NYT Pips Hints & Answers for August 8, 2026

Aug 8, 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, by Ian Livengood, is a compact grid where the deduction chain starts with the top-left equals region. That region forces all its three cells to share a value, immediately confining the double‑zero domino to [0,0] and [0,1], which sets [1,1] to zero as well. From there, the sum‑6 region on row 2 picks up a forced 5 from the adjacent cell, locking the second pip to 1, and the rest of the board falls into place via a greater‑5 cell that demands 6 and a sum‑6 pairing of 4 and 2, neatly under a less‑5 cap.

Rodolfo Kurchan’s medium puzzle introduces an unequal region and an equals pair, shaping a constrained flow. A less‑2 squeeze on [3,1] eliminates most candidates, while the equals region on [1,1] and [2,1] forces both to 1. Meanwhile, a greater‑6 double‑cell region on [2,2] and [2,3] is only satisfied by a 6‑6 domino, cascading into a sum‑6 trio that demands two 2s and a 2, and a complementary sum‑6 pair of 0 and 6. The logic threads from the bottom‑up constraints, with the double‑zero domino anchoring the left side.

Kurchan’s hard scales up with a massive equals region spanning six cells, all forced to 6 by intersecting sum and greater constraints. An equals region of three cells on the top row demands a run of 5s, and a sum‑0 pair at [0,3]–[1,3] drags in zeros. The bottom row features a less‑3 cascade (cells [7,0] and [7,6] both below 3) and a sum‑11 trio that must be 5,3,3, resolved by a 3‑3 domino. The entire grid is a tightly wound graph where equals clusters propagate pip values across rows and columns, with only one viable domino sequence.

