What is the difference between GPT-5.2 and Amazon Nova 2 Lite?

GPT-5.2 is developed by OpenAI while Amazon Nova 2 Lite is developed by Amazon. GPT-5.2 has a 400K token context window vs Amazon Nova 2 Lite's 1.0M. You can compare their actual outputs across 35 challenges on RIVAL to see how they differ in practice.

Which is better, GPT-5.2 or Amazon Nova 2 Lite?

It depends on your use case. GPT-5.2 and Amazon Nova 2 Lite each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 35 challenges so you can judge which fits your needs best.

How much does GPT-5.2 cost compared to Amazon Nova 2 Lite?

GPT-5.2 costs $1.75/M input tokens and Amazon Nova 2 Lite costs $0.5/M input tokens. Amazon Nova 2 Lite is $1.25/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 GPT-5.2 and Amazon Nova 2 Lite on RIVAL?

This page shows a side-by-side comparison of GPT-5.2 and Amazon Nova 2 Lite 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

GPT-5.2vsAmazon Nova 2 Lite

35 challenges

Why GPT-5.2?

No standout advantages

Why Amazon Nova 2 Lite?

8.3x cheaper overall$0.50/M in · $1.50/M out

2.5x more context1.0M

Favorites

GPT-5.2Amazon Nova 2 Lite

Movie

Spirited Away

20018.5

—

Album

Kind of Blue

Miles Davis

—

Book

To Kill a Mockingbird

Harper Lee

—

City

Kyoto

Japan

—

Game

Outer Wilds

Indie, Adventure4.4

—

12 / 35

Estimate Complexityreasoning

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

GPT-5.2

Openai · Dec 2025text

Below is a back-of-the-envelope estimate for the total training compute (in FLOPs) for GPT‑3, done from first principles and common Transformer training heuristics.

1) What we’re estimating

We want total FLOPs over the whole training run:

[ \text{Total FLOPs} \approx (\text{FLOPs per token}) \times (\text{# training tokens}) ]

So we need:

# parameters (P) (GPT‑3 “175B” ⇒ (P \approx 1.75\times 10^{11}))
# training tokens (T) (we’ll estimate)
A way to relate FLOPs/token to (P)

2) FLOPs per token ≈ a small multiple of parameter count

A commonly used rule for dense Transformers is:

[ \text{Training FLOPs} \approx 6 , P , T ]

Why “6”?

A forward pass through a big dense network costs on the order of ~2 FLOPs per parameter (roughly one multiply-add per weight, counted as ~2 FLOPs).
Backward pass (gradients wrt activations + wrt weights) costs about ~2× forward for dense layers.
So forward + backward ≈ ~3× forward.
That yields about: [ \text{FLOPs/token} \approx 3 \times (2P) = 6P ]

This ignores some extras (attention score computations, softmax, layernorm, embedding lookups), but for very large models the big matrix multiplies dominate, and the (6PT) scaling is usually the right order of magnitude.

So we’ll use: [ \text{Total FLOPs} \approx 6PT ]

3) Estimate the number of training tokens (T)

We need an assumption. GPT‑3 is widely described as trained on hundreds of billions of tokens. Without looking it up, a reasonable guess range is:

Low: (T \approx 200\text{B} = 2\times 10^{11})
Mid: (T \approx 300\text{B} = 3\times 10^{11})
High: (T \approx 500\text{B} = 5\times 10^{11})

I’ll compute the mid estimate and show the range.

4) Compute total FLOPs

Given:

(P = 1.75\times 10^{11})
(T = 3\times 10^{11})

[ \text{Total FLOPs} \approx 6 \times (1.75\times 10^{11}) \times (3\times 10^{11}) ]

Multiply constants:

(6 \times 1.75 \times 3 = 31.5)

Multiply powers of ten:

(10^{11}\times 10^{11} = 10^{22})

So: [ \text{Total FLOPs} \approx 31.5 \times 10^{22} = 3.15\times 10^{23}\ \text{FLOPs} ]

Range using different token assumptions

If (T=2\times 10^{11}): [ 6 \cdot 1.75 \cdot 2 \times 10^{22} = 21 \times 10^{22} = 2.1\times 10^{23} ]
If (T=5\times 10^{11}): [ 6 \cdot 1.75 \cdot 5 \times 10^{22} = 52.5 \times 10^{22} = 5.25\times 10^{23} ]

5) Final estimate

Estimated total training compute for GPT‑3 (175B): [ \boxed{\text{On the order of }10^{23}\text{ FLOPs, roughly }3\times 10^{23}\text{ FLOPs (with a plausible range }2\text{–}5\times 10^{23}\text{).}} ]

If you tell me what you want to assume for the training tokens (e.g., 300B vs 500B) or whether you want to include optimizer overhead (Adam adds extra memory ops; FLOP impact is smaller but nonzero), I can refine the estimate accordingly.

Amazon Nova 2 Lite

Amazon · Dec 2025text

Estimating the Total FLOPs for Training GPT-3

To estimate the total number of floating-point operations (FLOPs) required to train GPT-3, we’ll break the problem down step by step. The key factors are:

Number of parameters in GPT-3.
Total number of tokens processed during training.
FLOPs per token (computation per token during training).

We’ll derive reasonable estimates without searching online, using logical reasoning based on known transformer architecture principles and typical training practices.

Step 1: Number of Parameters (N)

GPT-3 has 175 billion parameters.

This is a well-publicized fact about GPT-3, so we take it as a given:
[ N = 175 \times 10^9 = 1.75 \times 10^{11} \text{ parameters} ]

Step 2: Total Tokens Processed During Training (T)

Training FLOPs depend on how much data the model sees. We need to estimate the total number of tokens ((T)) processed during training.

Reasoning:

Dataset size: GPT-3 was trained on a massive text corpus (e.g., books, web data).
Token definition: In transformer models, a "token" is typically ~1–2 characters (e.g., subword units like BPE).
Rough token count per byte:
- Text averages ~5–6 characters per word.
- A word ≈ 1 token (after tokenization).
- Thus, ≈5–6 bytes per token (since ASCII/UTF-8 uses 1–4 bytes per character, but tokenization groups characters).
- Conservative estimate: 5 bytes per token.
Dataset size in bytes:
GPT-3’s training data is often cited as " Hundreds of gigabytes to a few terabytes".
- Let’s assume ~300 billion tokens (a common ballpark for large LLM training).
  Why?
  - Smaller models (e.g., GPT-2) used ~40 GB of text ≈ 40 billion tokens.
  - GPT-3 is ~4,000× larger in parameters than GPT-2 (175B vs. 110M).
  - Training data often scales less than linearly with model size (due to data saturation), but for estimation, we’ll use:
    [ T \approx 300 \times 10^9 = 3 \times 10^{11} \text{ tokens} ]

Step 3: FLOPs per Token (F)

Now we estimate FLOPs required to process one token during training (both forward and backward passes).

Key Operations per Token:

For a transformer model like GPT-3, processing one token involves:

Self-attention mechanism:
- Query, Key, Value projections: 3 matrix multiplications.
- Output projection: 1 matrix multiplication.
- Softmax (cheaper than matrix multiplies).
Feed-forward network (FFN): 2 matrix multiplications (input → hidden → output).
Residual additions and layer norms: Minor compared to matrix multiplies.
Backward pass: Doubles the FLOPs of the forward pass (gradients, weight updates).

FLOPs per Layer:

A single transformer layer with (d_{\text{model}}) dimensions processes a token with ~(6d_{\text{model}}^2) FLOPs** (forward + backward).
- Why? Each matrix multiply of size (d_{\text{model}} \times d_{\text{model}}) costs (2d_{\text{model}}^2) FLOPs (1 multiply + 1 add). With ~3–4 such operations per layer (attention + FFN), forward pass ≈ (4d_{\text{model}}^2) FLOPs. Backward pass is similar, so total ≈ (8d_{\text{model}}^2).
- However, parameters (N) relate to (d_{\text{model}}):
  Total parameters (N \approx 2 \times \text{layers} \times d_{\text{model}}^2) (for attention + FFN weights).
  Thus, (d_{\text{model}}^2 \approx N / (2 \times \text{layers})).
  For simplicity, we use an empirical rule of thumb:
  [ \text{FLOPs per token} \approx 6N ] This is widely used in literature for transformer models (accounts for all layers, forward/backward, and overhead).

Justification:

If (N = 1.75 \times 10^{11}), then:
[ F \approx 6 \times N = 6 \times 1.75 \times 10^{11} = 1.05 \times 10^{12} \text{ FLOPs per token} ]

Step 4: Total FLOPs for Training

Now combine all factors:
[ \text{Total FLOPs} = T \times F = T \times 6N ]

Plugging in the values:

(T = 3 \times 10^{11}) tokens
(N = 1.75 \times 10^{11}) parameters

[ \text{Total FLOPs} = (3 \times 10^{11}) \times 6 \times (1.75 \times 10^{11}) ]

Calculation:

Compute (6 \times N):
[ 6 \times 1.75 \times 10^{11} = 10.5 \times 10^{11} = 1.05 \times 10^{12} ]
Multiply by (T):
[ (3 \times 10^{11}) \times (1.05 \times 10^{12}) = 3.15 \times 10^{23} \text{ FLOPs} ]

Final Estimate

[ \boxed{3 \times 10^{23} \text{ FLOPs}} ]

Summary of Assumptions:

Factor	Value	Reasoning
Parameters ((N))	(1.75 \times 10^{11})	Publicly stated for GPT-3.
Total tokens ((T))	(3 \times 10^{11})	Estimated from typical LLM training data (scaled from smaller models).
FLOPs/token ((F))	(6N)	Empirical rule for transformers (forward + backward passes).
Total FLOPs	(\approx 3 \times 10^{23})	(T \times 6N).

Context:

This is consistent with known estimates for large LLM training (e.g., training GPT-3 was reported to use ~10^23 – 10^24 FLOPs).
Real-world training uses optimizations (mixed precision, gradient accumulation), but our estimate captures the theoretical lower bound.

This calculation shows why training GPT-3 required massive computational resources (e.g., thousands of GPUs/TPUs for weeks).

Vote

2 / 35

AI Board Game Logicreasoning

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

GPT-5.2

Openai · Dec 2025svg

Amazon · Dec 2025text

No content available for this output.

Vote

Minimalist Landing Pageweb design

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

GPT-5.2

Openai · Dec 2025website

Amazon Nova 2 Lite

Amazon · Dec 2025website

Vote

12 of 35

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Estimating the Total FLOPs for Training GPT-3

To estimate the total number of floating-point operations (FLOPs) required to train GPT-3, we’ll break the problem down step by step. The key factors are:

Number of parameters in GPT-3.
Total number of tokens processed during training.
FLOPs per token (computation per token during training).

We’ll derive reasonable estimates without searching online, using logical reasoning based on known transformer architecture principles and typical training practices.

Step 1: Number of Parameters (N)

GPT-3 has 175 billion parameters.

This is a well-publicized fact about GPT-3, so we take it as a given:
[ N = 175 \times 10^9 = 1.75 \times 10^{11} \text{ parameters} ]

Step 2: Total Tokens Processed During Training (T)

Training FLOPs depend on how much data the model sees. We need to estimate the total number of tokens ((T)) processed during training.

Reasoning:

Dataset size: GPT-3 was trained on a massive text corpus (e.g., books, web data).
Token definition: In transformer models, a "token" is typically ~1–2 characters (e.g., subword units like BPE).
Rough token count per byte:
- Text averages ~5–6 characters per word.
- A word ≈ 1 token (after tokenization).
- Thus, ≈5–6 bytes per token (since ASCII/UTF-8 uses 1–4 bytes per character, but tokenization groups characters).
- Conservative estimate: 5 bytes per token.
Dataset size in bytes:
GPT-3’s training data is often cited as " Hundreds of gigabytes to a few terabytes".
- Let’s assume ~300 billion tokens (a common ballpark for large LLM training).
  Why?
  - Smaller models (e.g., GPT-2) used ~40 GB of text ≈ 40 billion tokens.
  - GPT-3 is ~4,000× larger in parameters than GPT-2 (175B vs. 110M).
  - Training data often scales less than linearly with model size (due to data saturation), but for estimation, we’ll use:
    [ T \approx 300 \times 10^9 = 3 \times 10^{11} \text{ tokens} ]

Step 3: FLOPs per Token (F)

Now we estimate FLOPs required to process one token during training (both forward and backward passes).

Key Operations per Token:

For a transformer model like GPT-3, processing one token involves:

Self-attention mechanism:
- Query, Key, Value projections: 3 matrix multiplications.
- Output projection: 1 matrix multiplication.
- Softmax (cheaper than matrix multiplies).
Feed-forward network (FFN): 2 matrix multiplications (input → hidden → output).
Residual additions and layer norms: Minor compared to matrix multiplies.
Backward pass: Doubles the FLOPs of the forward pass (gradients, weight updates).

FLOPs per Layer:

A single transformer layer with (d_{\text{model}}) dimensions processes a token with ~(6d_{\text{model}}^2) FLOPs** (forward + backward).
- Why? Each matrix multiply of size (d_{\text{model}} \times d_{\text{model}}) costs (2d_{\text{model}}^2) FLOPs (1 multiply + 1 add). With ~3–4 such operations per layer (attention + FFN), forward pass ≈ (4d_{\text{model}}^2) FLOPs. Backward pass is similar, so total ≈ (8d_{\text{model}}^2).
- However, parameters (N) relate to (d_{\text{model}}):
  Total parameters (N \approx 2 \times \text{layers} \times d_{\text{model}}^2) (for attention + FFN weights).
  Thus, (d_{\text{model}}^2 \approx N / (2 \times \text{layers})).
  For simplicity, we use an empirical rule of thumb:
  [ \text{FLOPs per token} \approx 6N ] This is widely used in literature for transformer models (accounts for all layers, forward/backward, and overhead).

Justification:

If (N = 1.75 \times 10^{11}), then:
[ F \approx 6 \times N = 6 \times 1.75 \times 10^{11} = 1.05 \times 10^{12} \text{ FLOPs per token} ]

Step 4: Total FLOPs for Training

Now combine all factors:
[ \text{Total FLOPs} = T \times F = T \times 6N ]

Plugging in the values:

(T = 3 \times 10^{11}) tokens
(N = 1.75 \times 10^{11}) parameters

[ \text{Total FLOPs} = (3 \times 10^{11}) \times 6 \times (1.75 \times 10^{11}) ]

Calculation:

Compute (6 \times N):
[ 6 \times 1.75 \times 10^{11} = 10.5 \times 10^{11} = 1.05 \times 10^{12} ]
Multiply by (T):
[ (3 \times 10^{11}) \times (1.05 \times 10^{12}) = 3.15 \times 10^{23} \text{ FLOPs} ]

Final Estimate

[ \boxed{3 \times 10^{23} \text{ FLOPs}} ]

Summary of Assumptions:

Factor	Value	Reasoning
Parameters ((N))	(1.75 \times 10^{11})	Publicly stated for GPT-3.
Total tokens ((T))	(3 \times 10^{11})	Estimated from typical LLM training data (scaled from smaller models).
FLOPs/token ((F))	(6N)	Empirical rule for transformers (forward + backward passes).
Total FLOPs	(\approx 3 \times 10^{23})	(T \times 6N).

Context:

This is consistent with known estimates for large LLM training (e.g., training GPT-3 was reported to use ~10^23 – 10^24 FLOPs).
Real-world training uses optimizations (mixed precision, gradient accumulation), but our estimate captures the theoretical lower bound.

This calculation shows why training GPT-3 required massive computational resources (e.g., thousands of GPUs/TPUs for weeks).