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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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Want hints instead? Scroll down for progressive clues that won't spoil the fun.
🎲 Today's Puzzle Overview
Today's Pips for March 30, 2026, edited by Ian Livengood, opens with a pyramid-shaped Easy grid: six dominoes arranged across three tiers, widest at the bottom and narrowing to just two cells at the peak. That shape isn't just decorative — it concentrates the most constraining region at the base, where a single sum rule immediately tells you which domino belongs there.
Livengood constructed both the Easy and Medium grids himself. The Medium board is shaped like a picture frame: a full top row, a full bottom row, and two short vertical spines connecting them at the sides, leaving the interior open. This layout isolates the spine cells in a way that makes their constraints feel especially decisive — solve the spines, and the top and bottom rows unravel in sequence. The Hard puzzle comes from constructor Rodolfo Kurchan, whose board uses 14 dominoes across a 7×6 boundary, with large stretches intentionally left empty to give each region more leverage.
What ties today's three puzzles together is the value of commitment. In each grid, once you identify the one move that is truly forced — not just plausible — the board begins to close itself. If you find yourself weighing two options simultaneously, there's almost always a prior constraint that eliminates one of them. Today is a good day to practice trusting the logic.
💡 Progressive Hints
Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!
💡 Start with the most restrictive sum
One region in the bottom row has a sum target high enough that only a single domino in today's set can satisfy it. That's your entry point — no guessing required.
💡 A sum of 10 across two cells
The two-cell region on the far left of the bottom row must add up to 10. Think about which domino from today's six is the only one that produces that exact total across a pair of pips.
💡 Full solution
The bottom-left pair (sum 10) must hold [5|5] — it's the only domino that reaches that total. With the left side of the bottom row settled, look at the three cells stacked vertically through the center: they sum to 13, which means 6+4+3 from top to bottom. [6|0] goes at the top with its 6 in the column and 0 pointing right; [4|4] fills the middle row with one 4 in the column and the other extending left; [3|1] sits in the bottom row with 3 in the column and 1 extending right. The two-cell equals region at the top of the pyramid forces both cells to show 0 — [1|0] places its 0 there to match the 0 already in place, with its 1 extending right into the four-cell low-sum region. [1|1] fills the last two bottom-right cells, and the four-cell sum of 4 is complete: 1+1+1+1.
💡 Two cells demand a high pip
Two isolated cells on opposite sides of the board both carry a constraint that rules out most of today's dominoes. Identify what those cells need, then see which dominoes in the set can actually deliver it.
💡 Only two dominoes qualify
Both isolated cells require a pip value above 4. Scan the eight dominoes: only two contain a pip that high. Each claims one of those cells — but which goes where? Look at the lower-right corner to find the answer.
💡 Full solution
The left-spine lower cell (greater than 4) takes [6|0] with its 6 there and 0 rising to the cell above. The right-spine upper cell (greater than 4) takes [6|2] with its 6 there and 2 falling below — this is forced because [6|0] in the right spine would leave a 0 at its lower end that can't be tripled in the bottom-right equals region without a [0|0] tile (which isn't in today's set). With [6|2] in the right spine, [2|2] fills the bottom-right equals trio. Working the top row right to left: the far-right less-than-3 cell takes [1|2]'s 1-end (1 < 3 ✓), its 2 landing in the next cell; the equals pair to its left needs 2 as well, so [2|0] sits there with 2 at the inner cell and 0 pointing outward; the next equals pair copies that 0, placing [0|3] with 0 at the inner cell and 3 at the far-left top corner (3 < 4 ✓). The bottom row closes with [2|3] placing its 2 at the left single-cell (sum 2 ✓) and its 3 beside it, then [4|1] delivering 3+4=7 for the sum-of-7 pair.
💡 Two regions need very high sums
Two separate two-cell regions both carry a sum constraint with a target above 8. Before placing anything, consider which domino pairs could possibly produce a sum that large.
💡 The all-zero region is your real starting point
There is one region in today's Hard grid where every pip must be exactly 0. That constraint is far more restrictive than the greater-than constraints — start there, not with the high-sum pairs.
💡 Chain outward from the zeros
Once the zero region is filled, its neighbors each have tight sum constraints that become immediately solvable. Follow the chain: zero region, then the adjacent pair, then the column below, then the bottom edge, and up the left side.
💡 The left side solves bottom-up, the right side top-down
After locking the left column chain from bottom to top, the top row's sum-of-3 across four cells becomes solvable. From there the right spine drops into place, and the three-cell equals column closes the grid.
