Topic: Pollard's kangaroo ECDLP solver - page 21. (Read 59389 times)

There seems to be some confusion here. the largest possible private key is 115792089237316195423570985008687907852837564279074904382605163141518161494336 or in hex FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140

Zero cannot be a private key therefor there are 115,792,089,237,316,195,423,570,985,008,687,907,852,837,564,279,074,904,382,605,163,141,518,161,494,336 valid private keys
It turns out there are only                                     57,896,044,618,658,097,711,785,492,504,343,953,926,418,782,139,537,452,191,302,581,570,759,080,747,168 valid public x coordinates

It really is not straight forward to grasp as it took me time though it happens at half of the largest key which is 57896044618658097711785492504343953926418782139537452191302581570759080747168 or in hex 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681B20A0
 
you see 57.........168 and 57.........169 have both the same public x coordinate and opposite y parity (fancy word for even/odd)

removing the lead 03 or 02 It turns out that every public x coord from 1 to 57.........168 is exactly equal to the public x coord at the same spot in the sequence from 115........336 to 57.........169
finally, if you search every private key from 1 to 57,896,044,618,658,097,711,785,492,504,343,953,926,418,782,139,537,452,191,302,581,570,759,080,747,168 (exactly half the normal range) for a public x coord (the majority of a compressed public key) you would find it whether or not the original private key was larger or smaller than 57.........169 however a leading 03 would be 02 and vice versa therefor the limitation doesn't apply to the resulting addresses. In a sense the only thing we care about a private key with this equation is the position of the xcoord's "foundkey" 1 + (115........336-foundkey) or 115........336 - (foundkey-1) depending on which side of 57.........168.5 the foundkey is as we can infer the other and deduce the private key.

In other words this would cut the time of brute forcing any public key in half provided you are searching the full range of 1 to 115........336.
57.........168 is still a lot of private keys so bitcoin is not exactly broken.....yet.

do you mean the last "6" ,
115792089237316195423570985008687907852837564279074904382605163141518161494337 ?

Half way of n
n//2 is wrong, check in above posts, mention clearly formula for div
thankx

yes I am still here, this was the only thing I have found so far.

paniker this is meaning that you can change the n's

modular elliptic curve

Total of all the wallets n is the last number. n= 115792089237316195423570985008687907852837564279074904382605163141518161494337 (In Dec)

n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (In HEX)

Half way of n n//2 = 57896044618658097711785492504343953926418782139537452191302581570759080747169

57896044618658097711785492504343953926418782139537452191302581570759080747169 Lenght Bits = 255

very nice and thanks to boris.. you still here..

Hi still not understand how it works and whats doing, donot understand big numbers, not all of them

I would have to know the key for 0237aff652dd11e2870636b0c4ce4fb324f4b0df45e70f7d8c77d15fcc9ae73525 as it would we below ~2^255 if 0337aff652dd11e2870636b0c4ce4fb324f4b0df45e70f7d8c77d15fcc9ae73525 is above.

Absolutely, so it seems that the amount of valid public x-coords is exactly half the amount of private keys.
So as the theory is any key in the full range (1-115792089237316195423570985008687907852837564279074904382605163141518161494336 ~2^256) either it is less than half (57896044618658097711785492504343953926418782139537452191302581570759080747168) or has the same x coordinate as one that does (above 57896044618658097711785492504343953926418782139537452191302581570759080747169).

I have a mathematical function to find the resulting twin on either side.
I really do like the work that has been done here on the Kangaroo software so I will provide my python function for reference of finding said twin so long as you know 1 of 2 private keys you will know both.

~2^255 is still a very large number.
In the case of uncompressed keys you still have to compute the y coord after but it is trivial.
(using the bit library for python as the function is not intensive)

