Why do GANs minimize -log D(G(z)) instead of the minimax form log(1 - D(G(z)))?

## The two candidate generator losses The value function is `V(D, G) = E[log D(x)] + E[log(1 - D(G(z)))]`, and the game says the generator minimizes V. Since only the second term contains the generator, the **saturating** (minimax) generator loss is `L_G_sat = E_z[ log(1 - D(G(z))) ]`, minimized. The **non-saturating** (heuristic) alternative is `L_G_ns = E_z[ -log D(G(z)) ]`, minimized. Both are minimized by driving `D(G(z))` toward 1, so they agree about what the generator wants. They disagree about how hard they push at each level of the discriminator's confidence, and that difference decides whether training starts at all. ## Where the saturation comes from The discriminator ends in a sigmoid: `D = sigmoid(s)`, where `s` is the pre-sigmoid logit the network computes. Everything the generator learns arrives through `s`, so the honest way to compare the two losses is to differentiate each with respect to `s`. For the saturating form: `d/ds log(1 - sigmoid(s)) = -1/(1 - D) * D(1 - D) = -D`. For the non-saturating form: `d/ds [-log sigmoid(s)] = -1/D * D(1 - D) = -(1 - D)`. So the two gradient magnitudes are `D(G(z))` and `1 - D(G(z))` — mirror images. Now think about early training. The generator is random, the discriminator learns within a few hundred steps to reject its output, and `D(G(z))` falls to something like 0.02. The saturating loss then supplies a gradient of magnitude 0.02: essentially nothing. The non-saturating loss supplies 0.98: nearly full strength. Late in training, if the generator ever fools the discriminator completely, the two swap roles — but that regime is not where training gets stuck, so the asymmetry is the one you want. Notice the mechanism carefully, because this is where candidates fumble. It is *not* that `log(1 - D)` has a small derivative with respect to `D` — `d/dD log(1 - D) = -1/(1 - D)`, which is about -1 near D = 0. The flatness comes from the **sigmoid saturating**: a confident discriminator sits far out on the logit axis, where `dD/ds = D(1 - D)` is tiny, and the chain rule kills the signal before it reaches the generator. ## What the swap costs you theoretically The elegance of the minimax form is that, against an optimal discriminator, the generator is minimizing a symmetric divergence between its distribution and the data's. The non-saturating loss gives that up. It is not the same function, not a monotone transform of it, and not a constant offset — only the fixed point (the generator wants to be indistinguishable) survives. A published analysis of the non-saturating gradient under an optimal discriminator shows it corresponds to a combination of a reverse KL term and a *negatively* weighted Jensen-Shannon term, so it is not descending any single well-behaved divergence. In practice this is a trade everyone takes: an objective with a beautiful interpretation and no usable gradient at initialization is worse than a heuristic that actually moves. The discriminator's own objective is untouched. It still maximizes the original value function on real and generated batches; only the generator's loss changes. That asymmetry — different loss for each player, no longer a single shared value function — is why some people prefer to describe modern adversarial training as two coupled objectives rather than as a minimax game. ## When the saturating form is not actually a problem If the discriminator is weak or deliberately handicapped, `D(G(z))` hovers near 0.5, both losses give gradients around 0.5 in magnitude, and the saturating form trains fine. The pathology is specific to a *confident* discriminator, which is the normal state at initialization because distinguishing random noise output from real data is trivial. So the failure is a start-of-training failure above all, and it can recur later whenever the discriminator pulls decisively ahead. ## Interview framing A complete answer has four beats: state both losses; explain that the discriminator wins early so `D(G(z))` is near 0; show that the saturating loss's gradient through the logit scales as `D(G(z))` while the non-saturating one scales as `1 - D(G(z))`; and concede that the swap preserves the generator's preferred direction but breaks the exact divergence interpretation of the game. If you can add that the discriminator's update is unchanged, you have covered everything the question is testing.