Pages:
Author

Topic: Pollard's kangaroo ECDLP solver - page 33. (Read 60037 times)

jr. member
Activity: 81
Merit: 2
September 28, 2021, 07:40:58 AM
guys big lol here plus some explanation from my side.

the script shared by NotATether is just doing opposite of divisor it is increasing 5 bit and giving you last key 5 bit uper range. only difference is while doing divisor you will get 1 key from down 5 bit and other all from defined uper bits range on exact same distance from reference value of this formula

Code:
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
for i in range(32):
    print(i, hex((((N-1)//32)*i+1)%N))

lets say we have key 10  and we do divisor of 10 so what we get is like this (just example)

upper reference range
210 ,  200   , 190 , 180 .................. until 10
assume after divisor each key will have 5 bit down from reference ranges based of exact same distance
205  ,  195 ,  185 , 175 .................  until 5

here is problem , position in 32 keys list is not always same so you cant guess which key is in which range at which position. but if you guess one key correctly which range that key is then you can find easily which is correct 5 bit down key at what position as all other keys ranges are always in sequence. if you do increment of one key all the keys will get increment as well and perhaps they will switch position also in divisor output.


now talk about NotATether script,

the script he posted is doing mod inverses and it is just multiplying value until reach 5 uper bit. (no one can get 120 how can they will get 125 lolololo)

exapmle

key is in 120 range and you used his script output will be

120, 121 , 122, 123, 124, 125 thats it . all 32 keys will be from 120 to 125 range and some range will have 2 or more keys. but guaranteed all keys will be between 120 ~ 125 range at known position as 125 bit will be at 32 position and divisor will have only one key from lower 5 bit guaranteed (not known position) and all other keys from exact same distance with upper reference range. hope you get the point.


now talk about brainless theory -


NotATether and brainless are misunderstanding each other brainless maybe joked that he reduced keys 720 by doing multiplication , addition and subtraction bla bla bla until 90 or 100 bit but NotATether  is insisting what he explained inside his posts is not a way & there is also no way to achieve that and perhaps he never achieved that one and just keep lying.

now what i think is brainless have to explain this to community

Code:
" I got it down to 104 bits today, but with 32,000 pubkeys; better than the normal 2^16 normally required, but I can't figure out a way to shrink it down to one key... "
for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip

as how he claimed this one and plus dont forget he claimed before that he found the 120 key but no plan to cash it but same time he asked .75 bitcoin to provide 115 range one key to buy 3090 (WTF)

well i think i got headache now time to drink coffee  Roll Eyes
.75 bitcoin to provide 115 range one key to buy 3090
this statment is about 4july2021 to 5july2021
and claim for have key could be very next minut, but not same time Smiley

btw you are writing a very interesting topics, and defiantly not boring and same time you are getting headache and need coffee, you are writing seems upset
read your above post, and if you feel more headache, then next time dont try coffee, try some wine, special red wine, Smiley 
dont mind stay happy and healthy and wealthy



brainless  Kiss
member
Activity: 348
Merit: 34
September 28, 2021, 07:38:51 AM
How about I give you a 120 bit public key and you give me exactly one 115 bit public key.
read back my post i need for process that required 3090 gpu's, i dont have, for base on cpu its required multiple days, and i am systems are busy already to process some other projects

btw for all you 1 Q, do you know those digits which can be divide without floating point in results from 1 to 10
thos digit multiplywhat ever, result also could be div without float from 1 to 10

remember i am brainless but my talks are not useless, in past i post magic numbers for division and multiply
you all never make any tests on use of those magic numbers, and talking like higly math experts about div/mul

above all my post have lot of clue for your new thinking levels, but i see you all still bound in basics

waiting your research for those numbers who div from 1 to 10 without float in results
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 07:27:36 AM
How about I give you a 120 bit public key and you give me exactly one 115 bit public key.
member
Activity: 348
Merit: 34
September 28, 2021, 07:17:44 AM
guys big lol here plus some explanation from my side.

the script shared by NotATether is just doing opposite of divisor it is increasing 5 bit and giving you last key 5 bit uper range. only difference is while doing divisor you will get 1 key from down 5 bit and other all from defined uper bits range on exact same distance from reference value of this formula

