Here is another example of #40
#40 sqr(2^(bit-1))
python pollard-kangaroo-multi.py 40
//privkey = 1003651412950
#40 sqr(2^(bit-2))
python pollard-kangaroo-multi.py 8:3FFFFFFFFF 028dfd0e801ed8d495b6a0b68b38fba4f32d7423af363c717cca6c2ebd1e11a399
//privkey = 179017692118
Addfactor: 824633720832
//result1 = 1003651412950 true
824633720832 is just 0xC000000000
179017692118 is 0x29AE4933D6
1003651412950 is 0xE9AE4933D6
So 0xC000000000 is just the mid point of the normal #40 range 8000000000:FFFFFFFFFF
in other words the #40 sqr(2^(bit-2)) range you are mentioning is just the position of the private key from the mid-point.
You are not doing less work, you've just changed the origin.
Instead of working with the tame kangaroo matching the private key solution, you are working with the corresponding wild kangaroo.
Exactly that was the plan > wraping the space W with the middle of the space to point infinity.
So that actually all x values get halfed.
I thought possibly that when a wild kangaroo jumps, a jump from the tame kangaroo in the negative space in the different direction could also lead to the same result in less time(cause it would be the same distance).
The key is definitely to find in the space(bit-2), however if you say regarding the kangaroo method, that there is no speed gain, the method is useless.
i have to test #35. i would not understand if the key is not findable. like #50 if the key is on the negative side it will find the privatekey and publickey on the positiv side,there will be just no publickey match> Watch it in the console im not sure if gets to the result.txt
Thanks for testing!