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Topic: Pollard's kangaroo ECDLP solver - page 12. (Read 59436 times)

copper member
Activity: 205
Merit: 1
August 02, 2023, 02:59:44 PM
hi nomachine, doing a nice 250 300 but it is single core version is there a multicore version out
or can we edit this to do that , thanks mate for sharing.

https://github.com/Telariust/pollard-kangaroo/blob/master/pollard-kangaroo-multi.py

There is a link in his post.
full member
Activity: 431
Merit: 105
August 02, 2023, 02:44:51 PM
hi nomachine, doing a nice 250 300 but it is single core version is there a multicore version out
or can we edit this to do that , thanks mate for sharing.
member
Activity: 499
Merit: 38
August 02, 2023, 10:56:35 AM

That link no longer exists. I barely managed to find the original pollard_kangaroo.txt on some Russian site. But it is deprecated for Python 2.x. I updated to 3.x and it's still slow, but if you want to play, here you go.


Code:
import time
import random
import gmpy2
from functools import lru_cache

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, infinite in real x,y-plane

def mul2(P, p=modulo):
    c = (3 * P.x * P.x * pow(2 * P.y, -1, p)) % p
    R = Point()
    R.x = (c * c - 2 * P.x) % p
    R.y = (c * (P.x - R.x) - P.y) % p
    return R

def add(P, Q, p=modulo):
    dx = Q.x - P.x
    dy = Q.y - P.y
    #c = (dy * pow(dx, -1, p)) % p
    c = dy * gmpy2.invert(dx, p) % p
    R = Point()
    R.x = (c * c - P.x - Q.x) % p
    R.y = (c * (P.x - R.x) - P.y) % p
    return R

@lru_cache(maxsize=None)
def X2Y(X, p=modulo):
    if p % 4 != 3:
        print('prime must be 3 modulo 4')
        return 0
    X = (X ** 3 + 7) % p
    pw = (p + 1) // 4
    Y = 1
    tmp = X
    for w in range(256):
        if (pw >> w) & 1 == 1:
            Y = (Y * tmp) % p
        tmp = pow(tmp, 2, p)
    return Y

def compute_P_table():
    P = [PG]
    for k in range(255):
        P.append(mul2(P[k]))
    return P

P = compute_P_table()

print('P-table prepared')

def comparator(A, Ak, B, Bk):
    result = set(A).intersection(set(B))
    if result:
        sol_kt = A.index(next(iter(result)))
        sol_kw = B.index(next(iter(result)))
        print('total time: %.2f sec' % (time.time() - starttime))
        d = Ak[sol_kt] - Bk[sol_kw]
        print('SOLVED:', d)
        with open("results.txt", 'a') as file:
            file.write(('%d' % (Ak[sol_kt] - Bk[sol_kw])) + "\n")
            file.write("---------------\n")
        return True
    else:
        return False

def check(P, Pindex, DP_rarity, file2save, A, Ak, B, Bk):
    if P.x % DP_rarity == 0:
        A.append(P.x)
        Ak.append(Pindex)
        with open(file2save, 'a') as file:
            file.write(('%064x %d' % (P.x, Pindex)) + "\n")
        return comparator(A, Ak, B, Bk)
    else:
        return False

