标签：+= 大于 ble 不同 test testing 第三方 技术分享 数字签名
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实 验 目 的 |
1. 熟悉RSA算法，理解其原理 2.网上找相关资料实现RSA算法 |
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实 验 环 境 |
Python2+pycharm |
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实 验 步 骤 |
算法基本思路： 1.公钥与私钥的生成： (1)随机挑选两个大质数 p 和 q，构造N = p*q； (2)计算欧拉函数φ(N) = (p-1) * (q-1)； (3)随机挑选e，使得gcd(e, φ(N)) = 1，即 e 与 φ(N) 互素； (4)计算d，使得 e*d ≡ 1 (mod φ(N))，即d 是e 的乘法逆元。 此时，公钥为（e, N），私钥为（d, N），公钥公开，私钥自己保管。 2.加密信息： (1)待加密信息（明文）为 M，M < N；（因为要做模运算，若M大于N，则后面的运算不会成立，因此当信息比N要大时，应该分块加密） (2)密文C = Me mod N (3)解密Cd mod N = (Me)d mod N = Md*e mod N ； 要理解为什么能解密？要用到欧拉定理aφ(n) ≡ 1 (mod n)，再推广：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)。 因此，Md*e mod N = M1 + k*φ(N) mod N = M 我的理解就是：服务器用客户的公钥加密信息发给我，然后客户用私钥解密。 3.数字签名： (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） 我的理解就是：我用自己的密钥加密签名，别人用我的公钥解密可以看到这是我的签名。注意，这个不具有隐私性，即任何人都可以解密此签名。
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实 验 结 果 |
PlainText: 10855225086173939930078735827643414599362568936 5889113498649228757882549939690 Encryption of plainText: 9643670328277461591386781304632351444421180914158605020517 8025562797860283631742532185008096175459168035060137521984 5956515264084827140511733851353574504602300562727743205958 6030439105507942610235389151100878860062259683721684449352 6995565035371405430097017917181968434081359786502175905113 26578152573016111 Decryption of cipherText: 1085522508617393993007873582764341459936256893658891134986 49228757882549939690 The algorithm is correct: True |
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实 验 总 结 |
本实验可用老师MixCS软件生成素数p，q，从而生成公钥和私钥，鉴于网上有相关介绍，本人还是坚持用python试一试。
加载python强大的第三方库实现RSA算法，如：RSA模块，Pycrypto模块等，这些模块都能自动生成极大素数p，q，进而生成公钥，私钥，因此与老师的所给的参考文件中的公钥,私钥不同，我还是希望一直用python做完实验的
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附录
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
标签：+= 大于 ble 不同 test testing 第三方 技术分享 数字签名
原文地址：https://www.cnblogs.com/WhiteHatKevil/p/10133123.html
