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In the paper 'Intriguing Properties of Neural Networks', the process of finding adversarial examples is set up as follows (section 4.1):

We denote by $f : \mathbb{R}^m → \{1 . . . k\}$ a classifier mapping image pixel value vectors to a discrete label set. We also assume that $f$ has an associated continuous loss function denoted by $\text{loss}_f$ : $\mathbb{R}^m × \{1 . . . k\} → \mathbb{R}^+$. For a given $x ∈ \mathbb{R}^m$ image and target label $l ∈ \{1 . . . k\}$, we aim to solve the following box-constrained optimization problem:

Minimize $\lVert r \rVert_2$ subject to:

  1. $f(x + r) = l$
  2. $x + r ∈ [0, 1]^m$

The minimizer $r$ might not be unique, but we denote one such $x + r$ for an arbitrarily chosen minimizer by $D(x, l)$. Informally, $x + r$ is the closest image to $x$ classified as $l$ by $f$. Obviously, $D(x, f(x)) = f(x)$, so this task is non-trivial only if $f(x) \neq l$. In general, the exact computation of $D(x, l)$ is a hard problem, so we approximate it by using a box-constrained L-BFGS. Concretely, we find an approximation of $D(x, l)$ by performing line-search to find the minimum $c > 0$ for which the minimizer $r$ of the following problem satisfies $f(x + r) = l$.

Minimize $c|r| + \text{loss}_f (x + r, l)$ subject to $x + r ∈ [0, 1]^m$

I have two questions about this passage.

  1. Why is $D(x,f(x))=f(x)$? My interpretation of the definition of $D(x,l)$ from the sentence before is that it is equal to $x+r$ where $r$ is the minimum magnitude vector such that $f(x+r)=l$. It appears that $D(x,f(x))=x$. Am I misunderstanding something here?

  2. Why are we looking for the minimum such $c$? My intuition is that the $c|r|$ term of the problem serves to pull $r$ towards the $0$ vector, while the $loss_f$ term serves to pull $r$ towards the "perfect" input image representing label $l$, which will likely be away from the $0$ vector. If this intuition is true, then increasing $c$ should pull the minimizer $r$ towards the $0$ vector and reduce its magnitude. So I would think we would want to find the maximum $c$ for which the minimizer $r$ of that expression satisfies $f(x+r)=l$, as this would lead to a smaller perturbation that still causes an adversarial example. What is wrong with this logic?

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For your first question -- probably just a typo. For your second question, the lagrangian dual of the first formulation would be

$$\max_{\lambda > 0} \min_{r} |r|+ \lambda \text{loss}(x+r,l)$$

If we set $c = 1/\lambda$ and multiply through, then we have

$$\min_{c,r} c|r|+\text{loss}(x+r,l)$$

To gain better intuition on this part, it might help to search for some geometric illustrations of duality (i think math stack exchange has some nice posts about this).

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