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ttbrunner avatar ttbrunner commented on July 18, 2024

Hi - sorry for the late answer. That's right, the images are flattened to vectors and the dot product is taken. This is equivalent:

np.vdot(a, b) == np.dot(a.flatten(), b.flatten())

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htZhang25 avatar htZhang25 commented on July 18, 2024

By the way, in your work, if the computed gradient of the current sample (corresponding to the label of target sample ) has positive values in certain element, you will project the vector '-gradient' which has negative values in those elments onto the orthogonal plane, to keep the prediction result away from the target label. Is it right?

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ttbrunner avatar ttbrunner commented on July 18, 2024

Well, the idea is simply to make sure the distance to the image under attack stays the same, no matter in which (orthogonal) direction we move. This is the premise of the original Boundary Attack (the Figures in the original paper by Brendel et al. https://arxiv.org/abs/1712.04248 show it nicely).

If we followed the gradient without projection, we might significantly increase the distance to the original image, and the attack might diverge. The projection makes sure that the Boundary Attack will only generate orthogonal samples (keeping the distance constant), and from there a small step towards the original image is taken. So, in the worst case (the label is not as desired), nothing will happen except that we need to draw more and more samples. But the attack cannot diverge/fail.

Projecting the gradient makes sure we don't violate the assumption of the Boundary Attack. Still, maybe it's not necessary and we could make a stronger attack without this projection. Unfortunately I haven't followed the newest papers on this.

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