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Topic: Pollard's kangaroo ECDLP solver - page 23. (Read 58537 times)

sr. member
Activity: 443
Merit: 350
November 20, 2021, 05:09:01 PM
-snip-
                        
input two point
privatekey    3     -    2      =   1
                 point1 - point2 = point3
                                               X                                                                                                                         Y
point1 f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9      388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
point2 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5   1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a


i need 3rd point subtract value

please write easy understand make python or explain mathematicaly


Actually the scalar subtraction is the addition where you change y value of the second point from y to modulo - y
Please find the python code with the sub function which subtracts one Point from another:

Demonstration of the function result:
Code:
>>> Point1 = Point(0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9, 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672)
>>> Point2 = Point(0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5, 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a)
>>> Point3 = sub(Point1,Point2)
>>> hex(Point3.x),hex(Point3.y)
('0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798', '0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8')

Python code:
Code:
import gmpy2

modulo = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
order  = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

class Point:
    def __init__(self, x=0, y=0):
        self.x = x
        self.y = y

PG = Point(Gx,Gy)
Z = Point(0,0) # zero-point

# returns (g, x, y) a*x + b*y = gcd(x, y)
def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, x, y = egcd(b % a, a)
        return (g, y - (b // a) * x, x)

def modinv(m, n = modulo):
    while m < 0:
        m += n
    g, x, _ = egcd(m, n)
    if g == 1:
        return x % n

    else: print (' no inverse exist')

def mul2(Pmul2, p = modulo):
    R = Point(0,0)
    #c = 3*Pmul2.x*Pmul2.x*modinv(2*Pmul2.y, p) % p
    c = 3*Pmul2.x*Pmul2.x*gmpy2.invert(2*Pmul2.y, p) % p
    R.x = (c*c-2*Pmul2.x) % p
    R.y = (c*(Pmul2.x - R.x)-Pmul2.y) % p
    return R

def add(Padd, Q, p = modulo):
    if Padd.x == Padd.y == 0: return Q
    if Q.x == Q.y == 0: return Padd
    if Padd == Q: return mul2(Q)
    R = Point()
    dx = (Q.x - Padd.x) % p
    dy = (Q.y - Padd.y) % p
    c = dy * gmpy2.invert(dx, p) % p
    #c = dy * modinv(dx, p) % p
    R.x = (c*c - Padd.x - Q.x) % p
    R.y = (c*(Padd.x - R.x) - Padd.y) % p
    return R

def mulk(k, Pmulk, p = modulo):
    if k == 0: return Z
    if k == 1: return Pmulk
    if (k % 2 == 0): return mulk(k//2, mul2(Pmulk, p), p)
    return add(Pmulk, mulk((k-1)//2, mul2(Pmulk, p), p), p)

def sub(P1, P2, p = modulo): #scalar subtraction P1-P2
    if P1 == P2: return Z
    if P2.x == P2.y == 0: return P1
    return add (P1, Point(P2.x, modulo - P2.y))

If you do not have gmpy2 I recommend to install it (it is 50x faster than self implemented inverse), but you can just uncomment c calculation in mul2 and add functions and comment the formula with gmpy2.

EDIT: corrected typo mistakes
legendary
Activity: 1974
Merit: 1077
^ Will code for Bitcoins
November 20, 2021, 02:02:25 PM


yes i know this git code
i am not understand c language so,
                        
input two point
privatekey    3     -    2      =   1
                 point1 - point2 = point3
                                               X                                                                                                                         Y
point1 f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9      388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
point2 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5   1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a


i need 3rd point subtract value

please write easy understand make python or explain mathematicaly


Code:
class Point:
    b = 7
    def __init__(self, x=float('inf'), y=float('inf')):
        self.x = x
        self.y = y
 
    def copy(self):
        return Point(self.x, self.y)
 
    def is_zero(self):
        return self.x > 1e20 or self.x < -1e20
 
    def neg(self):
        return Point(self.x, -self.y)
 
    def dbl(self):
        if self.is_zero():
            return self.copy()
        try:
            L = (3 * self.x * self.x) / (2 * self.y)
        except ZeroDivisionError:
            return Point()
        x = L * L - 2 * self.x
        return Point(x, L * (self.x - x) - self.y)
 
