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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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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Ian Livengood built today's easy, and it's a tight little five-domino puzzle that's really about noticing two things at once. Both top corners carry sum=0 constraints — they each need pip 0. Once you realize the double-blank [0|0] can handle one corner slot all by itself (filling both cells with zeros), the second corner snaps into place, and the rest of the board follows in a single chain without any decisions.
Rodolfo Kurchan's medium is a step up. Seven dominoes, a pair of equals cells running down column 1, and two separate sum=11 regions that both need a 5+6 split. The equals column is your entry point — there's only one domino in your set where both faces are identical, and it belongs right there. Once it's down, the less-than constraint and the two sum=11 regions hand you the rest of the board one piece at a time.
Hard is also Kurchan's, and it's 16 dominoes across a wide grid. The headline constraint is a sum=18 region — three cells that can only hold 6+6+6. That one region locks three dominoes and confirms two inequality constraints at the same time. After those three 6s are placed, a five-cell equals block becomes solvable, a chain of sum constraints falls out cleanly, and the puzzle stops feeling as large as it looks.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Hint 1: Two corners share the same constraint
Look at the top-left and top-right corners — they both carry sum=0. That means both need pip 0. Figure out which domino can take care of both those spots, and you've got your starting move.
💡 Hint 2: [0|0] handles one corner, [1|0] handles the other
The double-blank [0|0] goes vertically at col 3, rows 0–1: pip 0 fills [0,3] (sum=0 ✓), and the second pip 0 drops into the empty cell [1,3]. That frees [1|0] to place its 0-pip at [0,0] (sum=0 ✓) with its 1-pip at [1,0], feeding the sum=3 region. From there the board finishes in a straight line.
💡 Full answer
① [0|0] vertical col 3 rows 0–1: 0-pip at [0,3] (sum=0 ✓), 0-pip at [1,3] (empty). ② [1|0] vertical col 0 rows 1–0: 1-pip at [1,0], 0-pip at [0,0] (sum=0 ✓). ③ [4|2] horizontal row 1 cols 2–1: 4-pip at [1,2], 2-pip at [1,1] (1+2=3 ✓). ④ [5|1] vertical col 2 rows 3–2: 5-pip at [3,2] (equals), 1-pip at [2,2] (4+1=5 ✓). ⑤ [5|0] vertical col 1 rows 3–2: 5-pip at [3,1] (equals: 5=5 ✓), 0-pip at [2,1] (empty).
💡 Hint 1: One constraint nearly solves itself
There's an equals region running down column 1 — two cells that must hold the same pip value. You've got one domino in your set where both faces are identical. That's the piece that belongs here, and once it's down the puzzle opens up fast.
💡 Hint 2: The double-one kicks off a chain
[1|1] goes vertically at col 1, rows 2–3: both cells pip 1 (equals ✓). From there, the less-than cell at [4,1] can only take pip 0, 1, or 2, and the sum=11 constraint at [4,2]+[4,3] needs a 5+6 split — only one domino threads both conditions at once.
💡 Full answer
① [1|1] vertical col 1 rows 2–3: pip 1 at both (equals ✓). ② [0|6] horizontal row 4 cols 1–2: 0-pip at [4,1] (<3 ✓), 6-pip at [4,2]. ③ [5|2] vertical col 3 rows 4–3: 5-pip at [4,3] (6+5=11 ✓), 2-pip at [3,3] (<3 ✓). ④ [0|3] horizontal row 2 cols 3–4: 0-pip at [2,3] (<3 ✓), 3-pip at [2,4] (empty). ⑤ [6|5] vertical col 0 rows 2–3: 6-pip at [2,0], 5-pip at [3,0] (6+5=11 ✓). ⑥ [3|6] vertical col 3 rows 1–0: 3-pip at [1,3], 6-pip at [0,3] (>3 ✓). ⑦ [2|6] horizontal row 1 cols 1–2: 2-pip at [1,1] (<3 ✓), 6-pip at [1,2] (6≠3 ✓).
