What is the difference between Kimi K2 and Qwen3 Coder Next?

Kimi K2 is developed by Moonshot AI while Qwen3 Coder Next is developed by Qwen. You can compare their actual outputs across 32 challenges on RIVAL to see how they differ in practice.

Which is better, Kimi K2 or Qwen3 Coder Next?

It depends on your use case. Kimi K2 and Qwen3 Coder Next each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 32 challenges so you can judge which fits your needs best.

How much does Kimi K2 cost compared to Qwen3 Coder Next?

Kimi K2 costs $0.57/M input tokens and Qwen3 Coder Next costs $0.2/M input tokens. Qwen3 Coder Next is $0.37/M cheaper per input. Check their side-by-side outputs on RIVAL to see if the price difference is justified by quality.

How can I compare Kimi K2 and Qwen3 Coder Next on RIVAL?

This page shows a side-by-side comparison of Kimi K2 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

Kimi K2vsQwen3 Coder Next

32 challenges

Why Kimi K2?

No standout advantages

Why Qwen3 Coder Next?

1.6x cheaper overall$0.20/M in · $1.50/M out

Leads 1 of 1 benchmarks

Stronger on SWE-bench Verified70.6% vs 65.8%

7 months newerFeb 2026

Favorites

Kimi K2Qwen3 Coder Next

Movie

Blade Runner

19827.9

The Shawshank Redemption

19948.7

Album

Dark Side Of The Moon

suisside

OK Computer

Radiohead

Book

The Left Hand of Darkness

Ursula K. Le Guin

Братья Карамазовы

Фёдор Михайлович Достоевский

City

Kyoto

Japan

Kyoto

Japan

Game

Outer Wilds

Indie, Adventure4.4

The Legend of Zelda: Ocarina of Time

Adventure, Action4.4

12 / 32

Estimate Complexityreasoning

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

Kimi K2

Moonshotai · Jul 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.

✅ Final Answer:

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

Vote

2 / 32

Logic Puzzlereasoning

Tests an AI's ability to solve a simple but potentially confusing logic puzzle

Kimi K2

Moonshotai · Jul 2025text

No content available for this output.

Qwen3 Coder Next

Qwen · Feb 2026text

No content available for this output.

Vote

3 / 32

Stochastic Consistency Testanalysis

Tests an AI's randomness and creativity

Kimi K2

Moonshotai · Jul 2025text

Here are five freshly-minted jokes:

I tried to start a procrastinators’ club, but nobody showed up… yet.
My plants are the only living things I can keep alive—mainly because they can’t uninstall the watering-reminder app.
I asked the cloud for my data back; it just rained SQL on my parade.
My smart fridge just texted me, “Stop staring, your calories are counting themselves.”
I told my therapist I’m afraid of negative numbers. She said, “Don’t worry, we’ll address that on the minus couch.”

Qwen3 Coder Next

Qwen · Feb 2026website

Vote

9 / 32

Pokémon Battle UI Recreationweb design

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

Kimi K2

Moonshotai · Jul 2025website

No content available for this output.

Qwen3 Coder Next

Qwen · Feb 2026website

No content available for this output.

Moonshotai · Jul 2025website

Qwen3 Coder Next

Qwen · Feb 2026website

No content available for this output.

Vote

12 of 32

FAQ

Continue exploring

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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.