comSec作业四

1.

x 0 1 2 3 4 5 6
y 1,6 2,5

可得椭圆曲线中的点有

(0,1), (0,6), (1,2), (1,5)

2.

-P = (3, -5) = (3 ,2); -Q = (2,-5) = (2,2); -R = (5, -0) = (5,0)

3.

y = x 3 + x + 7 y = x^3 + x + 7 y=x3+x+7

如果 P = Q, 则 l a m d a = ( 3 x 1 2 + 1 ) / 2 y 1 lamda = (3x_1^2 + 1)/ 2y_1 lamda=(3x12+1)/2y1

否则, l a m d a = ( y 2 − y 1 ) / ( x 2 − x 1 ) lamda = (y_2-y_1)/(x_2-x_1) lamda=(y2y1)/(x2x1)

经过计算

- G 2G 3G 4G 5G 6G 7G 8G 9G 10G 11G 12G 13G
Point (3,2) (10,4) (1,8) (5,4) (4,8) (7,7) (6,8) (6,3) (7,4) (4,3) (5.7) (1,3) (10,7)
lamda 7 5 8 1 6 4 2 4 6 1 8 5

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