What is the difference between DeepSeek V3.1 and Gemini 3 Flash Preview?

DeepSeek V3.1 is developed by DeepSeek while Gemini 3 Flash Preview is developed by Google AI. DeepSeek V3.1 has a 164K token context window vs Gemini 3 Flash Preview's 1.0M. You can compare their actual outputs across 51 challenges on RIVAL to see how they differ in practice.

Which is better, DeepSeek V3.1 or Gemini 3 Flash Preview?

It depends on your use case. DeepSeek V3.1 and Gemini 3 Flash Preview each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 51 challenges so you can judge which fits your needs best.

How much does DeepSeek V3.1 cost compared to Gemini 3 Flash Preview?

DeepSeek V3.1 costs $0.2/M input tokens and Gemini 3 Flash Preview costs $0.5/M input tokens. DeepSeek V3.1 is $0.30/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 DeepSeek V3.1 and Gemini 3 Flash Preview on RIVAL?

This page shows a side-by-side comparison of DeepSeek V3.1 and Gemini 3 Flash Preview 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

DeepSeek V3.1vsGemini 3 Flash Preview

46 challenges

Why DeepSeek V3.1?

3.6x cheaper overall$0.20/M in · $0.80/M out

Why Gemini 3 Flash Preview?

6.4x more context1.0M

Leads 1 of 1 benchmarks

Stronger on SWE-bench Verified78.0% vs 66.0%

4 months newerDec 2025

Favorites

DeepSeek V3.1Gemini 3 Flash Preview

Movie

The Princess Bride

19877.7

Interstellar

20148.5

Album

The Rise and Fall of Ziggy Stardust and the Spiders From Mars

David Bowie

Kind of Blue

Miles Davis

Book

Gödel, Escher, Bach

Douglas R. Hofstadter

La biblioteca de babel

Jorge Luis Borges

City

Kyoto

Japan

Kyoto

Japan

Game

The Legend of Zelda: Breath of the Wild

Adventure, Action4.5

Outer Wilds

Indie, Adventure4.4

12 / 46

Estimate Complexityreasoning

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

DeepSeek V3.1

Deepseek · Aug 2025text

No content available for this output.

Gemini 3 Flash Preview

Google · Dec 2025text

To estimate the total floating-point operations (FLOPs) required to train GPT-3, we can use a standard heuristic derived from the architecture of Transformer models.

Here is the step-by-step reasoning.

1. Identify Key Constants

First, we need the two primary variables that determine training compute:

Number of Parameters ($N$): GPT-3 (the full version) is known to have 175 billion parameters.
Number of Training Tokens ($D$): GPT-3 was trained on a massive dataset. Based on the original research paper, the 175B model was trained on approximately 300 billion tokens.

2. The Heuristic Formula ($6ND$)

In deep learning, specifically for Transformers, there is a well-established "rule of thumb" for estimating training compute: $$\text{Compute (FLOPs)} \approx 6 \times N \times D$$

Why 6? The factor of 6 comes from the breakdown of the forward and backward passes:

Forward Pass ($2N$): For every parameter in the model, there is roughly one multiply-accumulate operation (which counts as 2 FLOPs) per token.
Backward Pass ($4N$): The backward pass (calculating gradients) is computationally more expensive. It typically requires roughly twice the compute of the forward pass ($2 \times 2N = 4N$).
Total: $2N (\text{forward}) + 4N (\text{backward}) = 6N$ FLOPs per token.

3. The Calculation

Now, we plug in the values for GPT-3:

$N = 1.75 \times 10^{11}$ (175 billion)
$D = 3.0 \times 10^{11}$ (300 billion)

$$\text{Total FLOPs} = 6 \times (1.75 \times 10^{11}) \times (3.0 \times 10^{11})$$

Step-by-step multiplication:

Multiply the coefficients: $6 \times 1.75 = 10.5$
Multiply by the tokens: $10.5 \times 3.0 = 31.5$
Combine the powers of ten: $10^{11} \times 10^{11} = 10^{22}$

$$\text{Total FLOPs} = 31.5 \times 10^{22}$$ $$\text{Total FLOPs} = 3.15 \times 10^{23}$$

4. Final Estimate

The total estimated compute for training GPT-3 is $3.15 \times 10^{23}$ FLOPs.

