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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's easy is built around three sum constraints that each lock an entire group of cells at once — one forces all-zeros, one forces all-threes, one forces all-sixes. Each group has a single candidate domino, and placing them in order resolves the whole grid. Medium, also by Livengood, opens on a three-cell equals region that immediately points to a double domino, then cascades through two equals chains before closing on a pair of sum-of-7 pairs. Hard, by Rodolfo Kurchan, is the standout: a 7-cell unequal region in the center forces all seven distinct pip values (0 through 6) to appear exactly once across those positions. The outer constraints — a sum-of-12 pair in the bottom row, a sum-of-0 corner, and a three-cell equals column — anchor the perimeter first, leaving the interior to be resolved through a chain of sum-of-2 constraints and the unequal region's own logic.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Three sum constraints, three groups
Today's easy has no equals or less-than constraints — just three sum regions, each covering multiple cells. One of them has a target of zero, which is a very strong restriction. Start there.
💡 Zero locks a domino; the rest follows
The sum-of-0 region spans three cells (row 1 col 1, row 2 cols 0–1). All three must be zero pips. One double in today's set is the blank domino — it fills two of those cells. The third cell's zero then pins a second domino. From there, the sum-of-12 and sum-of-24 regions each have only one way to reach their targets.
💡 Full answer
The [0-0] blank double fills row 2 cols 0–1 (both 0). The [0-3] domino goes horizontally in row 1 cols 1–2: 0 at col 1, 3 at col 2. The [3-6] fills row 1 cols 3–4: 3 at col 3, 6 at col 4. The [3-3] double fills row 2 cols 2–3 (both 3). The [6-6] double fills row 2 cols 4–5 (both 6). The [6-0] sits horizontally at row 0 cols 3–4: 6 at col 4 (wait — actually [6-0] at row 0: (0,4)=6, (0,3)=0).
💡 The three-cell equals region is your entry
Three cells in an equals region must all hold the same pip value. Two of those cells belong to the same domino — find the double that places equal values automatically, and you've locked the first piece.
💡 Doubles cascade downward
The [5-5] double anchors rows 1–2 on the right side. With that in place, the sum-of-12 region below (three cells) and the two-cell equals in column 1 both resolve cleanly. Each placement limits the candidates for the sum-of-7 pairs.
💡 Full answer
The [5-5] double goes in row 1 cols 2–3 (both 5). The [6-6] double fills row 3 cols 2–3 (both 6), forcing the remaining sum-of-12 cell to 0: [0-5] at row 2 cols 1–2 (0 at col 2, 5 at col 1). The [5-4] fills row 3 cols 0–1: 5 at col 1, 4 at col 0. The [1-6] goes vertically in col 0 rows 1–2: 6 at row 1, 1 at row 2. The [2-5] fills row 2 cols 3–4: 5 at col 3, 2 at col 4. The [2-0] goes vertically in col 4 rows 0–1: 0 at row 0, 2 at row 1.
💡 Seal the bottom row first
The sum-of-12 constraint on the two rightmost cells of the bottom row is extremely restrictive — only one pip combination reaches 12. That forces both cells to the same value and identifies the domino immediately.
💡 Sum-of-0 corners and the right edge
Two cells in the bottom-right corner must sum to 0, and a single cell in the top-right column must also be 0. These empty-pip anchors lock the dominoes running up the right side and reveal a chain through the sum-of-2 constraints.
💡 Column 0 three-cell equals
Cells (rows 1, 2, 3) in column 0 are all equal. Once you know the value from the left-side sum-of-7 and sum-of-2 chain, a blank double fills two of those cells and the third follows. The top-left sum-of-1 then closes the corner.
💡 The unequal region in the center
Seven cells in the middle of the grid must all hold different pip values — every value from 0 to 6 appears exactly once in that region. By the time you reach this step, several of those cells are already determined; the remaining slots fill in by elimination.
💡 Full answer
Bottom row: 3,4,5,6,6,0 (cols 0–5). The [0-0] blank double sits at row 3 cols 1–2 (both 0). Row 0: 1,0,3,1,0,1. The [4,4] double sits at row 3 cols 3–4 (both 4), completing the sum-of-12 triple with (2,5)=4. Row 2: 2,2,3,6,2,4. The unequal region holds 2,3,6,0,4,5,1 — all seven pip values exactly once.
🎨 Pips Solver
Apr 20, 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: Sum-of-0 forces blank cells
The sum-of-0 region covers three cells: (row 1, col 1), (row 2, col 0), and (row 2, col 1). All three must be zero. Place the [0-0] blank double horizontally in row 2 cols 0–1: both cells = 0. Then (row 1, col 1) must also be 0, which belongs to domino [0-3] — place it horizontally in row 1 cols 1–2: 0 at col 1, 3 at col 2.
2
Step 2: Sum-of-12 fills with threes
The sum-of-12 region covers (row 1, cols 2–3) and (row 2, cols 2–3). We already know (row 1, col 2) = 3. Remaining three cells must sum to 9. The [3-6] domino goes horizontally in row 1 cols 3–4: 3 at col 3, 6 at col 4. The [3-3] double fills row 2 cols 2–3: both = 3. Check: 3+3+3+3 = 12 ✓.
