In the research paper titled Conditional Generative Adversarial Nets by Mehdi Mirza and Simon Osindero, there is a notion of conditioning a neural network on class label.

It has been mentioned in abstract that we need to simply feed extra input $y$ to the generator and discriminator of an unconditional GAN.

Generative Adversarial Nets were recently introduced as a novel way to train generative models. In this work we introduce the conditional version of generative adversarial nets, which can be constructed by simply feeding the data, $y$, we wish to condition on to both the generator and discriminator. We show that this model can generate MNIST digits conditioned on class labels. We also illustrate how this model could be used to learn a multi-modal model, and provide preliminary examples of an application to image tagging in which we demonstrate how this approach can generate descriptive tags which are not part of training labels.

So, I cannot see whether there is any special treatment for input $y$.

If there is no special treatment for the data $y$ then why do they call $y$ a condition and follow the notation of conditional probability such as $G(z|y), D(x|y)$ instead of $G(z,y), D(x,y)$?

If there is a special treatment to input $y$ then what is that special? Don't they pass $y$ same as $x$ to the neural networks?

  • $\begingroup$ Normally data is not fed to both the generator and the discriminator, so isn't that already a special treatment? $\endgroup$
    – user253751
    Aug 5 at 10:35
  • $\begingroup$ @user253751 yeah true. In general $z$ is the only input to generator and $x$ or $G(z)$ is the actual input to discriminator. Please answer it if you are sure that $y$ has no difference in treatment as $z$ for generator and $y$ has no difference in treatment as $x$ or $G(z)$ for discriminator. But, how would they pass $y$ is another doubt in my question. Will they pass normally? Suppose $z=[1, 2, 3, 4]$ and $y = [1, 0, 0, 0, 0]$ then what can be the actual input for $G$? $\endgroup$
    – hanugm
    Aug 7 at 22:32
  • $\begingroup$ @user253751 why can't they use $G(z, y)$ instead of $G(z/y)$ etc.,? $\endgroup$
    – hanugm
    Aug 7 at 22:33
  • $\begingroup$ they can, or they can not. It's their choice since they are the ones writing the letters. $\endgroup$
    – user253751
    Aug 8 at 20:33

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