List of torsion subgroups
A list of elliptic curves with each possible torsion subgroup can be found here:
And PARI code that lists infinitely many elliptic curve with each torsion subgroup is here.
A list of elliptic curves with each possible torsion subgroup can be found here:
And PARI code that lists infinitely many elliptic curve with each torsion subgroup is here.
From Torsion Points of Elliptic Curves Over Number Fields by Christine Croll :
so ours is finite order. Then
hah!
More about torsion points:
[head ready to explode]
for :
(00:37) gp > a = 1
%1 = 1
(00:37) gp > b = 1
%2 = 1
(00:37) gp > E = ellinit([0,0,0,a,b])
%3 = [0, 0, 0, 1, 1, 0, 2, 4, -1, -48, -864, -496, 6912/31, [-0.6823278038280193273694837397, 0.3411639019140096636847418698 - 1.161541399997251936087917687*I, 0.3411639019140096636847418698 + 1.161541399997251936087917687*I]~, 3.749942978094342855851406868, -1.874971489047171427925703434 + 1.321720533565204538833995727*I, -1.256789871861911570289134735 + 0.E-29*I, 0.6283949359309557851445673678 - 1.280744177088026904445230577*I, 4.956376633845946955308257251]
(00:38) gp > elltors(E)
%4 = [1, [], []]
for
(00:38) gp > a = 8
%5 = 8
(00:39) gp > b = 9
%6 = 9
(00:39) gp > E=ellinit([0,0,0,a,b])
%7 = [0, 0, 0, 8, 9, 0, 16, 36, -64, -384, -7776, -67760, 3538944/4235, [-1.000000000000000000000000000, 0.5000000000000000000000000000 - 2.958039891549808021283664145*I, 0.5000000000000000000000000000 + 2.958039891549808021283664145*I]~, 2.323573124298217095517745754, -1.161786562149108547758872877 + 0.9328742056162391756323628615*I, -1.773647591593647783280373514 + 0.E-28*I, 0.8868237957968238916401867572 - 2.064141081460241175088749935*I, 2.167601432520942242537573241]
(00:39) gp > elltors(E)
%8 = [2, [2], [[-1, 0]]]
hmm… still trying to digest this
me too. still don’t understand what is the torsion and the generators π¦
Now experimenting on a very small curve, taken from Guide to Elliptic Curve Cryptography #27, with reduction polynomial , (, ).
Have checked that the points on #81 are on curve.
Next to do is to perform curve operation
Notes:
This curve is not a Koblitz curve. Going compare this one with Koblitz (by changing a =1 or a = 0 and b = 1). To generate points on curve look at P1363.
Still thinking how to make a flexible ecc system to calculate all of these.
and what it’s gonna be called? a simulator? platform?
what do you mean by ‘small curve’? if we only have 4 bit, it means that we only have a little bit combinations. then we have a rough curve, rather than a smooth curve if we have more bit. is it right?
@soni: i’m afraid it’s not that simple. yes that small curves will have only a small number of points but higher bits doesn’t not determine the smoothness of the curve. the curve is called “smooth” only if it is defined in real numbers, not in finite fields.
ha ha ha … koblitz guy?
@BR: good guess π how do you know it’s not the koblitz girl? :))
because he’s not wearing skirt, lipstick, eye lashes, or those girly stuff π
@BR: smart answer π hei but how do you know koblitz is not wearing eyelashes? :))
I keep forget how to paste latex equation in WordPress, so I post it here (I might have been but don’t remember).
Oh, finally. The code proves that the point (x, y) from the sample parameter is on the Koblitz-163.