Code:
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
for i in range(32):
    print(i, hex((((N-1)//32)*i+1)%N))

lets say we have key 10  and we do divisor of 10 so what we get is like this (just example)

upper reference range
210 ,  200   , 190 , 180 .................. until 10
assume after divisor each key will have 5 bit down from reference ranges based of exact same distance
205  ,  195 ,  185 , 175 .................  until 5

here is problem , position in 32 keys list is not always same so you cant guess which key is in which range at which position. but if you guess one key correctly which range that key is then you can find easily which is correct 5 bit down key at what position as all other keys ranges are always in sequence. if you do increment of one key all the keys will get increment as well and perhaps they will switch position also in divisor output.


now talk about NotATether script,

the script he posted is doing mod inverses and it is just multiplying value until reach 5 uper bit. (no one can get 120 how can they will get 125 lolololo)

exapmle

key is in 120 range and you used his script output will be

120, 121 , 122, 123, 124, 125 thats it . all 32 keys will be from 120 to 125 range and some range will have 2 or more keys. but guaranteed all keys will be between 120 ~ 125 range at known position as 125 bit will be at 32 position and divisor will have only one key from lower 5 bit guaranteed (not known position) and all other keys from exact same distance with upper reference range. hope you get the point.


now talk about brainless theory -


NotATether and brainless are misunderstanding each other brainless maybe joked that he reduced keys 720 by doing multiplication , addition and subtraction bla bla bla until 90 or 100 bit but NotATether  is insisting what he explained inside his posts is not a way & there is also no way to achieve that and perhaps he never achieved that one and just keep lying.

now what i think is brainless have to explain this to community

Code:
" I got it down to 104 bits today, but with 32,000 pubkeys; better than the normal 2^16 normally required, but I can't figure out a way to shrink it down to one key... "
for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip

as how he claimed this one and plus dont forget he claimed before that he found the 120 key but no plan to cash it but same time he asked .75 bitcoin to provide 115 range one key to buy 3090 (WTF)

well i think i got headache now time to drink coffee  Roll Eyes
.75 bitcoin to provide 115 range one key to buy 3090
this statment is about 4july2021 to 5july2021
and claim for have key could be very next minut, but not same time Smiley

btw you are writing a very interesting topics, and defiantly not boring and same time you are getting headache and need coffee, you are writing seems upset
read your above post, and if you feel more headache, then next time dont try coffee, try some wine, special red wine, Smiley 
dont mind stay happy and healthy and wealthy
sr. member
Activity: 652
Merit: 316
September 28, 2021, 06:36:58 AM
Here is no magic, here is script to shiftdown pubkey:
Code:
import random
import math
import hashlib
import base58
def inverse(x, p):
    """
    Calculate the modular inverse of x ( mod p )    
    """
    inv1 = 1
    inv2 = 0
    n=1
    while p != 1 and p!=0:        
        quotient = x // p
        
        inv1, inv2 = inv2, inv1 - inv2 * quotient
        x, p = p, x % p        
        n = n+1
    
    return inv2

def dblpt(pt, p):
    """
    Calculate pt+pt = 2*pt
    """
    if pt is None:
        return None
    (x,y)= pt
    if y==0:
        return None
    
    slope= 3*pow(x,2,p)*pow(2*y,p-2,p)
    
    
    xsum= pow(slope,2,p)-2*x
    
    ysum= slope*(x-xsum)-y  
    
    return (xsum%p, ysum%p)

def addpt(p1,p2, p):
    """
    Calculate p1+p2
    """
    if p1 is None or p2 is None:
        return None
    (x1,y1)= p1
    (x2,y2)= p2
    if x1==x2:
        return dblpt(p1, p)
        
