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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
Easy feels like a gift today: Ian Livengood has built the entire puzzle around a single forced cell. One constraint tells you exactly which pip goes there before anything else, and a two-step chain follows with zero decision points. If you find the entry, you're done in under two minutes.
Medium is where Rodolfo Kurchan asks you to look past the obvious. Sum constraints dominate the board, but the real key is a three-cell equals region at the bottom — a rarer constraint type that only a handful of domino combinations can satisfy. Spot that first and the sum chains above it collapse in sequence.
The hard grid is Kurchan at his most architecturally precise. The opening challenge is a four-cell region where the total must be less than 2 — a brutal ceiling that forces near-zero pips across the board's center-left. That cluster of zeros feeds directly into two sum-2 constraints stacked in the same column, and once the left side resolves, a sum-9 pair on the right (the theoretical maximum for two cells) provides the opposite anchor. The bottom row then closes on its own.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 One cell tells you everything
There's a region in this puzzle that contains exactly one cell and carries a sum constraint. That single cell's pip value is fully determined before you touch anything else.
💡 Sum of 1 means pip of 1
The single-cell sum target is 1. That means exactly one pip goes in that cell. Scan your domino set for any piece with a 1-end — only one qualifies.
💡 4–1 starts the cascade
The single-cell sum forces a 1 at row 1, column 3. The 4–1 domino runs vertically there: 4 at row 2, 1 at row 1. Then the 3–3 double drops vertically down column 2 — both ends are 3, satisfying the equals pair at rows 0–1. The 4 from the first placement creates an equals pair with the 4-end of 4–2, placing that domino horizontal across the bottom-right (row 3). The 2–0 follows vertically at column 1, rows 2–3 (2 below, 0 above), and 6–0 runs horizontal at row 1, columns 0–1 (6 left, 0 right) — confirming the final equals constraint where both 0s meet at column 1.
💡 One constraint type is harder to satisfy
Scan the grid for the equals regions. One of them contains three cells instead of the usual two — that's the unusual constraint that opens this puzzle.
💡 Triple equals means one shared pip across all three cells
All three cells in the bottom row must show the same pip. Look for a double domino in the set — a piece with both ends identical can cover two of those cells and lock the shared value.
💡 6–6 anchors the bottom, everything else follows upward
The 6–6 double fills the bottom two cells (both equal 6), and the 5–6 domino runs vertically at the rightmost active column — its 6-end snaps into the third equals cell, its 5-end settles below, satisfying the greater-than-1 constraint at row 4. The sum-7 pair at column 1 resolves next: 3 at the top (from 0–3, laid horizontal with 0 at column 2) plus 4 at row 1 (from 4–5, laid horizontal) = 7. The 4–5 places a 5 into the equals pair at column 0, rows 1–2 — matched by the 5-end of 1–5. That domino's 1-end feeds the sum-7 pair in the center row, paired with the 6-end of 1–6. Finally, 0–3 and 0–5 each contribute a 0 to the equals pair at row 0, columns 2–3.
💡 Find the most restrictive region first
One region has a sum ceiling so low it nearly forces every cell to zero. That's your opening — not the sum constraints scattered elsewhere.
💡 Less-than-2 across four cells means almost no pips allowed
A four-cell region where the total must be less than 2 can only be satisfied if the sum is 0 or 1. With four cells to fill, practically every pip must be 0. That cluster immediately rules out most of the domino set for that area.
💡 The zero cluster connects two sum-2 columns
Once you fill the four-cell region with zeros, you'll notice it overlaps two separate sum-2 constraints running down the left column. Work those sum-2 pairs in order from top to bottom — each one propagates a pip value to the next cell below it.
💡 Sum-9 on the right is the maximum possible
Six plus three equals nine — the highest achievable sum for two cells. On the right side, a pair of cells must hit exactly 9. Only one domino combination in the set reaches that total. Place it and the sum-5 constraints above it resolve immediately.
