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RSA 算法科普及简单 Python 实现

June 22, 2016 • Read: 3470 • CTF

最近做几个ctf的crypto题，好几道RSA都没法用以前无脑工具破搞定了，所以深入研究了一下原理和一些小trick，把自己的一点体会分享出来，共同学习。

1 算法基本思路

RSA算法是一种非对称加密算法，是现在广泛使用的公钥加密算法，主要应用是加密信息和数字签名。详情请看维基

1.1 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。

1.2 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（其实是费马小定理的推广）

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
简单来说，别人用我的公钥加密信息发给我，然后我用私钥解密。

1.3 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）

简单来说，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。

算法的安全性：基于大整数N难以分解出p和q，构造φ(N)；或由N直接构造φ(N)同样难。

2 Python实现基本原理

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

3 实例

3.1 veryeasyRSA

已知RSA公钥生成参数：
p = 3487583947589437589237958723892346254777
q = 8767867843568934765983476584376578389
e = 65537
求d =
请提交PCTF{d}

对上述代码稍作修改，取有用的扩展欧拉和计算d的两个函数，构造如下代码，即可解出答案。

# coding = utf-8
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 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)
p = 3487583947589437589237958723892346254777
q = 8767867843568934765983476584376578389
e = 65537
n = p * q
fn = (p - 1) * (q - 1)
d = computeD(fn, e)
print d

3.2 easyRSA

还记得veryeasy RSA吗？是不是不难？那继续来看看这题吧，这题也不难。
已知一段RSA加密的信息为：0xdc2eeeb2782c且已知加密所用的公钥：
(N=322831561921859 e = 23)
请解密出明文，提交时请将数字转化为ascii码提交
比如你解出的明文是0x6162，那么请提交字符串ab
提交格式:PCTF{明文字符串}

依然是一样的思路，取有用的函数，稍作修改。

# coding = utf-8
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 computeD(fn, e):
    (x, y, r) = extendedGCD(fn, e)
    #y maybe < 0, so convert it
    if y < 0:
        return fn + y
    return y
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 decryption(C, d, n):
    #RSA M = C^d mod n
    return fastExpMod(C, d, n)
p = 13574881
q = 23781539
n = p * q
fn = (p - 1) * (q - 1)
e = 23
d = computeD(fn, e)
C = int('0xdc2eeeb2782c', 16)
M = decryption(C, d, n)
flag = str(hex(M))[2:-1]
print d
print flag.decode('hex')
ctfcrypto
Last Modified: February 14, 2017
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