NYT Pips Hints & Answers for August 12, 2026

Aug 12, 2026

🚨 SPOILER WARNING

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.

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🎲 Today's Puzzle Overview

Today's NYT Pips easy puzzle by Ian Livengood unfolds on a compact 3x4 grid with a clear deduction graph. The chain begins at a less-2 singleton at [2,0], which forces the sole zero-bearing domino [0,3] to lock in vertically; that immediately seeds the sum-5 region on the top edge. From there, the sum-11 anchor at [0,1]-[1,1] dictates a 6-5 split, directing the remaining dominos into a tidy cascade. The empty cell at [1,0] functions as a neutral bridge, and the final sum-6 and sum-5 regions click into place with no ambiguity. Livengood's medium 3x5 grid weaves equals regions with greater-than columns. A solitary sum-4 cell at [2,1] forces a 4-2 domino, which sets a three-cell equals zone to all 2s. That equality then ripples upward, forcing the top-row equals to all 1s via the [2,1] and [3,1] dominos. Two greater-than columns on the right are resolved by high-pip dominos [5,4] and [3,6], with every placement flowing from the central sum-4 hub. Rodolfo Kurchan's hard grid expands to 9x9 and is dominated by zero-heavy equals regions. A solo sum-3 at [0,1] forces a 3-0 domino, which injects a zero into a three-cell vertical equals column. That zero then propagates into a second equal-sized region at [2,3]-[3,4], setting off a domino effect of zero placements. The puzzle's core is a long equals column of 3s running from [5,5] to [8,5], which intertwines with sum-14 and sum-12 targets on the right and bottom edges. The architecture is a radial deduction web: single cells with less/greater constraints sculpt the low end, while high-pip dominos [5,5], [5,6], [6,6] resolve the distant corners.

💡 Progressive Hints

Try these hints one at a time. Each hint becomes more specific to help you solve it yourself!

💡 Hint 1: Spot the restrictive single cell
There is a cell with a less-than-2 constraint—it can only hold a 0 or a 1. Scour the available dominos and see which one can deliver that tiny pip.
💡 Hint 2: Follow the zero to the top
That forced domino also places a 3 in the empty cell above it. Now look at the sum-11 region at [0,1] and [1,1]; the required total demands a 6 and a 5. Consider which two dominos carry those values and where their partners must land.
💡 Hint 3: The full unraveling
Place the [0,3] domino vertically at [2,0]-[1,0] (0 below, 3 above). Then the [2,6] domino sits horizontally at [0,1]-[0,2] with 6 left and 2 right; the [3,3] domino goes vertically at [0,3]-[1,3] (both 3s). Finally, the [4,5] domino runs vertically at [2,1]-[1,1] (4 below, 5 above) and the [2,2] domino finishes horizontally at [2,2]-[2,3] (2 and 2).
💡 Hint 1: Find the equality chain
A region forces all its cells to share the same pip value—it's a horizontal stripe of three cells on the top row. That equality will lock in a low number early on.
💡 Hint 2: Listen to the lone sum-4
The singleton sum-4 at the bottom left, [2,1], demands an exact 4. That 4 must pair with a 2, which then lands in the equals region directly to its right, forcing that whole zone to 2. This triggers a domino that slides a 1 upward.
💡 Hint 3: The complete medium chain
Place the [4,2] domino horizontally at [2,1]-[2,2] (4 left, 2 right). The equals region forces [1,1]=2 and [1,2]=2, so the [2,1] domino goes vertically at [1,1]-[0,1] (2 below, 1 above) and the [5,2] domino horizontally at [1,2]-[1,3] (2 left, 5 right). For the top row, [3,1] runs vertically at [1,0]-[0,0] (3 below, 1 above), and [1,4] horizontally at [0,2]-[0,3] (1 left, 4 right). The right-side columns are resolved by [5,4] vertically at [2,3]-[2,4] (5 above, 4 below) and [3,6] vertically at [0,4]-[1,4] (3 above, 6 below).
💡 Hint 1: Target the lone sum-3
Search for a single-cell region that must sum to 3—it will force a domino containing a 3 onto the board, and its partner will seed a cascade in the column directly below.
💡 Hint 2: Lock the zero column
The equals region spanning [1,1], [2,1], and [3,1] must all hold the same value. With the abundance of zero-dominos, aim to fill that column entirely with zeros; the [0,3] and [0,6] dominos are your tools.
💡 Hint 3: Expand the zero web
A second equals region at [2,3], [3,3], and [3,4] also needs three equal cells. The double-zero domino [0,0] can cover two of them, while another zero from a different domino completes the trio. Meanwhile, a sum-5 singleton awaits at [2,4].
💡 Hint 4: Erect the 3‑column
A long vertical equals region from [5,5] down to [8,5] demands four identical pips—they will all be 3. Use the [3,3] domino to anchor it, then weave in the [3,4] and [3,5] dominos to fill the rest. Don't forget the neighboring equals-4 region at [7,3]-[8,4].
💡 Hint 5: Full hard solution
Place [0,3] vertically at [1,1]-[0,1] (0 below, 3 above). [0,6] goes horizontally at [2,1]-[2,0] (0 left, 6 right); [0,1] vertically at [3,1]-[3,0] (0 above, 1 below). [0,0] horizontally at [3,3]-[3,4] (0,0); [0,5] horizontally at [2,3]-[2,4] (0 left, 5 right). Equals-3 column: [3,3] vertically at [5,5]-[6,5]; [3,4] horizontally at [8,5]-[8,4] (3 left, 4 right); [3,5] horizontally at [7,5]-[7,4] (3 left, 5 right); [4,4] vertically at [7,3]-[8,3] (4,4). Right side: [5,5] vertically at [1,6]-[2,6] (5,5); [1,6] horizontally at [3,7]-[3,6] (1 left, 6 right). Bottom left: [1,2] vertically at [7,0]-[6,0] (1 above, 2 below); [5,6] horizontally at [8,0]-[8,1] (5,6). Top right: [6,6] horizontally at [6,7]-[6,8] (6,6); [0,2] vertically at [8,7]-[7,7] (0 above, 2 below). Finally, [1,4] horizontally at [5,3]-[5,4] (1 left, 4 right).

