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