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Step 3: Sum-of-24 fills with sixes
The sum-of-24 region covers (row 0, col 4), (row 1, col 4), (row 2, cols 4–5). We know (row 1, col 4) = 6. Remaining three cells must sum to 18. Place the [6-6] double in row 2 cols 4–5: both = 6. Then (row 0, col 4) = 6. The [6-0] domino fills row 0 cols 3–4: 6 at col 4, 0 at col 3. Check: 6+6+6+6 = 24 ✓.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Three-cell equals points to a double
The equals region spans (row 1, cols 2–3) and (row 2, col 3). Two of those cells — (1,2) and (1,3) — share a domino. For a single domino to place equal values in two cells, it must be a double. [5-5] is the only applicable double: place it horizontally in row 1 cols 2–3 (both 5). This forces (row 2, col 3) = 5 as well.
2
Step 2: Sum-of-12 and the second double
The sum-of-12 region covers (row 2, col 2) and (row 3, cols 2–3). [6-6] placed at row 3 cols 2–3 contributes 12 on its own, forcing (row 2, col 2) = 0. That places domino [0-5] horizontally in row 2 cols 1–2: 0 at col 2, 5 at col 1.
3
Step 3: Column 1 equals leads to column 0 sum
The equals region on (row 2, col 1) and (row 3, col 1) forces (row 3, col 1) = 5. Domino [5-4] at row 3 cols 0–1: 5 at col 1, 4 at col 0. Then the sum-of-5 region on (row 2, col 0) and (row 3, col 0): (3,0) = 4, so (2,0) = 1. Domino [1-6] vertically in col 0 rows 1–2: 6 at row 1, 1 at row 2.
4
Step 4: Right-side equals and less-than close the grid
With (row 2, col 3) = 5, domino [2-5] fills row 2 cols 3–4: 5 at col 3, 2 at col 4. The equals region on (row 1, col 4) and (row 2, col 4) forces (row 1, col 4) = 2. Domino [2-0] goes vertically in col 4 rows 0–1: 0 at row 0, 2 at row 1. The less-than-2 constraint at (row 0, col 4): 0 < 2 ✓. And greater-than-5 at (row 1, col 0): (1,0) = 6 > 5 ✓.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Bottom-row sum-of-12 forces two sixes
The sum-of-12 region covers (row 5, cols 3–4). The only way two pip values can reach 12 is 6+6. Both cells must be 6. Domino [0-6] at (5,5)–(5,4): (5,5)=0, (5,4)=6. Then (5,3)=6 is covered by [6-1] at (5,3)–(4,3): (5,3)=6, (4,3)=1.
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Step 2: Sum-of-0 corners and right edge
The sum-of-0 region on (4,5) and (5,5): (5,5)=0 already; (4,5)=0. Domino [0-4] at (4,5)–(3,5): (4,5)=0, (3,5)=4. Single-cell sum-of-0 at (0,4): (0,4)=0. Domino [0-1] at (0,4)–(0,5): (0,4)=0, (0,5)=1. Sum-of-2 at (0,5)+(1,5): (1,5)=1. Domino [4-1] at (2,5)–(1,5): (2,5)=4, (1,5)=1. Single-cell sum-of-2 at (2,4): (2,4)=2. Domino [6-2] at (2,3)–(2,4): (2,3)=6, (2,4)=2.
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Step 3: Bottom-left chain and three-cell equals
Domino [5-5] double at (4,2)–(5,2): both=5 (satisfies >4 at (5,2) and >4 check). Domino [4-3] at (5,1)–(5,0): (5,1)=4 (>3 ✓), (5,0)=3. Sum-of-7 at (4,0)+(5,0): (4,0)=4. Domino [2-4] at (3,0)–(4,0): (3,0)=2, (4,0)=4. Equals region (1,0)=(2,0)=(3,0): all must equal 2. Domino [2-2] double at (2,0)–(2,1): (2,0)=2, (2,1)=2. Domino [1-2] at (0,0)–(1,0): (0,0)=1, (1,0)=2.
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Step 4: Top row and sum-of-12 triple
Sum-of-1 at (0,0)+(0,1): (0,0)=1, so (0,1)=0. Domino [0-3] at (0,1)–(0,2): (0,1)=0, (0,2)=3 (>1 ✓). Sum-of-12 triple at (2,5)+(3,4)+(3,5): (2,5)=4, (3,5)=4; so (3,4)=4. Domino [4-4] double at (3,3)–(3,4): (3,3)=4, (3,4)=4. Sum-of-0 at (3,1)=0. Domino [0-0] double at (3,1)–(3,2): (3,1)=0, (3,2)=0.
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Step 5: Remaining dominos and unequal verification
Domino [3-3] double at (1,2)–(2,2): (1,2)=3, (2,2)=3. Both satisfy >1 ✓. Domino [1-3] at (0,3)–(1,3): (0,3)=1, (1,3)=3 (>1 ✓). The unequal region now holds: (2,1)=2, (2,2)=3, (2,3)=6, (3,2)=0, (3,3)=4, (4,2)=5, (4,3)=1 — all seven values 0–6, each exactly once ✓. Grid complete.
💡 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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