number of tries to generate a specific key with a specific probability

bruteforce_table.py

# Prints out a table of the number of tries to generate a specific key with a
# specific probability
#
# More info at https://forum.safedev.org/t/secure-random-relocation/1321/41
#
# It's messy but the maths is correct.

import math

def tidyNum(n):
    if n < 1000:
        return str(n)
    if n < 1000000:
        return str(round(n/1000)) + "K" 
    if n < 1000000000:
        return str(round(n/1000000)) + "M" 
    return str(round(n/1000000000)) + "B" 

# Initiate the results
probabilities = [0.1, 0.5, 0.9, 0.99, 0.9999]
probabilityHeading = [""]
probabilityHeading.extend(probabilities)
probabilityHeading.extend(["", ""])
rows = [ 
    probabilityHeading,
    ["Prefix Length", "", "", "", "", "", "Sections", "Vaults"],
]

# calculate the values
for prefixLength in range(3,41):
    row = [prefixLength]
    for probability in probabilities:
        totalGuesses = round(math.log(1-probability) / math.log(1-1/(2**prefixLength)))
        tidyGuesses = tidyNum(totalGuesses)
        row.append(tidyGuesses)
    numSections = 2**prefixLength
    numVaults = numSections * 50
    row.extend([tidyNum(numSections), tidyNum(numVaults)])
    rows.append(row)

# print the output to stdout
rowFmt = "{:>13} | {:>8} {:>8} {:>6} {:>6} {:>6} | {:>6} {:>6}"

print("Number of guesses to find a key matching a prefix\n")
print(" "*13 + " | probability of generating a valid key  |")
for row in rows:
    print(rowFmt.format(*row))