RSA算法是一種非對稱加密算法,是如今普遍使用的公鑰加密算法,主要應用是加密信息和數字簽名。詳情請看維基:http://zh.wikipedia.org/wiki/RSA%E5%8A%A0%E5%AF%86%E6%BC%94%E7%AE%97%E6%B3%95html
算法基本思路:算法
1.公鑰與私鑰的生成:安全
(1)隨機挑選兩個大質數 p 和 q,構造N = p*q;dom
(2)計算歐拉函數φ(N) = (p-1) * (q-1);ide
(3)隨機挑選e,使得gcd(e, φ(N)) = 1,即 e 與 φ(N) 互素;函數
(4)計算d,使得 e*d ≡ 1 (mod φ(N)),即d 是e 的乘法逆元。測試
此時,公鑰爲(e, N),私鑰爲(d, N),公鑰公開,私鑰本身保管。加密
2.加密信息:spa
(1)待加密信息(明文)爲 M,M < N;(由於要作模運算,若M大於N,則後面的運算不會成立,所以當信息比N要大時,應該分塊加密)code
(2)密文C = Me mod N
(3)解密Cd mod N = (Me)d mod N = Md*e mod N ;
要理解爲何能解密?要用到歐拉定理(實際上是費馬小定理的推廣)aφ(n) ≡ 1 (mod n),再推廣:aφ(n)*k ≡ 1 (mod n),得:aφ(n)*k+1 ≡ a (mod n)
注意到 e*d ≡ 1 mod φ(N),即:e*d = 1 + k*φ(N)。
所以,Md*e mod N = M1 + k*φ(N) mod N = M
簡單來講,別人用個人公鑰加密信息發給我,而後我用私鑰解密。
3.數字簽名:
(1)密文C = Md mod N
(2)解密M = Ce mod N = (Md)e mod N = Md*e mod N = M ;(原理同上)
簡單來講,我用本身的密鑰加密簽名,別人用個人公鑰解密能夠看到這是個人簽名。注意,這個不具備隱私性,即任何人均可以解密此簽名。
算法的安全性:基於大整數N難以分解出p和q,構造φ(N);或由N直接構造φ(N)一樣難。
算法的實現:
1.快速冪取模;http://www.cnblogs.com/7hat/p/3398394.html
2.素性測試;http://www.cnblogs.com/7hat/p/3400831.html
3.擴展歐幾里得求乘法逆元和最大公約數;http://www.cnblogs.com/7hat/p/3406494.html
實現代碼:
import random def fastExpMod(b, e, m): """ e = e0*(2^0) + e1*(2^1) + e2*(2^2) + ... + en * (2^n) b^e = b^(e0*(2^0) + e1*(2^1) + e2*(2^2) + ... + en * (2^n)) = b^(e0*(2^0)) * b^(e1*(2^1)) * b^(e2*(2^2)) * ... * b^(en*(2^n)) b^e mod m = ((b^(e0*(2^0)) mod m) * (b^(e1*(2^1)) mod m) * (b^(e2*(2^2)) mod m) * ... * (b^(en*(2^n)) mod m) mod m """ result = 1 while e != 0: if (e&1) == 1: # ei = 1, then mul result = (result * b) % m e >>= 1 # b, b^2, b^4, b^8, ... , b^(2^n) b = (b*b) % m return result def primeTest(n): q = n - 1 k = 0 #Find k, q, satisfied 2^k * q = n - 1 while q % 2 == 0: k += 1; q /= 2 a = random.randint(2, n-2); #If a^q mod n= 1, n maybe is a prime number if fastExpMod(a, q, n) == 1: return "inconclusive" #If there exists j satisfy a ^ ((2 ^ j) * q) mod n == n-1, n maybe is a prime number for j in range(0, k): if fastExpMod(a, (2**j)*q, n) == n - 1: return "inconclusive" #a is not a prime number return "composite" def findPrime(halfkeyLength): while True: #Select a random number n n = random.randint(0, 1<<halfkeyLength) if n % 2 != 0: found = True #If n satisfy primeTest 10 times, then n should be a prime number for i in range(0, 10): if primeTest(n) == "composite": found = False break if found: return n def extendedGCD(a, b): #a*xi + b*yi = ri if b == 0: return (1, 0, a) #a*x1 + b*y1 = a x1 = 1 y1 = 0 #a*x2 + b*y2 = b x2 = 0 y2 = 1 while b != 0: q = a / b #ri = r(i-2) % r(i-1) r = a % b a = b b = r #xi = x(i-2) - q*x(i-1) x = x1 - q*x2 x1 = x2 x2 = x #yi = y(i-2) - q*y(i-1) y = y1 - q*y2 y1 = y2 y2 = y return(x1, y1, a) def selectE(fn, halfkeyLength): while True: #e and fn are relatively prime e = random.randint(0, 1<<halfkeyLength) (x, y, r) = extendedGCD(e, fn) if r == 1: return e def computeD(fn, e): (x, y, r) = extendedGCD(fn, e) #y maybe < 0, so convert it if y < 0: return fn + y return y def keyGeneration(keyLength): #generate public key and private key p = findPrime(keyLength/2) q = findPrime(keyLength/2) n = p * q fn = (p-1) * (q-1) e = selectE(fn, keyLength/2) d = computeD(fn, e) return (n, e, d) def encryption(M, e, n): #RSA C = M^e mod n return fastExpMod(M, e, n) def decryption(C, d, n): #RSA M = C^d mod n return fastExpMod(C, d, n) #Unit Testing (n, e, d) = keyGeneration(1024) #AES keyLength = 256 X = random.randint(0, 1<<256) C = encryption(X, e, n) M = decryption(C, d, n) print "PlainText:", X print "Encryption of plainText:", C print "Decryption of cipherText:", M print "The algorithm is correct:", X == M