💡 Hint 1: One region tells you a lot
Somewhere in this puzzle there's a three-cell region with a sum so high that there's only one possible combination of pip values. Find it before you touch anything else — it places three dominoes and confirms several other constraints at the same time.
💡 Hint 2: Sum=18 forces three sixes
The region at [3,1], [3,2], and [4,1] has to sum to 18. The only way to get there with three cells is 6+6+6. All three must be pip 6. That immediately places [3|6], [6|2], and [6|0], confirms [3,0]>2 and [4,0]<2 for free, and feeds a 2-pip into the sum=2 constraint at [3,3].
💡 Hint 3: Five cells, one pip value
The equals block at [2,2][2,3][2,4][2,5][3,4] spans five cells — all must match. The less-than constraint at [2,1] forces a low pip there, and the domino covering [2,1] is [5|0] reversed: 0-pip at [2,1] (<2 ✓), 5-pip at [2,2]. That sets the whole equals block to pip 5. With [3,4]=5 locked in, sum=6 at [3,5]+[4,5] needs two 3s.
💡 Hint 4: Solo sum cells give away four more dominoes
Sum=2 at [5,1] means [5,1]=2. [2|5] covers [5,1][5,2]: 2 ✓, 5-pip at [5,2]. That plus sum=10 at [4,2]+[5,2] gives [4,2]=5, and sum=8 at [4,3]+[4,4] gives two fours. Sum=1 at [6,3] forces [6,3]=1. Sum>9 at [7,3]+[7,4] can only come from [6|6].
💡 Full answer
① [3|6] horizontal row 3 cols 0–1: 3-pip at [3,0] (>2 ✓), 6-pip at [3,1] (sum=18). ② [6|2] horizontal row 3 cols 2–3: 6-pip at [3,2] (sum=18 ✓ with 6+6+6), 2-pip at [3,3] (sum=2 ✓). ③ [6|0] horizontal row 4 cols 1–0: 6-pip at [4,1] (sum=18 ✓), 0-pip at [4,0] (<2 ✓). ④ [5|0] horizontal row 2 cols 2–1: 5-pip at [2,2] (equals), 0-pip at [2,1] (<2 ✓). ⑤ [5|5] horizontal row 2 cols 3–4: 5-pip at both (equals: 5=5 ✓). ⑥ [5|3] horizontal row 3 cols 4–5: 5-pip at [3,4] (equals ✓), 3-pip at [3,5] (sum=6 starts). ⑦ [2|3] vertical col 5 rows 5–4: 2-pip at [5,5] (empty), 3-pip at [4,5] (3+3=6 ✓). ⑧ [5|6] horizontal row 2 cols 5–6: 5-pip at [2,5] (equals ✓), 6-pip at [2,6] (sum=6 ✓). ⑨ [2|5] horizontal row 5 cols 1–2: 2-pip at [5,1] (sum=2 ✓), 5-pip at [5,2]. ⑩ [4|5] horizontal row 4 cols 3–2: 4-pip at [4,3], 5-pip at [4,2] (5+5=10 ✓). ⑪ [0|4] vertical col 4 rows 5–4: 0-pip at [5,4] (empty), 4-pip at [4,4] (4+4=8 ✓). ⑫ [4|1] vertical col 3 rows 5–6: 4-pip at [5,3] (empty), 1-pip at [6,3] (sum=1 ✓). ⑬ [6|6] horizontal row 7 cols 3–4: both 6-pips (12>9 ✓). ⑭ [1|1] horizontal row 1 cols 7–8: both 1-pips. ⑮ [1|2] horizontal row 1 cols 6–5: 1-pip at [1,6] (1+1+1=3 ✓), 2-pip at [1,5]. ⑯ [2|2] horizontal row 0 cols 5–6: both 2-pips (2+2+2=6 ✓).
