椭圆曲线密码算法一
作者：互联网
一、关于椭圆曲线密码算法中点加、点乘的例子
As an example of the encryption process (taken from [KOBL94]), take p=751, Ep(1,188), which is equivalent to the curve y2=x3-x+188; and G=(0,376). Suppose the A wishes to send a message to B that is encoded in the elliptic point Pm=(562,201) and that A selects the random number k=386. B’s public key is Pb=(201,5). We have 386(0,376)=(676,558), and (562,201)+386(201,5)=(385,328). Thus, A sends the cipher text {(676,558), (385,328}.
1、计算386(0,376)
386(0,376)
=(256 + 128 + 2)(0,376)
=256(0,376) + 128(0,376) + 2(0,376)
1) 2(0,376)即为2G
相同点相加，故
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 02 - 1)/(2 x 376) (mod p)
=-752-1 (mod p)
因1 x 752 = 1 mod p，故752-1 = 1
=-1 (mod p)
= p - 1
= 751 - 1
= 750
2G=(0,376)+(0,376)
=(7502 - 0 - 0 (mod p), 750(0 - xr) - 376 (mod p))
=(1, -750-376 (mod p))
=(1, -375 (mod p))
=(1, p - 375)
=(1, 376) // 2G
2) 4G = 2(2G) = (1,376) + (1,376)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 12 - 1)/(2 x 376) (mod p)
=376-1 (mod p)
因2 x 376 = 1 mod p，故376-1 = 2
=2 (mod p)
=2
4G=(1,376)+(1,376)
=(22 - 1 - 1 (mod p), 2(1 - xr) - 376 (mod p))
=(2, -378 (mod p))
=(2, p - 378)
=(2, 373) // 4G
3) 8G = 2(4G) = (2,373) + (2,373)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 22 - 1)/(2 x 373) (mod p)
=11 x 746-1 (mod p)
因150 x 746 = 1 mod p，故746-1 = 150
=11 x 150 (mod p)
=148
8G=(2,373) + (2,373)
=(1482 - 2 - 2 (mod p), 148(2 - xr) - 373 (mod p))
=(121, -148 x 119 - 373 (mod p))
=(121, -712 (mod p))
=(121, p - 712)
=(121, 39) // 8G
4) 16G = 2(8G) = (121,39) + (121,39)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 1212 - 1)/(2 x 39) (mod p)
=364 x 78-1 (mod p)
因337 x 78 = 1 mod p，故78-1 = 337
=364 x 337 (mod p)
=255
16G=(121,39) + (121,39)
=(2552 - 121 - 121 (mod p), 255(121 - xr) - 39 (mod p))
=(197, 255 x (-76) - 39 (mod p))
=(197, -19419 (mod p))
=(197, -644 (mod p))
=(197, p - 644)
=(197, 107) // 16G
5) 32G = 2(16G) = (197,107) + (197,107)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 1972 - 1)/(2 x 107) (mod p)
=21 x 214-1 (mod p)
因186 x 214 = 1 mod p，故214-1 = 186
=21 x 186 (mod p)
=151
32G=(197,107) + (197,107)
=(1512 - 197 - 197 (mod p), 151(197 - xr) - 107 (mod p))
=(628, -602 (mod p))
=(628, p - 602)
=(628, 149) // 32G
6) 64G = 2(32G) = (628,149) + (628,149)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 6282 - 1)/(2 x 149) (mod p)
=326 x 298-1 (mod p)
因688 x 298 = 1 mod p，故298-1 = 688
=326 x 688 (mod p)
=490
64G= (628, 149) + (628, 149)
=(4902 - 628 - 628 (mod p), 490(628 - xr) - 149 (mod p))
=(26, 439) // 64G
7) 128G = 2(64G) = (26,439) + (26,439)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 262 - 1)/(2 x 439) (mod p)
=525 x 127-1 (mod p)
因615 x 127 = 1 mod p，故127-1 = 615
=525 x 615 (mod p)
=696
128G= (26, 439) + (26, 439)
=(6962 - 26 - 26 (mod p), 696(26 - xr) - 439 (mod p))
=(720,-570 (mod p))
=(720, 181) // 128G
8) 256G = 2(128G) = (720,181) + (720,181)
t = (3xp2 + a)/(2yp) (mod p)
=(3 x 7202 - 1)/(2 x 181) (mod p)
=629 x 362-1 (mod p)
因139 x 362 = 1 mod p，故362-1 = 139
=629 x 139 (mod p)
=315
256G= (720,181) + (720,181)
=(3152 - 720 - 720 (mod p), 315(720 - xr) - 181 (mod p))
=(155,558) // 256G
2、计算386(0,376)
386(0,376) =256(0,376) + 128(0,376) + 2(0,376)
386(0,376) =(155,558) + (720, 181) + (1, 376)
因 P != Q，故 t = (yQ - yP) / (xQ - xP) (mod p)
t =(181 - 558)/(720 - 155) (mod p)
t =-377 x 565-1 (mod p)
因537 x 565 = 1 (mod p)，故565-1 = 537
t =-377 x 537 (mod p)
t =-430 (mod p)
t =p - 430
t =321
则
386(0,376) =(155,558) + (720, 181) + (1, 376)
386(0,376) =(3212 - 155 - 720, 321(155 - xr) - 558) (mod p) + (1,376) (mod p)
386(0,376) =(30, 515) + (1, 376) (mod p)
再因 P != Q，故 t = (yQ - yP) / (xQ - xP) (mod p)
t =(376 - 515) / (1 - 30) (mod p)
t =(-139) / (-29) (mod p)
t = 139 x 29 -1 (mod p)
因259 x 29 = 1 (mod p)，故29-1 = 259
t = 139 x 259 (mod p)
t =704 (mod p)
t =704
则
386(0,376) =(30, 515) + (1, 376) (mod p)
386(0,376) =(7042 - 30 - 1, 704(30 - xr) - 515) (mod p)
386(0,376) =(676, -193) (mod p)
386(0,376) =(676, 558) (mod p)
3、一个求逆的简单程序
#include <stdio.h>
int main(void)
{
int i;
int value = 29;
for (i = 1; i < 751; ++i) {
if (i * value % 751 == 1) {
printf("iv value of [%d] is [%d]\n", value, i);
break;
}
}
if (i == 751) {
printf("no value\n");
}
return 0;
}
【参考文献】
标签：椭圆,720,121,197,密码,算法,386,mod,376 来源： https://www.cnblogs.com/luop/p/15134285.html