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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
You scan the easy grid and a row of zeros jumps out immediately — the sum-of-0 constraint locks all four cells in the third row to blanks, and that one observation hands you three domino placements at once. The left column's sum-of-17 follows automatically once you see which domino's six got pushed upward, and the last two pieces settle into the top row without any guessing.
Medium's entry feels like a gift: a single cell constrained to exactly 1 points you straight at the right domino. Place it and you notice one of its cells falls inside a three-way equals region — all three must match. The double-six locks in, and from there the grid peels apart in two clean passes, one across the top row and one across the bottom.
Rodolfo Kurchan's hard opens with three freebies — three single-cell constraints that immediately name three doubles and hand you your first three placements. With those anchors set, the equals and sum chains cascade from the top corners down and from the bottom-right corner up. By the time you reach the seven-cell unequal region in the center, most of it is already solved; the last two slots fill by elimination.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 The row of zeros is your door in
One constraint forces an entire row to sum to zero. That means every single cell in that row must be a blank face. Find that row and ask yourself: which dominoes in today's set have a zero pip, and where must they go?
💡 Three dominoes share that zero row
The [0-0] blank double fills two of the four zero cells. The remaining two zeros each belong to a different domino, pushing their non-zero pips either up or down into adjacent cells. Once those three dominoes lock in, the left column's sum-of-17 has only one valid fill — and the less-than-6 cell in the bottom-right takes care of itself.
💡 Full answer
The [0-0] blank double sits horizontally in row 3 cols 1–2 (both 0). The [0-6] runs vertically: 0 at (row 3, col 0), 6 at (row 2, col 0). The [5-0] runs vertically: 5 at (row 4, col 3), 0 at (row 3, col 3). The [6-5] fills the left column vertically: 6 at (row 0, col 0), 5 at (row 1, col 0) — sum 6+5+6=17 ✓. The [4-4] double sits horizontally in row 0 cols 1–2 (both 4). The [1-2] caps vertically: 1 at (row 0, col 3), 2 at (row 1, col 3) — row 0 sum 4+4+1=9 ✓.
💡 One cell tells you everything
A single cell is constrained to sum to exactly 1. There's only one pip value that satisfies that: the cell must be 1. Find that cell, identify which domino it belongs to, and place it — you've just made your first deduction without any cross-checking at all.
💡 A triple equals cascade
Once the first domino is placed, one of its cells sits inside a three-way equals region. All three cells in that region must hold the same value. You already know one of them equals 6 — so all three must be 6. The only domino that can fill the other two adjacent positions with 6s is the double-six. Place it and watch the rest of the grid open up from both the top row and the bottom row.
💡 Full answer
The [6-1] sits horizontally in row 2: 6 at col 3, 1 at col 2. The [6-6] double sits vertically in col 4 rows 2–3 (both 6). The [2-5] sits horizontally in row 0: 5 at col 0, 2 at col 1. The [4-5] sits vertically in col 0 rows 1–2: 5 at row 1, 4 at row 2. The [2-2] double sits horizontally in row 0 cols 2–3 (both 2). The [2-0] sits horizontally in row 0 cols 4–5: 2 at col 4, 0 at col 5. The [4-2] sits vertically in col 0 rows 3–4: 4 at row 3, 2 at row 4. The [3-0] sits horizontally in row 4 cols 1–2: 3 at col 1, 0 at col 2. The [0-1] sits horizontally in row 4 cols 3–4: 0 at col 3, 1 at col 4.
💡 Three freebies right away
Three cells in this grid each have a single-cell constraint that states the pip value directly. All three happen to be doubles. Start there — three placements with zero ambiguity — and you'll have a solid framework before the real deductions begin.
💡 Top row resolves from the anchors
With the three doubles in place, the equals constraint at the top of the grid (two adjacent cells must match) and the sum-of-2 pair each resolve in a single step. Follow those into the bottom-right corner: a sum-of-12 pair has only one possible solution.
💡 Bottom row chains work inward
The sum-of-12 pair in the bottom-right forces two sixes. From there, two more sum constraints in the bottom row cascade leftward, placing three additional dominoes. Then a sum-of-10 pair in the lower-left corner locks the double-five.
