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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 grid hands you two freebies before you've even started thinking — two sum cells spell out their values on sight, and once those dominoes land, each subsequent constraint just unlocks the next one. You'll reach the end before you realize you never had to pause.
The medium feels more spread out. The same two-anchor opening is here, but the anchors sit on opposite ends of the grid and set off separate chains that don't meet until the final few cells. There's a moment midway where you need to think about what values are still available — Ian keeps it honest but asks you to stay organized.
Rodolfo Kurchan's hard grid opens with two forced cells that feel deceptively easy, but they kick off long chains across a sprawling irregular layout. The real challenge is the unequal region at the center — four cells that must all differ, fed from multiple directions. You'll need to track three or four values at once before that region resolves. Methodical solvers will get there; it's a satisfying close.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Two free anchors
Two cells in this grid have sum constraints that force their values immediately — no cross-checking needed. Find both and you've got solid footholds to build from.
💡 Top-right and mid-bottom
One forced cell is in the top-right corner of the grid; the other is near the bottom of the second column. Once both dominoes land, an equals pair in the middle row connects directly to one of your placed values, handing you the third domino for free.
💡 Full answer
Top-right corner has sum=2 → value 2 there. Its domino [2-3] sits vertically — 3 lands in the cell below, satisfying the less-than-4 constraint. Second column bottom has sum=2 → value 2. Its domino [6-2] sits vertically, placing 6 in the cell above. The equals pair in row 1 (cols 1–2) must match that 6, so the adjacent cell is also 6 — domino [0-6] vertical in col 2, with 0 at the top. The equals pair in col 3 (rows 1–2) can only be satisfied by [5-5], placed vertically. The remaining [3-5] fills col 0 vertically with 5 on top, satisfying the greater-than-4 constraint.
💡 Start with the zeros
Two single-cell sum constraints anchor this grid — one forces a value of zero, the other forces a small number you can spot quickly. Both give you complete domino placements with no deduction required.
💡 Opposite corners
The zero-sum cell sits near the bottom of a column; once its domino is placed, the value above it feeds directly into a two-cell sum at the top of that column — giving you a third domino. The other forced cell is on the far right of row 1; its domino runs left and triggers an equals pair that unlocks the middle of the grid.
💡 Full answer
Row 2, col 1 has sum=0 → value 0. Domino [2-0] sits vertically, placing 2 at row 1 col 1. Row 1 col 1 + row 0 col 1 must equal 7; 7−2=5, so row 0 col 1 = 5. Domino [5-1] runs horizontally leftward, placing 1 at row 0 col 0. Row 1, col 5 has sum=3 → value 3. Domino [3-5] runs left, placing 5 at row 1 col 4. The equals pair at row 1 cols 3–4 forces row 1 col 3 = 5 as well; domino [5-4] vertical places 4 at row 2 col 3. The equals pair at row 2 cols 2–3 forces row 2 col 2 = 4; domino [4-6] vertical places 6 at row 3 col 2, satisfying the greater-than-4 constraint. The equals pair at row 0 cols 2 and row 1 col 2 forces both = 3 — domino [3-3] vertical. The only domino left is [0-0], which fills the top-right two cells and satisfies the less-than-2 constraint with a sum of zero.
💡 Two corners, two gifts
Two single-cell sum constraints in the left portion of the grid hand you complete domino placements immediately. Find both and you have two starting points on opposite sides of a long chain.
💡 Top-left and mid-left
One forced cell is at the very top-left corner; the other is several rows down in the leftmost column. Each anchors a horizontal domino, and each then feeds a constraint that determines a third cell — so two placements become four quickly.
💡 The left chain opens up
Top-left has sum=5 → value 5. Domino [5-2] horizontal places 2 at col 1. The equals pair at row 0 cols 1–2 forces col 2 = 2 as well; domino [4-2] runs horizontally and places 4 at col 3. The two-cell sum at rows 0–1 col 3 must equal 5; with 4 placed at row 0, row 1 col 3 = 1. Domino [1-4] vertical places 4 at row 2 col 3. Meanwhile, the mid-left forced cell at row 3 col 0 has sum=4 → value 4; domino [4-5] horizontal places 5 at row 3 col 1.