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

def search(Nt, Nw, problem, kangoo_power, starttime):
    DP_rarity = 1 << ((problem - 2 * kangoo_power) // 2 - 2)
    hop_modulo = ((problem - 1) // 2) + kangoo_power
    T, t, dt = [], [], []
    W, w, dw = [], [], []
    for k in range(Nt):
        t.append((3 << (problem - 2)) + random.randint(1, (1 << (problem - 1))))
        T.append(mulk(t[k]))
        dt.append(0)
    for k in range(Nw):
        w.append(random.randint(1, (1 << (problem - 1))))
        W.append(add(W0, mulk(w[k])))
        dw.append(0)
    print('tame and wild herds are prepared')
    oldtime = time.time()
    Hops, Hops_old = 0, 0
    t0 = time.time()
    oldtime = time.time()
    starttime = oldtime
    while True:
        for k in range(Nt):
            Hops += 1
            pw = T[k].x % hop_modulo
            dt[k] = 1 << pw
            solved = check(T[k], t[k], DP_rarity, "tame.txt", T, t, W, w)
            if solved:
                return 'sol. time: %.2f sec' % (time.time() - starttime)
            t[k] += dt[k]
            T[k] = add(P[pw], T[k])
        for k in range(Nw):
            Hops += 1
            pw = W[k].x % hop_modulo
            dw[k] = 1 << pw
            solved = check(W[k], w[k], DP_rarity, "wild.txt", W, w, T, t)
            if solved:
                return 'sol. time: %.2f sec' % (time.time() - starttime)
            w[k] += dw[k]
            W[k] = add(P[pw], W[k])
        t1 = time.time()
        if (t1 - t0) > 5:
            print('%.3f h/s' % ((Hops - Hops_old) / (t1 - t0)))
            t0 = t1
            Hops_old = Hops

start = 2147483647
end = 4294967295
search_range = end - start + 1
problem = search_range.bit_length()

compreessed_public_key = "0209c58240e50e3ba3f833c82655e8725c037a2294e14cf5d73a5df8d56159de69" #Puzzle 32
kangoo_power = 3
Nt = Nw = 2 ** kangoo_power
X = int(compreessed_public_key, 16)
Y = X2Y(X % (2 ** 256))
if Y % 2 != (X >> 256) % 2:
    Y = modulo - Y
X = X % (2 ** 256)
W0 = Point(X, Y)
starttime = oldtime = time.time()

Hops = 0
random.seed()

hops_list = []
N_tests = 3

for k in range(N_tests):
    with open("tame.txt", 'w') as tame_file, open("wild.txt", 'w') as wild_file:
        tame_file.write('')
        wild_file.write('')
    search_result = search(Nt, Nw, problem, kangoo_power, starttime)
    print(search_result)
    M, D = 0, 0
    if len(hops_list) > 0:
        M = sum(hops_list) * 1.0 / len(hops_list)
        D = sum((xi - M) ** 2 for xi in hops_list) * 1.0 / len(hops_list)
    print(M, '+/-', (D / (len(hops_list) - 1)) ** 0.5)
    print('Average time to solve: %.2f sec' % ((time.time() - starttime) / N_tests))



It is about 142827.704 h/s here....
member
Activity: 348
Merit: 34
July 18, 2023, 05:04:38 AM
Who can solve
 p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765

in pubkey

p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73

substract formula
p = int(2**256 - 2**32 - 977)
x1 =  # fill pubkey1-x
y1=  # fill pubkey1-y
x2=  # fill pubkey2-x
y2=  # fill pubkey2-y



dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))


if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey

Would you plz be specific, like solve for what? What is expected result conditions? How result should look like, can explain how the successful result should be some how
Above Dec calculate with mod P 2 digit less, same all Dec figures taken to create pubkey
Substraction formula original ecc with original mod P
Adjust mod P to 2 digit less for get correct pubkey as mention in pubkey result
newbie
Activity: 18
Merit: 0
July 18, 2023, 04:47:18 AM
Who can solve
 p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765

in pubkey

p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73

substract formula
p = int(2**256 - 2**32 - 977)
x1 =  # fill pubkey1-x
y1=  # fill pubkey1-y
x2=  # fill pubkey2-x
y2=  # fill pubkey2-y



dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))


if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey

Would you plz be specific, like solve for what? What is expected result conditions? How result should look like, can explain how the successful result should be some how
member
Activity: 348
Merit: 34
July 18, 2023, 01:03:02 AM
 Who can solve
 p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 1099511627776
b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
c = (a-b) %p
result = 1612236468765

in pubkey

p =115792089237316195423570985008687907852837564279074904382605163141518161494335
a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
c = (a-b) %p
result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73