💡 The center unequal region fills by elimination
Seven cells in the middle of the grid must each hold a different pip value — every value from 0 to 6 appears exactly once. By the time you reach this step, five of those seven cells are already determined. The remaining two slots belong to a single domino, and the only candidate that fills the two missing values is [2-3].
💡 Full answer
The [4-4] double: vertically col 0 rows 1–2 (both 4). The [6-6] double: vertically col 2 rows 1–2 (both 6). The [1-0]: vertically col 4 rows 0–1 (1 at row 0, 0 at row 1). The [4-1]: horizontally row 0 cols 2–3 (4 at col 2, 1 at col 3). The [4-0]: vertically col 1 rows 0–1 (4 at row 0, 0 at row 1). The [1-1] double: vertically col 1 rows 2–3 (both 1). The [5-6]: vertically col 4 rows 4–5 (5 at row 4, 6 at row 5). The [6-3]: horizontally row 6 (6 at col 4, 3 at col 3). The [3-5]: horizontally row 6 (3 at col 2, 5 at col 1). The [4-5]: vertically col 0 rows 6–5 (4 at row 6, 5 at row 5). The [5-5] double: vertically col 0 rows 3–4 (both 5). The [4-3]: vertically col 4 rows 3–2 (4 at row 3, 3 at row 2). The [2-3]: vertically col 2 rows 3–4 (2 at row 3, 3 at row 4). Unequal region: 0,4,1,6,5,2,3 — all seven values ✓.
🎨 Pips Solver
Apr 21, 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 anchors the entire third row
The sum-of-0 region spans four cells: (row 3, cols 0–3). Every cell must be a blank pip. Place the [0-0] double horizontally at row 3 cols 1–2 (both 0). The [0-6] domino must orient vertically with its 0 at (3,0) — that pushes the 6 up to (2,0). The [5-0] must orient vertically with its 0 at (3,3) — that pushes the 5 down to (4,3). Three dominoes placed, zero ambiguity.
2
Step 2: Column sum-of-17 places the last heavy domino
The left column covers three cells: (row 0, col 0), (row 1, col 0), and (row 2, col 0). We know (2,0)=6. The remaining two cells must sum to 11. Only [6-5] reaches 11 with two pips — place it vertically: 6 at (0,0), 5 at (1,0). Check: 6+5+6=17 ✓. As a bonus, (4,3)=5 satisfies the less-than-6 constraint ✓.
3
Step 3: Row 0 sum-of-9 closes the grid
Three cells in row 0 (cols 1–3) must sum to 9. Two dominoes remain: [4-4] and [1-2]. Place [4-4] horizontally at row 0 cols 1–2 (both 4): that's 8. Then (0,3) must be 1 to reach 9. The [1-2] orients vertically: 1 at (0,3), 2 at (1,3). Final check: 4+4+1=9 ✓. Grid complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Single-cell sum-of-1 identifies the first domino
The region at (row 2, col 2) must equal exactly 1. That cell belongs to a domino — and only [6-1] contains a pip of 1. Place [6-1] horizontally at row 2 cols 2–3: 1 at col 2, 6 at col 3. No other arrangement is possible; the adjacent cell (2,3) is the only valid neighbor, and the domino's 6 goes there.
2
Step 2: Three-cell equals forces the double-six
The equals region spans (2,3), (2,4), and (3,4). We just placed (2,3)=6, so all three cells must be 6. The only domino that places two consecutive 6s is [6-6]. Put it vertically at col 4 rows 2–3: both 6. ✓ All three equals cells now hold 6.
3
Step 3: Column 0 equals and top row
The equals region at (0,0) and (1,0) requires both to match. Domino [2-5] goes horizontally in row 0 at cols 1–0: 2 at (0,1), 5 at (0,0). Domino [4-5] goes vertically in col 0 rows 1–2: 5 at (1,0), 4 at (2,0). Both cells equal 5 ✓. Sum-of-8 in row 0 cols 1–4: (0,1)=2; need cols 2–4 to sum to 6. Domino [2-2] double at (0,2)–(0,3): both 2. Then (0,4)=2. Domino [2-0] at (0,4)–(0,5): 2 at col 4, 0 at col 5. Less-than-2 at (0,5): 0<2 ✓.