💡 Progressive Hints

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

💡 Hint 1: Spot the equalizer
Look for the equals region that covers three cells in the top‑left corner; it forces them all to share the same pip value, which drastically narrows down the domino pool.
💡 Hint 2: Zero in on the anchor
The equals region includes [0,0], [0,1], and [1,1]. Notice that a domino covering [0,0]‑[0,1] can only be [0,0] to satisfy the zero‑sum elsewhere, locking [1,1] to 0 as well.
💡 Hint 3: Full solve
Place domino [0,0] on [0,0]‑[0,1]. Then the [0,5] domino must go on [1,1]‑[2,1] with 0 on [1,1] and 5 on [2,1]. That forces the sum‑6 region at [2,1]‑[2,2] to take 1 at [2,2] from [5,1] domino, placing 5 at [2,3]. The greater‑5 cell [3,0] demands 6 from [6,2] domino, so [3,1] gets 2, and sum‑6 at [3,1]‑[3,2] completes with [4,3] (4 at [3,2], 3 at [3,3]).
💡 Hint 1: Look left
Focus on the less‑2 and equals constraints on the left side; they force a very restricted set of small pips, and the greater‑6 region only has one possible domino.
💡 Hint 2: The zero‑cell squeeze
The less‑2 cell at [3,1] can only be 0 or 1. However, the equals pair [1,1] and [2,1] ties them together, and [2,0] is less‑3. After the double‑zero domino, look at the unequal pair [0,3]‑[1,3] that must hold a 1 and a 3.
💡 Hint 3: Full solve
Domino [0,1] (0,1) goes on [2,0]‑[2,1], with 0 on [2,0] and 1 on [2,1] to satisfy equals. Domino [0,0] (0,0) on [3,1]‑[3,2] covers the less‑2 cell and the sum‑6 region. The unequal pair [1,3]‑[0,3] takes [1,3] with 1 at [1,3] and 3 at [0,3]. Domino [1,5] (1,5) on [1,1]‑[1,2], giving 1 to the equals region and 5 to greater‑3. The greater‑6 region [2,2]‑[2,3] uses [6,6] (6,6). Finally, sum‑6 region [2,4]‑[2,5]‑[3,4] takes [2,2] (2,2) on [2,4]‑[2,5] and [2,6] (2,6) on [3,4]‑[3,3] with 2 on [3,4] and 6 on [3,3] to complete sum‑6 pair.
💡 Hint 1: Cluster spotlight
The grid is dominated by large equals clusters and sum constraints. Look for the massive equals region of six cells, and the sum‑0 pair at the top, which will force zeros early.
💡 Hint 2: Zeros and fives
The sum‑0 region at [0,3] and [1,3] forces both cells to be 0, but no single domino provides two zeros; they will be placed by separate dominos that each carry a zero. Next, the three‑cell equals region at top (cells [0,1], [0,2], [1,2]) forces all to the same pip, likely a 5 given the available dominos.
💡 Hint 3: Splitting the equal 5s
The three‑cell equals region can be covered by a [5,5] domino on two cells and a [5,6] domino placing 5 on the third, leaving 6 on the greater‑3 cell at [0,0].
💡 Hint 4: The 6‑cluster and sum‑11
The massive equals region of six cells (spanning [2,2],[2,3],[3,2],[4,2],[4,3],[5,3]) all share the same pip, and the only pip that can appear six times across dominos is 6, using [6,6] and other 6‑bearing dominos. This forces [2,2] and [3,2] to be 6, dragging in the [6,6] domino there. Below, the sum‑11 region on row 7 includes three cells that must sum to 11; the only viable combination with available pips is 5,3,3, using a [3,3] domino. That also satisfies the less‑3 cell [7,6] with 0 from an adjacent domino.
💡 Hint 5: Full solve
Place [5,5] domino vertically on [0,2]‑[1,2] (both 5), and [5,6] horizontally on [0,1]‑[0,0] (5 at [0,1], 6 at [0,0] to satisfy greater‑3). For sum‑0: [0,4] on [0,3]‑[0,4] (0,4) and [0,6] on [1,3]‑[2,3] (0,6). Equals region [2,1]‑[3,1] gets 4 via [3,4] on [1,1]‑[2,1] (3,4) and later [2,4] on [4,1]‑[3,1] (2,4). The large equals cluster (cells [2,2],[2,3],[3,2],[4,2],[4,3],[5,3]) all become 6: place [6,6] on [2,2]‑[3,2]; [4,6] on [3,3]‑[4,3] (4 at [3,3] for sum‑4, 6 at [4,3]); [2,6] on [5,2]‑[4,2] (2 at [5,2], 6 at [4,2]); [1,6] on [6,3]‑[5,3] (1 at [6,3] for less‑3, 6 at [5,3]). The equals cluster on the left ([4,1],[5,1],[5,2],[6,1],[6,2],[7,1]) all become 2: use [2,2] on [5,1]‑[6,1]; [2,5] on [6,2]‑[7,2] (2,5); [2,4] on [4,1]‑[3,1] (2,4); [0,2] on [7,0]‑[7,1] (0,2). Bottom row sum‑11: [7,3]‑[7,4] take [3,3] (3,3) joining 5 at [7,2]. Finally [2,3] on [5,6]‑[4,6] (2 at [5,6], 3 at [4,6]) and [0,1] on [7,6]‑[6,6] (0 at [7,6], 1 at [6,6]) satisfy sum‑3 goals.

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Aug 8, 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 8, 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 8, 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: Top‑left equals
The equals region covers [0,0], [0,1], and [1,1]. The only way to place a domino with identical pips on [0,0]‑[0,1] is domino [0,0]. Place it there; then [1,1] is forced to 0.
2
Step 2: Zero propels a 5
With [1,1]=0, the only domino that can deliver a 0 there is [0,5]. Place it horizontally on [1,1]‑[2,1] (0 at [1,1], 5 at [2,1]). Now the sum‑6 region [2,1]‑[2,2] has 5, so [2,2] must be 1.
3
Step 3: The 1‑5 domino completes row 2
Domino [5,1] supplies the 1 at [2,2] and the companion 5 must go to the empty [2,3]. Place it vertically: [2,2]=1, [2,3]=5.
4
Step 4: Greater‑5 and the final sum
[3,0] demands a pip >5, so it must be 6. Domino [6,2] covers [3,0]‑[3,1] (6,2). Then sum‑6 region [3,1]‑[3,2] needs 4 at [3,2]; domino [4,3] fits perfectly, placing 4 there and 3 at [3,3] (satisfying less‑5).

