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Title: BTC Cryptography formular Post by: rddsd7 on December 03, 2017, 07:10:59 AM Hey,
Anyone knows the crypto formula for generating BTC address?? is it base on date+time...?? Any possibility of repeating an address?? Title: Re: BTC Cryptography formular Post by: aplistir on December 03, 2017, 10:47:40 AM Hey, Anyone knows the crypto formula for generating BTC address?? is it base on date+time...?? Any possibility of repeating an address?? Yep. First you need to have a 256 bit random number, and then you convert it to an address. The random number is supposed to be fully random, so it is not based on anything. By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. So you are better off playing lottery than trying to find an address collision :) Title: Re: BTC Cryptography formular Post by: rddsd7 on December 03, 2017, 03:20:39 PM Hey, Anyone knows the crypto formula for generating BTC address?? is it base on date+time...?? Any possibility of repeating an address?? Yep. First you need to have a 256 bit random number, and then you convert it to an address. The random number is supposed to be fully random, so it is not based on anything. By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. So you are better off playing lottery than trying to find an address collision :) Thanks. That implies a lot.. Title: Re: BTC Cryptography formular Post by: tromp on December 03, 2017, 03:50:07 PM By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. Assuming 10^7 addresses in use, the odds of address collision are 10^7/2^256, or roughly 10^-70. Assuming a lottery with 10^-8 odds of winning, the odds of winning a lottery 100 times in a row are 10^-800. You're WAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYYYYYYYYY off :-( Title: Re: BTC Cryptography formular Post by: rddsd7 on December 04, 2017, 05:31:57 AM By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. Assuming 10^7 addresses in use, the odds of address collision are 10^7/2^256, or roughly 10^-70. Assuming a lottery with 10^-8 odds of winning, the odds of winning a lottery 100 times in a row are 10^-800. You're WAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYYYYYYYYY off :-( Yeah.. I'm waaaaaaaaaaaayyyyyyyy off. but however there's a little possiblilty i guess.. so, i suggest if they would create the address by including the DATE+TIME, that would never repeat.. Title: Re: BTC Cryptography formular Post by: Dr.Z on December 04, 2017, 07:40:57 AM Address = RIPEMD160(SHA256(Public Key))
The public key is calculated from the private key using elliptic curve multiplication, which is irreversible: K = k * G where k is the private key, G is a constant point called the generator point and K is the resulting public key. Title: Re: BTC Cryptography formular Post by: BurtW on December 05, 2017, 08:53:51 PM By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. Assuming 10^7 addresses in use, the odds of address collision are 10^7/2^256, or roughly 10^-70. Assuming a lottery with 10^-8 odds of winning, the odds of winning a lottery 100 times in a row are 10^-800. You're WAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYYYYYYYYY off :-( Title: Re: BTC Cryptography formular Post by: haltingprobability on December 06, 2017, 02:44:18 AM By repeating an address, you mean address collision, where you generate an address that is already in use. That is possible, but the changes are so small that winning a lottery 100 times in a row is much more likely. Assuming 10^7 addresses in use, the odds of address collision are 10^7/2^256, or roughly 10^-70. Assuming a lottery with 10^-8 odds of winning, the odds of winning a lottery 100 times in a row are 10^-800. You're WAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYYYYYYYYY off :-( Last I checked, 0 = 0. Title: Re: BTC Cryptography formular Post by: shrivelfire on December 17, 2017, 07:58:47 PM You need to learn what are the possible addresses first. There are at least 256 types of addresses, and these are randomized.
Title: Re: BTC Cryptography formular Post by: BurtW on December 17, 2017, 08:38:44 PM You need to learn what are the possible addresses first. There are at least 256 types of addresses, and these are randomized. Noob knows not what they are talking about.There are 2160 possible Bitcoin addresses. There are 2256 possible public/private key pairs. So, there are approximately 2256 / 2160 = 2(256 - 160) = 296 public/private key pairs per Bitcoin address. |