4
Step 4: Bottom rows close through sums
Sum-of-8 at (2,0)+(3,0): (2,0)=4, so (3,0)=4. Domino [4-2] vertically at col 0 rows 3–4: 4 at row 3, 2 at row 4. Sum-of-5 at (4,0)+(4,1): (4,0)=2, so (4,1)=3. Domino [3-0] horizontally at row 4 cols 1–2: 3 at col 1, 0 at col 2. Equals at (4,2)=(4,3): (4,2)=0, so (4,3)=0. Domino [0-1] horizontally at row 4 cols 3–4: 0 at col 3, 1 at col 4. Grid complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Three single-cell constraints lock three doubles
Single-cell sum-of-4 at (1,0): (1,0)=4. The [4-4] double covers (1,0)–(2,0) vertically: both 4. Single-cell sum-of-6 at (1,2): (1,2)=6. The [6-6] double covers (1,2)–(2,2) vertically: both 6. Single-cell sum-of-0 at (1,4): (1,4)=0. The [1-0] domino covers (0,4)–(1,4) vertically: 1 at row 0, 0 at row 1. Three placements, all forced, no decisions needed.
2
Step 2: Top row resolves from equals and sum
Sum-of-2 at (0,3)+(0,4): (0,4)=1, so (0,3)=1. Domino [4-1] horizontally at row 0 cols 2–3: 4 at col 2, 1 at col 3. Equals at (0,1)=(0,2): (0,2)=4, so (0,1)=4. Domino [4-0] vertically at col 1 rows 0–1: 4 at row 0, 0 at row 1. Top row complete.
3
Step 3: Less-than-2 and the unequal region's left edge
Less-than-2 at (3,1) forces (3,1)=0 or 1. Domino [1-1] double covers (2,1)–(3,1) vertically: both 1. (3,1)=1<2 ✓. The unequal region now has (1,1)=0, (2,0)=4, (2,1)=1, (2,2)=6 determined. Values 0,1,4,6 are used; still needed among (3,0), (3,2), (4,2): values 2, 3, and 5.
4
Step 4: Bottom-right corner and bottom row chain
Sum-of-12 at (5,4)+(6,4): only 6+6=12. Domino [5-6] vertically at col 4 rows 4–5: 5 at row 4, 6 at row 5. Domino [6-3] horizontally at row 6 cols 3–4: 3 at col 3, 6 at col 4. Sum-of-6 at (6,2)+(6,3): (6,3)=3, so (6,2)=3. Domino [3-5] horizontally at row 6 cols 1–2: 5 at col 1, 3 at col 2. Sum-of-9 at (6,0)+(6,1): (6,1)=5, so (6,0)=4. Domino [4-5] vertically at col 0 rows 5–6: 5 at row 5, 4 at row 6. Sum-of-10 at (4,0)+(5,0): (5,0)=5, so (4,0)=5. Domino [5-5] double vertically at col 0 rows 3–4: both 5. (3,0)=5 now known for the unequal region.
5
Step 5: Sum-of-12 triple and unequal region completes
Sum-of-12 at (2,4)+(3,4)+(4,4): (4,4)=5, so (2,4)+(3,4)=7. Domino [4-3] vertically at col 4 rows 2–3: 3 at row 2, 4 at row 3. Unequal region remaining: (3,2) and (4,2) must use values 2 and 3 (the two not yet placed among 0–6: used are 0,1,4,5,6 from other cells; 3 placed at (3,4) which is outside the unequal region; so from the unequal region values 0,1,4,5,6 are placed, leaving 2 and 3). Domino [2-3] vertically at col 2 rows 3–4: 2 at row 3, 3 at row 4. Unequal region final values: 0,4,1,6,5,2,3 — all seven distinct ✓. 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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