Differential Privacy Epsilon

Problem

Return the privacy parameter epsilon of the Laplace mechanism given the query sensitivity and the Laplace noise scale:

epsilon = sensitivity / scale

Hint

Larger noise means smaller epsilon and stronger privacy.

case 1

in: [1,2]

expected: 0.5

case 2

in: [2,1]

expected: 2

case 3

in: [5,10]

expected: 0.5

case 4

in: [1,0.5]

expected: 2

Starter codePython

def laplace_epsilon(sensitivity, scale):
    # Your code here
    pass

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