🎨 Pips Solver
Mar 17, 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: Two sum=0 corners need pip 0
Both [0,0] and [0,3] carry sum=0 — each must hold pip 0. Three dominoes have a 0-face: [0|0], [5|0], and [1|0]. The double-blank [0|0] can fill two cells with zeros, so it goes vertically at col 3 rows 0–1: pip 0 at [0,3] ✓, pip 0 at [1,3] (empty region, no problem).
2
Step 2: [1|0] claims the left corner
[0,0] still needs pip 0. From what's left, only [1|0] has a 0-face. It goes vertically at col 0 rows 1–0: pip[1]=0 at [0,0] (sum=0 ✓), pip[0]=1 at [1,0], which drops straight into the sum=3 region.
3
Step 3: Sum=3 at [1,0]+[1,1] forces the horizontal pair
[1,0]=1 is locked. So [1,1] must be 2. The horizontal domino at row 1 cols 1–2 needs 2-pip at col 1. [4|2] reversed: 4-pip at [1,2], 2-pip at [1,1] (1+2=3 ✓).
4
Step 4: Sum=5 at [1,2]+[2,2] pulls in [5|1]
[1,2]=4 from the previous step. So [2,2] must be 1. The vertical domino at col 2 rows 2–3 needs 1-pip at the top. [5|1] reversed: 5-pip at [3,2] (bottom, entering the equals region), 1-pip at [2,2] (4+1=5 ✓).
5
Step 5: [5|0] finishes the board
Only [5|0] is left. It goes vertically at col 1 rows 3–2: 5-pip at [3,1], 0-pip at [2,1] (empty, fine). Equals at [3,1][3,2]: 5=5 ✓. Done.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Equals column locked by the double-one
The equals region at [2,1][3,1] needs both cells to match. Two separate dominoes must show the same pip face there. The only piece that guarantees identical faces is [1|1] — drop it vertically at col 1 rows 2–3: both cells pip 1.
2
Step 2: Less-than at [4,1] and sum=11 combine
[4,1]<3 means pip 0, 1, or 2. The region [4,2]+[4,3] must hit 11 — only a 5+6 split works. A horizontal domino at row 4 covering [4,1][4,2] with a low pip at col 1 and 6 at col 2: [0|6] puts 0 at [4,1] (<3 ✓) and 6 at [4,2].
3
Step 3: [4,3]=5, then [5|2] confirms the less-than pair
[4,2]=6 is set. Sum=11 means [4,3]=5. A vertical domino at col 3 rows 3–4 must show 5 at row 4. [5|2] reversed: 5-pip at [4,3] (6+5=11 ✓), 2-pip at [3,3]. Both cells [2,3] and [3,3] must be <3 — [3,3]=2<3 ✓.
4
Step 4: Less-than pair at rows 2–3 gives you [0|3]
[2,3] also needs to be <3. A horizontal domino at row 2 covering [2,3][2,4] must place a low pip at col 3. [0|3]: 0-pip at [2,3] (<3 ✓), 3-pip at [2,4] (empty region).
5
Step 5: Sum=11 at [2,0]+[3,0] — one more 5+6 split
The left column pair needs to hit 11 again. [6|5] is the match: vertical at col 0 rows 2–3, 6-pip at [2,0], 5-pip at [3,0] (6+5=11 ✓).
6
Step 6: Greater>3 at [0,3] points to [3|6]
[0,3] must be pip 4, 5, or 6. A vertical domino at col 3 rows 0–1 must put a high value at [0,3]. [3|6] reversed: 6-pip at [0,3] (>3 ✓), 3-pip at [1,3].
7
Step 7: [2|6] closes everything
Only [2|6] is left. Horizontal at row 1 cols 1–2: 2-pip at [1,1] (<3 ✓), 6-pip at [1,2]. Unequal [1,2]≠[1,3]: 6≠3 ✓. Puzzle done.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Sum=18 means 6+6+6 — no other option
The region at [3,1], [3,2], and [4,1] must sum to 18. Max pip is 6, and 6+6+6=18 is the only combination that works. All three cells are pip 6. That tells you exactly which dominoes go there.