    # calculate (y1-y2)/(x1-x2)  modulus p
    
    slope=(y1-y2)*pow(x1-x2,p-2,p)
    
    
    xsum= pow(slope,2,p)-(x1+x2)
  
    ysum= slope*(x1-xsum)-y1
    
    
    return (xsum%p, ysum%p)

def ptmul(pt,a, p):
    """
    Calculate pt*a
    """
    scale= pt    
    acc=None
  
    
    while a:
        
        if a&1:
            if acc is None:
                acc= scale
                
            else:    
                acc= addpt(acc,scale, p)                
              
        scale= dblpt(scale, p)
        a >>= 1
        
            
  
    return acc

def ptdiv(pt,a,p,n):  
    """
    Calculate pt/a
    """  
    divpt=inverse(a, n)%n    
    return ptmul(pt, divpt, p)


def isoncurve(pt,p):
    """
    returns True when pt is on the secp256k1 curve
    """
    (x,y)= pt
    return (y**2 - x**3 - 7)%p == 0


def getuncompressedpub(compressed_key):
    """
    returns uncompressed public key
    """
    y_parity = int(compressed_key[:2]) - 2    
    x = int(compressed_key[2:], 16)
    a = (pow(x, 3, p) + 7) % p
    y = pow(a, (p+1)//4, p)    
    if y % 2 != y_parity:
        y = -y % p        
    return (x,y)

def compresspub(uncompressed_key):
    """
    returns uncompressed public key
    """
    (x,y)=uncompressed_key
    y_parity = y&1
    head='02'
    if y_parity ==1:
        head='03'    
    compressed_key = head+'{:064x}'.format(x)      
    return compressed_key

def hash160(hex_str):
    sha = hashlib.sha256()
    rip = hashlib.new('ripemd160')
    sha.update(hex_str)
    rip.update( sha.digest() )    
    return rip.hexdigest()  # .hexdigest() is hex ASCII
    
def getbtcaddr(pubkeyst):
    
    hex_str = bytearray.fromhex(pubkeyst)
    # Obtain key:
    key_hash = '00' + hash160(hex_str)

    # Obtain signature:

    sha = hashlib.sha256()
    sha.update( bytearray.fromhex(key_hash) )
    checksum = sha.digest()
    sha = hashlib.sha256()
    sha.update(checksum)
    checksum = sha.hexdigest()[0:8]

    return (base58.b58encode( bytes(bytearray.fromhex(key_hash + checksum)) )).decode('utf-8')

def checkpub(realpub, temppub, id):
    localpt = ptmul(temppub, 1024, p)
    localaddpt = ptmul(g, id, p)
    respub= addpt(localpt,localaddpt, p)
    print ("respub-> ", compresspub(respub))
    
#secp256k1 constants
Gx=0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
Gy=0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8
n=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
p = 2**256 - 2**32 - 977
g= (Gx,Gy)


compressed_key='0234c1fd04d301be89b31c0442d3e6ac24883928b45a9340781867d4232ec2dbdf'
point=getuncompressedpub(compressed_key)

print(getbtcaddr("04%064x%064x"%point))
print(getbtcaddr(compressed_key))
divisor = 2**3
newpub=ptdiv(point,divisor,p,n)

(partGx,partGy)=ptdiv(g,divisor,p,n)
print ("1 Fraction part-> (%x,%064x)" % (partGx,partGy))

with open('pub.txt', 'w') as f:
    f.write("04%064x%064x"%newpub)
    f.write('\n')
    print ("Compressed NewPUB (",0,")-> ", compresspub(newpub),"addr",getbtcaddr(compresspub(newpub)))
    i=1
    (pointx,pointy)=(partGx,partGy)
    while i        (newpubtempx,newpubtempy) = addpt(newpub,(pointx,p-pointy), p)
        f.write("04%064x%064x"%(newpubtempx,newpubtempy))
        f.write('\n')
        print ("Compressed NewPUB (",i,")-> ", compresspub((newpubtempx,newpubtempy)),"addr",getbtcaddr(compresspub((newpubtempx,newpubtempy))))
        (pointx,pointy) = addpt((pointx,pointy),(partGx,partGy), p)  
        i=i+1

  



