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Claude 3.7 Thinking SonnetvsGPT-4o (Omni)

42 challenges

Why Claude 3.7 Thinking Sonnet?

1.6x more context200K
Stronger on MATH96.2% vs 75.9%
10 months newerFeb 2025

Why GPT-4o (Omni)?

2.9x cheaper overall$2.50/M in · $10.00/M out
Stronger on MMLU88.7% vs 77.1%
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Estimate Complexity

reasoning·hard

Tests an AI's ability to make educated estimates based on technical knowledge

Claude 3.7 Thinking Sonnet

Claude 3.7 Thinking Sonnet

Estimating GPT-3 Training FLOPs

I'll estimate the total number of floating point operations (FLOPs) required to train GPT-3 by breaking down the calculation step by step.

Model Architecture Facts

  • GPT-3 has approximately 175 billion parameters
  • Uses a transformer architecture with multiple layers
  • Context length of 2048 tokens
  • Trained on approximately 300 billion tokens

Calculating FLOPs per Token

For transformer models, we need to account for operations in:

  1. Self-attention mechanisms
  2. Feed-forward networks
  3. Layer normalization and other operations

For GPT-3, let's use these parameters:

  • Embedding dimension (d): ~12,288
  • Number of layers (L): ~96
  • Context length (s): 2,048

Forward Pass Calculation

For each token in each layer:

  • Self-attention: ~4 × d² + 2 × d × s FLOPs
  • Feed-forward network: ~8 × d² FLOPs
  • Other operations: ~d FLOPs

Total FLOPs per token (forward pass): ≈ L × (12 × d² + 2 × d × s) ≈ 96 × (12 × 12,288² + 2 × 12,288 × 2,048) ≈ 96 × (1.8 × 10⁹ + 5 × 10⁷) ≈ 1.8 × 10¹¹ FLOPs

Backward Pass

The backward pass typically requires about 2× the computation of the forward pass: ≈ 2 × 1.8 × 10¹¹ ≈ 3.6 × 10¹¹ FLOPs

Total FLOPs per token (forward + backward): ≈ 1.8 × 10¹¹ + 3.6 × 10¹¹ ≈ 5.4 × 10¹¹ FLOPs

Total Training FLOPs

Multiply by the total number of tokens processed: ≈ 5.4 × 10¹¹ × 300 × 10⁹ ≈ 1.62 × 10²³ FLOPs

Alternative Calculation

As a sanity check, using a simpler heuristic:

  • Each parameter requires ~2-4 FLOPs per token (including forward, backward, and weight update)
  • 175B parameters × 4 FLOPs × 300B tokens ≈ 2.1 × 10²³ FLOPs

Conclusion

The total training computation for GPT-3 was approximately 1.6-2.1 × 10²³ FLOPs (160-210 zettaFLOPs).

This massive computational requirement explains why training such models costs millions of dollars and requires specialized supercomputing infrastructure.

GPT-4o (Omni)

GPT-4o (Omni)

To estimate the total number of FLOPs required to train GPT-3, let's break it down into key components.

1. Model Size and Parameters

GPT-3 has 175 billion parameters.

2. FLOPs per Forward and Backward Pass

In deep learning, the FLOPs required for a forward pass and backward pass are estimated as:

  • Forward pass: ( 2 imes ) the number of parameters
  • Backward pass: ( 2 imes ) the FLOPs of the forward pass

Thus, total FLOPs per training step:

[ ext{Total FLOPs per step} = 6 imes ext{number of parameters} imes ext{tokens per batch} ]

For GPT-3, let's assume 175B parameters and a batch size of 2048 tokens (common for large-scale training).

[ ext{FLOPs per training step} = 6 imes 175 imes 10^9 imes 2048 ]

[ = 2.15 imes 10^{15} ext{ FLOPs per step} ]

3. Total Training Steps

GPT-3 was trained with 300 billion tokens (as per OpenAI's paper).

If the batch size is 2048 tokens, the total number of steps is:

[ ext{Total steps} = rac{300 imes 10^9}{2048} approx 1.46 imes 10^8 ]

4. Total FLOPs Required

Multiplying the FLOPs per step by the total number of steps:

[ ext{Total FLOPs} = (2.15 imes 10^{15}) imes (1.46 imes 10^8) ]

[ approx 3.14 imes 10^{23} ext{ FLOPs} ]

Final Answer:

The estimated total number of FLOPs required to train GPT-3 is about ( 3.14 imes 10^{23} ) floating point operations.

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2 / 42

AI Board Game Logic

reasoning·medium

Tests an AI's ability to understand game rules and strategy

Claude 3.7 Thinking Sonnet

Claude 3.7 Thinking Sonnet

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GPT-4o (Omni)

GPT-4o (Omni)

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