Laplace Noise Scale

Problem

Return the Laplace noise scale needed to achieve epsilon-differential privacy for a query with the given sensitivity:

scale = sensitivity / epsilon

Hint

This inverts epsilon = sensitivity / scale.

case 1

in: [1,1]

expected: 1

case 2

in: [2,0.5]

expected: 4

case 3

in: [5,10]

expected: 0.5

case 4

in: [1,2]

expected: 0.5

Starter codePython

def laplace_scale(sensitivity, epsilon):
    # Your code here
    pass

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