Deep Learning: Difference between revisions
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For l2 p>=2, <math>p \geq 2</math>, <math>vol(A(\epsilon g d p)) \geq 1 - \frac{\exp(-2 \pi \epsilon^2)}{2 \pi \epsilon}</math> | For l2 p>=2, <math>p \geq 2</math>, <math>vol(A(\epsilon g d p)) \geq 1 - \frac{\exp(-2 \pi \epsilon^2)}{2 \pi \epsilon}</math> | ||
For p=2, the diameter of the hypercube is <math>O(\sqrt{d})</math>. | For p=2, the diameter of the hypercube is <math>O(\sqrt{d})</math> so we should use <math>\epsilon \approx O(\sqrt{d})</math>. | ||
For p=infinity, the diameter of the cube is 1. | For p=infinity, the diameter of the cube is 1 so we should pick a constant <math>\epsilon</math>. | ||
This shows if you pick a random sample, there is a high probability of it being misclassified or there being an adversarial example within epsilon. | This shows if you pick a random sample, there is a high probability of it being misclassified or there being an adversarial example within epsilon. | ||