💡 Full solution
Start with the sum=0 region (three cells, lower center): [0|0] fills the bottom two, [0|1] places its 0 in the third cell with its 1 rising upward. That 1 is now the right cell of a sum=3 pair — its partner needs 2, so [2|2] goes there vertically, placing a second 2 below it. The sum=2 column below takes that 2 plus a 0 from [4|0], whose 4 lands to the left in the bottom row. The sum=6 pair in the bottom-left corner now has 4, needing 2 — [5|2] delivers with 2 at the bottom and 5 rising upward. That 5 feeds the sum=9 column above, forcing the upper cell to be 4 — [1|4] places its 4 there and its 1 one step higher. The sum=2 pair above forces the lower cell to be 1, so [1|6] goes vertically with 1 at the bottom and 6 at the top. Now the greater-than-8 left pair has 6 at its lower cell; its upper cell gets 5 from [1|5] (5+6=11 ✓), with that domino's 1 starting the top row. The sum=3 top-row region of four cells then resolves: 1 is already there, [1|1] fills the next two with 1+1, and [2|0] places its 0 at the fourth cell and its 2 at the far right. The sum=5 right-column pair has 2 at top, so the lower cell needs 3 — [3|6] goes there with 3 at top and 6 below. That 6 anchors the greater-than-8 right pair; [4|5] places 5 below it (6+5=11 ✓) and 4 in the bottom-right corner. The three-cell equals column below must all show 4 — [4|4] fills the remaining two cells. Finally, [5|3] closes the sum=8 interior pair with 5 and 3.
🎨 Pips Solver
Mar 30, 2026
Click a domino to place it on the board. You can also click the board, and the correct domino will appear.
✅ 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 bottom-left sum forces the first domino
Two cells on the far left of the bottom row carry a sum constraint of 10. Look through the six available dominoes and ask which one — as a pair of pips — adds up to exactly 10. Only [5|5] does. Place it horizontally across those two cells. The left side of the bottom row is now fixed.
2
Step 2: The central column sum identifies three more
Three cells run vertically through the center of the pyramid — one in each row — and their pips must sum to 13. One of those cells (the bottom one) is already adjacent to the placed [5|5], but belongs to a different domino. Which three pip values from today's remaining dominoes can reach 13? The combination is 6+4+3. Place [6|0] at the top of this column with its 6 pointing into the column and its 0 extending right. Place [4|4] in the middle row with one 4 in the column and the other extending left. Place [3|1] in the bottom row with its 3 at the column cell and its 1 extending right.
3
Step 3: The equals region at the top forces the next placement
The pyramid's top tier holds just two cells under an equals constraint — both must show the same pip. The [6|0] domino already placed its 0 at the right cell of that pair (top row, right position). So the left cell must also show 0. [1|0] delivers this: place it with its 0 at the inner top-row position and its 1 extending right and downward into the four-cell region below.
4
Step 4: The four-cell low-sum region closes the puzzle
The four-cell region sums to 4. Three of its cells are already determined by the dominoes placed so far — they each show a pip of 1 (from the tails of [3|1], [1|0], and the slot waiting for [1|1]). One cell remains: the bottom-right corner pair. [1|1] is the only domino left, and placing it there gives 1+1+1+1=4. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Medium Level
1
Step 1: Greater-than constraints identify the two six-pip dominoes
Two single cells — one at the lower end of the left column, one at the upper end of the right column — each require a pip value greater than 4. Scanning today's eight dominoes, only [6|0] and [6|2] carry a pip above 4. Each gets exactly one of those spine cells. The question is which goes where.
2
Step 2: The lower-right equals trio settles the assignment
The lower-right corner region requires three cells to all show the same pip: the bottom of the right column plus the right-end pair of the bottom row. If [6|0] occupied the right column, its lower cell would show 0, and you'd need two more 0s in the bottom-right corner — but no [0|0] domino exists in today's set. So [6|2] must go in the right column, with its 2 at the lower cell, and [6|0] goes in the left column, with its 6 at the lower cell and its 0 rising to the upper cell.
3
Step 3: The right-column placement locks the bottom-right corner
The right column's lower cell now shows 2. The three-cell equals region connecting it to the bottom row's rightmost pair must all be 2. Place [2|2] horizontally across those two bottom-right cells — both showing 2 — completing the equals trio.
4
Step 4: The top row resolves right to left
The far-right cell of the top row must be less than 3. Of the remaining dominoes, only [1|2] can place a value below 3 there — put its 1 at the far right and its 2 in the adjacent cell. Those two cells are also part of an equals pair, but [1|2]'s 2 lands one cell to the left of the equals boundary. The actual equals pair is the next two cells: the one just filled (showing 2 from [1|2]) and its neighbor. That neighbor must also be 2, so [2|0] goes there with 2 pointing right and 0 pointing left.