substract formula
p = int(2**256 - 2**32 - 977)
x1 =  # fill pubkey1-x
y1=  # fill pubkey1-y
x2=  # fill pubkey2-x
y2=  # fill pubkey2-y



dx = (x1 - x2) % p
dy = (y1 - (-y2)) % p
c = dy * gmpy2.invert(dx, p) % p
Rx = (c*c - x2 - x1) % p
Ry = (c*(x2 - Rx) - y2) % p
print (Rx , Ry)
print (hex(Rx) , hex(Ry))


if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
copper member
Activity: 205
Merit: 1
July 10, 2023, 04:53:18 PM

In the 30+ minutes that I wasted hammering (and subsequently trashing) a reply to this, I just updated the pkarith script to spit out the correct end points... took me just two minutes and 2 lines of code change. https://gist.github.com/ZenulAbidin/e8687d9e16189c99d192e97d37e71dbe

See how more effectively and faster I can implement stuff when I have all the info I need, clearly (emphasis on that), beforehand?

I can only work with information I have at hand (without guessing random stuff). Nowhere was it said that you had to divide max value by 32 and add that to start value to get the end value... take a look back at them yourself.

Otherwise you get stuff like a broken Kangaroo-256 with a borked hashtable lookup function.

There wouldn't have even been a flamewar if I had that info. Instead I got "0.01325 to 0.0625" which I understood to mean that the zones themselves are the start and end points. With this info, making an assumption about division at that stage would've been a guess.

And none of this was resolved until Counselor explained this important point a few posts up.


Hello.
  I read the topic from the end and saw this script and discussion.
I checked how it works, but no matches are found when searching for public keys with the ranges that go in the same block in the keyhunt.
Do the public keys in the output of this script match the ranges?
member
Activity: 77
Merit: 19
July 09, 2023, 05:49:24 AM
It is the same person or group which solve 120 bit.

So it is not kangaroo and not bsgs.

They must know something which we do not know.
The questions is -> what kind of math they use.   And second questions : I'm afraid BTC is not secure any more.
jr. member
Activity: 47
Merit: 13
July 09, 2023, 05:35:34 AM
Hi there  Smiley

Congratulations to the solver (or solvers) of the puzzle #125   Cheesy

full member
Activity: 1162
Merit: 237
Shooters Shoot...
May 19, 2023, 07:37:46 AM
Hey, guys! somebody can explain me why ive got error :
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing

this is -winfo
Code:
Loading: 1252.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 2238135375233024 2^50.991
Time      : 6.7d
DP Size   : 10.0/17.0MB
DP Count  : 261003 2^17.994
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 1.00
HT SDev   : 1.00
Kangaroos : 0 2^-inf

Loading: 125.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 1982479057551360 2^50.816
Time      : 6.0d
DP Size   : 9.0/29.8MB
DP Count  : 231003 2^17.818
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 0.88
HT SDev   : 0.94
Kangaroos : 67108864 2^26.000

MergeWork: destination argument missing = you are missing the 3rd argument.

If you are trying to merge 2 different work files, you have to tell it what to save it as; merge file1 file2 file3 = merge file1 and file2 and save it as file3
after merge result file have 15mb, but! work file 6gb each other
Yeah I don’t know. The winfo says that one file is only 9 MB.

newbie
Activity: 22
Merit: 1
May 19, 2023, 07:33:22 AM
Hey, guys! somebody can explain me why ive got error :
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing

this is -winfo
Code:
Loading: 1252.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 2238135375233024 2^50.991
Time      : 6.7d
DP Size   : 10.0/17.0MB
DP Count  : 261003 2^17.994
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 1.00
HT SDev   : 1.00
Kangaroos : 0 2^-inf

Loading: 125.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 1982479057551360 2^50.816
Time      : 6.0d
DP Size   : 9.0/29.8MB
DP Count  : 231003 2^17.818
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 0.88
HT SDev   : 0.94
Kangaroos : 67108864 2^26.000

MergeWork: destination argument missing = you are missing the 3rd argument.

