What is the difference between Claude Sonnet 4.5 and MiniMax M2?

Claude Sonnet 4.5 is developed by Anthropic while MiniMax M2 is developed by MiniMax. Claude Sonnet 4.5 has a 200K token context window vs MiniMax M2's 205K. in 54 community votes on RIVAL, Claude Sonnet 4.5 wins 59% of head-to-head matchups. These results are based on blind head-to-head voting across 27 challenges.

Which is better, Claude Sonnet 4.5 or MiniMax M2?

Based on 54 community votes on RIVAL, Claude Sonnet 4.5 wins 59% of head-to-head matchups against MiniMax M2. Claude Sonnet 4.5 is strongest in Reasoning, Image Generation, Conversation. However, MiniMax M2 leads in Web Design, Analysis.

How much does Claude Sonnet 4.5 cost compared to MiniMax M2?

Claude Sonnet 4.5 costs $3/M input tokens and MiniMax M2 costs $0/M input tokens. MiniMax M2 is $3.00/M cheaper per input. The more expensive model wins 59% of duels, so the premium may be justified by quality.

How are Claude Sonnet 4.5 vs MiniMax M2 votes collected on RIVAL?

RIVAL presents both models' outputs side-by-side in blind duels — voters see the responses but don't know which model produced each one. This eliminates brand bias. 54 votes have been collected for this pair across 27 challenges. All vote data is part of RIVAL's open dataset.

Rival

Claude Sonnet 4.5vsMiniMax M2

27 challenges

Why Claude Sonnet 4.5?

Leads 1 of 1 benchmarks

Stronger on SWE-bench Verified77.2% vs 69.4%

Why MiniMax M2?

No standout advantages

Favorites

Claude Sonnet 4.5MiniMax M2

Movie

2001: A Space Odyssey

19688.1

—

Album

Kind of Blue

Miles Davis

—

Book

Die Verwandlung

Franz Kafka

—

City

Amsterdam

Netherlands

—

Game

Portal 2

Shooter, Puzzle4.6

—

12 / 27

Estimate Complexityreasoning

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

Claude Sonnet 4.5

Anthropic · Sep 2025text

I'll work through this estimation step by step, using what I know about GPT-3's architecture and training.

Known/Estimated Parameters

Model size:

GPT-3 (175B version) has ~175 billion parameters

Training data:

Trained on ~300 billion tokens (this was mentioned in the paper)

Architecture:

Transformer decoder-only model

FLOPs Calculation

For transformer models, there's an approximate rule for FLOPs per token:

FLOPs per token ≈ 6P (for forward pass) + 6P (for backward pass) = 12P total

Where P = number of parameters

This factor of 6 comes from:

2 FLOPs per multiply-accumulate operation
Each parameter is involved in approximately 3 operations per forward pass (attention, FFN layers, etc.)

Calculation:

Parameters (P): 175 × 10⁹
Tokens (T): 300 × 10⁹
FLOPs per token: 12P = 12 × 175 × 10⁹ = 2.1 × 10¹²

Total training FLOPs:

Total = FLOPs per token × number of tokens
Total = 2.1 × 10¹² × 300 × 10⁹
Total = 6.3 × 10²³ FLOPs

Final Estimate

~3-6 × 10²³ FLOPs (or ~300-600 zettaFLOPs)

This is roughly equivalent to running a GPU capable of 100 TFLOPS continuously for about 75,000 years, or more practically, running thousands of GPUs for several weeks.

MiniMax M2

Minimax · Oct 2025text

To estimate the total FLOPs required to train GPT-3 (175B parameters) on 300 billion tokens (one epoch), we break down the process step by step:

Step 1: Understand GPT-3's Specifications

Parameters: 175 billion (175e9).
Training Tokens: 300 billion (300e9).
Context Length: 2048 tokens per sequence.
Training Epochs: 1 (as per the paper).

Step 2: FLOPs for Forward Pass

For a transformer model, the forward pass cost per token per parameter is approximately 2 FLOPs (one multiplication and one addition in matrix operations).
Total FLOPs for forward pass:
[ \text{Total FLOPs}_{\text{forward}} = \text{Tokens} \times \text{Parameters} \times 2 ] [ = 300 \times 10^9 \times 175 \times 10^9 \times 2 = 1.05 \times 10^{23} \text{ FLOPs} ]

Step 3: FLOPs for Backward Pass

The backward pass (gradient computation) costs approximately 2× the forward pass due to additional operations (e.g., chain rule applications).
Total FLOPs for backward pass:
[ \text{Total FLOPs}{\text{backward}} = 2 \times \text{Total FLOPs}{\text{forward}} = 2 \times 1.05 \times 10^{23} = 2.10 \times 10^{23} \text{ FLOPs} ]

Step 4: Quadratic Attention Overhead (Minor Correction)

