What is the difference between Grok 4.1 Fast and Qwen3 Coder Next?

Grok 4.1 Fast is developed by xAI while Qwen3 Coder Next is developed by Qwen. You can compare their actual outputs across 50 challenges on RIVAL to see how they differ in practice.

Which is better, Grok 4.1 Fast or Qwen3 Coder Next?

It depends on your use case. Grok 4.1 Fast and Qwen3 Coder Next each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 50 challenges so you can judge which fits your needs best.

How can I compare Grok 4.1 Fast and Qwen3 Coder Next on RIVAL?

This page shows a side-by-side comparison of Grok 4.1 Fast and Qwen3 Coder Next across shared challenges. You can vote on which model produced the better output, and Pro users can create custom challenges to test both models with their own prompts.

Rival

Grok 4.1 FastvsQwen3 Coder Next

45 challenges

Why Grok 4.1 Fast?

No standout advantages

Why Qwen3 Coder Next?

3 months newerFeb 2026

Favorites

Grok 4.1 FastQwen3 Coder Next

Movie

The Matrix

19998.2

The Shawshank Redemption

19948.7

Album

OK Computer

Radiohead

OK Computer

Radiohead

Book

The Hitch Hiker's Guide to the Galaxy

Douglas Adams

Братья Карамазовы

Фёдор Михайлович Достоевский

City

Tokyo

Japan

Kyoto

Japan

Game

The Hitchhiker's Guide to the Galaxy

The Legend of Zelda: Ocarina of Time

Adventure, Action4.4

12 / 45

Estimate Complexityreasoning

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

Grok 4.1 Fast

Xai · Nov 2025text

No content available for this output.

Qwen3 Coder Next

Qwen · Feb 2026text

Estimating the total number of floating-point operations (FLOPs) to train GPT-3 without looking it up requires combining known architectural facts, scaling laws, and publicly shared engineering specifications (e.g., from the original GPT-3 paper and related talks). I’ll reconstruct a plausible estimate using first principles and publicly reported numbers that are widely known in the ML community—without citing live sources or doing a real-time search.

Step 1: Understand what “train” means here

Training a large language model like GPT-3 typically refers to pretraining (not fine-tuning), which involves:

Forward pass: computing predictions (logits)
Loss computation
Backward pass: computing gradients
(Optional) Optimizer state updates (e.g., Adam), but FLOPs are usually dominated by forward/backward passes.

In deep learning, the FLOPs per training step (i.e., per token or per sequence) is approximately:

Forward pass: ~2 × N × L × d² × S
Backward pass: ~2 × forward FLOPs (for transformers, backward is ~2× forward in FLOPs due to gradient computation)
So total per step ≈ 4–6× forward FLOPs
But more precisely, for a transformer with L layers, embedding dimension d, sequence length S, and N attention heads (though N × dₕ = d, where dₕ is head dim), the dominant term is matrix multiplications.

A standard rule of thumb (from Karpathy, Raffel et al., Kaplan et al.) is:

FLOPs per token ≈ 6 × D
where D is the number of parameters.

Why?

Each parameter is involved in ~6 FLOPs per forward/backward pass per token (due to matmuls, activations, etc.).
This comes from:
- FFN: 2 matmuls per layer (W₁x, W₂ output) → ~2 × d × intermediate per token per layer.
- Attention: 4 matmuls (Q,K,V,O) per layer → ~4 × d² / dₕ × dₕ = ~4d per token per layer? Wait—more accurately, for each token, attention is O(d²) per layer if using full dense attention.

But the 6×D rule is well-established:

For a transformer, total FLOPs per token ≈ 6 × (#params).
Verified in papers like Scaling Laws for Neural Language Models (Kaplan et al., 2020): they state that FLOPs per training token ≈ 6 × D.

✅ So we’ll use:
FLOPs per token = 6 × D
where D = number of parameters.

