RSA算法初探
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RSA算法
最近在学习支付宝的支付方式,在支付宝接口开发中,用到了RSA加密的方式来加密,于是就对其感兴趣,于是参考网上的RSA算法,结合密码学,于是写出了这篇文章,本文不涉及到任何数学知识,只教你会用。
1. 公钥 私钥的概念
公钥就是大家都能看得到的密钥,也就是我们使用这个密钥来对文本进行加密,假设喜洋洋将hello
发给灰太狼,首先喜洋洋用公钥对hello
进行加密,一般对ascii
码来进行操作,设加密后为¥%#@!&……%
,当然了,有人会截取到这段加密后的文字,但不知道什么意思,当灰太狼拿到这个文字,使用私钥进行解密,因为私钥只有灰太狼有,所以只有灰太狼才能知道这段文字是什么意思!
2. 实战
首先我们要构造公钥和私钥
import random,sys,os,rabinMillerfrom rabinMiller import gcd,findModInversedef generateKey(keysize=2048): ############################# p = rabinMiller.generateLargePrime(keysize) q = rabinMiller.generateLargePrime(keysize) n = p*q ################################ while True: e = random.randrange(2**(keysize-1),2**keysize) if gcd(e,(p-1)*(q-1)) == 1: break ################################ d = findModInverse(e,(p-1)*(q-1)) ################################ public_key = (n,e) private_key = (n,d) return (public_key,private_key)
其中所需要的文件为:
# Primality Testing with the Rabin-Miller Algorithm# http://inventwithpython.com/hacking (BSD Licensed)import randomdef rabinMiller(num): # Returns True if num is a prime number. s = num - 1 t = 0 while s % 2 == 0: # keep halving s while it is even (and use t # to count how many times we halve s) s = s // 2 t += 1 for trials in range(5): # try to falsify num's primality 5 times a = random.randrange(2, num - 1) v = pow(a, s, num) if v != 1: # this test does not apply if v is 1. i = 0 while v != (num - 1): if i == t - 1: return False else: i = i + 1 v = (v ** 2) % num return Truedef isPrime(num): # Return True if num is a prime number. This function does a quicker # prime number check before calling rabinMiller(). if (num < 2): return False # 0, 1, and negative numbers are not prime # About 1/3 of the time we can quickly determine if num is not prime # by dividing by the first few dozen prime numbers. This is quicker # than rabinMiller(), but unlike rabinMiller() is not guaranteed to # prove that a number is prime. lowPrimes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997] if num in lowPrimes: return True # See if any of the low prime numbers can divide num for prime in lowPrimes: if (num % prime == 0): return False # If all else fails, call rabinMiller() to determine if num is a prime. return rabinMiller(num)def generateLargePrime(keysize=1024): # Return a random prime number of keysize bits in size. while True: num = random.randrange(2**(keysize-1), 2**(keysize)) if isPrime(num): return numdef gcd(a, b): while a != 0: a, b = b % a, a return bdef findModInverse(a, m): # Returns the modular inverse of a % m, which is # the number x such that a*x % m = 1 if gcd(a, m) != 1: return None # no mod inverse if a & m aren't relatively prime # Calculate using the Extended Euclidean Algorithm: u1, u2, u3 = 1, 0, a v1, v2, v3 = 0, 1, m while v3 != 0: q = u3 // v3 # // is the integer division operator v1, v2, v3, u1, u2, u3 = (u1 - q * v1), (u2 - q * v2), (u3 - q * v3), v1, v2, v3 return u1 % m
学过基本数学的话,这段代码还是能够看得懂得。
下面我们就来加密解密了。
message = 'hello' #我要加密的文字"""plaintext^e mod n"""public_key, private_key = generateKey() #构造私钥公钥n,e = public_keyn,d = private_key"""encrypt = plaintext ^ e % n"""encrytext = [pow(ord(x),e,n) for x in message]print(encrytext)"""plaintext = encrypt ^ d % n"""dencrypt = [chr(pow(x,d,n)) for x in encrytext]dencrypt = ''.join(dencrypt)print(dencrypt)
我们就简单地使用一下这个RSA算法,具体细节并没有深究,需要了解的同学可以网上了解一下。
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