💡 Center and right side
Row 3 cols 1–2 must sum to 8; with col 1 = 5, col 2 must be 3. Domino [3-3] horizontal places 3 at cols 2 and 3. Now the unequal region has two values known: row 2 col 3 = 4 and row 3 col 3 = 3. On the right: row 2 cols 5–6 must sum to 10. Domino [6-5] horizontal fills row 2 cols 4–5 with 6 and 5. Domino [5-1] vertical in col 6 places 5 at row 2 and 1 at row 3, making the sum 5+5=10 for cols 5–6 in row 2. Rows 3–4 in col 6 must sum to 2; with row 3 = 1, row 4 col 6 = 1. Domino [2-1] vertical places 2 at row 5 col 6 and 1 at row 4 col 6.
💡 Full answer
Unequal region (rows 2–4, col 3–4): values so far are 4 (row 2 col 3), 6 (row 2 col 4), and 3 (row 3 col 3). The last cell, row 4 col 3, must differ from all three — so not 3, 4, or 6. Domino [0-5] vertical places 0 at row 5 col 3 and 5 at row 4 col 3; 5 satisfies the unequal constraint. Row 5 cols 3–4 must sum to less than 2; with col 3 = 0, col 4 must be 0 or 1. Domino [4-0] horizontal places 4 at row 5 col 5 and 0 at col 4 — sum 0+0=0, less than 2. Row 5 cols 5–6 must sum to 6; 4+2=6, and col 6 = 2 is already placed — confirmed. Last domino [1-1] vertical fills rows 1–2 col 0; sum 1+1=2, satisfying less-than-3.
🎨 Pips Solver
Apr 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 forced anchors
The cell at top-right (row 0, col 4) has a sum=2 constraint — it can only hold the value 2. Separately, the cell at row 2, col 1 also has sum=2, so it holds value 2 as well. Both of these are single-cell constraints, so no cross-checking is needed; the values are immediately determined. You now know where two of the five dominoes go.
2
Step 2: Place both anchor dominoes
The cell at row 0 col 4 is value 2. This is part of domino [2-3]. The grid only allows it to sit vertically, placing 2 at top and 3 in the cell below (row 1 col 4). That 3 satisfies the less-than-4 constraint on row 1 col 4. The cell at row 2 col 1 is value 2, part of domino [6-2]. It sits vertically with 6 above at row 1 col 1, and 2 below at row 2 col 1.
3
Step 3: Equals pair in row 1 resolves col 2
The equals constraint says row 1 col 1 and row 1 col 2 must have the same value. Row 1 col 1 is now 6 (from the previous step), so row 1 col 2 must also be 6. The only unplaced domino with a 6 half is [0-6]. It sits vertically in col 2 with 0 at the top (row 0 col 2) and 6 at row 1 col 2.
4
Step 4: Equals pair in col 3 resolves [5-5]
The equals constraint on rows 1–2 in col 3 means both cells must have the same value. Looking at the remaining unplaced dominoes — [3-5] and [5-5] — only [5-5] has two equal halves. It must go here, sitting vertically with 5 at both row 1 col 3 and row 2 col 3.
5
Step 5: Last domino closes the puzzle
Only [3-5] remains. It must fill col 0 vertically: row 0 col 0 and row 1 col 0. The greater-than-4 constraint on row 0 col 0 means it must hold a value above 4 — either 5 or 6. With 5 and 3 as the two halves of [3-5], placing 5 at row 0 and 3 at row 1 satisfies the constraint. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Two forced single cells
Row 2, col 1 has a sum=0 constraint — the only pip value that equals zero is 0, so that cell is determined immediately. Separately, row 1, col 5 has sum=3, placing the value 3 there. Two anchors, no deduction required.
2
Step 2: Place the zero domino
Cell row 2 col 1 = 0 is part of domino [2-0]. The grid forces it to sit vertically: 2 at row 1 col 1, 0 at row 2 col 1. That 0 satisfies the sum=0 region. Row 1 col 1 is now known to be 2.
