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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 keeps today's easy accessible — the bottom-left sum target is small enough that you'll know exactly where two dominoes land before you've placed anything else. From there it's a single follow-up move and a clean finish. Give yourself two minutes, maybe three.
Medium, also from Ian Livengood, pivots on a four-cell equals block anchoring the right side. All four cells must share the same value, and once you figure out what that value is, the rest of the grid chains left without backtracking. The real challenge is spotting that anchor first.
Hard is a Rodolfo Kurchan construction — denser grid, 12 dominoes, and one standout constraint: a three-cell sum of 17 in the lower middle. That's near-maximum, and it tells you almost exactly what's there. The difficulty comes from a web of single-cell clues that you'll need to cross-reference before everything clicks.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Start with a sum constraint
One region in the bottom-left adds up to a very small number. That alone rules out almost every possible pip combination for those two cells.
💡 The sum forces both values
The bottom-left sum of 2 requires each of the two cells there to hold a 1 — they can't split the total any other way given the available dominoes. That pins down the orientation of two dominoes and also satisfies the greater-than constraint one row up.
💡 Full answer
The [5–1] domino sits vertically in column 0, rows 4–5, with 5 on top (satisfying greater-than-3) and 1 at the bottom. The [0–1] domino runs horizontally in row 5, cols 1–2, with 1 at col 1 and 0 at col 2 — completing the sum. The [3–3] double fills column 2, rows 2–3, satisfying the equals constraint there. The [4–3] domino sits vertically in column 4, rows 0–1, with 3 on top and 4 below (less than 5). Finally, the [6–3] domino runs horizontally in row 0, cols 2–3, with 6 left and 3 right — matching the 3 at col 4 to satisfy the top equals region.
💡 Find the large equals region
Four cells on the right side of the grid all share an equals constraint — every single one must show the same pip value.
💡 One double anchors the right side
Within that four-cell equals block, a double domino locks in two of the cells immediately. Once you identify which double fits and what value it forces, the other two cells in the region must match — pinning down two more domino placements.
💡 Full answer
The [3–3] double runs vertically in column 5, rows 1–2, filling the equals region core with 3s. The [4–3] domino goes horizontal in row 0, cols 4–5 (4 left, 3 right), and the [5–3] domino fills row 2, cols 3–4 (5 left, 3 right) — all four equals cells land on 3. For the three-cell equals region in the lower-middle, the [4–4] double goes vertically in column 2, rows 3–4, and the [2–4] domino runs horizontal in row 3, cols 0–1 (2 left, 4 right). The sum-of-3 trio at bottom-left is solved by the [1–1] double vertically at column 0, rows 4–5, and the [3–1] domino horizontal in row 5, cols 1–2 (1 at col 1). The [6–4] domino completes the grid vertically at column 3, rows 0–1 (6 on top, 4 below).
💡 Look for the high-sum row
A three-cell sum constraint somewhere in the grid has an unusually large target — one that leaves very little room for variation.
💡 The sum of 17 nearly solves itself
Three cells summing to 17 can only be achieved with values near the maximum of 6. There's essentially one combination that works, and it locks the orientation of three dominoes at once. Use the single-cell constraints nearby to confirm which domino goes where.
💡 Full answer
Row 5, cols 1–3 sum to 17 as 6+5+6: the [5–6] domino sits vertically at col 1, rows 4–5 (5 on top, 6 at bottom); the [5–2] domino sits vertically at col 2, rows 5–6 (5 at row 5, 2 at row 6); the [6–3] domino sits vertically at col 3, rows 4–5 (6 at bottom, satisfying sum; 3 at row 4, satisfying less-than-4). The [1–1] double runs horizontal in row 3, cols 1–2, anchoring the three-cell equals region (all 1s) — the [1–5] domino horizontal in row 2, cols 0–1 contributes 1 at col 1. The [6–6] double runs horizontal in row 4, cols 5–6. The [5–0] domino sits vertically at col 5, rows 1–2 (0 at top, 5 below — satisfying greater-than-2). The [2–0] domino sits vertically at col 4, rows 0–1 (0 at top, 2 at row 1). The [3–5] domino runs horizontal in row 1, cols 1–2 (3 at col 1, 5 at col 2). The [5–4] domino sits vertically at col 3, rows 1–2 (5 at top, 4 at bottom). The [6–4] domino runs horizontal in row 3, cols 4–5 (6 left, 4 right).
🎨 Pips Solver
Apr 23, 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: Lock the middle column
The equals constraint on column 2, rows 2–3 demands that both cells share the same pip value. That means the domino placed there must show the same number on each end — a double. Among the five dominoes in today's easy puzzle, the only double is [3–3]. Place it vertically in column 2, rows 2–3. Both cells are now fixed at 3.
2
Step 2: Solve the bottom-left sum
The sum-of-2 region covers cells at row 5, col 0 and row 5, col 1. The [5–1] domino must cover the left column, rows 4–5. The greater-than-3 constraint on row 4, col 0 forces the 5 to sit at the top, leaving 1 at the bottom (row 5, col 0). To hit the sum target of 2, row 5 col 1 must also equal 1. The [0–1] domino runs horizontally across row 5, cols 1–2, with 1 at col 1 and 0 at col 2. Both constraints are now satisfied.
