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  1. Home
  2. Models
  3. Claude Sonnet 4.6
  4. Estimate Complexity

See how Claude Sonnet 4.6 compares on this challenge:

  • Claude Sonnet 4.6 vs Gemini 3.1 Pro Preview
  • Claude Sonnet 4.6 vs Z.ai: GLM 5

3 models have responded to the "Estimate Complexity" challenge. Compare their approaches side-by-side on RIVAL. This response is part of RIVAL's open dataset of 5,600+ AI model responses.

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Back to Claude Sonnet 4.6

Estimate Complexity

Claude Sonnet 4.6's response to the 'Estimate Complexity' challenge.

Claude Sonnet 4.6
February 17, 2026
Text

?
Prompt

Estimate the total number of FLOPs required to train GPT-3 without searching online. Explain your reasoning step by step.

Claude Sonnet 4.6
Response

claude-sonnet-4.6 logo
Claude Sonnet 4.6Estimate Complexity
claude-sonnet-4.6 logo
Claude Sonnet 4.6

Estimating FLOPs to Train GPT-3

Step 1: Recall GPT-3's Architecture

GPT-3 has approximately:

  • 175 billion parameters (this is well-known)
  • 96 transformer layers
  • 96 attention heads
  • Hidden dimension: 12,288
  • Context length: 2,048 tokens

Step 2: FLOPs per Token (Forward Pass)

For a transformer, the dominant operations are matrix multiplications. There's a useful rule of thumb:

FLOPs per token ≈ 2 × N (for the forward pass)

where N = number of parameters.

Why 2×N?

  • Each parameter is involved in roughly one multiply-add operation per token
  • A multiply-add = 2 FLOPs
  • This holds when sequence length is much smaller than model dimension (which is roughly true here)

So forward pass FLOPs per token: $$2 \times 175 \times 10^9 = 3.5 \times 10^{11} \text{ FLOPs/token}$$


Step 3: Account for Backward Pass

Backpropagation costs approximately 2× the forward pass (computing gradients for weights + activations).

So total per token: $$3 \times 3.5 \times 10^{11} = 1.05 \times 10^{12} \text{ FLOPs/token}$$


Step 4: Training Tokens

GPT-3 was trained on approximately 300 billion tokens (from the paper's description of ~300B tokens from filtered Common Crawl, WebText, books, Wikipedia).


Step 5: Total FLOPs

$$\text{Total FLOPs} = 1.05 \times 10^{12} \times 3 \times 10^{11}$$

$$= 3.15 \times 10^{23} \text{ FLOPs}$$

$$\approx \boxed{3 \times 10^{23} \text{ FLOPs}}$$


Sanity Check

  • This is ~300 zettaFLOPs
  • At 1,000 A100s running at ~300 TFLOPs each (with ~50% utilization → ~150 TFLOPs effective):
    • Total compute = 1,000 × 1.5×10¹⁴ = 1.5×10¹⁷ FLOPs/second
    • Time = 3×10²³ / 1.5×10¹⁷ ≈ 2×10⁶ seconds ≈ 23 days
    • OpenAI reportedly used thousands of GPUs for weeks — this is plausible ✓

Summary

ComponentValue
Parameters175B
FLOPs/token (forward)~3.5 × 10¹¹
Training multiplier (fwd+bwd)×3
Training tokens~300B
Total FLOPs~3 × 10²³

The key formula to remember: C ≈ 6ND where N = parameters, D = training tokens.

Estimate Complexity

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About Claude Sonnet 4.6

Capabilities

ConversationReasoningCode GenerationAnalysisAgentic Tool UseTool Use

Categories

TextCodeMultimodal

Specifications

Provider
Anthropic
Released
2026-02-17
Size
LARGE
Context
1,000,000 tokens

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SAME PROMPT

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