There are several mistakes:
The coding is:
#include <stdio.h> #include "field2n.h" #include "poly.h" extern FIELD2N poly_prime; int main(){ INDEX i; FIELD2N a = {0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000001}; FIELD2N b = {0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x00000001}; /*FIELD2N b = {0x00000002, 0x0a601907, 0xb8c953ca, 0x1481eb10, 0x512f7874, 0x4a3205fd}; */ FIELD2N x = {0x00000002, 0xFE13C053, 0x7BBC11AC, 0xAA07D793, 0xDE4E6D5E, 0x5C94EEE8}; FIELD2N y = {0x00000002, 0x89070FB0, 0x5D38FF58, 0x321F2E80, 0x0536D538, 0xCCDAA3D9}; FIELD2N yy, xy; FIELD2N left, right; FIELD2N c; FIELD2N x_2, x_3, ax; null(&yy); null(&xy); null(&c); null(&left); null(&right); null(&x_2); null(&x_3); null(&ax); if (!irreducible(&poly_prime)) return(0); print_field("poly_prime = ", &poly_prime); printf("\nNUMBITS = %d", NUMBITS); printf("\nNUMWORD = %d", NUMWORD); printf("\nMAXLONG = %d\n", MAXLONG); print_field("a = ", &a); print_field("b = ", &b); print_field("x = ", &x); print_field("y = ", &y); poly_mul(&y, &y, &yy); print_field("yy = ", &yy); poly_mul(&x, &y, &xy); print_field("xy = ", &xy); SUMLOOP(i) left.e[i] = yy.e[i] ^ xy.e[i]; print_field("left = ", &left); poly_mul(&x, &x, &x_2); print_field("x_2 = ", &x_2); poly_mul(&x, &x_2, &x_3); print_field("x_3 = ", &x_3); poly_mul(&a, &x, &ax); print_field("ax = ", &ax); print_field("b = ", &b); SUMLOOP(i) right.e[i] = x_3.e[i] ^ x_2.e[i]; SUMLOOP(i) right.e[i] = right.e[i] ^ b.e[i]; print_field("right = ", &right); return 0; }
this header file is also have to be updated for the NUMBITS:
/*** field2n.h ***/ #define WORDSIZE (sizeof(int)*8) #define NUMBITS 163 #define NUMWORD (NUMBITS/WORDSIZE) #define UPRSHIFT (NUMBITS%WORDSIZE) #define MAXLONG (NUMWORD+1) #define MAXBITS (MAXLONG*WORDSIZE) #define MAXSHIFT (WORDSIZE-1) #define MSB (1L<<MAXSHIFT) #define UPRBIT (1L<<(UPRSHIFT-1)) #define UPRMASK (~(-1L<<UPRSHIFT)) #define SUMLOOP(i) for(i=0; i<MAXLONG; i++) typedef short int INDEX; typedef unsigned long ELEMENT; typedef struct { ELEMENT e[MAXLONG]; } FIELD2N;
and also this line from polymain.c:
FIELD2N poly_prime = {0x00000008, 0x00000000, 0x00000000, 0x00000000, 0x00000000, 0x000000c9}; /*163*/
and the result is:
poly_prime =
8 0 0 0 0 c9
NUMBITS = 163
NUMWORD = 5
MAXLONG = 6
a =
0 0 0 0 0 1
b =
0 0 0 0 0 1
2 fe13c053 7bbc11ac aa07d793 de4e6d5e 5c94eee8
y =
2 89070fb0 5d38ff58 321f2e80 536d538 ccdaa3d9
yy =
7 ca0561ef a7b090b5 ddf25eaf f0567c2c 39c1cad7
xy =
4 d7418721 62b253d5 a381f1f6 80b47e5c ad3aa2a
left =
3 1d44e6ce c502c360 7e73af59 70e20270 331260fd
x_2 =
6 710bd85f 2b559b08 5dc2832e 86f4a4c 7ef8d0be
x_3 =
5 6c4f3e91 ee575868 23b12c77 788d483c 4deab042
right =
3 1d44e6ce c502c360 7e73af59 70e20270 331260fd
Now left and right is equal! YAY!!!
ha ha ha … salah equation toh π
iya! gawat ya? :))
Modifying Rosing’s codes to do this checking if a point is on curve using Pari,
with the sample parameters:
the result is like this (on binary fields):
poly_prime =
8 0 0 0 0 c9
NUMBITS = 163
NUMWORD = 5
MAXLONG = 6
a =
0 0 0 0 0 1
b =
2 a601907 b8c953ca 1481eb10 512f7874 4a3205fd
3 f0eba162 86a2d57e a0991168 d4994637 e8343e36
y =
0 d51fbc6c 71a0094f a2cdd545 b11c5c0c 797324f1
left =
1 393a5074 f973003b 4ab508ce 55cc184a 928293df
right =
1 cf775de5 a25942e6 33c8b050 97bf9375 d364fba2
left and right, is not equal!
The idea is to compare if the left side and the right side of y^2 + xy = x^3 + ax + b is equal, then the point (x, y) is on the curve.
Something is still very wrong. Now will do debugging…
let’s try with less bits first
sigh. manual calculation? why can’t we use PARI?
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