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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
Ian Livengood is back for easy and medium on March 16, and both puzzles are built on the same principle: constraints that chain together so tightly there's barely a decision to make. Easy is a five-domino board where the equals region in the bottom corner is your starting gun — figure out which double belongs there and the puzzle hands the rest of itself to you in sequence. Medium is bigger and uses a less-than cell to kick off a cascade that ripples all the way up the grid without a single branching moment.
Hard is Rodolfo Kurchan's, and it's a different kind of puzzle entirely — 16 dominoes, a five-cell equals region, a seven-cell unequal block, a sum-to-15 with three cells, and a handful of solo sum cells that pin exact pip values before you've even thought about the bigger constraints. It looks like a lot, but those solo cells are doing you a favor: several placements are completely free, and they knock dominoes into position that lock down the trickier regions.
All three today share the same lesson. Start with the constraint that gives you the least room to argue. In easy it's the equals corner, in medium it's the less-than cell, in hard it's the zeros. Follow the chain from there and the board fills in on its own.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint #1 - Start with what must match
There's an equals region in the bottom-right corner — two cells that need identical pip values. Before you touch anything else, figure out which of your five dominoes can satisfy that constraint. That's where the puzzle starts.
💡 Hint #2 - [6|6] doesn't fit there
[6|6] looks like an obvious pick for the equals region, but putting it there breaks the big sum-18 region above — you simply can't make 18 from what's left. [0|0] is the only double that works. Drop it in the bottom-right and the rest of the board follows without any guessing.
💡 Hint #3 - Full answer
① [0|0] horizontal in the bottom-right, row 2 cols 2–3 (equals ✓). ② [5|2] vertical at col 3: 5-pip at row 1, 2-pip at row 0 (sum-2 ✓). ③ [5|0] vertical at col 0: 5-pip at row 2, 0-pip at row 1. ④ [3|6] vertical at col 1: 3-pip at row 2, 6-pip at row 1 (3+5=8 ✓). ⑤ [6|6] horizontal at row 0 cols 1–2 (6+6+0+6=18 ✓).
💡 Hint #1 - One cell is almost completely locked
Look for the cell with an inequality constraint — it's tight enough that only one or two pip values are legal there. Find it, figure out what fits, and the board starts unraveling from that spot.
💡 Hint #2 - One domino, eight dominoes sorted
The < 2 cell must hold pip 0 or 1. [1|3] is the only domino that can place a 1 there without immediately breaking the equals chain running along the bottom. Once [1|3] is in, the equals constraints at rows 3–4 force every remaining placement one by one, all the way up to the top.
💡 Hint #3 - Full answer
① [1|3] horizontal row 2 cols 1–0: 1-pip (<2 ✓), 3-pip at col 0. ② [6|3] vertical col 0 rows 4–3: 6-pip at row 4, 3-pip at row 3 (equals=3 ✓). ③ [6|0] horizontal row 4 cols 1–2: 6-pip (equals=6 ✓), 0-pip at col 2. ④ [0|2] horizontal row 4 cols 3–4: 0-pip (equals=0 ✓), 2-pip at col 4. ⑤ [2|2] vertical col 5 rows 3–4: pip=2 at both (equals=2 ✓). ⑥ [4|4] vertical col 5 rows 1–2: 4-pip at row 2 (4+2=6 ✓), 4-pip at row 1. ⑦ [4|5] horizontal row 0 cols 5–4: 4-pip (equals=4 ✓), 5-pip at col 4. ⑧ [5|5] horizontal row 0 cols 2–3: 5-pip at both (5>4 ✓, 5=5 ✓).
💡 Hint #1 - The board gives you several free answers
Several cells in this puzzle carry solo sum constraints — just one cell with a fixed target pip. Those spots hand you an exact value with zero deduction. Find all of them and you'll have a solid foundation before the real reasoning begins.
💡 Hint #2 - Four freebies and a sum-9
Cell [2,7] must be pip 0, [4,4] must also be pip 0, [3,0] must be pip 2, and [4,5] must be pip 4. Each one forces exactly one domino. On top of those, the two-cell sum-9 region at [0,6]–[1,6] has just one domino that reaches 9 by itself. That's five placements before the puzzle makes you think.
💡 Hint #3 - Five cells all agree on pip 1
The large equals region spans [2,6], [3,4], [3,5], [3,6], and [4,6] — five cells, all the same value. You already know [2,6]=1 (it's paired with the zero cell from the sum-0 constraint) and [3,5]=1 (from the sum-4 placement). That pins all five cells to pip 1, and three more dominoes drop right into place.
💡 Hint #4 - Three fives make fifteen, and two fours make eight
Sum-15 at [5,6],[5,7],[6,6] needs three pip-5 values — the only way to hit 15 with three cells. The equals region [3,7],[3,8],[4,7] is all pip 2 (confirmed from the sum-3 and equals-2 cells nearby). These two regions together account for four more dominoes.
