Question to the smart people:
If one is using the kangaroo method, searching for 32 bit distinguished points, and searching in the 120 bit range, how many x coordinates (x coordinate is what the program finds of a given public key, which in kangaroo terminology means "distance traveled") would have a 32 bit distinguished point?
There are 2^120 possibilities, but we know not every private key in the 120 bit range has a public key x coordinate that has a DP of 32. Is there a good guesstimate or is it just anyone's guess?
I should clarify, how many tame points in a 120 bit range would/could have distinguished points of 32 bits?
If one is using the kangaroo method, searching for 32 bit distinguished points, and searching in the 120 bit range, how many x coordinates (x coordinate is what the program finds of a given public key, which in kangaroo terminology means "distance traveled") would have a 32 bit distinguished point?
There are 2^120 possibilities, but we know not every private key in the 120 bit range has a public key x coordinate that has a DP of 32. Is there a good guesstimate or is it just anyone's guess?
I should clarify, how many tame points in a 120 bit range would/could have distinguished points of 32 bits?
If you're in the 120 bit range it should be 120/2 = 60 - 32 = 28. Finally answer is 2^28 = 268,435,456 tame distinguished points of 32 bits, that's my guess.
I went through a small range 1-300FFF and didn't find any 32 DPs. I was closely looking at trailing, but I don't think there were any leading or in between either.
Anywho, if we use your 2^28, that would mean 2^28 for each type of DP, leading, trailing, and in between. So that would mean 2^29.5 (minimum, because the in between ones can be in between anywhere.)
You said you went through a small range 1-300FFF, that's 2^24 range with distinguished points of 32 bits. 32 bits is to big for that small space. That space 2^24 would require distinguished points of 6 bits.