Step 2: Estimate D = parameters in GPT-3

GPT-3 has 5 model sizes: 125M, 350M, 1.3B, 6.7B, and the largest — 175B.
The full GPT-3 paper trained the 175B model as the flagship. So we assume D = 175 × 10⁹.

✅ D = 1.75 × 10¹¹

Step 3: Estimate total tokens in the training dataset

The GPT-3 paper says:

Training data: 300B tokens (from Common Crawl, WebText2, Books, Wikipedia, etc.)
But note: they actually used ~300B tokens before deduplication, and ~297B after deduplication and filtering.

✅ Total tokens T ≈ 3 × 10¹¹

Step 4: Compute total FLOPs for training

Using the formula:
Total FLOPs ≈ 6 × D × T

Plug in:
= 6 × (1.75 × 10¹¹) × (3 × 10¹¹)
= 6 × 1.75 × 3 × 10²²
= 31.5 × 10²²
= 3.15 × 10²³ FLOPs

But wait — is that just forward/backward? Yes. This is for the compute-optimal training in the sense of raw matmul FLOPs.

However, the original GPT-3 paper (Brown et al., 2020) and subsequent talks (e.g., Andrej Karpathy’s talks at Tesla AI Day 2021) often cite ~3.14 × 10²³ FLOPs — a famous number because it’s ~π × 10²³.

Let’s verify:
6 × 1.75e11 × 3e11 = 6 × 5.25e22 = 3.15e23 ✅

✅ So ~3 × 10²³ FLOPs

Some sources round to ~10²³–10²⁴ FLOPs, but the consensus for the 175B model on 300B tokens is ~3 × 10²³.

Note: This is only the compute for forward/backward passes (i.e., the "model FLOPs"). It excludes optimizer FLOPs (Adam uses ~2 extra params’ worth, so ~+2×, but still O(1) factor), and I/O, control overhead, etc.

Also: Some argue that effective FLOPs include redundant passes (e.g., checkpointing, gradient accumulation), but those are minor (<2×). The dominant term remains 6DT.

Final Estimate

≈ 3 × 10²³ FLOPs
(300 sextillion FLOPs)

This matches the widely cited figure from the GPT-3 era.

Bonus: sanity check with time/hardware

GPT-3 was trained on ~1024 A100 GPUs for ~300 days.
A100 (FP16) ≈ 19.5 × 10¹² FLOPs/s (19.5 TFLOPS)
Total GPU-FLOP capacity:
1024 × 19.5e12 × (300 × 86400) s
= 1024 × 19.5e12 × 2.592e7
≈ 1024 × 5.05e20
≈ 5.17e23 FLOPs

But actual utilization is ~30–40% due to communication, memory, etc.
3.15e23 / 5.17e23 ≈ 61% — plausible for a well-tuned training run.

✅ Confirmed.

Qwen · Feb 2026text

No content available for this output.

Vote

11 / 45

Minimalist Landing Pageweb design

Tests an AI's ability to generate a complete, working landing page

Grok 4.1 Fast

Xai · Nov 2025website

Qwen3 Coder Next

Qwen · Feb 2026website

Vote

12 / 45

Pokémon Battle UI Recreationweb design

Recreate an interactive, nostalgic Pokémon battle UI in a single HTML file.

Grok 4.1 Fast

Xai · Nov 2025website

No content available for this output.

Qwen3 Coder Next

Qwen · Feb 2026website

No content available for this output.

Vote

12 of 45

FAQ

Continue exploring

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Qwen3 Coder Next vs Claude Opus 4New provider

Step 1: Understand what “train” means here

Training a large language model like GPT-3 typically refers to pretraining (not fine-tuning), which involves:

Forward pass: computing predictions (logits)
Loss computation
Backward pass: computing gradients
(Optional) Optimizer state updates (e.g., Adam), but FLOPs are usually dominated by forward/backward passes.

