What is the difference between DeepSeek V3.2 Speciale and Gemini 3 Flash Preview?

DeepSeek V3.2 Speciale is developed by DeepSeek while Gemini 3 Flash Preview is developed by Google AI. DeepSeek V3.2 Speciale has a 131K token context window vs Gemini 3 Flash Preview's 1.0M. You can compare their actual outputs across 48 challenges on RIVAL to see how they differ in practice.

Which is better, DeepSeek V3.2 Speciale or Gemini 3 Flash Preview?

It depends on your use case. DeepSeek V3.2 Speciale and Gemini 3 Flash Preview each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 48 challenges so you can judge which fits your needs best.

How much does DeepSeek V3.2 Speciale cost compared to Gemini 3 Flash Preview?

DeepSeek V3.2 Speciale costs $0.28/M input tokens and Gemini 3 Flash Preview costs $0.5/M input tokens. DeepSeek V3.2 Speciale is $0.22/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 V3.2 Speciale and Gemini 3 Flash Preview on RIVAL?

This page shows a side-by-side comparison of DeepSeek V3.2 Speciale 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 V3.2 SpecialevsGemini 3 Flash Preview

43 challenges

Why DeepSeek V3.2 Speciale?

5.5x cheaper overall$0.28/M in · $0.42/M out

Why Gemini 3 Flash Preview?

8x more context1.0M

Favorites

DeepSeek V3.2 SpecialeGemini 3 Flash Preview

Movie

—

Interstellar

20148.5

Album

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Kind of Blue

Miles Davis

Book

—

La biblioteca de babel

Jorge Luis Borges

City

—

Kyoto

Japan

Game

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Outer Wilds

Indie, Adventure4.4

12 / 43

Estimate Complexityreasoning

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

DeepSeek V3.2 Speciale

Deepseek · Dec 2025text

Estimating the total number of Floating Point Operations (FLOPs) required to train GPT-3 involves understanding the model's architecture, training process, and computational requirements. Here's a step-by-step breakdown of the reasoning:

1. Understand GPT-3's Architecture

GPT-3 is a transformer-based language model with the following key parameters:

Number of parameters: 175 billion (175 × 10⁹).
Number of layers: 96.
Hidden size: 12,288.
Attention heads: 96.

2. Compute FLOPs per Forward Pass

In a transformer model, the FLOPs per forward pass can be approximated using the formula: [ \text{FLOPs} \approx 2 \times \text{Parameters} \times \text{Sequence Length} ]

Sequence length: GPT-3 uses a sequence length of 2048 tokens.
Parameters: 175 × 10⁹.

So, the FLOPs per forward pass are: [ \text{FLOPs}_{\text{forward}} \approx 2 \times 175 \times 10^9 \times 2048 = 7.168 \times 10^{14} ]

3. Compute FLOPs per Backward Pass

Backpropagation typically requires about twice the FLOPs of a forward pass. Therefore: [ \text{FLOPs}{\text{backward}} \approx 2 \times \text{FLOPs}{\text{forward}} = 2 \times 7.168 \times 10^{14} = 1.4336 \times 10^{15} ]

4. Total FLOPs per Training Step

Each training step consists of one forward pass and one backward pass. Thus: [ \text{FLOPs}{\text{step}} = \text{FLOPs}{\text{forward}} + \text{FLOPs}_{\text{backward}} = 7.168 \times 10^{14} + 1.4336 \times 10^{15} = 2.1504 \times 10^{15} ]

5. Total Training Steps

GPT-3 was trained on 300 billion tokens. With a batch size of 3.2 million tokens per step, the total number of training steps is: [ \text{Steps} = \frac{\text{Total Tokens}}{\text{Batch Size}} = \frac{300 \times 10^9}{3.2 \times 10^6} = 93,750 ]

6. Total FLOPs for Training

Multiply the FLOPs per step by the total number of steps: [ \text{Total FLOPs} = \text{FLOPs}_{\text{step}} \times \text{Steps} = 2.1504 \times 10^{15} \times 93,750 \approx 2.016 \times 10^{20} ]

Final Estimate

The total number of FLOPs required to train GPT-3 is approximately: [ \boxed{2 \times 10^{20} \text{ FLOPs}} ]

This estimate aligns with publicly reported figures and provides a reasonable approximation based on the model's architecture and training process.

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 / 43

AI Board Game Logicreasoning

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

DeepSeek V3.2 Speciale

Deepseek · Dec 2025svg

Google · Dec 2025text

No content available for this output.

Vote

12 / 43

Minimalist Landing Pageweb design

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

DeepSeek V3.2 Speciale

Deepseek · Dec 2025website

Gemini 3 Flash Preview

Google · Dec 2025website

Vote

12 of 43

FAQ

Continue exploring

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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).