Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
unpluggedcoin
on 18/07/2023, 09:47:18 UTC
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