    def add(self, q):
        if self.x == q.x and self.y == q.y:
            return self.dbl()
        if self.is_zero():
            return q.copy()
        if q.is_zero():
            return self.copy()
        try:
            L = (q.y - self.y) / (q.x - self.x)
        except ZeroDivisionError:
            return Point()
        x = L * L - self.x - q.x
        return Point(x, L * (self.x - x) - self.y)
 
    def mul(self, n):
        p = self.copy()
        r = Point()
        i = 1
        while i <= n:
            if i&n:
                r = r.add(p)
            p = p.dbl()
            i <<= 1
        return r
 
    def __str__(self):
        return "({:.3f}, {:.3f})".format(self.x, self.y)
 
def show(s, p):
    print(s, "Zero" if p.is_zero() else p)
 
def from_y(y):
    n = y * y - Point.b
    x = n**(1./3) if n>=0 else -((-n)**(1./3))
    return Point(x, y)
 
# demonstrate
a = from_y(1)
b = from_y(2)
show("a =", a)
show("b =", b)
c = a.add(b)
show("c = a + b =", c)
d = c.neg()
show("d = -c =", d)
show("c + d =", c.add(d))
show("a + b + d =", a.add(b.add(d)))
show("a * 12345 =", a.mul(12345))
Output:
Code:
a = (-1.817, 1.000)
b = (-1.442, 2.000)
c = a + b = (10.375, -33.525)
d = -c = (10.375, 33.525)
c + d = Zero
a + b + d = Zero
a * 12345 = (10.759, 35.387)
jr. member
Activity: 70
Merit: 1
November 20, 2021, 10:23:08 AM


yes i know this git code
i am not understand c language so,
                         
input two point
privatekey    3     -    2      =   1
                 point1 - point2 = point3
                                               X                                                                                                                         Y
point1 f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9      388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
point2 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5   1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a


i need 3rd point subtract value

please write easy understand make python or explain mathematicaly
full member
Activity: 706
Merit: 111
November 20, 2021, 07:57:47 AM
please eny update this post https://bitcointalksearch.org/topic/m.58481226

i need how to work to point subtract  Cry

any idea make subtract code in python  Sad
Python 3

https://github.com/WanderingPhilosopher/Windows-KeySubtractor/releases/tag/v1.0
jr. member
Activity: 70
Merit: 1
November 20, 2021, 06:12:52 AM
please eny update this post https://bitcointalksearch.org/topic/m.58481226

i need how to work to point subtract  Cry

any idea make subtract code in python  Sad
jr. member
Activity: 70
Merit: 1
November 20, 2021, 06:04:33 AM
When I look at this picture I see that this one value for x and y of R
has millions and billions different pairs of x and y of Q and P
I mean just change the slope of the line

Of course it has. Equation like x+y = 10 has infinite number of solutions.


i know but i am reasontly watch
read full please yes its real no scam only education purpos
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i am watch youtube video today 13/11/21 >>> how to find bitcoin private key >>search


important notice this hacker use only python languagae ,,
1000 BTC stolen
step 1:  hacker crack adderss to  public key convert >> note python code use it
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step 3:past Uncompressed public key 2nd python code $ run 7hr find private key

 yes its real no scam only education purpos video upload but youtube remove this video so no data get


share this message
please python code use find bitcoin private keys
you can find private key make video share me  Smiley Smiley

so i need subtract code please
member
Activity: 110
Merit: 61
November 20, 2021, 12:47:48 AM
When I look at this picture I see that this one value for x and y of R
has millions and billions different pairs of x and y of Q and P
I mean just change the slope of the line

Of course it has. Equation like x+y = 10 has infinite number of solutions.
jr. member
Activity: 70
Merit: 1
November 19, 2021, 11:19:47 PM
And also one more thing.
https://i.imgur.com/RU7sYqO.png

When I look at this picture I see that this one value for x and y of R
has millions and billions different pairs of x and y of Q and P
I mean just change the slope of the line


this code write for c in keysubtracter i need python >>>>  https://github.com/albertobsd/keysubtracter

i know tow point add,multiple in python code, but subtracter not understand https://github.com/albertobsd/keysubtracter/blob/main/keysubtracter.c please

modifiy this code c to python convert
another c code https://github.com/albertobsd/ecctools  subtract
please only subtract code get write to python

atleast  this c code https://github.com/albertobsd/keysubtracter/blob/main/keysubtracter.c decode to mathematical explain
for example :

c = K(qy-py)/(qx-px)
rx1 = K(c^2-px-qx)
ry2 = K(c*(px-rx)) #sorrymyBADenglish

----------------------------------------------------------i use this tow point multiple in python
P = 2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1
N = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
Gx = 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
Gy = 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
Acurve = 0

Cx,Cy = multiple(Gx,Gy)
print (hex(Cx)[2:66],hex(Cy)[2:66])
jr. member
Activity: 42
Merit: 11
November 19, 2021, 10:02:52 PM
And also one more thing.


