] and 2 at [5,4] (2 is <4).
5
Step 5: The row-8 sum-4 and the equals-1 chain
Cells [8,1] and [8,2] must sum to 4. The only remaining domino that can do that is [2,2] — place it horizontally for 2 and 2. Cells [7,5], [7,6], [8,6] are equals; the only unused equal-pip domino is [1,1]. Place it horizontally at [7,5]-[7,6] for two 1s. Then [8,6] must also be 1. To cover [9,6] and [8,6], use [3,1] vertically with 3 at [9,6] and 1 at [8,6].
6
Step 6: The final sum-11 and remaining pieces
Cells [9,4], [9,5], [9,6] sum to 11. We already placed 3 at [9,6] in Step 5. So [9,4]+[9,5] need to add to 8. Domino [6,2] gives exactly that: place it horizontally with 6 at [9,4] and 2 at [9,5] (6+2+3=11). The remaining dominoes [5,6] and [4,4] fill the last gaps: [5,6] horizontally at [6,2]-[6,1] with 5 at [6,2] (>4) and 6 at [6,1] (>5); then [4,4] horizontally at [2,4]-[2,5] with two 4s in the equals-4 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.
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