🎨 Pips Solver

Aug 12, 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

Pips hint for August 12, 2026 – hard level puzzle grid with critical first placements and strategy

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

NYT Pips August 12, 2026 hard puzzle full solution grid showing final answer with hints

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: Exploit the less‑2 cell at [2,0]
The less-2 constraint means [2,0] can only be 0 or 1. Of the five available dominos, only [0,3] contains a 0. Therefore [2,0] must be 0, and its partner 3 must occupy the only adjacent cell above, [1,0]. Place [0,3] vertically at [2,0]-[1,0]; this also sets the empty cell [1,0]=3.
2
Step 2: Resolve the top‑edge sum‑5
The region [0,2]-[0,3] requires a sum of 5. With the only 0 already used, [0,3] needs a 3 (from the [3,3] domino) and [0,2] must be a 2 (from [2,6]). Place [3,3] vertically at [0,3]-[1,3] (both 3s) and [2,6] horizontally at [0,1]-[0,2] with 6 at [0,1] and 2 at [0,2]. The sum-11 region now has its 6.
3
Step 3: Satisfy the sum‑11 anchor
The sum-11 region [0,1]-[1,1] already has [0,1]=6, so [1,1] must be 5. The only remaining domino with a 5 is [4,5]; its partner 4 must go to the adjacent cell below, [2,1]. Place [4,5] vertically at [2,1]-[1,1] with 4 below and 5 above.
4
Step 4: Complete the right column and bottom sum‑6
The sum-5 region [1,3]-[2,3] has [1,3]=3, so [2,3] must be 2. The unused [2,2] domino provides that 2; place it horizontally at [2,2]-[2,3] (both 2s). Now [2,1]=4 and [2,2]=2 sum to 6, satisfying the sum-6 constraint. The grid is finished.