When I look at this picture I see that this one value for x and y of R
has millions and billions different pairs of x and y of Q and P
I mean just change the slope of the line
member
Activity: 406
Merit: 47
November 19, 2021, 08:52:09 PM
I am not yet understand about subtract method I follow come later
May be subtract method it is not works. for now can not solve
What is thing make subtract method stuck, power of GPU not help right
jr. member
Activity: 42
Merit: 11
November 19, 2021, 07:36:01 PM
@mausuv

I am searching for this too... It looks to me that if | public key = private key * G point | then some parts of the whole process is the same for all keys... It does not need to be calculated again... Also what I understood is I have 2 points P and Q and I am searching for  x and y of R... So what does generator points have to do with all process... And how do I get values of P and Q if I know the private key and that private key has a value "0x2"... Somebody could explain which variable means what?

jr. member
Activity: 70
Merit: 1
November 19, 2021, 09:09:09 AM
guys please tell two point subtract example i need

tell easy understand because, i know tow point add and multipliction
but subtract not understand please write example easy math

for example :

c = K(qy-py)/(qx-px)
rx1 = K(c^2-px-qx)
ry2 = K(c*(px-rx))

your example write this method #sorrymyBADenglish
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
November 17, 2021, 11:04:02 PM
Are there any Rho programs?

Not that I know of.

I not yet deep understand about ECDLP and Pollard's kangaroo Algorithm and about Collision
I just know overview that not enough for knowledge can solve puzzle

I am not sure I understand it correctly
This Collision method meaning it have Collision every where on range around 2*32 will be found Collision of every key right (but size large 2**256 it is large enough for can not find it)

It is considering as a collision any key that has the same Most Significant Bytes as some other key in the hash table. How many MSBs that have to be identical depends on your DP setting. So if your DP size is 10, all keys with the same 10 beginning bits are considered a collision for hashtable purposes.
member
Activity: 406
Merit: 47
November 17, 2021, 07:57:15 PM

I not yet deep understand about ECDLP and Pollard's kangaroo Algorithm and about Collision
I just know overview that not enough for knowledge can solve puzzle

I am not sure I understand it correctly
This Collision method meaning it have Collision every where on range around 2*32 will be found Collision of every key right (but size large 2**256 it is large enough for can not find it)
full member
Activity: 706
Merit: 111
November 16, 2021, 08:22:43 AM
I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?

You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for.

Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no?

Rho is specifically designed for Discrete Logarithm (factoring) problem.

Also I read the Wikipedia page for Kangaroo which said that it's applicable for any group. In fact the paper for Pollard's Kangaroo originally mentioned it's use for multiplicative [again, RSA/factoring] groups.

Are there any Rho programs?
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
November 16, 2021, 07:54:01 AM
I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?

You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for.

Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no?

Rho is specifically designed for Discrete Logarithm (factoring) problem.

Also I read the Wikipedia page for Kangaroo which said that it's applicable for any group. In fact the paper for Pollard's Kangaroo originally mentioned it's use for multiplicative [again, RSA/factoring] groups.
jr. member
Activity: 48
Merit: 11
November 16, 2021, 04:48:50 AM

Can possible Pollard's kangaroo can search multiple pubkey in same time?
I mean check multiple pubkey at once

No, one at a time only
newbie
Activity: 6
Merit: 0
November 16, 2021, 04:45:02 AM

Can possible Pollard's kangaroo can search multiple pubkey in same time?
I mean check multiple pubkey at once
newbie
Activity: 5
Merit: 2
November 10, 2021, 05:51:45 AM
I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?

You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for.

Not ECDLP but DLP so Pollard Rho and Pollard Kangaroo should be applicable no?
legendary
Activity: 1568
Merit: 6660
bitcoincleanup.com / bitmixlist.org
November 10, 2021, 02:47:44 AM
I've shared the basic pseudo code for RSA Bruteforce below. Can you please write a basic pseudo-code for this method?

You cannot solve RSA problems with the Kangaroo method because the ECDLP problem is not applicable for RSA, as there are no logarithms to solve for.
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