In this example i use pubkey  0234c1fd04d301be89b31c0442d3e6ac24883928b45a9340781867d4232ec2dbdf, privkey is 0x67 and upper range is 2^7
Divisor is 2^3, so new upper range is 2^7-2^3=2^4
In file pub.txt you will find all pubkeys and their number is equil to divisor.
if you try to find all this keys in range 0x1:0xf you will see that only one pubkey will be lie in range
And this pubkey is 03d01115d548e7561b15c38f004d734633687cf4419620095bc5b0f47070afe85a with privkey 0xC
To produce real privkey need multiply privkey by divisor
0xC*0x8 = 0x60
After this need add to result founded public key number (7)
Totaly privekey = 0x60 +0x7=0x67
jr. member
Activity: 81
Merit: 2
September 28, 2021, 05:28:31 AM
Can somebody please explain what the "brainless"-method is?

What is he adding and subtracting? WTF.

chit chat he is adding and subtracting ~ dont worry he got no working method and by knowing private key and than start doing addition or subtraction is bad idea as you will know each stage where keys are going and you can modify the values to get the desire output to say wooowww to your self "i did it" is some high level shit  Roll Eyes

as i told you working blindly on public key is real skill but you will never know where you will land even if you know where you landed still range number will stab you in back so chill and drink soda while BSGS are running hahahaha  Kiss
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 05:22:36 AM
Can somebody please explain what the "brainless"-method is?

What is he adding and subtracting? WTF. Can somebody provide an example?

jr. member
Activity: 81
Merit: 2
September 28, 2021, 05:03:14 AM
ssxb

Your post makes not so much sense to me?!

How does brainless approach reduce the amount of public keys? Did you understand what he meant? Can you elaborate it further?

ok which part doesn't make sense , if you asking about 1/720 keys part ~ that what brainless said in his previous posts. check his last posts and you will find these comments.

and if you ask me this is possible  or not? , yes this is possible but not the way he explained.

but there is one problem with my yes. it is very difficult to guess specific range where you can start searching your calculated keys. perhaps you will guess 10 keys are in 110 range but they are not there. Kiss

is that right ssxb?

this is possible but not the way he explained.

which way you got?

it is very difficult to guess specific range where you can start searching your calculated keys. perhaps you will guess 10 keys are in 110 range but they are not there


correct
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 04:43:45 AM
Code:
from fastecdsa import curve
from fastecdsa.point import Point
import bit

G = curve.secp256k1.G
N = curve.secp256k1.q

def pub2point(pub_hex):
    x = int(pub_hex[2:66], 16)
    if len(pub_hex) < 70:
        y = bit.format.x_to_y(x, int(pub_hex[:2], 16) % 2)
    else:
        y = int(pub_hex[66:], 16)
    return Point(x, y, curve=curve.secp256k1)

# This function makes all the downscaled pubkeys obtained from subtracting
# numbers between 0 and divisor, before dividing the pubkeys by divisor.
def shiftdown(pubkey, divisor, file, convert=True):
    Q = pub2point(pubkey) if convert else pubkey
    # k = 1/divisor
    k = pow(divisor, N - 2, N)
    for i in range(divisor):
        P = Q - (i * G)
        P = k * P
        if (P.y % 2 == 0):
            prefix = "02"
        else:
            prefix = "03"
        hx = hex(P.x)[2:].zfill(64)
        hy = hex(P.y)[2:].zfill(64)
        file.write(prefix+ " " + hx+"\n") # Writes compressed key to file
    file.write("\n") # Writes compressed key to file

with open("input.txt", "r") as f, open("output.txt", "w") as outf:
    line = f.readline().strip()
    while line != '':
        outf.write("original: " +line + "\n")
        shiftdown(line, pow(2,1), outf)
        line = f.readline().strip()

input:
Code:
02e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
022f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4

..d13 is 4 and ...fe4 is 5

output:
Quote
original: 02e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
02 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
02 c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413

original: 022f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
03 5699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5
02 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5


02c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 is 2

Because i know 4 is even, i know it has to be the first pubkey. Because i know 5 is odd, I know it has to be the second key.

The problem is, that in bigger numbers, you can not determine if it is odd or even.