5. Contextualizing the Number

To put $3.15 \times 10^{23}$ into perspective:

In scientific notation, this is 315 Sextillion FLOPs.
In terms of "Zettaflops" (10²¹), it is 315 ZettaFLOPs.
If you used a single high-end NVIDIA A100 GPU (which performs at roughly 312 TFLOPS of FP16 with Tensor Cores), and assuming perfect efficiency, it would take roughly 32 years to complete the training. (In reality, OpenAI used thousands of GPUs in parallel to finish in a few months).

Note on accuracy: This estimate ignores the overhead of the "attention" mechanism (which scales with sequence length), but for models as large as GPT-3, the feed-forward and projection layers (represented by $N$) dominate the compute cost, making $6ND$ a very accurate approximation.

Vote

2 / 46

AI Board Game Logicreasoning

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

DeepSeek V3.1

Deepseek · Aug 2025svg

No content available for this output.

Google · Dec 2025text

No content available for this output.

Vote

12 / 46

Minimalist Landing Pageweb design

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

DeepSeek V3.1

Deepseek · Aug 2025website

Gemini 3 Flash Preview

Google · Dec 2025website

Vote

12 of 46

FAQ

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To estimate the total floating-point operations (FLOPs) required to train GPT-3, we can use a standard heuristic derived from the architecture of Transformer models.

Here is the step-by-step reasoning.

1. Identify Key Constants

First, we need the two primary variables that determine training compute:

Number of Parameters ($N$): GPT-3 (the full version) is known to have 175 billion parameters.
Number of Training Tokens ($D$): GPT-3 was trained on a massive dataset. Based on the original research paper, the 175B model was trained on approximately 300 billion tokens.

2. The Heuristic Formula ($6ND$)

In deep learning, specifically for Transformers, there is a well-established "rule of thumb" for estimating training compute: $$\text{Compute (FLOPs)} \approx 6 \times N \times D$$

Why 6? The factor of 6 comes from the breakdown of the forward and backward passes:

Forward Pass ($2N$): For every parameter in the model, there is roughly one multiply-accumulate operation (which counts as 2 FLOPs) per token.
Backward Pass ($4N$): The backward pass (calculating gradients) is computationally more expensive. It typically requires roughly twice the compute of the forward pass ($2 \times 2N = 4N$).
Total: $2N (\text{forward}) + 4N (\text{backward}) = 6N$ FLOPs per token.

3. The Calculation

Now, we plug in the values for GPT-3:

$N = 1.75 \times 10^{11}$ (175 billion)
$D = 3.0 \times 10^{11}$ (300 billion)

$$\text{Total FLOPs} = 6 \times (1.75 \times 10^{11}) \times (3.0 \times 10^{11})$$

Step-by-step multiplication:

Multiply the coefficients: $6 \times 1.75 = 10.5$
Multiply by the tokens: $10.5 \times 3.0 = 31.5$
Combine the powers of ten: $10^{11} \times 10^{11} = 10^{22}$

$$\text{Total FLOPs} = 31.5 \times 10^{22}$$ $$\text{Total FLOPs} = 3.15 \times 10^{23}$$

4. Final Estimate

The total estimated compute for training GPT-3 is $3.15 \times 10^{23}$ FLOPs.

5. Contextualizing the Number

To put $3.15 \times 10^{23}$ into perspective:

In scientific notation, this is 315 Sextillion FLOPs.
In terms of "Zettaflops" (10²¹), it is 315 ZettaFLOPs.
If you used a single high-end NVIDIA A100 GPU (which performs at roughly 312 TFLOPS of FP16 with Tensor Cores), and assuming perfect efficiency, it would take roughly 32 years to complete the training. (In reality, OpenAI used thousands of GPUs in parallel to finish in a few months).