3
Step 3: Sum constraint resolves the top-left
The two-cell sum region covering row 0 col 1 and row 1 col 1 must equal 7. Row 1 col 1 = 2, so row 0 col 1 must be 5. Domino [5-1] sits horizontally, placing 5 at row 0 col 1 and 1 at row 0 col 0.
4
Step 4: Right-side anchor triggers an equals chain
Row 1 col 5 = 3 is part of domino [3-5]. It sits horizontally leftward, placing 5 at row 1 col 4. The equals pair at row 1 cols 3–4 forces row 1 col 3 = 5 as well. That 5 is part of domino [5-4], which sits vertically, placing 4 at row 2 col 3.
5
Step 5: Equals pair in row 2 resolves the bottom
The equals region covering row 2 cols 2–3 forces row 2 col 2 to also be 4. Domino [4-6] sits vertically, placing 4 at row 2 col 2 and 6 at row 3 col 2. That 6 satisfies the greater-than-4 constraint on row 3 col 2.
6
Step 6: Last two dominoes fall into place
The equals pair at row 0 col 2 and row 1 col 2 forces both to the same value. The only remaining domino with two equal halves is [3-3], so both cells = 3. Finally, domino [0-0] fills row 0 cols 3–4 horizontally — both halves are 0, and their sum is 0, which satisfies the less-than-2 constraint. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: Two forced anchors in the left column
Row 0, col 0 has a sum=5 constraint — value 5 is placed immediately. Row 3, col 0 has sum=4 — value 4 placed immediately. These two cells are independent anchors and kick off separate chains.
2
Step 2: Top-left chain — equals and sum
Cell (0,0) = 5 is part of domino [5-2], placed horizontally: (0,0)=5, (0,1)=2. The equals pair at row 0 cols 1–2 forces (0,2) = 2. Domino [4-2] sits horizontally placing 4 at (0,3) and 2 at (0,2). The two-cell sum at rows 0–1 col 3 must equal 5; with (0,3)=4, row 1 col 3 must be 1. Domino [1-4] sits vertically: (1,3)=1, (2,3)=4.
3
Step 3: Mid-left chain — sum unlocks col 2
Cell (3,0) = 4 is part of domino [4-5], placed horizontally: (3,0)=4, (3,1)=5. The two-cell sum at row 3 cols 1–2 must equal 8; with (3,1)=5, (3,2) must be 3. Domino [3-3] sits horizontally: (3,2)=3, (3,3)=3.
4
Step 4: Right side — sum=10 and the column-6 chain
The two-cell sum at row 2 cols 5–6 must equal 10 — the only way to reach 10 with pip values is 5+5 or 6+4. Domino [6-5] placed horizontally fills row 2 cols 4–5: (2,4)=6, (2,5)=5. Domino [5-1] placed vertically in col 6 fills (2,6)=5 and (3,6)=1 — making the sum 5+5=10. Rows 3–4 col 6 must sum to 2; (3,6)=1, so (4,6)=1. Domino [2-1] vertical places (5,6)=2 and (4,6)=1.
5
Step 5: Unequal region and the bottom row
The unequal region (rows 2–4, col 3–4) requires all four cells to have different values. Known so far: (2,3)=4, (2,4)=6, (3,3)=3. The fourth cell (4,3) must differ from 4, 6, and 3. Domino [0-5] placed vertically: (5,3)=0, (4,3)=5 — satisfies the unequal constraint (values: 4, 6, 3, 5, all different). Row 5 cols 3–4 must sum to less than 2; (5,3)=0, so (5,4) must be 0 or 1. Domino [4-0] horizontal places (5,5)=4 and (5,4)=0 — sum 0+0=0, confirmed less than 2. Row 5 cols 5–6 must sum to 6: 4+2=6 ✓. Last, domino [1-1] vertical fills col 0 rows 1–2: (1,0)=1, (2,0)=1 — sum 1+1=2, satisfying the less-than-3 constraint. Puzzle 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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