Re: Pollard's kangaroo ECDLP solver

So there are 2^96 different private keys in the entire 2^256 space that can "unlock" for example this address: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (Puzzle 64)? Can anyone confirm this?

Yes exactly and for a simple fact :

16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (and every other wallet address of this type) is encoded on 160 bits (and not on 256 bits like the majority of other crypto parameters in bitcoin protocol)
16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN is base58 encoded and is 3ee4133d991f52fdf6a25c9834e0745ac74248a4 (20*8bytes = 160bits) in hexadecimal
(
Unlike public keys that are encoded on 256 bits

a simple wallet address (p2pkh) is simply obtained by the function hash160(public_key)

So if you have the possibility to browse the entire 1-2^256 space and to compute the hash160 function for every hash160(public_key) derived from 1-2^256 private keys you will find an average of 2^(256-160) = 2^96 public key with hash160=16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN .

But even if 2^96 seems big its far away smaller from 2^256. And it's pretty impossible to have a collision between two random public key in the using age of bitcoin.

But theoretically if you find any public key derived from a private key with hash160(public_key)= '16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN' or = to any other non zero wallet address you will able to unlock the coins.

Because the verification in the bitcoin protocol to prove that you are the owner of the private key is simply "have you signed the transaction with the private key associated to a public key that gives 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (or any other target address) with hash160 function

Thanks for the reply!

I'm missing something...So these addresses will have different public keys or not?

Can one private key control 2^96 different wallets?

For example when I import some random private key in Electrum wallet and click on "sweep funds" am I sweeping one wallet or 2^96 different wallets?

You got it backwards, one private key does NOT open 2^96 addresses .. instead, every address can be unlocked by 2^96 different private keys ..

because: number of private keys is 2^256 .. but number of addresses is only 2^160 .. which in verbal language means that for each address there's 2^96 extra private keys that can unlock it.

Number of all private keys: 2^256
Number of all public keys: 2^256
Number of all addresses: 2^160
Number of all Hash160: 2^160 (duh 🙄)

Quote from: BHWallet on Today at 07:13:31 PM

Quote from: CrunchyF on August 26, 2022, 05:55:01 PM

Quote from: BHWallet on August 26, 2022, 04:14:41 PM

So there are 2^96 different private keys in the entire 2^256 space that can "unlock" for example this address: 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN (Puzzle 64)? Can anyone confirm this?