position of two secp256k1 public keys

vjudeu

hero member

Activity: 667

Merit: 1529

Quote

Btw, how did you generate your table? I want to do that with different powers.

You can try using different power than "2" in Sage:

Code:

p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
K = GF(p)
a = K(0x0000000000000000000000000000000000000000000000000000000000000000)
b = K(0x0000000000000000000000000000000000000000000000000000000000000007)
E = EllipticCurve(K, (a, b))
G = E(0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8)
E.set_order(n * 0x1)
index = 0
while(index<256):
private_key = 2^index % n
public_key = private_key * G
print(hex(index),hex(private_key),hex(public_key[0]),hex(public_key[1]))
index += 1

Also note that you don't have to end with 2^255, you can go on ad infinitum, and after a lot of operations you will reach the base point again.

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: hexan123 on October 25, 2023, 12:57:16 AM

Quote from: digaran on October 24, 2023, 06:23:35 PM

Here, repeat this script with the same points several times and compare each time result.

# Public keys in hexadecimal format
hex_pubkey1 = "02b23790a42be63e1b251ad6c94fdef07271ec0aada31db6c3e8bd32043f8be384"
hex_pubkey2 = "034a4a6dc97ac7c8b8ad795dbebcb9dcff7290b68a5ef74e56ab5edde01bced775"

privkey1 = 2^255
privkey2 = 2^15

easy i have table of 2^* key

I hope you realize the script I posted is based on a script either posted by OP or someone else with the same question, the script doesn't do anything other than checking if the 2 keys are the same or not and just gives you the time it took to compare, as pointed out several times, knowing the time it takes to go from A to B, is easy, you just need to count each step, like 1=A, 50=B, then you count by 2, 3, 4 until you get to 50, the time depends on how fast you can count.

Btw, how did you generate your table? I want to do that with different powers.

hexan123

newbie

Activity: 17

Merit: 0

Quote from: digaran on October 24, 2023, 06:23:35 PM

Here, repeat this script with the same points several times and compare each time result.

# Public keys in hexadecimal format
hex_pubkey1 = "02b23790a42be63e1b251ad6c94fdef07271ec0aada31db6c3e8bd32043f8be384"
hex_pubkey2 = "034a4a6dc97ac7c8b8ad795dbebcb9dcff7290b68a5ef74e56ab5edde01bced775"

privkey1 = 2^255
privkey2 = 2^15

easy i have table of 2^* key

albert0bsd

hero member

Activity: 828

Merit: 657

Quote from: shelby0930 on October 24, 2023, 04:41:29 PM

i want to develop a way to measure distance between 2 public keys

There is no way to know the distance between 2 publickey that would break ECDSA

Quote from: shelby0930 on October 24, 2023, 04:41:29 PM

what i mean by distance is the time it takes to reach from point B to point A

What? that doesn't have sense, what you mean by time? running time of a program? or what? Number of steps?

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Here, repeat this script with the same points several times and compare each time result.

Code:

import time
import gmpy2

# Define function to convert SEC-encoded public key to coordinates
def sec_to_public_pair(sec):
if sec[0] != 2 and sec[0] != 3:
raise ValueError("Invalid SEC format")
x = gmpy2.mpz(int.from_bytes(sec[1:], 'big'))
y_squared = x**3 + 7
y = gmpy2.mpz(gmpy2.sqrt(y_squared))
if sec[0] == 3 and y % 2 == 0 or sec[0] == 2 and y % 2 == 1:
y = -y
return (x, y)

# Public keys in hexadecimal format
hex_pubkey1 = "02b23790a42be63e1b251ad6c94fdef07271ec0aada31db6c3e8bd32043f8be384"
hex_pubkey2 = "034a4a6dc97ac7c8b8ad795dbebcb9dcff7290b68a5ef74e56ab5edde01bced775"

# Convert public keys to coordinates
point1 = sec_to_public_pair(bytes.fromhex(hex_pubkey1))
point2 = sec_to_public_pair(bytes.fromhex(hex_pubkey2))

# Measure execution time
start_time = time.time()

# Compare the x-coordinates of the points
if point1[0] == point2[0]:
print("The public keys are the same.")
else:
print("The public keys are different.")

end_time = time.time()
execution_time = end_time - start_time
print(f"Execution time: {execution_time} seconds")

Calculating public keys take around the same amount of time, uncompressed public keys take double the time of compressed keys.
As I told you before, the distance between 50 and 100 would take e.g, 25 seconds if your key rate is 2 keys/s.
10keys/s = estimated time 5 seconds.
50keys/s = around 1 second etc.

Public key generation difficulty is not based on the distance between 2 points, because 1 point addition will always take the same time doesn't matter if you are adding 2+1 or 5000+2^250, it's not like a point representing a 256 bit key weighs more computationally and a point representing 1 bit key weighs less, no they all are the same computationally.

ymgve2

full member

Activity: 161

Merit: 230

What do you mean by "point negation"?

And what is your purpose? "time" between points is an irrelevant metric and a bad proxy for the distance between two points. And if you're trying to see how far two arbitrary points are from each other, time is irrelevant because even if you had billions of the fastest computers on Earth that could go stepwise from A to B, it would still use more time than the lifetime of the universe to go between two arbitrary points in secp256k1.

shelby0930

newbie

Activity: 24

Merit: 4

i want to develop a way to measure distance between 2 public keys what i mean by distance is the time it takes to reach from point B to point A lets say the private key of A is 50 and B is 100 lets assume that it takes about 2 minuets to go from B to A using point negation.. how do i get the estimate time it takes to go from B to A ?

Topic: position of two secp256k1 public keys (Read 132 times)