3
Step 3: Place the remaining two dominoes
Two dominoes remain: [4–3] and [6–3]. Cell row 1, col 4 carries a less-than-5 constraint. Placing [4–3] vertically at column 4, rows 0–1 puts 3 at the top (row 0) and 4 at row 1 — the value 4 satisfies less-than-5. The final domino [6–3] runs horizontally in row 0, cols 2–3, with 6 at col 2 and 3 at col 3. The equals region at row 0, cols 3–4 confirms both cells show 3. All constraints satisfied.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Anchor the four-cell equals region
The equals constraint covers four cells on the right side: row 0 col 5, row 1 col 5, row 2 col 4, and row 2 col 5. All four must share the same value. The [3–3] double placed vertically at column 5, rows 1–2 fills two of them with 3. That forces the remaining two cells — row 0 col 5 and row 2 col 4 — to also equal 3. The [4–3] domino placed horizontally in row 0, cols 4–5 (4 left, 3 right) gives row 0 col 5 = 3. The [5–3] domino placed horizontally in row 2, cols 3–4 (5 left, 3 right) gives row 2 col 4 = 3. Both greater-than constraints at row 1 col 3 and row 2 col 3 are checked: row 1 col 3 is not yet placed, row 2 col 3 = 5 > 4 ✓.
2
Step 2: Solve the three-cell equals region
Cells row 3 col 1, row 3 col 2, and row 4 col 2 must all be equal. The [4–4] double placed vertically at column 2, rows 3–4 sets both cells in column 2 to 4. So row 3 col 1 must also be 4. The [2–4] domino placed horizontally in row 3, cols 0–1 (2 at col 0, 4 at col 1) satisfies this. Cell row 3 col 0 is an empty constraint, so the value 2 there is unconstrained — that works.
3
Step 3: Resolve the bottom-left sum
The sum-of-3 region covers row 4 col 0, row 5 col 0, and row 5 col 1. The [1–1] double placed vertically at column 0, rows 4–5 contributes 1 at each cell. Sum so far: 1+1=2, so row 5 col 1 must equal 1. The [3–1] domino placed horizontally in row 5, cols 1–2 (1 at col 1, 3 at col 2) delivers exactly that. The single-cell sum-of-3 constraint at row 5 col 2 is also satisfied: that cell equals 3 ✓.
4
Step 4: Complete with the remaining domino
Only the [6–4] domino remains. It fits vertically at column 3, rows 0–1 (6 on top at row 0, 4 at row 1). Check: the sum-of-10 region at row 0, cols 3–4 reads 6+4=10 ✓. The greater-than-3 constraint at row 1 col 3 reads 4>3 ✓. All regions satisfied — grid complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Crack the sum-of-17 row
The sum constraint at row 5, cols 1–3 has a target of 17. With each cell capped at 6, the only combination that reaches 17 is 6+5+6. The [5–6] domino must cover row 5 col 1 (value 6) — placed vertically at col 1, rows 4–5, with 5 at row 4. The single-cell sum-of-5 constraint at row 4, col 1 confirms (5 = 5) ✓. The [5–2] domino placed vertically at col 2, rows 5–6 puts 5 at row 5 and 2 at row 6 — satisfying less-than-3 at row 6 col 2 (2 < 3) ✓. The [6–3] domino placed vertically at col 3, rows 4–5 puts 6 at row 5 (completing the sum) and 3 at row 4 — satisfying less-than-4 at row 4 col 3 (3 < 4) ✓.
2
Step 2: Work through the single-cell constraints
Cell row 1 col 1 must be less than 5. The [3–5] domino placed horizontally at row 1, cols 1–2 puts 3 at col 1 (satisfying < 5) and 5 at col 2. Cells row 1 cols 2–3 must sum to 10: (1,2)=5, so (1,3) must equal 5. The [5–4] domino placed vertically at col 3, rows 1–2 puts 5 at row 1 and 4 at row 2. Single-cell sum-of-4 at row 2 col 3: 4 = 4 ✓. Cell row 4 col 6 must be greater than 3 — the [6–6] double placed horizontally at row 4, cols 5–6 puts 6 at both cells, satisfying that ✓.
3
Step 3: Resolve the three-cell equals region
Cells row 2 col 1, row 3 col 1, and row 3 col 2 must all be equal. The [1–1] double placed horizontally at row 3, cols 1–2 sets both to 1. So row 2 col 1 must also be 1. The [1–5] domino placed horizontally at row 2, cols 0–1 puts 1 at col 1 and 5 at col 0 — satisfying the equals region ✓. Cell row 2 col 0 is an empty constraint, so value 5 is fine.
4
Step 4: Place the final three dominoes
The [5–0] domino placed vertically at col 5, rows 1–2 puts 0 at row 1 and 5 at row 2. Greater-than-2 at row 2 col 5: 5 > 2 ✓. The [2–0] domino placed vertically at col 4, rows 0–1 puts 0 at row 0 and 2 at row 1. Sum-of-2 region at rows 0–1, col 4 and row 1 col 5: (0,4)=0 + (1,4)=2 + (1,5)=0 = 2 ✓. The [6–4] domino placed horizontally at row 3, cols 4–5 puts 6 at col 4 and 4 at col 5. Greater-than-3 at row 3 col 4: 6 > 3 ✓. Unequal constraint at row 3 col 5 and row 4 col 5: 4 ≠ 6 ✓. 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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