💡 Full solution: zeros down the left, maximum on the right, bottom arithmetic closes everything
The four-cell less-than-2 region forces all four cells to 0: the 0–0 double fills two vertically, the 0–1 domino contributes its 0-end to a third cell and its 1-end to the corner above, and the 2–0 domino contributes its 0-end to the fourth cell (with 2 extending right). The 1 at the corner feeds sum-2 at the top-left pair — forcing 1 below it from the 2–1 domino, whose 2-end then drives the lower sum-2 to 0 (from the 6–0 domino). On the right: the 6–1 domino runs vertically with 6 at row 3 and 1 at row 2, providing the 6 for the sum-9 pair — matched by 3 from the 6–3 domino at row 4. The 3–4 domino fills the top-right corner (3 at row 0, 4 at row 1) confirming both sum-5 pairs. The 3–3 double fills the central equals pair. Bottom row: 6–4 and 4–2 each place a 4 at columns 0–1 (4+4=8 ✓), and 4–2's 2 plus 2–6's 2 at columns 2–3 (2+2=4 ✓).
🎨 Pips Solver
Apr 19, 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: The single-cell sum is the forced entry
Locate the constraint that covers exactly one cell. Its sum target is 1, which means the pip placed on that cell must equal exactly 1. No other value is possible — a one-cell region with sum 1 is fully determined before you know anything else about the board.
2
Step 2: The 4–1 domino is the only candidate
Only one piece in the set has a 1-pip end: the 4–1 domino. It must land here, with its 1-end in the sum-1 cell at row 1, column 3, and its 4-end extending downward to row 2, column 3. No other orientation or domino can satisfy this constraint.
3
Step 3: The equals pair in column 2 resolves by symmetry
Two cells in column 2 — rows 0 and 1 — must show equal pips. The 3–3 double is the natural fit: both ends are identical, so the constraint is satisfied regardless of orientation. Place it vertically in column 2, rows 0–1.
4
Step 4: Trace the equals chain down from the 4
The 4-end of the 4–1 domino sits at row 2, column 3. The region covering rows 2–3 in column 3 requires those two cells to be equal — so row 3, column 3 must also be 4. The 4–2 domino carries a 4: place it horizontal at row 3, columns 2–3, with 4 on the right and 2 on the left. Row 3 column 2 is now 2.
5
Step 5: The final two equals regions close the grid
Rows 3, columns 1–2 must be equal. Row 3, column 2 = 2, so row 3, column 1 = 2 as well. The 2–0 domino runs vertically at column 1, rows 2–3: 2 at row 3, 0 at row 2. That sets column 1, row 2 = 0. The equals pair at rows 1–2, column 1 needs both cells equal: row 2, column 1 = 0, so row 1, column 1 = 0. The 6–0 finishes the grid, running horizontal at row 1: 6 at column 0 (an empty-type cell with no constraint), 0 at column 1 ✓.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Target the three-cell equals region
Three cells at the bottom of the grid must all show the same pip value. This is far more restrictive than a standard two-cell equals pair. Before working any sum arithmetic, ask: which pip value can appear three times simultaneously, given the dominoes in the set?
2
Step 2: The 6–6 double anchors at value 6
The only double in the set with high identical pips is 6–6. Place it horizontal across the bottom two cells (row 3, columns 0–1) — both equal 6. The third equals cell (row 3, column 2) must also be 6. Find a domino with a 6-end that can reach column 2 from above.
3
Step 3: 5–6 runs vertically at column 2
The 5–6 domino covers the third equals cell with its 6-end at row 3 and its 5-end descending to row 4. The row 4 cell carries a greater-than-1 constraint: 5 > 1 ✓. The three-cell equals region is fully resolved at value 6.
