Compare Claude Sonnet 3.6 (2022-10-22) by Anthropic against INTELLECT-3 by OpenRouter, tested across 43 shared challenges. Updated March 2026.
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Tests an AI's ability to make educated estimates based on technical knowledge
To estimate the total number of FLOPs required to train GPT-3, we need to consider the model architecture, training process, and dataset size. GPT-3 has 175 billion parameters, a context length of 2048 tokens, 96 layers, a hidden dimension of 12,288, and a feed-forward dimension of 49,152 (4 × hidden dimension). The vocabulary size is approximately 50,257. The training dataset consists of 300 billion tokens.
The forward pass involves computations for each transformer layer and the output projection. For one layer:
Summing these for one layer:
Per-token cost (dividing by context length 2048):
Additionally, the output projection per token: (\text{hidden dimension} \times \text{vocabulary size} = 12,288 \times 50,257) FLOPs.
Total forward FLOPs per token: [ \text{FLOPs}_{\text{forward}} = 96 \times 3 \times 12,288 \times 49,152 + 12,288 \times 50,257 ] Calculating:
Training involves a forward pass and a backward pass. The backward pass typically requires twice the FLOPs of the forward pass due to gradient computation: [ \text{FLOPs}{\text{train}} = 3 \times \text{FLOPs}{\text{forward}} = 3 \times 1.7456 \times 10^{11} = 5.2368 \times 10^{11} \text{ FLOPs per token}. ]
Multiply by the total number of training tokens (300 billion): [ \text{Total FLOPs} = 300 \times 10^9 \times 5.2368 \times 10^{11} = 1.57104 \times 10^{23} \text{ FLOPs}. ]
The initial calculation includes only major operations. Layer normalization, residual connections, and other minor operations add a small overhead (approximately 5-10%). Accounting for this, we adjust:
However, given the complexity and approximations, a reasonable estimate is (1.6 \times 10^{23}) FLOPs. This aligns with known estimates for GPT-3 training, which range from (10^{23}) to (10^{24}) FLOPs.
The total FLOPs required to train GPT-3 is approximately (1.6 \times 10^{23}).
[ \boxed{1.6 \times 10^{23}} ]
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