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Board Bitcoin Discussion
Re: Bitcoin challenge transaction: ~100 BTC total bounty to solvers!
by
Darmont33
on 09/09/2019, 06:03:43 UTC
Here is the code with my changes
Code:

import time
import random
import gmpy2
import math
import sys

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

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

def rev(b, n = modulo):
    while b < 0:
        b += modulo
    g, x, _ = egcd(b, n)
    if g == 1:
        return x % n
       
def mul2(P, p = modulo):
    R = Point()
#    c = 3*P.x*P.x*rev(2*P.y, p) % p
    c = 3*P.x*P.x*gmpy2.invert(2*P.y, p) % p
    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):
    R = Point()
    dx = Q.x - P.x
    dy = Q.y - P.y   
    c = dy * gmpy2.invert(dx, p) % p     
    #c = dy * rev(dx, p) % p     
    R.x = (c*c - P.x - Q.x) % p
    R.y = (c*(P.x - R.x) - P.y) % p
    return R # 6 sub, 3 mul, 1 inv

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 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
    for w in range(256):
        if (pw >> w) & 1 == 1:
            tmp = X
            for k in range(w):
                tmp = (tmp**2)%p
            Y *= tmp
            Y %= p
    return Y

def comparator():
    A, Ak, B, Bk = [], [], [], []
    with open('tame.txt') as f:
        for line in f:
            L = line.split()
            a = int(L[0],16)
            b = int(L[1],16)
            A.append(a)
            Ak.append(b)
    with open('wild.txt') as f:
        for line in f:
            L = line.split()
            a = int(L[0],16)
            b = int(L[1],16)
            B.append(a)
            Bk.append(b)
    result = list(set(A) & set(B))
    if len(result) > 0:
        sol_kt = A.index(result[0])
        sol_kw = B.index(result[0])
        print ('total time: %.2f sec' % (time.time()-starttime))
        d = Ak[sol_kt] - Bk[sol_kw]
        print ('SOLVED: %64X' % d + '\n')
        file = open("results.txt",'a')
        file.write(('%X'%(Ak[sol_kt] - Bk[sol_kw])) + "\n")
        file.write("---------------\n")
        file.close()
        return True
    else:
        return False

def check(P, Pindex, DP_rarity, file2save):
    if P.x % (DP_rarity) == 0:
        file = open(file2save,'a')
        file.write(('%064X %064X'%(P.x,Pindex)) + "\n")
        file.close()
        return comparator()
    else:
        return False
   
P = [PG]
for k in range(255): P.append(mul2(P[k]))   
print ('P-table prepared')   

def search(a,b):
    global solved
    s=(a+b)>>1
    d=(b-a)
    problem=int(math.log(d,2))
#    print(a,b,s,d,'\n')
    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):
        qtf= s
        qtr= random.randint(1,d)
 #       print('tame\n',qtf,qtr)
        qt=qtf+qtr
        t.append(qt) 
        T.append(mulk(t[k]))
        dt.append(0)
    for k in range(Nw):
        qw=(random.randint(1, d))
  #      print('wild\n',qw)
        w.append(qw)
        W.append(add(W0,mulk(w[k])))
        dw.append(0)
    print ('tame and wild herds are prepared')
    oldtime = time.time()
    starttime = oldtime
    Hops, Hops_old = 0, 0
    t0 = time.time()
    oldtime = time.time()
    starttime = oldtime
    while (1):
        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")
            if solved: break
            t[k] += dt[k]
            T[k] = add(P[pw], T[k])
        if solved: break           
        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")
            if solved: break
            w[k] += dw[k]
            W[k] = add(P[pw], W[k])
        if solved: break
        t1 = time.time()
        if (t1-t0) > 5:
            print ('%.3f h/s'%((Hops-Hops_old)/(t1-t0)))
            t0 = t1
            Hops_old = Hops
    hops_list.append(Hops)       
    print ('Hops:', Hops)       
    return 'sol. time: %.2f sec' % (time.time()-starttime)   

s=sys.argv[1]
sa = sys.argv[2]
sb = sys.argv[3]
sk = sys.argv[4]
a = int(sa, 16)
b = int(sb, 16)
kangoo_power = int(sk, 10)
Nt = Nw = 2**kangoo_power
X = int(s, 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 = []

solved = False
open("tame.txt",'w').close()
open("wild.txt",'w').close()
search(a,b)

Example (case 32)

python kang.py 0387dc70db1806cd9a9a76637412ec11dd998be666584849b3185f7f9313c8fd28 80000000 FFFFFFFF 3

(the last number 3 is the kangooro_power)

P-table prepared
tame and wild herds are prepared
total time: 0.21 sec
SOLVED:                                                         7D4FE747

('Hops:', 23072)



Hey, nice work! and congratulations to the key opener 62!

Please clarify the following situation .. File tame.txt .. What is the significance of the intensity of filling this file? For example, in one range it is not recorded at all, and the range is slightly different and the file has a size of 2MB in 1 day... thx