If you are trying to merge 2 different work files, you have to tell it what to save it as; merge file1 file2 file3 = merge file1 and file2 and save it as file3
after merge result file have 15mb, but! work file 6gb each other
full member
Activity: 1162
Merit: 237
Shooters Shoot...
May 19, 2023, 06:43:55 AM
Hey, guys! somebody can explain me why ive got error :
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing

this is -winfo
Code:
Loading: 1252.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 2238135375233024 2^50.991
Time      : 6.7d
DP Size   : 10.0/17.0MB
DP Count  : 261003 2^17.994
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 1.00
HT SDev   : 1.00
Kangaroos : 0 2^-inf

Loading: 125.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 1982479057551360 2^50.816
Time      : 6.0d
DP Size   : 9.0/29.8MB
DP Count  : 231003 2^17.818
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 0.88
HT SDev   : 0.94
Kangaroos : 67108864 2^26.000

MergeWork: destination argument missing = you are missing the 3rd argument.

If you are trying to merge 2 different work files, you have to tell it what to save it as; merge file1 file2 file3 = merge file1 and file2 and save it as file3
newbie
Activity: 22
Merit: 1
May 19, 2023, 05:53:51 AM
Hey, guys! somebody can explain me why ive got error :
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing

this is -winfo
Code:
Loading: 1252.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 2238135375233024 2^50.991
Time      : 6.7d
DP Size   : 10.0/17.0MB
DP Count  : 261003 2^17.994
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 1.00
HT SDev   : 1.00
Kangaroos : 0 2^-inf

Loading: 125.save
Version   : 0
DP bits   : 33
Start     : 10000000000000000000000000000000
Stop      : 1FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key       : 0233709EB11E0D4439A729F21C2C443DEDB727528229713F0065721BA8FA46F00E
Count     : 1982479057551360 2^50.816
Time      : 6.0d
DP Size   : 9.0/29.8MB
DP Count  : 231003 2^17.818
HT Max    : 8 [@ 0074C5]
HT Min    : 0 [@ 000003]
HT Avg    : 0.88
HT SDev   : 0.94
Kangaroos : 67108864 2^26.000
member
Activity: 185
Merit: 15
Two things you should never abandon: Family & BTC
May 14, 2023, 09:56:16 PM

Anyway to answer your PM I'm 80% certain about the range. I created the wallet years ago manually and was dumb enough to use it!!

That said, I think I prefer searching for it myself. If you would like to help with the setup, where it's best to rent GPUs, what version of Kangaroo to use, etc happy to pay a fee. I hope you understand, it's too much BTC to just give away the public key on a forum!

Thanks!!
Yeah, somehow I doubt it, being a newbie years ago knowing how to manually generate a very low range key and somehow depositing a large amount in it and then after years now you remembered that you have these coins , yet the funny thing is that you have the public key not the private key, having the public key could mean that you have spent from the address, so the address has an exposed p in a low range, one could use all public keys with balance on blockchain and use kangaroo to find your imaginary key, but we all know there is no such a key and nobody would manually generate low range key to deposit and then forgets the key. Lol

You're implying that he's lying?
Do you know what i should say to you?
Do you?

I agree with you loool
copper member
Activity: 1330
Merit: 899
🖤😏
May 14, 2023, 07:58:29 PM

Anyway to answer your PM I'm 80% certain about the range. I created the wallet years ago manually and was dumb enough to use it!!

That said, I think I prefer searching for it myself. If you would like to help with the setup, where it's best to rent GPUs, what version of Kangaroo to use, etc happy to pay a fee. I hope you understand, it's too much BTC to just give away the public key on a forum!

