CRT优化RSA

转载：利用中国剩余定理加速 RSA

RSA 作为世界上使用最为流行的公钥密码算法，被广泛应用在数据加密和数字签名上。

为了提高加密和签名验证的效率，一般会将RSA的加密指数（一般是公钥位数）设置的较小，一般为 65537 ，而解密或签名效率却不能够简单的通过减小私钥的长度来实现，因为过短的私钥将直接导致安全问题。

于是乎，基于中国剩余定理（Chinese Remainder Theorem，简称 CRT）的 RSA 加速方案被提出。以RSA加解密为例，本文将首先讲解 RSA 基本原理，再介绍中国剩余理论和费马小定理，最后介绍 RSA-CRT算法。

RSA算法

RSA 包括密钥生成算法、加密算法和签名算法。

密钥生成

加解密

签名与验签

CRT

Garner's formulax(x1, x2)（参考：H. Garner. The Residue Number System. IRE Transactions on Electronic Computers, EC-8 (6), pp. 140 – 147, June 1959.），对于\(x_1 = x(mod p)\)和\(x_2=x(mod q)\)：

\(x=x_2 + h.q\)

\(h=(x_1-x_2)(q^{-1}mod p) mod p\)

欧拉定理和费马小定理

RSA-CRT

• \(d = d(mod p-1)+k(p-1)\)，这个如何实现？
• 求出\(m_q\)和\(m_p\)后，就可以基于CRT求出\(m\)。

举例

## RSA加解密
p = 137, q = 131, n = 137.131 = 17947, e = 3, d = 11787.
私钥（n，d），公钥（n，e）
m = 513
加密：c = 5133 mod n = 8363
解密：m'=c^d mod n = 513

## RSA-CRT加解密
dP = d mod (p-1) = 11787 mod 136 = 91
dQ = d mod (q-1) = 11787 mod 130 = 87

qInv = q^{-1} mod p = 131-1 mod 137 = 114
m1 = cdP mod p = 836391 mod 137 = 102
m2 = cdQ mod q = 836387 mod 131 = 120

h = qInv.(m1 - m2) mod p = 114.(102-120+137) mod 137 = 3 
m = m2 + h.q = 120 + 3.131 = 513.

程序

import time

def chinese_remainder_theorem(c, d, p, q):
    dp = d % (p - 1)
    dq = d % (q - 1)
    q_inv = modinv(q, p)  # calculates the inverse
    m1 = pow(c % p, dp, p)
    m2 = pow(c % q, dq, q)
    h = q_inv * (m1 - m2) % p
    m = m2 + h * q
    return m


def modinv(e, phi):  # function used to calculate modular inverse
    d = 0
    x1 = 0
    x2 = 1
    y1 = 1
    temp_phi = phi

    while e > 0:  # extended euclidean algorithm
        temp1 = temp_phi // e
        temp2 = temp_phi - temp1 * e
        temp_phi = e
        e = temp2

        x = x2 - temp1 * x1
        y = d - temp1 * y1

        x2 = x1
        x1 = x
        d = y1
        y1 = y

    if temp_phi == 1:
        return d + phi


def gcd(a, h):  # function used to calculate the GCD
    temp = 0
    while (1):
        temp = a % h
        if (temp == 0):
            return h
        a = h
        h = temp


start_time = time.time()

# p and q are 1024 bit primes. Tested using Miller Rabbin algorithm from Question 2
p = 137
q = 131
n = p*q
e = 3
phi = (p-1)*(q-1)
d = modinv(e, phi)

msg = 513
print("Message data = ", msg)
c = pow(msg, e, n)  # encryption c = (msg ^ e) % n
print("Encrypted data = ", c)

# decryption using chinese remainder theorem
decrypted_msg = chinese_remainder_theorem(c, d, p, q)
print("Original Message Sent = ", decrypted_msg)

end_time = time.time()

elapsed_time = end_time - start_time
print("Time taken for RSA with CRT: {:.6f} seconds".format(elapsed_time))

参考

1. 第二十一个知识点：CRT算法如何提高RSA的性能?
2. Using the CRT with RSA