Paillier，同态加密算法。Java版。
直接上链接。
http://www.csee.umbc.edu/~kunliu1/research/Paillier.html
稍微修改一下，代码如下。
[java] view plain copy /** * This program is free software: you can redistribute it and/or modify it * under the terms of the GNU General Public License as published by the Free * Software Foundation, either version 3 of the License, or (at your option) * any later version. * * This program is distributed in the hope that it will be useful, but WITHOUT * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License for * more details. * * You should have received a copy of the GNU General Public License along with * this program. If not, see <http://www.gnu.org/licenses/>. */ import java.math.*; import java.util.*; /** * Paillier Cryptosystem <br> * <br> * References: <br> * [1] Pascal Paillier, * "Public-Key Cryptosystems Based on Composite Degree Residuosity Classes," * EUROCRYPT'99. URL: * <a href="http://www.gemplus.com/smart/rd/publications/pdf/Pai99pai.pdf">http: * //www.gemplus.com/smart/rd/publications/pdf/Pai99pai.pdf</a><br> * * [2] Paillier cryptosystem from Wikipedia. URL: * <a href="http://en.wikipedia.org/wiki/Paillier_cryptosystem">http://en. * wikipedia.org/wiki/Paillier_cryptosystem</a> * * @author Kun Liu (kunliu1@cs.umbc.edu) * @version 1.0 */ public class Paillier { /** * p and q are two large primes. lambda = lcm(p-1, q-1) = * (p-1)*(q-1)/gcd(p-1, q-1). */ private BigInteger p, q, lambda; /** * n = p*q, where p and q are two large primes. */ public BigInteger n; /** * nsquare = n*n */ public BigInteger nsquare; /** * a random integer in Z*_{n^2} where gcd (L(g^lambda mod n^2), n) = 1. */ private BigInteger g; /** * number of bits of modulus */ private int bitLength; /** * Constructs an instance of the Paillier cryptosystem. * * @param bitLengthVal * number of bits of modulus * @param certainty * The probability that the new BigInteger represents a prime * number will exceed (1 - 2^(-certainty)). The execution time of * this constructor is proportional to the value of this * parameter. */ public Paillier(int bitLengthVal, int certainty) { KeyGeneration(bitLengthVal, certainty); } /** * Constructs an instance of the Paillier cryptosystem with 512 bits of * modulus and at least 1-2^(-64) certainty of primes generation. */ public Paillier() { KeyGeneration(512, 64); } /** * Sets up the public key and private key. * * @param bitLengthVal * number of bits of modulus. * @param certainty * The probability that the new BigInteger represents a prime * number will exceed (1 - 2^(-certainty)). The execution time of * this constructor is proportional to the value of this * parameter. */ public void KeyGeneration(int bitLengthVal, int certainty) { bitLength = bitLengthVal; /* * Constructs two randomly generated positive BigIntegers that are * probably prime, with the specified bitLength and certainty. */ p = new BigInteger(bitLength / 2, certainty, new Random()); q = new BigInteger(bitLength / 2, certainty, new Random()); n = p.multiply(q); nsquare = n.multiply(n); g = new BigInteger("2"); lambda = p.subtract(BigInteger.ONE).multiply(q.subtract(BigInteger.ONE)) .divide(p.subtract(BigInteger.ONE).gcd(q.subtract(BigInteger.ONE))); /* check whether g is good. */ if (g.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).gcd(n).intValue() != 1) { System.out.println("g is not good. Choose g again."); System.exit(1); } } /** * Encrypts plaintext m. ciphertext c = g^m * r^n mod n^2. This function * explicitly requires random input r to help with encryption. * * @param m * plaintext as a BigInteger * @param r * random plaintext to help with encryption * @return ciphertext as a BigInteger */ public BigInteger Encryption(BigInteger m, BigInteger r) { return g.modPow(m, nsquare).multiply(r.modPow(n, nsquare)).mod(nsquare); } /** * Encrypts plaintext m. ciphertext c = g^m * r^n mod n^2. This function * automatically generates random input r (to help with encryption). * * @param m * plaintext as a BigInteger * @return ciphertext as a BigInteger */ public BigInteger Encryption(BigInteger m) { BigInteger r = new BigInteger(bitLength, new Random()); return g.modPow(m, nsquare).multiply(r.modPow(n, nsquare)).mod(nsquare); } /** * Decrypts ciphertext c. plaintext m = L(c^lambda mod n^2) * u mod n, where * u = (L(g^lambda mod n^2))^(-1) mod n. * * @param c * ciphertext as a BigInteger * @return plaintext as a BigInteger */ public BigInteger Decryption(BigInteger c) { BigInteger u = g.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).modInverse(n); return c.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).multiply(u).mod(n); } /** * sum of (cipher) em1 and em2 * * @param em1 * @param em2 * @return */ public BigInteger cipher_add(BigInteger em1, BigInteger em2) { return em1.multiply(em2).mod(nsquare); } /** * main function * * @param str * intput string */ public static void main(String[] str) { /* instantiating an object of Paillier cryptosystem */ Paillier paillier = new Paillier(); /* instantiating two plaintext msgs */ BigInteger m1 = new BigInteger("20"); BigInteger m2 = new BigInteger("60"); /* encryption */ BigInteger em1 = paillier.Encryption(m1); BigInteger em2 = paillier.Encryption(m2); /* printout encrypted text */ System.out.println(em1); System.out.println(em2); /* printout decrypted text */ System.out.println(paillier.Decryption(em1).toString()); System.out.println(paillier.Decryption(em2).toString()); /* * test homomorphic properties -> D(E(m1)*E(m2) mod n^2) = (m1 + m2) mod * n */ // m1+m2,求明文数值的和 BigInteger sum_m1m2 = m1.add(m2).mod(paillier.n); System.out.println("original sum: " + sum_m1m2.toString()); // em1+em2，求密文数值的乘 BigInteger product_em1em2 = em1.multiply(em2).mod(paillier.nsquare); System.out.println("encrypted sum: " + product_em1em2.toString()); System.out.println("decrypted sum: " + paillier.Decryption(product_em1em2).toString()); /* test homomorphic properties -> D(E(m1)^m2 mod n^2) = (m1*m2) mod n */ // m1*m2,求明文数值的乘 BigInteger prod_m1m2 = m1.multiply(m2).mod(paillier.n); System.out.println("original product: " + prod_m1m2.toString()); // em1的m2次方，再mod paillier.nsquare BigInteger expo_em1m2 = em1.modPow(m2, paillier.nsquare); System.out.println("encrypted product: " + expo_em1m2.toString()); System.out.println("decrypted product: " + paillier.Decryption(expo_em1m2).toString()); //sum test System.out.println("--------------------------------"); Paillier p = new Paillier(); BigInteger t1 = new BigInteger("21");System.out.println(t1.toString()); BigInteger t2 = new BigInteger("50");System.out.println(t2.toString()); BigInteger t3 = new BigInteger("50");System.out.println(t3.toString()); BigInteger et1 = p.Encryption(t1);System.out.println(et1.toString()); BigInteger et2 = p.Encryption(t2);System.out.println(et2.toString()); BigInteger et3 = p.Encryption(t3);System.out.println(et3.toString()); BigInteger sum = new BigInteger("1"); sum = p.cipher_add(sum, et1); sum = p.cipher_add(sum, et2); sum = p.cipher_add(sum, et3); System.out.println("sum: "+sum.toString()); System.out.println("decrypted sum: "+p.Decryption(sum).toString()); System.out.println("--------------------------------"); } } 顶 0