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.
Topic: Pollard's kangaroo ECDLP solver - page 9. (Read 55445 times)
two script python
http://bitchain.pl/100btc/pollard_kangaroo.txt
and
https://github.com/Telariust/pollard-kangaroo/blob/master/pollard-kangaroo-multi.py
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))
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....
Hi there,
i am playing around with kangaroo and i am not sure if it's working correctly. Why is it saying "Kangaroos: 0 2^-inf"?
Also, whats the meaning of HT Max/Min/...? Would be great if someone could explain it to me. I am running kangaroo in client/server mode with server command: ./kangaroo -w save.work -wsplit -wi 600 -o result.txt -d 24 -s in.txt
Here is an example of one of the workfiles, as it's saved every 10minutes there are severall workfiles now.
Do you merge them for example once a day and only keep the merged file or do we need the save.workTIMESTAMP too?
Thanks a lot
i am playing around with kangaroo and i am not sure if it's working correctly. Why is it saying "Kangaroos: 0 2^-inf"?
Also, whats the meaning of HT Max/Min/...? Would be great if someone could explain it to me. I am running kangaroo in client/server mode with server command: ./kangaroo -w save.work -wsplit -wi 600 -o result.txt -d 24 -s in.txt
Here is an example of one of the workfiles, as it's saved every 10minutes there are severall workfiles now.
Quote
Kangaroo v2.2
Loading: save.work_27Jul23_101320
Version : 0
DP bits : 24
Start : 200000000000000000000000000000000
Stop : 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key : 03633CBE3EC02B9401C5EFFA144C5B4D22F87940259634858FC7E59B1C09937852
Count : 0 2^-inf
Time : 00s
DP Size : 2.3/5.7MB
DP Count : 11186 2^13.449
HT Max : 3 [@ 00B207]
HT Min : 0 [@ 000000]
HT Avg : 0.04
HT SDev : 0.21
Kangaroos : 0 2^-inf
Loading: save.work_27Jul23_101320
Version : 0
DP bits : 24
Start : 200000000000000000000000000000000
Stop : 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Key : 03633CBE3EC02B9401C5EFFA144C5B4D22F87940259634858FC7E59B1C09937852
Count : 0 2^-inf
Time : 00s
DP Size : 2.3/5.7MB
DP Count : 11186 2^13.449
HT Max : 3 [@ 00B207]
HT Min : 0 [@ 000000]
HT Avg : 0.04
HT SDev : 0.21
Kangaroos : 0 2^-inf
Do you merge them for example once a day and only keep the merged file or do we need the save.workTIMESTAMP too?
Thanks a lot
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
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
Substraction formula original ecc with original mod P
Adjust mod P to 2 digit less for get correct pubkey as mention in pubkey result
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
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
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
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
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?
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.
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.
Hi there
Congratulations to the solver (or solvers) of the puzzle #125
Congratulations to the solver (or solvers) of the puzzle #125
Hey, guys! somebody can explain me why ive got error :
this is -winfo
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing
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
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
Hey, guys! somebody can explain me why ive got error :
this is -winfo
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing
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
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
Hey, guys! somebody can explain me why ive got error :
this is -winfo
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing
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
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
Hey, guys! somebody can explain me why ive got error :
this is -winfo
Code:
D:\kangaroo>kangaroo.exe -wm 125.save 1252.save
Kangaroo v2.2
MergeWork: destination argument missing
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
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
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!!
You're implying that he's lying?
Do you know what i should say to you?
Do you?
I agree with you loool
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!!
@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. 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!!
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.
@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!!
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!!
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!!!
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.
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!!!
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!!!
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.
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