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Solves problem 59
This commit is contained in:
Julien Lengrand-Lambert
@@ -71,11 +71,14 @@ So you may find some of the code here quite ugly. And this is the case :). Why o
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54 - How many hands does Player 1 win in radndom hands of Poker? - < 1 sec <br />
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55 - How many Lychrel numbers are there below ten-thousand? - < 1 sec <br />
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56 - Considering natural numbers of the form, a^b, finding the maximum digital sum. > 3 sec <br />
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59 - Decrypt XOR ASCII - 23.4s<br />
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67 - Using an efficient algorithm find the maximal sum in the triangle? - 0.027 <br />
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## In progress:
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47 - Find the first four consecutive integers to have four distinct primes factors. <br />
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57 - How many fractions contain a numerator with more digits than denominator? <br />
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58 - What is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%? <br />
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97 - Find the last ten digits of the non-Mersenne prime: 28433 <20> 2^7830457 + 1. <br />
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## Contact
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@@ -87,4 +90,4 @@ You can contact me at julien at lengrand dot fr, or on my [current website](http
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Last update : 22/05/2014
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Last update : 25/05/2014
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8
e_49.py
8
e_49.py
@@ -87,12 +87,4 @@ def findPrimePerms(prime):
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if __name__ == '__main__':
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primes = fourLengthPrime()
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print getPrimesPermutations(primes, 3)
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#itertools.combinations(lst, i)]
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#getPrimesPermutations(primes, 3)
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# try:
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# while primes.next():
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# print primes.next()
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# except StopIteration:
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# print "Reached end of list"
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#print "Answer : %d " % (last_ten())
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30
e_57.py
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30
e_57.py
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@@ -0,0 +1,30 @@
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#!/usr/bin/env python
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'''
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Created on 10 feb. 2012
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@author: Julien Lengrand-Lambert
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DESCRIPTION: Solves problem 57 of Project Euler
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It is possible to show that the square root of two can be expressed
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as an infinite continued fraction.
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sqrt 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
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By expanding this for the first four iterations, we get:
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1 + 1/2 = 3/2 = 1.5
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1 + 1/(2 + 1/2) = 7/5 = 1.4
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1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
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1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
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The next three expansions are 99/70, 239/169, and 577/408,
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but the eighth expansion, 1393/985, is the first example where the number
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of digits in the numerator exceeds the number of digits in the denominator.
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In the first one-thousand expansions, how many fractions contain a numerator
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with more digits than denominator?
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'''
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if __name__ == '__main__':
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print "plouf"
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#print "Answer : %d " % (last_ten())
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31
e_58.py
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31
e_58.py
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#!/usr/bin/env python
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'''
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Created on 10 feb. 2012
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@author: Julien Lengrand-Lambert
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DESCRIPTION: Solves problem 58 of Project Euler
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Starting with 1 and spiralling anticlockwise in the following way,
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a square spiral with side length 7 is formed.
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37 36 35 34 33 32 31
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38 17 16 15 14 13 30
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39 18 5 4 3 12 29
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40 19 6 1 2 11 28
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41 20 7 8 9 10 27
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42 21 22 23 24 25 26
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43 44 45 46 47 48 49
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It is interesting to note that the odd squares lie along the bottom right diagonal,
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but what is more interesting is that 8 out of the 13 numbers lying along both
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diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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If one complete new layer is wrapped around the spiral above, a square spiral
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with side length 9 will be formed. If this process is continued,
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what is the side length of the square spiral for which the ratio of primes
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along both diagonals first falls below 10%?
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'''
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if __name__ == '__main__':
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print "plouf"
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#print "Answer : %d " % (last_ten())
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255
e_59.py
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255
e_59.py
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#!/usr/bin/env python
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'''
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Created on 10 feb. 2012
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@author: Julien Lengrand-Lambert
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DESCRIPTION: Solves problem 59 of Project Euler
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Each character on a computer is assigned a unique code
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and the preferred standard is ASCII (American Standard Code for Information Interchange).