💡 Hint #5 - Full answer
① [0|1] horizontal [2,7]–[2,6]: 0 at [2,7] (sum=0 ✓), 1 at [2,6] (equals). ② [2|3] vertical [2,8]–[3,8]: 3 at [2,8] (sum=3 ✓), 2 at [3,8]. ③ [2|2] horizontal [3,0]–[3,1]: 2 at [3,0] (sum=2 ✓), 2 at [3,1]. ④ [0|0] horizontal [4,3]–[4,4]: 0 at both (sum=0 ✓). ⑤ [1|4] vertical [3,5]–[4,5]: 1 at [3,5] (equals), 4 at [4,5] (sum=4 ✓). ⑥ [6|3] vertical [0,6]–[1,6]: 6+3=9 ✓. ⑦ [2|1] horizontal [3,7]–[3,6]: 2 at [3,7], 1 at [3,6] (equals). ⑧ [1|1] horizontal [3,3]–[3,4]: 1 at both (equals). ⑨ [1|5] vertical [4,6]–[5,6]: 1 at [4,6] (equals), 5 at [5,6]. ⑩ [2|5] vertical [4,7]–[5,7]: 2 at [4,7] (equals=2 ✓), 5 at [5,7]. ⑪ [5|4] horizontal [6,6]–[6,7]: 5 at [6,6] (sum=15 ✓), 4 at [6,7]. ⑫ [4|6] horizontal [7,7]–[7,8]: 4 at [7,7] (4+4=8 ✓), 6 at [7,8] (sum=6 ✓). ⑬ [0|6] horizontal [7,4]–[7,3]: 0 at [7,4], 6 at [7,3] (0+6=6 ✓). ⑭ [4|4] vertical [5,2]–[6,2]: 4 at both (sum=4 ✓). ⑮ [3|3] vertical [5,3]–[6,3]: 3 at both (sum=3 ✓). ⑯ [6|5] vertical [4,2]–[3,2]: 6 at [4,2], 5 at [3,2] (unequal: 2,5,1,6,0,4,3 — all different ✓).
🎨 Pips Solver
Mar 16, 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: Double-blank claims the equals corner
The equals region at [2,2]–[2,3] needs two matching pips. [6|6] would work in isolation but leaves an unsolvable sum-18 — there's no combination from the remaining pieces that reaches 18 in those four cells. [0|0] is the only double that fits. It goes horizontally in the bottom-right.
2
Step 2: Sum-2 at the top-right pins the pip
Cell [0,3] is a solo sum constraint — it has to hold pip 2. Only [5|2] has a 2-pip face, so it goes vertically at col 3: pip 5 at row 1 (unconstrained empty region), pip 2 at row 0 (sum=2 ✓).
3
Step 3: Sum-8 draws from two vertical dominoes
The sum-8 region covers [2,0] and [2,1]. These two cells are each the bottom half of separate vertical dominoes — they're not part of the same piece. You need two pip values that add to 8. From what's left: [5|0] and [3|6] give 5+3=8.
4
Step 4: The vertical pair locks into place
[5|0] goes vertically at col 0: pip 5 at row 2, pip 0 at row 1. [3|6] goes vertically at col 1: pip 3 at row 2, pip 6 at row 1. Both pip-at-bottom values land in the sum-8 region — 5+3=8 ✓.
5
Step 5: [6|6] closes the big sum
Only [6|6] is left, and it fits perfectly at row 0 cols 1–2. The sum-18 region gets 6+6 from this piece and 0+6 from the bottoms of the two vertical dominoes — 6+6+0+6=18 ✓. Done.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: The less-than cell starts everything
Cell [2,1] has a < 2 constraint — pip must be 0 or 1. That cell is only reachable by a horizontal domino paired with [2,0]. From the available set, [1|3] places 1 at [2,1] ✓, with 3-pip landing at [2,0]. No other domino with a 0 or 1 face can go here without breaking what comes next.
2
Step 2: The 3-pip at col 0 must equal its neighbor below
The equals constraint at [2,0]–[3,0] means [3,0] also has to be 3. Cell [3,0] is covered by a vertical domino running down to [4,0]. From the remaining pieces, [6|3] is the one that places 3 at [3,0] — it goes vertically with 3 at row 3 and 6 at row 4.
3
Step 3: The 6-pip at [4,0] triggers the bottom-row equals chain
[4,0]=[4,1] by the equals constraint, so [4,1] must also be 6. [4,1] is the left cell of a horizontal domino. [6|0] is the match — 6-pip at col 1, 0-pip at col 2.
4
Step 4: Zero equals zero, two follows
[4,2]=[4,3] equals constraint, and [4,2]=0 just set. So [4,3]=0. [0|2] horizontal at [4,3]–[4,4] puts 0 at col 3 (equals=0 ✓) and 2-pip at col 4.
5
Step 5: Two-two closes the bottom row
[4,4]=[4,5] equals, and [4,4]=2. So [4,5]=2. [2|2] goes vertically at col 5, rows 3–4: both cells pip 2. [3,5]=2 and [4,5]=2 ✓.
