UCB Exploration Bonus

Problem

Compute the UCB exploration bonus c * sqrt(ln(t) / count).

count is how many times the action has been pulled and t is the total number of pulls so far. A non-positive count returns 0.0 to keep the result finite.

Hint

The bonus shrinks as the action is pulled more often.

Test cases

case 1

in: [5,10,1]

expected: 0.6786140424415112

case 2

in: [1,100,2]

expected: 4.291932052578694

case 3

in: [0,10,1]

expected: 0

case 4

in: [4,4,1]

expected: 0.5887050112577373

Starter codePython

import math
def ucb_exploration_bonus(count, t, c):
    # Your code here
    pass

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