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For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
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A modern encryption method is to take a text file, convert the bytes to ASCII,
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then XOR each byte with a given value, taken from a secret key.
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The advantage with the XOR function is that using the same encryption key
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on the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
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then 107 XOR 42 = 65.
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For unbreakable encryption, the key is the same length as the plain text message,
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and the key is made up of random bytes.
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The user would keep the encrypted message and the encryption key in different locations,
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and without both "halves", it is impossible to decrypt the message.
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Unfortunately, this method is impractical for most users,
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so the modified method is to use a password as a key. If the password is
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shorter than the message, which is likely, the key is repeated cyclically
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throughout the message. The balance for this method is using a sufficiently
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long password key for security, but short enough to be memorable.
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Your task has been made easy, as the encryption key consists of three lower
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case characters. Using cipher1.txt (right click and 'Save Link/Target As...'),
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a file containing the encrypted ASCII codes, and the knowledge that the plain
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text must contain common English words, decrypt the message and find
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the sum of the ASCII values in the original text.
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'''
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import itertools
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import string
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base = 16 # hey, its hex!
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printable_letters = string.printable # list of all printable characters
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#frequency of english letters
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en_freq = {'a': 0.08167,
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'b': 0.01492,
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'c':0.02782,
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'd':0.04253,
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'e':0.12702,
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'f':0.02228,
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'g':0.02015,
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'h':0.06094,
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'i':0.06966,
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'j':0.00153,
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'k':0.00772,
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'l':0.04025,
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'm':0.02406,
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'n':0.06749,
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'o':0.07507,
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'p':0.01929,
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'q':0.00095,
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'r':0.05987,
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's':0.06327,
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't':0.09056,
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'u':0.02758,
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'v':0.00978,
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'w':0.02360,
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'x':0.00150,
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'y':0.01974,
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'z':0.00074}
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def _xorhex(a, b):
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"""
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XORs a against b and returns the result
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a and b are expected to be int values
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XOR Truth Table
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Input Output
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A B
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0 0 0
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0 1 1
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1 0 1
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1 1 0
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"""
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long_res = a ^ b
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string_res = '%x' % long_res # returns a string version
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return long_res
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def load_data(filename):
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"Loads the data of the filename into a table"
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file = open(filename, "r")
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stuff = file.read()
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return [int(val) for val in stuff.split(',')]
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def all_passwords(range, length):
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'''
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Returns all possible passwords given a range of ascii_values, and the length of the password
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'''
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return list(set(itertools.combinations(range*3, length)))
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def apply_password(value, password):
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"""
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Applies a given password to the message.
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"""
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long_pass = repeat_password(password, len(value))
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return [_xorhex(a, b) for a, b in zip(value, long_pass)]
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def repeat_password(password, length):
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"""
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Repeats a password to match the requested size.
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Ex : [1, 2, 3], 7 => [1, 2, 3, 1, 2, 3, 1]
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"""
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repeat = length / len(password)
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rest = length % len(password)
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return password * repeat + password[:rest]
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def all_solutions(data, pass_range, length):
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"""
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Prints all possible solutions
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"""
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res = []
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all_pass = all_passwords(pass_range, length)
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total = len(all_pass)
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print total
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idx = 0
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for password in all_pass:
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idx += 1
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print "%d / %d" % (idx, total)
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sol = apply_password(data, password)
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res.append([to_ascii(sol), password])
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return res
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def to_ascii(data):
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"""
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Turns a list of ascii values into a proper string
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"""
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res = ""
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for el in data:
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res += chr(el)
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return res
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def discard_non_printable(solutions):
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"""
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We discard from a list of hex all the elements that cannot be printed.
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#FIXME: Do better
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"""
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res = []
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for val in solutions:
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is_printable = True
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for v in val[0]:
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if not(v in printable_letters):
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is_printable = False
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if is_printable:
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res.append(val)
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return res
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def letter_freq(work_val):
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"""
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Calculate the frequency of the elements in the string val
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Returns a dictionary of the form {element: frequency} for all elements of val.