6
Step 6: Sum-6 at col 5 pulls the next piece in
Sum-6 covers [2,5]+[3,5]. [3,5]=2 is locked. So [2,5]=4. The vertical domino at col 5 rows 1–2 must show 4 at row 2. [4|4] does it — both pips 4, and 4+2=6 ✓.
7
Step 7: The right-edge equals chain runs up
[0,5]=[1,5] equals, and [1,5]=4 just placed. So [0,5]=4. The horizontal domino at row 0 covering cols 4–5 must show 4 at col 5. [4|5] reversed: 4-pip at col 5, 5-pip at col 4.
8
Step 8: Double-five finishes the top
[0,3]=[0,4] equals, and [0,4]=5 from the previous step. So [0,3]=5. [5|5] at row 0 cols 2–3: both 5s. [0,2]=5>4 ✓, [0,3]=5 closes equals ✓. Puzzle done.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum-0 at [2,7] is a free placement
Cell [2,7] must hold pip 0 — no other value is legal. [0|1] is the domino that covers it: 0-pip at [2,7] ✓, 1-pip at [2,6], which happens to sit inside the five-cell equals region.
2
Step 2: Sum-3 at [2,8] gives you another freebie
Cell [2,8] must be pip 3. [2|3] covers [2,8]–[3,8]: 3-pip at [2,8] ✓, 2-pip at [3,8] — that 2 will matter shortly for the equals region below.
3
Step 3: Sum-2 at [3,0] forces the double-two
Cell [3,0] must be pip 2. [2|2] goes horizontally at [3,0]–[3,1]: 2-pip at [3,0] ✓, 2-pip at [3,1] (entering the unequal block).
4
Step 4: Sum-0 at [4,4] hands you [0|0]
Cell [4,4] must be pip 0. [0|0] spans [4,3]–[4,4] horizontally — both cells pip 0. [4,3] also lands in the unequal block.
5
Step 5: Sum-4 at [4,5] brings a 1 into the equals region
Cell [4,5] must be pip 4. [1|4] goes vertically [3,5]–[4,5]: 1-pip at [3,5] (entering the five-cell equals region) and 4-pip at [4,5] ✓.
6
Step 6: Sum-9 at the top right has one answer
Cells [0,6]+[1,6] need to sum to 9. [6|3] is the only domino that hits 9 on its own. It goes vertically: 6-pip at [0,6], 3-pip at [1,6].
7
Step 7: Five cells, all pip 1 — three more placements drop in
The equals region [2,6],[3,4],[3,5],[3,6],[4,6] forces one common value. [2,6]=1 from Step 1, [3,5]=1 from Step 5. All five must equal 1. Three cells remain: [3,6], [3,4], [4,6]. [2|1] horizontal [3,7]–[3,6] places 1 at [3,6]. [1|1] horizontal [3,3]–[3,4] places 1 at [3,4]. [1|5] vertical [4,6]–[5,6] places 1 at [4,6] and 5-pip at [5,6].
8
Step 8: Equals at [3,7],[3,8],[4,7] — pip 2 across the board
[3,7]=2 (from [2|1] in Step 7), [3,8]=2 (from Step 2). So [4,7] also must be 2. [2|5] vertical [4,7]–[5,7]: 2-pip at [4,7] ✓, 5-pip at [5,7].
9
Step 9: Sum-15 needs three fives
[5,6]+[5,7]+[6,6] has to reach 15. [5,6]=5 and [5,7]=5 are already set. So [6,6]=5. [5|4] goes horizontally [6,6]–[6,7]: 5-pip at [6,6] ✓, 4-pip at [6,7].
10
Step 10: Sum-8 at [6,7],[7,7] — four plus four
[6,7]=4 from Step 9. Sum-8 needs [7,7]=4. [4|6] horizontal [7,7]–[7,8]: 4-pip at [7,7] ✓, 6-pip at [7,8].
11
Step 11: Sum-6 at [7,8] and sum-6 at [7,3],[7,4]
[7,8]=6 from Step 10 — sum=6 ✓. Separately, [7,3]+[7,4] must sum to 6. [0|6] goes horizontal [7,4]–[7,3]: 0-pip at [7,4], 6-pip at [7,3] (0+6=6 ✓).
12
Step 12: Sum-4 at [6,2] pins a four to the unequal block
Cell [6,2] must be pip 4. [4|4] goes vertically [5,2]–[6,2]: 4-pip at both. [5,2]=4 enters the unequal block.
13
Step 13: Sum-3 at [6,3] pins a three to the unequal block
Cell [6,3] must be pip 3. [3|3] goes vertically [5,3]–[6,3]: 3-pip at both. [5,3]=3 enters the unequal block.
14
Step 14: The last domino clears the unequal block
The unequal block [3,1],[3,2],[3,3],[4,2],[4,3],[5,2],[5,3] needs seven distinct pip values. So far: [3,1]=2, [3,3]=1, [4,3]=0, [5,2]=4, [5,3]=3. Remaining cells [3,2] and [4,2] need the two missing values — 5 and 6. [6|5] vertical [4,2]–[3,2]: 6-pip at [4,2], 5-pip at [3,2]. All seven cells hold different values ✓. Puzzle solved.
💡 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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