我在Java中实施了Fiege Fiat Shamir识别方案,我非常确定它在数学方面很好。 (我已多次检查过)但它永远不会起作用(当调用check时,它几乎总是假的,即使用数字调用也应该有效)。我之前没有序列,(k
值为1),它已经让它工作,但现在它不起作用。救命啊!
public class ZKPTimeTrials {
public static int gcd(int p, int q) {
if (q == 0) return p;
else return gcd(q, p % q);
}
public static int randomR(int min, int max) {
Random randgen = new Random();
return randgen.nextInt((max - min) + 1) + min;
}
public static int getRandomCoprime(int n) {
int result = n;
while (gcd(n, result) != 1) {
result = randomR(2, n-1);
}
return result;
}
public static int[] makeSi(int k, int n) {
int[] result = new int[k];
for(int i = 0; i < result.length; i++) {
result[i] = getRandomCoprime(n);
}
return result;
}
public static int[] makeVi(int[] si, int n) {
int[] result = new int[si.length];
for(int i = 0; i < result.length; i++) {
result[i] = (si[i] * si[i]) % n;
}
return result;
}
public static int[] makeEi(int k) {
int[] result = new int[k];
for(int i = 0; i < k; i++) {
result[i] = randomR(0, 1);
}
return result;
}
public static int makeY(int r, int[] ei, int[] si, int n) {
int result = r;
for(int i = 0; i < si.length; i++) {
result *= (int) Math.pow(si[i], ei[i]);
}
return result % n;
}
public static boolean check(int n, int x, int y, int[] ei, int[] vi) {
int signBit = ZKPTimeTrials.randomR(0, 1);
if(signBit == 0) {
signBit = -1;
}
int shouldY = x * signBit;
for(int i = 0; i < vi.length; i++) {
shouldY *= (int) Math.pow(vi[i], ei[i]);
}
return ((y * y) % n) == shouldY % n;
}
public static void main(String args[]) {
int n = 71 * 7;
int t = 50;
int k = 10;
int[] si = makeSi(k, n);
int[] vi = makeVi(si, n);
int r = randomR(2, n-1);
int ei[] = makeEi(k);
int s = randomR(0, 1);
if(s == 0) {
s = -1;
}
int x = (s * r * r) % n;
int y = makeY(r, ei, si, n);
for(int i = 0; i < si.length; i++) System.out.print(ei[i] + " ");
System.out.println();
for(int i = 0; i < si.length; i++) System.out.print(si[i] + " ");
System.out.println(check(n, x, y, ei, vi));
}
}
答案 0 :(得分:0)
第一个问题是makeY中的整数溢出并检查:在两个函数&#39; result&#39;很可能会溢出,因为您先构建产品并在之后减少模数。尝试在每次乘法后减少mod n以保持&#39;结果&#39;小。
例如在makeY中,写:
int result = r % n;
for (int i = 0; i < si.length; i++) {
if (ei[i] == 1)
result = (result * si[i]) % n;
}
return result;
(我还删除了Math.pow()以使其更具可读性和效率,但这不是错误。)
第二个问题是检查功能的逻辑:不需要signBit变量,而是应该检查y * y是否等于shouldY 或 -shouldY。
public static boolean check(int n, int x, int y, int[] ei, int[] vi) {
int shouldY = x % n;
for (int i = 0; i < vi.length; i++) {
if (ei[i] == 1)
shouldY = (shouldY * vi[i]) % n;
}
return (y*y - shouldY) % n == 0 || (y*y + shouldY) % n == 0;
}
通过这些小修正,我设法让你的代码运行。希望它有所帮助...