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Only present elements will be present in the dictionary (frequency > 0)
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"""
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counts = map(work_val.count ,work_val) # gets count of each element of hex_val
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freqs = [(float(x) / len(work_val)) for x in counts]
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#freqs = [x for x in counts]
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return dict(zip(work_val,freqs)) # transforms into dict of unique values associated to frequencies
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def chi2(ref_dict, test_dict, length):
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"""
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Performs a chi2 adequation testing.
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Given a reference dictionary and a corresponding dctionary to be tested, returns a score corresponding to
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the similitude between the test data and the reference data.
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The lower the score, the higher the similitude.
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Dictionaries are expected to be under the {value:frequency} form.
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Example : chi2({a:0.08, b:0.92}, {a:0.80, b:0.20}, 12)
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length corresponds to the number of elements needed to acquire test_dict
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Returns a float value
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"""
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my_keys = ref_dict.keys()
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score = 0
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for m in my_keys:
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if test_dict.has_key(m): # testing if occurence happened
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freq_val = length * test_dict[m]
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else:
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freq_val = 0
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ref_val = length * ref_dict[m]
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score += pow(freq_val - ref_val, 2) / ref_val
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return score
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def strip_string_to_lowercase(s):
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"""
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Returns s in lowercase, with all its special characters stripped
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Taken from here: http://stackoverflow.com/questions/638893/what-is-the-most-efficient-way-in-python-to-convert-a-string-to-all-lowercase-st#639325
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"""
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tmpStr = s.lower().strip()
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retStrList = []
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for x in tmpStr:
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if x in string.ascii_lowercase:
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retStrList.append(x)
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return ''.join(retStrList)
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def close_to_english(element):
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"""
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Returns a score of likeliness for element to be english
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The computed score is the chi2 adequation test.
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The closest to 0, the most probable the sentence is to be in english
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Table can be found here, solutions at row 25 (english is 26 degrees of liberty).
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http://www.medcalc.org/manual/chi-square-table.php to get the values to compare to
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"""