See,if I add the pubkey of 3
Quote
original: 02f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 -> pubkey of 3
02 c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413 -> not the halve
02 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 -> pubkey of 1

original: 02e493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13 -> pubkey of 4
02 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5 -> odd pubkey of 5
02 c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413 -> even pubkey of 3

original: 022f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4 -> pubkey of 5
03 5699b93fc6e1bd29e09a328d657a607b4155b61a6b5fcbedd7c12df7c67df8f5 -> not the halve
02 c6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5-> pubkey of 2

You see... You can not determine from the halve if it is the correct one.
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 04:32:18 AM
Also dont forget:
The Problem we have is, that we don't know if the last bit is even or odd. If you know if the last bit is even or odd, you can crack any public key in no time.
How does this affect? Please explain in more detail..

You can always shift down a public key, aka halving a public key. You get two new public keys. But one is really halving and the other one is moving to the key middle range.

E.g.

You have the public key of 4 (binary represantation is 100). You halve it, and get the public key of 2 (10) and the public key of midrange - 2.
You have the public key of 5 (binary representation is 101). You halve it, and get the public key of public key of midrange - 2 or 3 and 2 (binary 10).

As you dont know the lowest bit, you can not determine which one of the pubkeys is the one you need. So you you have to take both and check them for the correct result.
jr. member
Activity: 40
Merit: 7
September 28, 2021, 04:19:05 AM
Also dont forget:
The Problem we have is, that we don't know if the last bit is even or odd. If you know if the last bit is even or odd, you can crack any public key in no time.
How does this affect? Please explain in more detail..

maybe SSXB can explain this
jr. member
Activity: 33
Merit: 7
September 28, 2021, 04:15:12 AM
Also dont forget:
The Problem we have is, that we don't know if the last bit is even or odd. If you know if the last bit is even or odd, you can crack any public key in no time.
How does this affect? Please explain in more detail..
jr. member
Activity: 40
Merit: 7
September 28, 2021, 04:09:34 AM
ssxb

Your post makes not so much sense to me?!

How does brainless approach reduce the amount of public keys? Did you understand what he meant? Can you elaborate it further?

ok which part doesn't make sense , if you asking about 1/720 keys part ~ that what brainless said in his previous posts. check his last posts and you will find these comments.

and if you ask me this is possible  or not? , yes this is possible but not the way he explained.

but there is one problem with my yes. it is very difficult to guess specific range where you can start searching your calculated keys. perhaps you will guess 10 keys are in 110 range but they are not there. Kiss

is that right ssxb?
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 03:47:03 AM
ssxb

Your post makes not so much sense to me?!

How does brainless approach reduce the amount of public keys? Did you understand what he meant? Can you elaborate it further?
jr. member
Activity: 81
Merit: 2
September 28, 2021, 02:46:13 AM
guys big lol here plus some explanation from my side.

the script shared by NotATether is just doing opposite of divisor it is increasing 5 bit and giving you last key 5 bit uper range. only difference is while doing divisor you will get 1 key from down 5 bit and other all from defined uper bits range on exact same distance from reference value of this formula

Code:
N=115792089237316195423570985008687907852837564279074904382605163141518161494337
for i in range(32):
    print(i, hex((((N-1)//32)*i+1)%N))

lets say we have key 10  and we do divisor of 10 so what we get is like this (just example)

upper reference range
210 ,  200   , 190 , 180 .................. until 10
assume after divisor each key will have 5 bit down from reference ranges based of exact same distance
205  ,  195 ,  185 , 175 .................  until 5

here is problem , position in 32 keys list is not always same so you cant guess which key is in which range at which position. but if you guess one key correctly which range that key is then you can find easily which is correct 5 bit down key at what position as all other keys ranges are always in sequence. if you do increment of one key all the keys will get increment as well and perhaps they will switch position also in divisor output.


now talk about NotATether script,

the script he posted is doing mod inverses and it is just multiplying value until reach 5 uper bit. (no one can get 120 how can they will get 125 lolololo)

exapmle

key is in 120 range and you used his script output will be

120, 121 , 122, 123, 124, 125 thats it . all 32 keys will be from 120 to 125 range and some range will have 2 or more keys. but guaranteed all keys will be between 120 ~ 125 range at known position as 125 bit will be at 32 position and divisor will have only one key from lower 5 bit guaranteed (not known position) and all other keys from exact same distance with upper reference range. hope you get the point.


