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Since the set of all private keys is denumerable, it is at least possible to create a searcheable record of all keypairs. Which, when finished, is a killer for the whole idea. Am I right or do I miss some essential part?
Is the denumerability of private keys a threat?
Answers
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There's around 2^256 possible private keys. However, there's only 2^160 valid wallet addresses. To store each key-address pair we need 160+256 bits = 25 bytes, so the searchable record will take several million mass of solar system if stored on 4TiB harddrives.
But first we need to create this database. If we equip each PC on Earth (around 2 billion) with 4 GPUs (each generating around 20Mkey/s), we will need much more time than the age of the universe. By the way, we most likely will also hit the hard drive write performance limit if we write each pair to disk.
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No, it is not a threat, because the number of possible private keys is about 2^256. Even if every atom on earth was a Petabyte hard drive, you would still not have nearly enough storage for a record of all keys.
(More interesting is the finite number of Bitcoin addresses, but that is still enormously large.)
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You're meaning something like this?
– aland – 2014-05-01T11:35:02.907@aland Clever! I think I just found the smallest vanity address: 19's compressed address starts with
– Tim S. – 2014-05-01T17:13:22.2471AmyIt's pretty hard to represent non-denumerable sets on a finite state digital computer. – Nate Eldredge – 2014-09-12T06:25:15.690
OP - It's not denumerable, it's finite. "Denumerable" in standard mathematical terminology means countably infinite. – Meni Rosenfeld – 2014-09-12T13:24:27.490