🔧 Step-by-Step Answer Walkthrough For Medium Level

1
Step 1: Less‑2 anchors double‑zero
The less‑2 cell at [3,1] restricts values to 0 or 1. Placing domino [0,0] (double‑zero) on [3,1]‑[3,2] satisfies both the less‑2 and initiates the sum‑6 region, forcing [3,3] to 6 later.
2
Step 2: Less‑3 and equals dictate 0 and 1
[2,0] is less‑3; the adjacent equals pair [1,1]‑[2,1] must match. Domino [0,1] placed on [2,0]‑[2,1] gives 0 at [2,0] and 1 at [2,1], locking the equals value to 1.
3
Step 3: The 1‑5 feeds the equals partner
[1,1] needs 1. Domino [1,5] fits horizontally on [1,1]‑[1,2], delivering 1 to [1,1] and 5 to the greater‑3 cell [1,2].
4
Step 4: Unequal pair sorted
The unequal region [0,3]‑[1,3] requires distinct pips. Domino [1,3] placed vertically sets [1,3]=1 and [0,3]=3, satisfying the constraint.
5
Step 5: Big 6s and final sums
The greater‑6 region [2,2]‑[2,3] demands 6 in both; domino [6,6] goes there horizontally. The triple sum‑6 region [2,4],[2,5],[3,4] uses domino [2,2] on [2,4]‑[2,5] (2,2) and domino [2,6] on [3,4]‑[3,3] (2 at [3,4], 6 at [3,3], already needed).

🔧 Step-by-Step Answer Walkthrough For Hard Level

1
Step 1: Sum‑0 splits zeros
The sum‑0 region [0,3] and [1,3] both must be 0. No domino holds two zeros, so use separate ones: [0,4] horizontally on [0,3]‑[0,4] (0,4) and [0,6] vertically on [1,3]‑[2,3] (0,6). The 4 at [0,4] satisfies greater‑3, and the 6 at [2,3] starts the big equals cluster.
2
Step 2: Triple equals fills with 5s
The top‑row equals region [0,1],[0,2],[1,2] forces all cells to 5. Domino [5,5] covers [0,2]‑[1,2] vertically, then [5,6] on [0,1]‑[0,0] gives 5 at [0,1] and 6 at [0,0] (greater‑3).
3
Step 3: A small equals pair gets 4
The equals pair [2,1]‑[3,1] needs both 4. Domino [3,4] placed on [1,1]‑[2,1] provides 3 at [1,1] (sum‑3) and 4 at [2,1]. Later, [2,4] will bring 4 to [3,1] from [4,1].
4
Step 4: Massive 6‑cluster coordinates
The six‑cell equals region must hold 6. Place [6,6] on [2,2]‑[3,2]; [4,6] on [3,3]‑[4,3] (4 at [3,3] for sum‑4, 6 at [4,3]); [1,6] on [6,3]‑[5,3] (1 at [6,3] less‑3, 6 at [5,3]); [2,6] on [5,2]‑[4,2] (2 at [5,2], 6 at [4,2]). Now the entire cluster is 6.
5
Step 5: Left‑side 2‑cluster and sum‑11 start
The second massive equals region (six cells) forces all to 2. Use [2,2] on [5,1]‑[6,1]; [2,5] on [6,2]‑[7,2] (2 at [6,2], 5 at [7,2] starting sum‑11); [2,4] on [4,1]‑[3,1] (2 at [4,1], 4 at [3,1] completing the 4‑pair); and [0,2] on [7,0]‑[7,1] (0 at [7,0] less‑3, 2 at [7,1]).
6
Step 6: Final sums and the bottom row
The sum‑11 region needs 5+3+3. With [7,2]=5, place [3,3] horizontally on [7,3]‑[7,4] (3,3). The remaining tidy‑up: [2,3] on [5,6]‑[4,6] (2 at [5,6], 3 at [4,6] sum‑3), and [0,1] on [7,6]‑[6,6] (0 at [7,6] less‑3, 1 at [6,6] completing sum‑3 with [5,6]=2). Grid solved.

💡 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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