5
Step 5: The leftmost top-row equals pair and the less-than constraint finish the top
The far-left pair of the top row has an equals constraint: both cells in it must match. The inner cell now shows 0 (from [2|0]'s left end), so the outer cell must also be 0. Place [0|3] with its 0 at the inner position and its 3 at the far-left corner. The less-than-4 constraint covering the far-left top cell and the left column's upper cell checks out: 3 < 4 and 0 < 4.
6
Step 6: The bottom row closes with two sum constraints
Three dominoes remain for the six bottom-row cells. The far-left single-cell constraint requires a pip of exactly 2 — place [2|3] with its 2 there and its 3 one step right. The sum-of-7 constraint on the next two cells has 3 as its left value; it needs a right value of 4. Place [4|1] with its 4 satisfying the sum and its 1 filling the last cell. Puzzle complete.
🔧 Step-by-Step Answer Walkthrough For Hard Level
1
Step 1: The sum=0 region forces three placements
Three cells in the lower-center area must all show 0 — their combined sum is 0, and no individual pip can be negative. The [0|0] double-blank fills two of them in one tile. The third cell has a 0 forced on it too, and the [0|1] domino satisfies this: place its 0 in that third cell, with its 1 rising upward to the cell above.
2
Step 2: The sum=3 pair above chains from the zero region
The 1 placed by [0|1] now sits at the right cell of a two-cell sum=3 region. Its partner to the left must show 2. The [2|2] domino fills that spot — one 2 at the constrained cell, the other extending downward into the column below.
3
Step 3: The sum=2 column below locks the next domino
Two cells stack vertically in this column: the lower 2 from [2|2] is the upper one, and the cell below it needs the total to reach 2. So the lower cell must be 0. The [4|0] domino places its 0 there and its 4 to the left in the bottom row.
4
Step 4: The sum=6 bottom pair and the left column chain upward
The two-cell constraint at the bottom-left of the board sums to 6. The right cell of that pair now shows 4 (from [4|0]), so the left cell must be 2. The [5|2] domino delivers: its 2 at the bottom-left, its 5 rising upward to the cell above.
5
Step 5: Sum=9 and sum=2 complete the left column
The two-cell column above the [5|2] placement must sum to 9. Its lower cell shows 5, so the upper cell must be 4. The [1|4] domino fits vertically — 4 at the lower cell, 1 at the cell above it. That 1 is now the lower cell of a sum=2 pair in the left interior, forcing the cell above it to be 1 as well. The [1|6] domino places its 1 there and its 6 one step higher.
6
Step 6: The greater-than-8 left pair and the top-row anchor
The two-cell column at the very top-left has a combined sum greater than 8. Its lower cell now shows 6 (from [1|6]). For 6 plus something to exceed 8, the upper cell needs at least 3 — and in practice the only remaining domino that fits is [1|5], which places its 5 at the top cell (5+6=11 > 8 ✓) and its 1 one step right along the top row.
7
Step 7: The four-cell top-row sum resolves with two dominoes
Four cells across the top row — including the 1 just placed by [1|5] — must sum to 3. That 1 accounts for one pip; the remaining three cells must contribute 2 more. The [1|1] domino fills the next two cells (1+1=2), and the [2|0] domino places its 0 in the fourth cell (total: 1+1+1+0=3 ✓), with its 2 extending right to the far end of the top row.
8
Step 8: The right-column chain drops from the top row
The far-right top cell shows 2 (from [2|0]). The sum=5 constraint on that cell and the one below it means the lower cell needs 3. The [3|6] domino fits vertically — 3 at the top, 6 dropping below. That 6 is now the upper cell of the greater-than-8 right pair.
9
Step 9: The right-side chain closes through the equals column
The greater-than-8 right pair has 6 at its top cell; its lower cell needs at least 3 for the sum to exceed 8. The [4|5] domino places its 5 there (6+5=11 ✓) and its 4 in the cell below that — which is the top of the three-cell equals column. All three cells in that column must match, so each must be 4. The [4|4] double-four fills the remaining two cells of the equals column.
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Step 10: The interior sum=8 pair closes the puzzle
One domino remains: [5|3]. One region remains: a two-cell horizontal pair in the interior that must sum to 8. Place [5|3] with its 5 on the right cell and its 3 on the left — 3+5=8. 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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