Attention layers introduce quadratic cost in context length ((O(S^2)) per token, where (S = 2048)).
Overhead per token: ( \sim 4% ) of the linear cost (empirically for GPT-3).
Adjusted forward/backward FLOPs:
[ \text{Total FLOPs}_{\text{linear}} = \text{Forward} + \text{Backward} = 1.05 \times 10^{23} + 2.10 \times 10^{23} = 3.15 \times 10^{23} \text{ FLOPs} ]
Attention overhead: (3.15 \times 10^{23} \times 0.04 = 1.26 \times 10^{22} \text{ FLOPs}).
Revised total:
[ 3.15 \times 10^{23} + 1.26 \times 10^{22} \approx 3.28 \times 10^{23} \text{ FLOPs} ]

Step 5: Final Adjustment for Practical Efficiency

Training uses mixed precision (FP16) and optimizations (e.g., gradient accumulation), reducing effective FLOPs by ~10–20%.
Adjusted total:
[ 3.28 \times 10^{23} \times 0.85 \approx 2.79 \times 10^{23} \text{ FLOPs} ] (Accounts for inefficiencies, data loading, etc.).

Conclusion

The estimated total FLOPs to train GPT-3 (175B) on 300B tokens is ~2.79 × 10²³ FLOPs.
This aligns with OpenAI's reported value of 3.14 × 10²³ FLOPs (accounting for minor rounding and approximation differences).

Key Assumptions:

1 training epoch on 300B tokens.
Dominance of linear matrix multiplications (attention overhead <5%).
Mixed precision reduces cost by ~15%.
No hyperparameter tuning or intermediate checkpoints.

Final Answer:
[ \boxed{2.79 \times 10^{23}} ]
FLOPs (rounded to two significant figures, approximately 2.8 × 10²³).

Vote

2 / 27

AI Board Game Logicreasoning

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

Claude Sonnet 4.5

Anthropic · Sep 2025svg

MiniMax M2

Minimax · Oct 2025svg

Vote

3 / 27

Logic Puzzlereasoning

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

Claude Sonnet 4.5

Anthropic · Sep 2025text

I need to figure out how many sisters Sally has.

Given information:

Sally is a girl
Sally has 3 brothers

Now, each brother has 2 sisters.

If each of Sally's brothers has 2 sisters, and Sally is one of those sisters, then there must be one more sister besides Sally.

Let me verify:

Total sisters in the family: Sally + 1 other sister = 2 sisters
Each of the 3 brothers would have these same 2 sisters
This checks out ✓

Therefore, Sally has 1 sister.

MiniMax M2

Minimax · Oct 2025text

Sally has 3 brothers, meaning there are 3 boys in the family.
Each brother has 2 sisters. Since the brothers' sisters are the girls in the family, and one of them is Sally, there must be one other sister to make it 2 sisters per brother.
Therefore, there are 2 girls in total: Sally and one other sister.
From Sally's perspective, she has 1 sister (the other girl).

Answer: 1

Vote

4 / 27

Math Misconception Testreasoning

Tests an AI's understanding of number representation

Claude Sonnet 4.5

Anthropic · Sep 2025text

No, 9.11 is not greater than 9.9.

9.9 is greater than 9.11.

When comparing decimals, you can think of them as:

9.11 = 9 + 0.11
9.9 = 9 + 0.9 (or 9.90)

Since 0.90 > 0.11, we have 9.9 > 9.11.

MiniMax M2

Minimax · Oct 2025text

No, 9.11 is not greater than 9.9. Here's why:

Decimal Comparison: When comparing decimals, look at the digits after the decimal point.
- 9.11 has digits "11" after the decimal.
- 9.9 (or 9.90) has digits "90" after the decimal.
  Since 11 is less than 90, 9.11 is less than 9.9.
Numerical Values:
- 9.9 is equivalent to 9.90.
- 9.90 - 9.11 = 0.79, which is positive, confirming 9.90 > 9.11.
Common Misconception: The confusion might stem from software version numbers (e.g., 9.11 might be considered newer than 9.9), but numerically, 9.11 is smaller.

In summary, 9.9 is greater than 9.11. If you're comparing versions, context matters, but numerically, 9.9 wins.

Vote

5 / 27

Stochastic Consistency Testanalysis

Tests an AI's randomness and creativity

Claude Sonnet 4.5

Anthropic · Sep 2025text

5 Jokes

Why don't scientists trust atoms? Because they make up everything!
I told my wife she was drawing her eyebrows too high. She looked surprised.
Why did the scarecrow win an award? He was outstanding in his field.
I'm reading a book about anti-gravity. It's impossible to put down.
Why don't eggs tell jokes? They'd crack each other up.

Answer: All 5 jokes are unique. Each one has a different setup and punchline, covering different topics (atoms, eyebrows, scarecrows, gravity, and eggs).

MiniMax M2

Minimax · Oct 2025text

No content available for this output.

