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Roughly there are 2^256 possible private key and 2^160 bitcoin address (excluding address with prefix 3 and bc1), so of course private key is possible.
P.S. please prove why Pollard's kangaroo algorithm is the most effective private key search tool
P.S. please prove why Pollard's kangaroo algorithm is the most effective private key search tool
Pollard's kangaroo algorithm is a mathematical calculation. Brute Force is not very effectiveWith the help of enumeration, much less keys are checked. When busting, you waste time and computer power. And in mathematical calculation, you are ahead of time. All this is described in detail in the documents:
How pollard-kangaroo works, the Tame and Wild kangaroos, is a simple explanation.
Suppose there is pubkeyX, unknow privkeyX, but privkeyX is in range w=[L..U]. The keys have a property - if we increase pubkey by S, then its privkey will also increase by S. We start step-by-step to increase pubkeyX by S(i), keeping sumX(S(i)). This is a Wild kangaroo. We select a random privkeyT from range [L..U], compute pubkeyT. We start step-by-step to increment pubkeyT by S(i) while maintaining sumT(S(i)). This is a Tame kangaroo. The size of the jump S(i) is determined by the x coordinate of the current point, so if a Wild or Tame kangaroo lands on one point, their paths will merge. (we are concerned with pseudo random walks whose next step is determined by the current position) Thanks to the Birthday Paradox (Kruskal's card trick), their paths will one day meet. Knowing each traveled path (sumX and sumT), privkeyX is calculated. The number of jumps is approximately 2w1/2 group operations, which is much less than a full search w.
Kangaroos, Monopoly and Discrete Logarithms --> https://web.northeastern.edu/seigen/11Magic/KruskalsCount/PollardKangarooMonopoly.pdf
Using Pollard's kangaroo algorithm, you can’t crack all the addresses, but you can partially determine the desired private key from a huge list of public keys. Theoretical is possible!
