Privacy Amplification Bits

Problem

Return the number of secure bits produced by privacy amplification:

max(0, min_entropy - leaked - 2 * log2(1 / epsilon))

where min_entropy is the min-entropy the eavesdropper does not know, leaked is the number of exposed bits, and epsilon the desired statistical security parameter.

Hint

A smaller epsilon costs two extra bits per decade of security.

Test cases

case 1

in: [10,2,0.001]

expected: 0

case 2

in: [20,2,0.01]

expected: 4.712287620450551

case 3

in: [8,0,0.5]

expected: 6

case 4

in: [5,4,0.1]

expected: 0

Starter codePython

import math
def privacy_amplification_bits(min_entropy, leaked, epsilon):
    # Your code here
    pass

4 test cases · run your code against them in the editor