Vote

6 / 27

SVG Layout Challengeimage generation

Tests an AI's ability to generate vector graphics

Claude Sonnet 4.5

Anthropic · Sep 2025svg

MiniMax M2

Minimax · Oct 2025svg

No content available for this output.

Vote

7 / 27

Xbox Controller SVG Artimage generation

Tests an AI's ability to create detailed SVG illustrations of gaming hardware

Claude Sonnet 4.5

Anthropic · Sep 2025svg

MiniMax M2

Minimax · Oct 2025svg

No content available for this output.

Vote

8 / 27

Generate a Stand-Up Routineconversation

Tests an AI's humor and creative writing ability

Claude Sonnet 4.5

Anthropic · Sep 2025text

No content available for this output.

MiniMax M2

Minimax · Oct 2025text

No content available for this output.

Vote

9 / 27

Character Voice Testconversation

Tests an AI's ability to write in distinct character voices

Claude Sonnet 4.5

Anthropic · Sep 2025text

No content available for this output.

MiniMax M2

Minimax · Oct 2025text

No content available for this output.

Vote

10 / 27

Pokémon Battle UI Recreationweb design

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

Claude Sonnet 4.5

Anthropic · Sep 2025website

MiniMax M2

Minimax · Oct 2025website

No content available for this output.

Vote

11 / 27

Linear App Cloneweb design

Tests an AI's ability to replicate an existing UI with Tailwind CSS

Claude Sonnet 4.5

Anthropic · Sep 2025website

MiniMax M2

Minimax · Oct 2025website

No content available for this output.

Vote

12 / 27

Framer-Style Animationweb design

Tests an AI's ability to create smooth web animations

Claude Sonnet 4.5

Anthropic · Sep 2025website

No content available for this output.

MiniMax M2

Minimax · Oct 2025website

No content available for this output.

Vote

12 of 27

FAQ

Continue exploring

Claude Sonnet 4.5 vs Pony AlphaNew provider

Claude Sonnet 4.5 vs GPT OSS 20BNew provider

Claude Sonnet 4.5 vs MiniMax M2.5Cross-provider

To estimate the total FLOPs required to train GPT-3 (175B parameters) on 300 billion tokens (one epoch), we break down the process step by step:

Step 1: Understand GPT-3's Specifications

Parameters: 175 billion (175e9).
Training Tokens: 300 billion (300e9).
Context Length: 2048 tokens per sequence.
Training Epochs: 1 (as per the paper).

Step 2: FLOPs for Forward Pass

For a transformer model, the forward pass cost per token per parameter is approximately 2 FLOPs (one multiplication and one addition in matrix operations).
Total FLOPs for forward pass:
[ \text{Total FLOPs}_{\text{forward}} = \text{Tokens} \times \text{Parameters} \times 2 ] [ = 300 \times 10^9 \times 175 \times 10^9 \times 2 = 1.05 \times 10^{23} \text{ FLOPs} ]

Step 3: FLOPs for Backward Pass

The backward pass (gradient computation) costs approximately 2× the forward pass due to additional operations (e.g., chain rule applications).
Total FLOPs for backward pass:
[ \text{Total FLOPs}{\text{backward}} = 2 \times \text{Total FLOPs}{\text{forward}} = 2 \times 1.05 \times 10^{23} = 2.10 \times 10^{23} \text{ FLOPs} ]

Step 4: Quadratic Attention Overhead (Minor Correction)

Attention layers introduce quadratic cost in context length ((O(S^2)) per token, where (S = 2048)).
Overhead per token: ( \sim 4% ) of the linear cost (empirically for GPT-3).
Adjusted forward/backward FLOPs:
[ \text{Total FLOPs}_{\text{linear}} = \text{Forward} + \text{Backward} = 1.05 \times 10^{23} + 2.10 \times 10^{23} = 3.15 \times 10^{23} \text{ FLOPs} ]
Attention overhead: (3.15 \times 10^{23} \times 0.04 = 1.26 \times 10^{22} \text{ FLOPs}).
Revised total:
[ 3.15 \times 10^{23} + 1.26 \times 10^{22} \approx 3.28 \times 10^{23} \text{ FLOPs} ]

Step 5: Final Adjustment for Practical Efficiency

Training uses mixed precision (FP16) and optimizations (e.g., gradient accumulation), reducing effective FLOPs by ~10–20%.
Adjusted total:
[ 3.28 \times 10^{23} \times 0.85 \approx 2.79 \times 10^{23} \text{ FLOPs} ] (Accounts for inefficiencies, data loading, etc.).

Conclusion

Key Assumptions:

1 training epoch on 300B tokens.
Dominance of linear matrix multiplications (attention overhead <5%).
Mixed precision reduces cost by ~15%.
No hyperparameter tuning or intermediate checkpoints.

Final Answer:
[ \boxed{2.79 \times 10^{23}} ]
FLOPs (rounded to two significant figures, approximately 2.8 × 10²³).