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Anchor on the sum‑4 singleton
The single-cell region [2,1] demands a pip of exactly 4. Among dominos with a 4, [4,2] is ideal: placing it horizontally at [2,1]-[2,2] sets [2,1]=4 and [2,2]=2. This 2 immediately forces the adjacent equals region [1,1],[1,2],[2,2] to all become 2.
2
Step 2: Equalize the middle band
With [2,2]=2, the equals region forces [1,1]=2 and [1,2]=2. To fill these, place the [2,1] domino vertically at [1,1]-[0,1] (2 below, 1 above) and the [5,2] domino horizontally at [1,2]-[1,3] (2 left, 5 right). The [0,1] now holds a 1, seeding the top‑row equality.
3
Step 3: Satisfy the top‑row equals
The region [0,0],[0,1],[0,2] must all be equal; [0,1]=1 forces [0,0]=1 and [0,2]=1. Use the [3,1] domino vertically at [1,0]-[0,0] (3 below, 1 above) to cover [0,0]. Then the [1,4] domino horizontally at [0,2]-[0,3] puts 1 at [0,2] and 4 at [0,3].
4
Step 4: Resolve the right‑side greater‑than columns
The column [1,3]-[2,3] already has [1,3]=5; place the [5,4] domino vertically at [2,3]-[2,4] (5 above, 4 below) to satisfy the constraint. For [0,3]-[0,4], [0,3]=4 is set; place the [3,6] domino vertically at [0,4]-[1,4] (3 above, 6 below) to complete the column— and [1,4]=6 meets its greater‑than rule.
5
Step 5: Verify all constraints
Every region now checks out: the equals bands are uniform, the sum‑4 singleton holds, the greater‑than columns are above their thresholds, and all dominos are placed without conflict.

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Fire the sum‑3 and zero‑column
The singleton sum‑3 at [0,1] forces [0,1]=3. The only domino that can pair a 3 with a zero is [0,3]; place it vertically at [1,1]-[0,1] (0 below, 3 above). This injects a 0 into the equals region [1,1],[2,1],[3,1], forcing all three cells to 0. To fill them, place [0,6] horizontally at [2,1]-[2,0] (0 left, 6 right)—the 6 satisfies the greater‑3 at [2,0]. Then place [0,1] vertically at [3,1]-[3,0] (0 above, 1 below); the 1 meets the less‑3 at [3,0].
2
Step 2: Build the second zero‑equals block
The equals region [2,3],[3,3],[3,4] must be uniform. The [0,0] domino covers two cells: place it horizontally at [3,3]-[3,4] (both 0). The third cell [2,3] gets its 0 from [0,5] placed horizontally at [2,3]-[2,4] (0 left, 5 right). The 5 at [2,4] now satisfies the sum‑5 singleton.
3
Step 3: Assemble the sum‑16 trio
The region [1,6],[2,6],[3,6] must sum to 16. With [0,1]... already placed, the only workable combination is 5+5+6. Place [5,5] vertically at [1,6]-[2,6] (both 5s) and [1,6] horizontally at [3,7]-[3,6] (1 left, 6 right). The 1 at [3,7] satisfies its greater‑0 constraint.
4
Step 4: Erect the central 3‑column
The long equals region [5,5],[6,5],[7,5],[8,5] demands all 3s. Anchor it with [3,3] vertically at [5,5]-[6,5] (both 3). The remaining two cells need 3s: place [3,4] horizontally at [8,5]-[8,4] (3 left, 4 right) and [3,5] horizontally at [7,5]-[7,4] (3 left, 5 right). This feeds the equals-4 region [7,3],[8,3],[8,4]: [8,4]=4, [7,3] and [8,3] become 4 via [4,4] vertically at [7,3]-[8,3] (both 4). The greater‑3 at [7,4] is met by the 5.
5
Step 5: Solve the bottom‑left sum‑12
The region [7,0],[8,0],[8,1] sums to 12. We still have [1,2] (1,2) and [5,6] (5,6). Place [1,2] vertically at [7,0]-[6,0] (1 above, 2 below), with the 2 satisfying greater‑0 at [6,0]. Then place [5,6] horizontally at [8,0]-[8,1] (5,6) to complete the sum 1+5+6=12.
6
Step 6: Finish top‑right and remaining regions
The sum‑14 region [6,7],[6,8],[7,7] needs 6+6+2. Place [6,6] horizontally at [6,7]-[6,8] (both 6) and [0,2] vertically at [8,7]-[7,7] (0 above, 2 below); the 0 at [8,7] is empty, and 2 at [7,7] completes 14. Finally, the less‑3 at [5,3] and greater‑3 at [5,4] are handled by [1,4] horizontally at [5,3]-[5,4] (1 left, 4 right). All constraints lock.

💡 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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