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# cleans the hex val of special characters, and puts to lowercase
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stripped_el = strip_string_to_lowercase(element)
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length_el = len(stripped_el)
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#calculates the letters frequency of element
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freqs = letter_freq(stripped_el)
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return chi2(en_freq, freqs, length_el)
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def find_most_probable(data, pass_range, length):
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solutions = all_solutions(data, pass_range, length)
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# discards easy fuck ups
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solutions = discard_non_printable(solutions)
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best_candidate = ""
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best_chi = 1000000
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for sol in solutions:
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score = close_to_english(sol[0])
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if score < best_chi:
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best_chi = score
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best_candidate = sol
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return best_candidate
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def get_solution(data, pass_range, length):
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solution = find_most_probable(data, ascii_range, 3)[0]
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sum = 0
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for val in solution:
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sum+=ord(val)
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return sum
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if __name__ == '__main__':
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data = load_data("e_59_cipher1.txt")
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ascii_range = range(ord('a'), ord('z') + 1 )
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print get_solution(data, ascii_range, 3)
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1
e_59_cipher1.txt
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1
e_59_cipher1.txt
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@@ -0,0 +1 @@
|
||||
79,59,12,2,79,35,8,28,20,2,3,68,8,9,68,45,0,12,9,67,68,4,7,5,23,27,1,21,79,85,78,79,85,71,38,10,71,27,12,2,79,6,2,8,13,9,1,13,9,8,68,19,7,1,71,56,11,21,11,68,6,3,22,2,14,0,30,79,1,31,6,23,19,10,0,73,79,44,2,79,19,6,28,68,16,6,16,15,79,35,8,11,72,71,14,10,3,79,12,2,79,19,6,28,68,32,0,0,73,79,86,71,39,1,71,24,5,20,79,13,9,79,16,15,10,68,5,10,3,14,1,10,14,1,3,71,24,13,19,7,68,32,0,0,73,79,87,71,39,1,71,12,22,2,14,16,2,11,68,2,25,1,21,22,16,15,6,10,0,79,16,15,10,22,2,79,13,20,65,68,41,0,16,15,6,10,0,79,1,31,6,23,19,28,68,19,7,5,19,79,12,2,79,0,14,11,10,64,27,68,10,14,15,2,65,68,83,79,40,14,9,1,71,6,16,20,10,8,1,79,19,6,28,68,14,1,68,15,6,9,75,79,5,9,11,68,19,7,13,20,79,8,14,9,1,71,8,13,17,10,23,71,3,13,0,7,16,71,27,11,71,10,18,2,29,29,8,1,1,73,79,81,71,59,12,2,79,8,14,8,12,19,79,23,15,6,10,2,28,68,19,7,22,8,26,3,15,79,16,15,10,68,3,14,22,12,1,1,20,28,72,71,14,10,3,79,16,15,10,68,3,14,22,12,1,1,20,28,68,4,14,10,71,1,1,17,10,22,71,10,28,19,6,10,0,26,13,20,7,68,14,27,74,71,89,68,32,0,0,71,28,1,9,27,68,45,0,12,9,79,16,15,10,68,37,14,20,19,6,23,19,79,83,71,27,11,71,27,1,11,3,68,2,25,1,21,22,11,9,10,68,6,13,11,18,27,68,19,7,1,71,3,13,0,7,16,71,28,11,71,27,12,6,27,68,2,25,1,21,22,11,9,10,68,10,6,3,15,27,68,5,10,8,14,10,18,2,79,6,2,12,5,18,28,1,71,0,2,71,7,13,20,79,16,2,28,16,14,2,11,9,22,74,71,87,68,45,0,12,9,79,12,14,2,23,2,3,2,71,24,5,20,79,10,8,27,68,19,7,1,71,3,13,0,7,16,92,79,12,2,79,19,6,28,68,8,1,8,30,79,5,71,24,13,19,1,1,20,28,68,19,0,68,19,7,1,71,3,13,0,7,16,73,79,93,71,59,12,2,79,11,9,10,68,16,7,11,71,6,23,71,27,12,2,79,16,21,26,1,71,3,13,0,7,16,75,79,19,15,0,68,0,6,18,2,28,68,11,6,3,15,27,68,19,0,68,2,25,1,21,22,11,9,10,72,71,24,5,20,79,3,8,6,10,0,79,16,8,79,7,8,2,1,71,6,10,19,0,68,19