2
Step 2: [3|6] horizontal at row 3 — first six into place
[3,1]=6 needs a domino that places 6 at col 1. [3|6] goes horizontal at row 3 cols 0–1: 3-pip at [3,0], 6-pip at [3,1] (sum=18). Greater>2 at [3,0]: 3>2 ✓.
3
Step 3: [6|2] fills the second sum=18 cell and confirms sum=2
[3,2]=6. [6|2] horizontal row 3 cols 2–3: 6-pip at [3,2] (all three 6+6+6=18 ✓), 2-pip at [3,3] (sum=2 ✓).
4
Step 4: [6|0] closes the sum=18 and checks [4,0]
[4,1]=6. [6|0] horizontal row 4 cols 1–0: 6-pip at [4,1] ✓, 0-pip at [4,0]. Less<2 at [4,0]: 0<2 ✓.
5
Step 5: Less-than at [2,1] opens the five-cell equals block
[2,1]<2 means 0 or 1. A horizontal domino at row 2 covering [2,1][2,2] must place a low pip at col 1. [5|0] reversed: 0-pip at [2,1] (<2 ✓), 5-pip at [2,2]. That cell is in the five-cell equals region — so all five cells [2,2][2,3][2,4][2,5][3,4] must be pip 5.
6
Step 6: [5|5] horizontal fills two more equals cells
[2,3] and [2,4] are both in the equals block, both must be pip 5. [5|5] horizontal row 2 cols 3–4: 5-pip at both ✓.
7
Step 7: [5|3] brings the equals block into row 3
[3,4] is the fifth equals cell — must be pip 5. [5|3] horizontal row 3 cols 4–5: 5-pip at [3,4] (equals ✓), 3-pip at [3,5]. Sum=6 at [3,5]+[4,5] needs [4,5] to also be 3.
8
Step 8: [2|3] sets [4,5]=3 and finishes the sum=6 pair
[4,5]=3 needed. [2|3] vertical col 5 rows 5–4: 2-pip at [5,5] (empty), 3-pip at [4,5] (3+3=6 ✓).
9
Step 9: [5|6] covers the last equals cell and sum=6 at [2,6]
[2,5] is the last equals cell — must be pip 5. [5|6] horizontal row 2 cols 5–6: 5-pip at [2,5] (equals ✓), 6-pip at [2,6] (sum=6 ✓).
10
Step 10: Sum=2 at [5,1] is a freebie
[5,1] must be pip 2. [2|5] horizontal row 5 cols 1–2: 2-pip at [5,1] (sum=2 ✓), 5-pip at [5,2].
11
Step 11: Sum=10 at [4,2]+[5,2] gives [4,2]=5
[5,2]=5 is locked. So [4,2]=5. [4|5] horizontal row 4 cols 3–2: 4-pip at [4,3], 5-pip at [4,2] (5+5=10 ✓).
12
Step 12: Sum=8 at [4,3]+[4,4] — two fours
[4,3]=4 from above. So [4,4]=4. [0|4] vertical col 4 rows 5–4: 0-pip at [5,4] (empty), 4-pip at [4,4] (4+4=8 ✓).
13
Step 13: Sum=1 at [6,3] and sum>9 at the bottom
[6,3] must be pip 1. [4|1] vertical col 3 rows 5–6: 4-pip at [5,3] (empty), 1-pip at [6,3] ✓. For [7,3]+[7,4]>9, only [6|6] hits 12>9 — horizontal row 7 cols 3–4.
14
Step 14: Three-cell sum=3 and sum=6 clear the top
Sum=3 at [1,6][1,7][1,8]: three 1s. [1|1] horizontal row 1 cols 7–8 puts both 1s. [1|2] horizontal row 1 cols 6–5: 1-pip at [1,6] (1+1+1=3 ✓), 2-pip at [1,5]. Sum=6 at [0,5][0,6][1,5]: [1,5]=2 is set, so [0,5]+[0,6] must be 2+2=4 to make 6. [2|2] horizontal row 0 cols 5–6: both 2-pips (2+2+2=6 ✓). Done.
💡 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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