Thanks!!
Yeah, somehow I doubt it, being a newbie years ago knowing how to manually generate a very low range key and somehow depositing a large amount in it and then after years now you remembered that you have these coins , yet the funny thing is that you have the public key not the private key, having the public key could mean that you have spent from the address, so the address has an exposed p in a low range, one could use all public keys with balance on blockchain and use kangaroo to find your imaginary key, but we all know there is no such a key and nobody would manually generate low range key to deposit and then forgets the key. Lol
full member
Activity: 1162
Merit: 237
Shooters Shoot...
May 14, 2023, 05:16:25 PM
@WanderingPhilospher it seems that I can only write 1 PM every hour, sorry. So that's not going to work.

Anyway to answer your PM I'm 80% certain about the range. I created the wallet years ago manually and was dumb enough to use it!!

That said, I think I prefer searching for it myself. If you would like to help with the setup, where it's best to rent GPUs, what version of Kangaroo to use, etc happy to pay a fee. I hope you understand, it's too much BTC to just give away the public key on a forum!

Thanks!!
Understood.

You’ll have to use JLPs version. Using 256 GPUs, it solved a 110 bit key in under 3 days.

So 128 GPUs should solve it in around 4.5 days, 64 GPUs should solve it in around 9 days.

I’m not the one to ask about renting GPUs as far as best prices because I rarely rent any and only rented my first one about a week ago.
newbie
Activity: 2
Merit: 0
May 14, 2023, 04:51:28 PM
@WanderingPhilospher it seems that I can only write 1 PM every hour, sorry. So that's not going to work.

Anyway to answer your PM I'm 80% certain about the range. I created the wallet years ago manually and was dumb enough to use it!!

That said, I think I prefer searching for it myself. If you would like to help with the setup, where it's best to rent GPUs, what version of Kangaroo to use, etc happy to pay a fee. I hope you understand, it's too much BTC to just give away the public key on a forum!

Thanks!!
full member
Activity: 1162
Merit: 237
Shooters Shoot...
May 14, 2023, 12:22:30 PM
Hello,

How long does it take to find a key in the 2^110 range? I know for this challenge it was found on 2020-05-30? But 3 years past and surely this range can be searched much quicker now? Can someone give me a time estimate and hardware if we were to search realistically this range? I am asking because I might have an old address where the key could be in that range...but not sure. And I don't want to spend thousands on hardware without knowing approx. how log it twill take to search the range with Kangaroo.

Thanks!!!

Honestly, it all depends on your hardware.

With my new mod/speed increase, I could solve a 110-bit range key, using 64 GPUs in less than 3 days.

Obviously if you use more GPUs, you can solve it quicker, if you have less than it would be longer.

I would help, depending on how much btc is in your wallet. I couldn't help unless I could reimburse my costs, plus a little extra.

If you want to throw some cash at it, let me know. DM and we can discuss and work out specifics.
newbie
Activity: 2
Merit: 0
May 14, 2023, 10:13:08 AM
Hello,

How long does it take to find a key in the 2^110 range? I know for this challenge it was found on 2020-05-30? But 3 years past and surely this range can be searched much quicker now? Can someone give me a time estimate and hardware if we were to search realistically this range? I am asking because I might have an old address where the key could be in that range...but not sure. And I don't want to spend thousands on hardware without knowing approx. how log it twill take to search the range with Kangaroo.

Thanks!!!
full member
Activity: 1162
Merit: 237
Shooters Shoot...
May 14, 2023, 01:48:17 AM
I am not sure how you perform elliptic curve cryptography (ECC) calculations on a GPU as CUDA GPUs do not currently support 128-bit integers. Could you please explain your implementation, preferably with some code?
First page, link to GitHub, go look at the code.

This program has already solved 2 keys, at 109 and 114 bits.
  2 steps folding 512bits to 256bits reduction using 64 bits digits. According to the document, I found out that this is not the secp256K1 algorithm. I want to implement GPU calculation in Python. Do you have any sample code?
So are you saying that if somebody wanted to find the private key of a known bitcoin public key, then this program will not work?!
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