一、關於橢圓曲線密碼演算法中點加、點乘的例子

As an example of the encryption process (taken from [KOBL94]), take p=751, E_p(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 P_m=(562,201) and that A selects the random number k=386

1、計算386(0,376)

386(0,376)

=(256 + 128 + 2)(0,376)

=256(0,376) + 128(0,376) + 2(0,376)

1)2(0,376)即為2G

相同點相加，故

t = (3x_p²+ a)/(2y_p) (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 - x_r) - 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 = (3x_p²+ a)/(2y_p) (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 - x_r) - 376 (mod p))

=(2, -378(mod p))

=(2, p - 378)

=(2, 373) // 4G

3) 8G = 2(4G) = (2,373) + (2,373)

t = (3x_p²+ a)/(2y_p) (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 - x_r) - 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 = (3x_p²+ a)/(2y_p) (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 - x_r) - 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 = (3x_p²+ a)/(2y_p) (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 - x_r) - 107 (mod p))

=(628, -602(mod p))

=(628, p - 602)

=(628, 149)// 32G

6) 64G = 2(32G) = (628,149) + (628,149)

t = (3x_p²+ a)/(2y_p) (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 - x_r) - 149 (mod p))

=(26, 439)// 64G

7) 128G = 2(64G) = (26,439) + (26,439)

t = (3x_p²+ a)/(2y_p) (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 - x_r) - 439 (mod p))

=(720,-570 (mod p))

=(720, 181)// 128G

8) 256G = 2(128G) = (720,181) + (720,181)

t = (3x_p²+ a)/(2y_p) (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 - x_r) - 181 (mod p))

=(155,558)// 256G

2、計算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 = (y_Q- y_P) / (x_Q- x_P) (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

則

386(0,376)=(155,558) +(720, 181)+ (1, 376)

386(0,376)=(321² - 155 - 720, 321(155 - x_r) - 558) (mod p) + (1,376) (mod p)

386(0,376)=(30, 515) + (1, 376) (mod p)

再因 P != Q，故 t = (y_Q- y_P) / (x_Q- x_P) (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

則

386(0,376)=(30, 515) + (1, 376) (mod p)

386(0,376)=(704²- 30 - 1, 704(30 - x_r) - 515) (mod p)

386(0,376)=(676, -193) (mod p)

386(0,376)=(676, 558) (mod p)

3、一個求逆的簡單程式

#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;
}

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