,7,1,71,24,11,21,3,0,73,79,85,87,79,38,18,27,68,6,3,16,15,0,17,0,7,68,19,7,1,71,24,11,21,3,0,71,24,5,20,79,9,6,11,1,71,27,12,21,0,17,0,7,68,15,6,9,75,79,16,15,10,68,16,0,22,11,11,68,3,6,0,9,72,16,71,29,1,4,0,3,9,6,30,2,79,12,14,2,68,16,7,1,9,79,12,2,79,7,6,2,1,73,79,85,86,79,33,17,10,10,71,6,10,71,7,13,20,79,11,16,1,68,11,14,10,3,79,5,9,11,68,6,2,11,9,8,68,15,6,23,71,0,19,9,79,20,2,0,20,11,10,72,71,7,1,71,24,5,20,79,10,8,27,68,6,12,7,2,31,16,2,11,74,71,94,86,71,45,17,19,79,16,8,79,5,11,3,68,16,7,11,71,13,1,11,6,1,17,10,0,71,7,13,10,79,5,9,11,68,6,12,7,2,31,16,2,11,68,15,6,9,75,79,12,2,79,3,6,25,1,71,27,12,2,79,22,14,8,12,19,79,16,8,79,6,2,12,11,10,10,68,4,7,13,11,11,22,2,1,68,8,9,68,32,0,0,73,79,85,84,79,48,15,10,29,71,14,22,2,79,22,2,13,11,21,1,69,71,59,12,14,28,68,14,28,68,9,0,16,71,14,68,23,7,29,20,6,7,6,3,68,5,6,22,19,7,68,21,10,23,18,3,16,14,1,3,71,9,22,8,2,68,15,26,9,6,1,68,23,14,23,20,6,11,9,79,11,21,79,20,11,14,10,75,79,16,15,6,23,71,29,1,5,6,22,19,7,68,4,0,9,2,28,68,1,29,11,10,79,35,8,11,74,86,91,68,52,0,68,19,7,1,71,56,11,21,11,68,5,10,7,6,2,1,71,7,17,10,14,10,71,14,10,3,79,8,14,25,1,3,79,12,2,29,1,71,0,10,71,10,5,21,27,12,71,14,9,8,1,3,71,26,23,73,79,44,2,79,19,6,28,68,1,26,8,11,79,11,1,79,17,9,9,5,14,3,13,9,8,68,11,0,18,2,79,5,9,11,68,1,14,13,19,7,2,18,3,10,2,28,23,73,79,37,9,11,68,16,10,68,15,14,18,2,79,23,2,10,10,71,7,13,20,79,3,11,0,22,30,67,68,19,7,1,71,8,8,8,29,29,71,0,2,71,27,12,2,79,11,9,3,29,71,60,11,9,79,11,1,79,16,15,10,68,33,14,16,15,10,22,73
|
||||
@@ -3,9 +3,33 @@
|
||||
{
|
||||
"selected_items":
|
||||
[
|
||||
[
|
||||
"sol",
|
||||
"solution"
|
||||
],
|
||||
[
|
||||
"all_",
|
||||
"all_pass"
|
||||
],
|
||||
[
|
||||
"asci",
|
||||
"ascii_range"
|
||||
],
|
||||
[
|
||||
"all",
|
||||
"all_permutations"
|
||||
"all_passwords"
|
||||
],
|
||||
[
|
||||
"find",
|
||||
"find_most_probable"
|
||||
],
|
||||
[
|
||||
"all_p",
|
||||
"all_pass"
|
||||
],
|
||||
[
|
||||
"rep",
|
||||
"repeat_password"
|
||||
],
|
||||
[
|
||||
"ele",
|
||||
@@ -35,10 +59,6 @@
|
||||
"ee",
|
||||
"election_id"
|
||||
],
|
||||
[
|
||||
"find",
|
||||
"find_by_uuid"
|
||||
],
|
||||
[
|
||||
"ca",
|
||||
"candidates"
|
||||
@@ -477,6 +497,14 @@
|
||||
},
|
||||
"file_history":
|
||||
[
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_59.py",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/README.markdown",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_59_cipher1.txt",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_18.py",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_13.data",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_18.data",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_22.data",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_22.py",
|
||||
"/home/jll/Documents/01_perso/03_project_euler/e_47.py",
|
||||
"/home/jll/Documents/01_perso/00_myelections/app/controllers/application_controller.rb",
|
||||
"/home/jll/Documents/01_perso/00_myelections/app/views/new_election_admin_mailer/new_election_admin_email.text.erb",
|
||||
@@ -596,15 +624,7 @@
|
||||
"/home/jll/Documents/01_perso/18_blog/welcome-in-my-world/index.html",
|
||||
"/home/jll/Documents/01_perso/18_blog/posts.md",
|
||||
"/home/jll/Documents/01_perso/18_blog/404.md",
|
||||
"/home/jll/Documents/01_perso/18_blog/_config.yml",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2012-03-05-computer-vision-companies.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2011-10-17-17.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2011-10-17-home.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2011-10-13-job-space.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2012-09-07-ivolution-development-status-36.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2012-06-11-remove-all-your-thumbs-files.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2012-06-04-is-descartes-the-father-of-agile-development.markdown",
|
||||
"/home/jll/Documents/01_perso/18_blog/_posts/2012-02-22-computer-vision-companies.markdown"
|
||||
"/home/jll/Documents/01_perso/18_blog/_config.yml"
|
||||
],
|
||||
"find":
|
||||
{
|
||||
@@ -821,7 +841,7 @@
|
||||
},
|
||||
"output.exec":
|
||||
{
|
||||
"height": 236.0
|
||||
"height": 253.0
|
||||
},
|
||||
"output.jenkinsIndicator":
|
||||
{
|
||||
|
||||
Reference in New Issue
Block a user