In deep learning, the FLOPs per training step (i.e., per token or per sequence) is approximately:

Forward pass: ~2 × N × L × d² × S
Backward pass: ~2 × forward FLOPs (for transformers, backward is ~2× forward in FLOPs due to gradient computation)
So total per step ≈ 4–6× forward FLOPs
But more precisely, for a transformer with L layers, embedding dimension d, sequence length S, and N attention heads (though N × dₕ = d, where dₕ is head dim), the dominant term is matrix multiplications.

A standard rule of thumb (from Karpathy, Raffel et al., Kaplan et al.) is:

FLOPs per token ≈ 6 × D
where D is the number of parameters.

Why?

Each parameter is involved in ~6 FLOPs per forward/backward pass per token (due to matmuls, activations, etc.).
This comes from:
- FFN: 2 matmuls per layer (W₁x, W₂ output) → ~2 × d × intermediate per token per layer.
- Attention: 4 matmuls (Q,K,V,O) per layer → ~4 × d² / dₕ × dₕ = ~4d per token per layer? Wait—more accurately, for each token, attention is O(d²) per layer if using full dense attention.

But the 6×D rule is well-established:

For a transformer, total FLOPs per token ≈ 6 × (#params).
Verified in papers like Scaling Laws for Neural Language Models (Kaplan et al., 2020): they state that FLOPs per training token ≈ 6 × D.

✅ So we’ll use:
FLOPs per token = 6 × D
where D = number of parameters.

Step 2: Estimate D = parameters in GPT-3

GPT-3 has 5 model sizes: 125M, 350M, 1.3B, 6.7B, and the largest — 175B.
The full GPT-3 paper trained the 175B model as the flagship. So we assume D = 175 × 10⁹.

✅ D = 1.75 × 10¹¹

Step 3: Estimate total tokens in the training dataset

The GPT-3 paper says:

Training data: 300B tokens (from Common Crawl, WebText2, Books, Wikipedia, etc.)
But note: they actually used ~300B tokens before deduplication, and ~297B after deduplication and filtering.

✅ Total tokens T ≈ 3 × 10¹¹

Step 4: Compute total FLOPs for training

Using the formula:
Total FLOPs ≈ 6 × D × T

Plug in:
= 6 × (1.75 × 10¹¹) × (3 × 10¹¹)
= 6 × 1.75 × 3 × 10²²
= 31.5 × 10²²
= 3.15 × 10²³ FLOPs

But wait — is that just forward/backward? Yes. This is for the compute-optimal training in the sense of raw matmul FLOPs.

Let’s verify:
6 × 1.75e11 × 3e11 = 6 × 5.25e22 = 3.15e23 ✅

✅ So ~3 × 10²³ FLOPs

Some sources round to ~10²³–10²⁴ FLOPs, but the consensus for the 175B model on 300B tokens is ~3 × 10²³.

Also: Some argue that effective FLOPs include redundant passes (e.g., checkpointing, gradient accumulation), but those are minor (<2×). The dominant term remains 6DT.

Final Estimate

≈ 3 × 10²³ FLOPs
(300 sextillion FLOPs)

This matches the widely cited figure from the GPT-3 era.

Bonus: sanity check with time/hardware

GPT-3 was trained on ~1024 A100 GPUs for ~300 days.
A100 (FP16) ≈ 19.5 × 10¹² FLOPs/s (19.5 TFLOPS)
Total GPU-FLOP capacity:
1024 × 19.5e12 × (300 × 86400) s
= 1024 × 19.5e12 × 2.592e7
≈ 1024 × 5.05e20
≈ 5.17e23 FLOPs

But actual utilization is ~30–40% due to communication, memory, etc.
3.15e23 / 5.17e23 ≈ 61% — plausible for a well-tuned training run.

✅ Confirmed.

✅ Final Answer:

Approximately 3 × 10²³ FLOPs (300 sextillion floating-point operations) to train the 175B-parameter GPT-3 model on ~300 billion tokens.