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 Kurapica 01-15-2022 06:41

Math Problem

.............

 chants 01-16-2022 15:42

The usual variable names for ECC points P and Q being points on a curve is a distraction. ECC involves modular arithmetic with addition, multiplication, inverses. Quite clear this is not ECC.

The good news about XOR is it is bitwise addition modulo 2. There are many nice properties. Used alone it cannot really be used to protect anything as it's linear mapping and easily reversible. x^x==0, x^0==x, x^~x==1, x^1==~x. The basic properties are sufficient to invert this or simplify it.

 Kurapica 01-16-2022 18:28

yes those names of variables were my idea, the code itself is not available to make sure it involves any ECC or other known functions.

still I can't find the relation with Q or Qy as explained.

 Git 01-18-2022 22:37

I've no idea if any of this helps, and you probably know them anyway, but...

using you example :

R = 0x0B42 CE12 CF5C 8C9B
A = 0x4CBD A859 22EF 6E40
B = 0x03EB E297 3C74 4952
D = 0xED77 B8C3
E = 0xDBFE 0085

F = 0x4F56 6DDB A11B A24F
G = 0x0021 883A AFFC 966E
H = 0x9314 5272 7148 BF10
J = 0x0000 2715 BF80 855D

Lets take all the XORs in hex for reference :

R^F = 0x4414 a3c9 6e47 2ed4
R^G = 0x9856 9c60 be14 338b
R^H = 0x9856 9c60 be14 338b
R^J = 0x0b42 e907 70dc 09c6

A^F = 0x03eb c582 83f4 cc0f
A^G = 0xdfa9 fa2b 53a7 d150
A^H = 0xdfa9 fa2b 53a7 d150
A^J = 0x4cbd 8f4c 9d6f eb1d

B^F = 0x4cbd 8f4c 9d6f eb1d
B^G = 0x90ff b0e5 4d3c f642
B^H = 0x90ff b0e5 4d3c f642
B^J = 0x03eb c582 83f4 cc0f

D^F = 0x4f56 6ddb 4c6c 1a8c
D^G = 0x9314 5272 9c3f 07d3
D^H = 0x9314 5272 9c3f 07d3
D^J = 0x0000 2715 52f7 3d9e

E^F = 0x4f56 6ddb 7ae5 a2ca
E^G = 0x9314 5272 aab6 bf95
E^H = 0x9314 5272 aab6 bf95
E^J = 0x0000 2715 647e 85d8

Again, no idea if they have any significance, but I noted the following relations, for this example anyway :

R^G = R^H
A^G = A^H
B^G = B^H
D^G = D^H
E^G = E^H
(D^G) & 0xFFFFFFFF00000000 = (E^G) & 0xFFFFFFFF00000000
(D^J) & 0xFFFFFFFF00000000 = (E^J) & 0xFFFFFFFF00000000
(A^J) & 0xFFFFFFFF00000000 = (B^F) & 0xFFFFFFFF00000000
(A^F) & 0xFFFFFFFF00000000 = (B^J) & 0xFFFFFFFF00000000
(D^F) & 0xFFFFFFFF00000000 = (E^F) & 0xFFFFFFFF00000000

Git

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