from bit import Key
import secrets
def twin(i, pubhex):
    max = 115792089237316195423570985008687907852837564279074904382605163141518161494336
    if len(pubhex) == 66:
        publichex = str(pubhex)[2:66]
        if i < 57896044618658097711785492504343953926418782139537452191302581570759080747169:
            twin = max - (i-1)
            if str(pubhex)[:2] == '02':
                twinprefix = '03'
                return [twin,f'{twinprefix}{publichex}']
            elif str(pubhex)[:2] == '03':
                twinprefix = '02'
                return [twin, f'{twinprefix}{publichex}']
        elif i > 57896044618658097711785492504343953926418782139537452191302581570759080747168:
            twin = 1 + (max-i)
            if str(pubhex)[:2] == '02':
                twinprefix = '03'
                return [twin,f'{twinprefix}{publichex}']
            elif str(pubhex)[:2] == '03':
                twinprefix = '02'
                return [twin,f'{twinprefix}{publichex}']
    elif len(pubhex) == 130:
        publichex = str(pubhex)[2:66]
        if i < 57896044618658097711785492504343953926418782139537452191302581570759080747169:
            twin = max - (i-1)
            return [twin,f'uncomp,{publichex}']
        elif i > 57896044618658097711785492504343953926418782139537452191302581570759080747168:
            twin = 1 + (max-i)
            return [twin, f'uncomp,{publichex}']

max = 115792089237316195423570985008687907852837564279074904382605163141518161494336
for x in range(100):
    q = secrets.randbelow(max)
    k = Key.from_int(x)
    t = twin(x,k.pub_to_hex())
    pt = t[0]
    ptpub = Key.from_int(pt).pub_to_hex()
    print(t[1],ptpub)

'''
# Or you can do this
for x in range(1000):
    x = secrets.randbelow(max)
    k = Key.from_int(x)
    t = twin(x,k.pub_to_hex())
    pt = t[0]
    ptpub = Key.from_int(pt).pub_to_hex()
    assert t[1] == ptpub
'''

'''
# for uncompressed
for x in range(100000):
    x = secrets.randbelow(max)
    k = Key.from_int(x)
    k._public_key = k._pk.public_key.format(compressed=False)
    t = twin(x,k.pub_to_hex())
    pt = Key.from_int(t[0])
    # this next line is not nessicary as we format the response without the leading '04'
    pt._public_key = pt._pk.public_key.format(compressed=False)
    ptpub = pt.pub_to_hex()
    assert t[1] == f'uncomp,{str(ptpub)[2:66]}'
'''

Tell us if you've already decided to brag  

Upd. Better yet, just prove it. I generated a random key. Here is the public key for it
0337aff652dd11e2870636b0c4ce4fb324f4b0df45e70f7d8c77d15fcc9ae73525

Fake! Don't trust anything like this.

Depends on what the method is, what is the method?

Can you tells us more about it?))

I found a method to find any private key within 2^255 bit space. Is that new?

Hello i was playing with reverse brut on 62-63? but not so fast as they found it it interesting
i try to change ranges in:
(for y in range(170,200)
 ( for cc in range(922,1845): )  

and got strange results.
only even numbers on end (...99679470) (..00198327750)

and it's going not random, searching 1 by1

I think he is just not interested anymore, as #120 seems to be unsolvable in reasonable time with the current software & hardware.
I do not know, maybe there were other tensions, maybe some users wanted to use him as a free source of tools to crack bitcoin. No idea, but reading his topics I see there were many demanding ignorant.
Or maybe he is still here, using other account. Who knows.

Jean_Luc made his last post on October 11, 2020, and he was last active July 28, 2021 (more than 6 months ago). His last commits on github were also a year ago (February 2021)

Is everything ok with TC? (Jean_Luc)

maybe he created and testing, but not for public, as last found puzzle 115 by joint vent.. and those version were not publicly available...

How come no one has attempted to put a bloom filter on kangaroo. Kangaroo hasn't been updated of modified.


JeanLucPons commented on May 7, 2020
Yes, I added load/save/work
The multi key will be done with the help of theses new features.
The idea is to allow to pre compute large tame kangaroo file and to solve multiple key using this file.
Then when a key is solved, all the wild will become tame and more keys are solved more chance to solve the others....


JeanLucPons commented on May 8, 2020
Yes,
I will add some note about this on the readme. It is a bit tricky.
Multi key support is not yet supported, for this you will need first to create a large tame array for a given range and then attack keys with it.

What DP and range do you use?
Maybe it is related to BOLD statement from ReadMe file:

Please, read this: https://github.com/JeanLucPons/Kangaroo#note-on-timememory-tradeoff-of-the-dp-method

Maybe your system is just too fast for problem you launch for ;-)