my test is this kangaroo python
puzzle #40 pubkey = 03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4
https://github.com/Telariust/pollard-kangaroo/blob/master/pollard-kangaroo-multi.pypython pollard-kangaroo-multi.py 40 03a2efa402fd5268400c77c20e574ba86409ededee7c4020e4b9f0edbee53de0d4
set verbose to 2 for show message it working process
flag_verbose = 2 # 0, 1, 2
use verbose 2 script will write file tame.txt and wild.txt and on work message
message show this
[childs][tame-2#15344] newDP: X=ec9d5049bd4509aa41e1f5c312de1900d3eac4adba0692a41286b2497e520000
[childs][tame-2#15344] newDP: dK=0x102840cf7d0
if flag_verbose > 1:
print("\n[childs][%s#%s] newDP: X=%064x " % (child_name,child_pid, Xcoord));
print("\n[childs][%s#%s] newDP: dK=0x%x " % (child_name,child_pid, dK));
two value on tame.txt and wild.txt is
Xcoord
dK
but python code is very complicate
I think code is work function Elliptic Curve on script not using library function
I can see code but don't understand ECC calculate
I use ruby script from this easy
https://learnmeabitcoin.com/technical/public-keyand test print x y to see
point = multiply(k, $g) # this point is the public key
# convert x and y values of this point to hexadecimal
x = point[:x].to_s(16).rjust(64, "0")
y = point[:y].to_s(16).rjust(64, "0")
puts "point[:x] = " + point[:x].to_s
puts "point[:y] =" + point[:y].to_s
this x is same value on tame.txt and wild.txt
I don't know much
but think may be same on python script done
Can any help to save file tame.txt and wild.txt from kangaroo.exe
kangaroo.exe save file in binary right I can not read it