Topic: How to define which coordinate is negative? (Read 929 times)

hi can you please tell me how to find the publickey of btc address

Since you know the private keys, and you want to sort by private keys, then you must compare the private keys.

You should only compare X points if you want to sort by public key. But since you want to compare by private keys, you have to compare the private keys themselves.

There is no way to order a list of private keys using only their public keys - it is like trying to order a list of strings based on their SHA256 hashes - that is impossible.

all points with lowest Y-coordinate are odd and with highest are even.

thanks sir!!

is there any way to find private key range
first half below this(57896044618658097711785492504343953926418782139537452191302581570759080747168) or
second half above this(57896044618658097711785492504343953926418782139537452191302581570759080747168) using public key

Since you know the private keys, and you want to sort by private keys, then you must compare the private keys.

You should only compare X points if you want to sort by public key. But since you want to compare by private keys, you have to compare the private keys themselves.

There is no way to order a list of private keys using only their public keys - it is like trying to order a list of strings based on their SHA256 hashes - that is impossible.

pk1 = '02C6047F9441ED7D6D3045406E95C07CD85C778E4B8CEF3CA7ABAC09B95C709EE5'#private key=2
pk2 = '022f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4'#private key=5


P = pub2point(pk1)
Q = pub2point(pk2)

eq = P.x == Q.x
lt = P.x < Q.x
R = eq or lt
print(R)

>>>>>>>.false


Q is big because it is private key 5 but im getting false output

Well you should split up this operation


You should make it:


You cannot directly compare two Point classes. Read this page to understand why: https://crypto.stackexchange.com/questions/98189/how-to-prove-that-an-elliptic-curve-point-is-smaller-or-greater-than-half-of-the


[1] They are talking about a similar problem, where you find if the point is on the smaller half or bigger half of the private key range. But this discussion is why you cannot check two Point classes for Less Than - because their private keys are unknown.

---

To work around this, you should compare the X coordinates directly.

This will tell you if the X points are smaller or bigger, and allows you to order the public keys by characteristic P (instead of by private key order N).

P.S. I would appreciate it if you post the questions about the code inside this thread only, posting the same thing in the thread *and* in my PM mailbox is a bit annoying because I am already watching this thread.

comparing two public key im trying to find which one big or small

is there any alternative way to do this ,already pm you my code

please help!

R = P <= Q

eq = P.x == Q.x
lt = P.x < Q.x
R = eq or lt

If there were a polynomial time solution[1] then this would provide a polynomial time solution to the elliptic curve discrete logarithm problem. We strongly believe this not to be the case.
...

The last part effectively rules out dividing the Y by 2 and discarding the fractional part, to attempt to discern the subranges of X (e.g. between 0,2^64-1 and 2^64,2^128-1)


Are you sure you are using this code smapl, because this error message looks funny - there are no point types in this code.

all points with lowest Y-coordinate are odd and with highest are even. and if we could calculate exact position of point in this group there could be a correlation
to secp256k1 subgroup S={G,2G,3G,...,(N-1)G}.
but secp256k1 P-N is 432420386565659656852420866390673177326 more than 2^128 since not all values of P can become X-coordinate.

hi is there any way to find private key range using public key

pk1 = '022f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4'

pk2 = '02C6047F9441ED7D6D3045406E95C07CD85C778E4B8CEF3CA7ABAC09B95C709EE5'

pk1<=pk2
false


any python code useful for me
im getting below error

typeerror ' ' not supported between instances of 'point' and 'point'

if we consider the entire points group like this:
from lowest possible X
(1)X:0000000000000000000000000000000000000000000000000000000000000001 Y:4218f20ae6c646b363db68605822fb14264ca8d2587fdd6fbc750d587e76a7ee
(2)X:0000000000000000000000000000000000000000000000000000000000000001 Y:bde70df51939b94c9c24979fa7dd04ebd9b3572da7802290438af2a681895441
(3)X:0000000000000000000000000000000000000000000000000000000000000002 Y:66fbe727b2ba09e09f5a98d70a5efce8424c5fa425bbda1c511f860657b8535e
(4)X:0000000000000000000000000000000000000000000000000000000000000002 Y:990418d84d45f61f60a56728f5a10317bdb3a05bda4425e3aee079f8a847a8d1
...
to highest possible X
(.)X:fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2c Y:0e994b14ea72f8c3eb95c71ef692575e775058332d7e52d0995cf8038871b67d
(.)X:fffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2c Y:f166b4eb158d073c146a38e1096da8a188afa7ccd281ad2f66a307fb778e45b2

all points with lowest Y-coordinate are odd and with highest are even. and if we could calculate exact position of point in this group there could be a correlation
to secp256k1 subgroup S={G,2G,3G,...,(N-1)G}.
but secp256k1 P-N is 432420386565659656852420866390673177326 more than 2^128 since not all values of P can become X-coordinate.

Gx = 55066263022277343669578718895168534326250603453777594175500187360389116729240
Gy = 32670510020758816978083085130507043184471273380659243275938904335757337482424
P = 115792089237316195423570985008687907853269984665640564039457584007908834671663
N = 115792089237316195423570985008687907852837564279074904382605163141518161494337
#print(P-N)432420386565659656852420866390673177326
    #2^128 340282366920938463463374607431768211456
#input()
starting_x = 0x0
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
beta  = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
beta2 = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40

def oncurve(x,y): # Checks if point satifies x^3 +7 = y^2 , mod P
  x = (x*x*x+7) % p
  y = (y*y) % p
  return x==y
  
while starting_x < p:
    x = starting_x
    ysquared = ((x*x*x+7) % p)      
    y1 = pow(ysquared, (p+1)//4, p)
    y2 = (y1 * -1) % p
    if (y1**2) % p == (x**3 + 7) % p:
        print(f"Secp256k1 True X Coordinate: {hex(x)[2:].zfill(64)} [{x}]")
        print(f'{hex(x)[2:].zfill(64)}:{hex(y1)[2:].zfill(64)}')
        print( oncurve( x,int(hex(y1)[2:].zfill(64), 16)) )
        print(f'{hex(x)[2:].zfill(64)}:{hex(y2)[2:].zfill(64)}')
        print( oncurve( x,int(hex(y2)[2:].zfill(64), 16)) )
        print(hex((x*beta) % p)[2:].zfill(64))
        print(hex((x*beta2) % p)[2:].zfill(64))
        print('---------------------------------------------------------------------------------------------------------------------------------')
    starting_x += 1

hi @brainless

is there any way to guess private key odd or even using public key

hi @brainless

is there any way to guess private key odd or even using public key

hi @brainless

is there any way to guess private key odd or even using public key

PK is 96, but i try to test how much participant , no one interested, as they loose hopes, maximum newbie apear and start trying brute force or kanagroo, vantisearch etc
Enjoy your time

how to run
what is input,explain please message me reply [email protected]

let me give it a try:
still open to correction

0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63

now if you start mapping points, you will get  

....
025699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5
03c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413
0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
02c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413
035699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5
....

like this points will shrink back toward their min or max value with filling of 02 or 03

until

0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
....
0300000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
0200000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63
....
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

and only known way to know which point is on which side is "non" hahaha.

chill and happy brute force while calling your luck.

PK is 96, but i try to test how much participant , no one interested, as they loose hopes, maximum newbie apear and start trying brute force or kanagroo, vantisearch etc
Enjoy your time

it was first what i checked.  It is not correlated (