4
Step 4: Sum-7 pairs resolve in order from top down
Row 0, column 1 and row 1, column 1 must sum to 7. The 0–3 domino runs horizontal at row 0 with 3 at column 1 and 0 at column 2, contributing 3. So row 1, column 1 must be 4. The 4–5 domino fits horizontal at row 1: 4 at column 1, 5 at column 0. That also locks the equals pair at column 0, rows 1–2: row 2, column 0 must be 5, filled by the 5-end of the 1–5 domino (5 at column 0, 1 at column 1).
5
Step 5: Center sum-7 and the final top pieces
Row 2, columns 1–2 must sum to 7. Row 2, column 1 = 1 (from 1–5). So row 2, column 2 = 6: the 1–6 domino runs vertically at column 2, rows 1–2 (1 at row 1, 6 at row 2), giving 1+6=7 ✓. The less-than-7 pair at row 1, columns 2–3: 1+5=6 < 7 ✓. The 0–5 domino fills the top-right area (0 at row 0 column 3, 5 at row 1 column 3), and 0 at row 0 column 2 (from 0–3) matches 0 at row 0 column 3, confirming the top equals pair ✓.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Locate the four-cell less-than-2 region
A region covering four cells — at rows 0–2, column 1–2 area — has a sum ceiling of less than 2. The total of all four pips must be 0 or 1. Since even a single 2-pip would push the sum to at least 2 if accompanied by anything nonzero, the only viable solution is all four cells equal 0. These four positions are your forced starting pips.
2
Step 2: Three dominoes fill the four zero cells
The 0–0 double covers two of those cells vertically (rows 1–2, column 2) — both 0. The 0–1 domino covers row 0: place it with 0 at column 1 and 1 at column 0. The 2–0 domino covers column 2, row 0: place it with 0 at column 2 and 2 extending right to column 3. All four cells in the less-than-2 region are now 0 ✓.
3
Step 3: Propagate through the sum-2 column constraints
Row 0, column 0 and row 1, column 0 must sum to 2. Row 0, column 0 = 1 (from the 0–1 domino). So row 1, column 0 = 1, filled by the 2–1 domino running vertically at column 0, rows 1–2 (1 at row 1, 2 at row 2). Next, rows 2–3 at column 0 must also sum to 2: row 2, column 0 = 2 (from 2–1). So row 3, column 0 = 0, filled by the 6–0 domino running vertically at column 0, rows 3–4 (0 at row 3, 6 at row 4).
4
Step 4: Sum-5 pairs resolve on the right side
Row 0, columns 3–4 must sum to 5. Row 0, column 3 = 2 (from the 2–0 domino). So row 0, column 4 = 3, from the 3–4 domino running vertically at column 4, rows 0–1 (3 at row 0, 4 at row 1). Then rows 1–2 at column 4 must also sum to 5: row 1, column 4 = 4 (just placed). So row 2, column 4 = 1, from the 6–1 domino running vertically at column 4, rows 2–3 (1 at row 2, 6 at row 3).
5
Step 5: Sum-9 confirms and extends the right column
Rows 3–4 at column 4 must sum to 9 — the maximum achievable. Row 3, column 4 = 6 (from 6–1 already placed). So row 4, column 4 = 3, from the 6–3 domino running vertically at column 4, rows 4–5 (3 at row 4, 6 at row 5). 6+3=9 ✓.
6
Step 6: The central equals pair and bottom row close the grid
Rows 3–4 at column 2 must show equal pips. The 3–3 double fits vertically here — both cells equal 3, no arithmetic required ✓. Finally, the bottom row: rows 6, columns 0–1 must sum to 8. The 6–4 domino at column 0, rows 5–6 places 4 at row 6. The 4–2 domino runs horizontal at row 6, columns 1–2 (4 at column 1, 2 at column 2): 4+4=8 ✓. Columns 2–3 at row 6 must sum to 4: row 6, column 2 = 2. The 2–6 domino runs horizontal at row 6, columns 3–4 (2 at column 3, 6 at column 4): 2+2=4 ✓. The 6 at column 4 carries an empty-type constraint — no restriction ✓.
💡 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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