now talk about brainless theory -


NotATether and brainless are misunderstanding each other brainless maybe joked that he reduced keys 720 by doing multiplication , addition and subtraction bla bla bla until 90 or 100 bit but NotATether  is insisting what he explained inside his posts is not a way & there is also no way to achieve that and perhaps he never achieved that one and just keep lying.

now what i think is brainless have to explain this to community

Code:
" I got it down to 104 bits today, but with 32,000 pubkeys; better than the normal 2^16 normally required, but I can't figure out a way to shrink it down to one key... "
for 10 bit down = 1024 pubkeys
for 20 bit down = 1024*1024 = 1048576 pubkeys
for 30 bit down = 1024*1024*1024 = 1073741824 pubkeys
1048576 and 1073741824 pubkeys with each other addition and mutiplication will return you 260 pubkeys apear where 16 pubkeys sure inside 10 bit down from main pubkey
these 260 pubkeys again played for get 30 bit down for 1/720 pubkeys
now you can start to find with above tip

as how he claimed this one and plus dont forget he claimed before that he found the 120 key but no plan to cash it but same time he asked .75 bitcoin to provide 115 range one key to buy 3090 (WTF)

well i think i got headache now time to drink coffee  Roll Eyes
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 02:38:26 AM
Also dont forget:
The Problem we have is, that we don't know if the last bit is even or odd. If you know if the last bit is even or odd, you can crack any public key in no time.
a.a
member
Activity: 126
Merit: 36
September 28, 2021, 01:48:18 AM
@WanderingPhilosopher
If you store it in a file, you can lookup the position. But is there a formula for shifting it up by knowing the position of the pubkey?

@fxsniper

Do you read? It is a memory tradeoff! If you bitshift a 256 bit key 128 bit you will have 2^128 keys to check in a range of 1-2^128. 2^128 is how many petabytes???

Please inform yourself about ECDLP. I invested last few weeks in my freetime to understand how it works and what to do.
member
Activity: 406
Merit: 47
September 28, 2021, 01:29:47 AM
this is just idea think out of box from "subtracted" methods 

I don't know much about in deep of ECDLP
I know only basic, simple and overview about ECDLP

Did have some video or document description for learning ECDLP  from scratch? (include how to multiply private key to ECDLP step by step, I know how to using function but can not calculate by manual?

Can ECDLP can be scale?
y2=x3+0x+7(mod p)
Y^2 = X^3 + ax + b
(what correct formula of algorithm)

Now ECDLP is 256 bit right?
Can ECDLP scale down to 128 bit?
Can ECDLP scale up to 512 bit?

if can not, please ignore.
Just silly question?

if ECDLP can scale
Can possible to test ECDLP 128 bit for fast found then up scan it
or
Can possible to test ECDLP 512 bit for possible hit one and down scale it

I forget knowledge a lot it like brainwasher myself all about crypto (now like a newbies)
member
Activity: 406
Merit: 47
September 28, 2021, 12:50:51 AM
How can test and proof "subtracted" methods  it is work?
I would like to try and testing.
full member
Activity: 1232
Merit: 242
Shooters Shoot...
September 27, 2021, 08:40:15 PM
Quote
So if you would have 65 bits, you would bitshift it by the amount of pubkeys you can effectively  store in your bloom filter. Lets say you can store 2 ^ 31 pubkeys in your bloom filter. Now you have only to generate the pubkeys and store them and crack them. Get the privatekey, shiftup and remove the first bits from the 65 bits of the pubkey and get a new shorter pubkey. Crack again like you did with the first bits. Shift up and remove the next bits. small amount of bits left, crack them. Now add privatekeys from the three rounds and voila you have the privatekey of the 65 bit in about an no time.
You just plug it back into your script that help you bitshift and you have the private key to the original 65 bit pub key. All that shifting back up, shorter pub key, etc. is over kill.
Pages:
Jump to: