<p>I want to build a model that when given two vectors, outputs the probability of one vector being the encoded form of the other. I have 2 strategies for this: (<strong>Dataset available</strong>)</p>
<ol>
<li><p>I can directly feed them in concatenation to a neural network and take the output as the probability.</p>
</li>
<li><p>I can train a conditional GAN with the conditional vector being the encoded vector and using the original vector as the generated one. In this case, the discriminator works as the network that I train in the first approach.</p>
</li>
</ol>
<p>Which approach is better? Am I thinking in the right direction?</p>

CC BY-SA 4.0
2021-04-27T04:46:14.003Z
I haven't dealt with vector encodings, but I would suggest changing the title of your post to be more relevant. Maybe something like "What model best suits vector encoding classification"
Recessive
CC BY-SA 4.0
2021-04-27T08:58:59.793Z
Okayy! I'll do that
Aditya Abhiram
<neural-networks><generative-adversarial-networks><models>
Which approach best suits vector encodings?
Aditya Abhiram
# Which approach best suits vector encodings?

(Asked by: username_0 on 2021-04-27)

I want to build a model that when given two vectors, outputs the probability of one vector being the encoded form of the other. I have 2 strategies for this: (**Dataset available**)

1. I can directly feed them in concatenation to a neural network and take the output as the probability.
2. I can train a conditional GAN with the conditional vector being the encoded vector and using the original vector as the generated one. In this case, the discriminator works as the network that I train in the first approach.

Which approach is better? Am I thinking in the right direction?

**Tags:** neural-networks, generative-adversarial-networks, models

> **Comments:**
> **username_1**: I haven't dealt with vector encodings, but I would suggest changing the title of your post to be more relevant. Maybe something like "What model best suits vector encoding classification"  
> **username_0**: Okayy! I'll do that  

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## 0 answers

