novacn


October 23, 2014, 12:33:45 PM 

I heard that there are some tools to generate bitcoin address with predefined patterns.
If I have a key pair with public address like this: 1abcdefghijklmn....
Any one know roughly how hard is it for someone to generate a key pair with pubkey like this: 1axcxexfxhxjxlxn..., where x means any letter or number? I think it should be pretty hard, but don't know exactly how hard it is.
Maybe its a weird question. I'm just very curious... Thought maybe some tech guru here knows the answer.
Thanks in advance!








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qdoop
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October 23, 2014, 01:01:02 PM 

Elementary probability theory.
Each position has 1/58 probability to be the desired symbol. So if you fix let say 3 positions the probability to find a matching pattern would be (1/58)*(1/58)*(1/58) It is independent of the position you fix in the address. 1ace....xxxx is as difficult as 1axcxex.......
The possibility vanishes very quick as a power of n(positions you fix)




novacn


October 23, 2014, 02:07:22 PM 

Elementary probability theory.
Each position has 1/58 probability to be the desired symbol. So if you fix let say 3 positions the probability to find a matching pattern would be (1/58)*(1/58)*(1/58) It is independent of the position you fix in the address. 1ace....xxxx is as difficult as 1axcxex.......
The possibility vanishes very quick as a power of n(positions you fix)
Thanks @qdoop! That's a quick and concise answer!




findftp
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October 23, 2014, 02:23:44 PM 

Elementary probability theory.
Each position has 1/58 probability to be the desired symbol. So if you fix let say 3 positions the probability to find a matching pattern would be (1/58)*(1/58)*(1/58) It is independent of the position you fix in the address. 1ace....xxxx is as difficult as 1axcxex.......
The possibility vanishes very quick as a power of n(positions you fix)
Thanks @qdoop! That's a quick and concise answer! FYI, the tool is called vanitygen. findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdefg Difficulty: 51529903411245 [195.63 Kkey/s][total 16791552][Prob 0.0%][50% in 5.8y] findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdef Difficulty: 888446610538 [221.12 Kkey/s][total 1468160][Prob 0.0%][50% in 32.2d] So just a "simple" prefix takes a long time on a intel xeon w3505 edit: must say I run this linux on software vmware.

1FdwDuV2qgsw2w1Lza8nsea7L6mQxsc7g3



novacn


October 23, 2014, 02:32:09 PM 

Elementary probability theory.
Each position has 1/58 probability to be the desired symbol. So if you fix let say 3 positions the probability to find a matching pattern would be (1/58)*(1/58)*(1/58) It is independent of the position you fix in the address. 1ace....xxxx is as difficult as 1axcxex.......
The possibility vanishes very quick as a power of n(positions you fix)
Thanks @qdoop! That's a quick and concise answer! FYI, the tool is called vanitygen. findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdefg Difficulty: 51529903411245 [195.63 Kkey/s][total 16791552][Prob 0.0%][50% in 5.8y] findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdef Difficulty: 888446610538 [221.12 Kkey/s][total 1468160][Prob 0.0%][50% in 32.2d] So just a "simple" prefix takes a long time on a intel xeon w3505 edit: must say I run this linux on software vmware. Thanks, findftp! I'll play around with that vanitygen. Seems a great tool, cheers!




findftp
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October 23, 2014, 02:48:47 PM 

FYI, the tool is called vanitygen. findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdefg Difficulty: 51529903411245 [195.63 Kkey/s][total 16791552][Prob 0.0%][50% in 5.8y] findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen 1abcdef Difficulty: 888446610538 [221.12 Kkey/s][total 1468160][Prob 0.0%][50% in 32.2d] So just a "simple" prefix takes a long time on a intel xeon w3505 edit: must say I run this linux on software vmware. Thanks, findftp! I'll play around with that vanitygen. Seems a great tool, cheers! FYI, if I do the same abcdefg case insensitive, it goes much faster because then 1Abcdefg and 1ABcdefg and 1aBcdefg ... etc is also valid. I don't really know about the math behind it. It'll probably be a smaller base number to calculate with. findftp@mint16 ~/Downloads/vanitygenmaster $ ./vanitygen i 1abcdefg Difficulty: 13419245680 [240.02 Kkey/s][total 1666560][Prob 0.0%][50% in 10.8h]

1FdwDuV2qgsw2w1Lza8nsea7L6mQxsc7g3



teukon
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October 24, 2014, 01:13:43 AM Last edit: October 24, 2014, 01:27:28 AM by teukon 

I heard that there are some tools to generate bitcoin address with predefined patterns.
If I have a key pair with public address like this: 1abcdefghijklmn....
Any one know roughly how hard is it for someone to generate a key pair with pubkey like this: 1axcxexfxhxjxlxn..., where x means any letter or number? I think it should be pretty hard, but don't know exactly how hard it is.
Maybe its a weird question. I'm just very curious... Thought maybe some tech guru here knows the answer.
Thanks in advance!
An alternative approach. It is believed that there are roughly 2^160 different possible Bitcoin addresses (RIPEMD160). If you want to specify half of the address, you must generate roughly 2^80 different random addresses for a reasonable chance of finding one (probability about e ^{1}). Assuming for simplicity that all hash types are the same, this is approximately the total work done by the Bitcoin network to date (the total work in the main chain up to block 326696 is 2 ^{81.2} hashes to 3.s.f). So to make such an address, you'd probably have to compute as many hashes as all Bitcoin miners have ever computed. That's about one hash for every atom in a 1kg gold bar.




novacn


October 24, 2014, 02:24:42 AM 

I heard that there are some tools to generate bitcoin address with predefined patterns.
If I have a key pair with public address like this: 1abcdefghijklmn....
Any one know roughly how hard is it for someone to generate a key pair with pubkey like this: 1axcxexfxhxjxlxn..., where x means any letter or number? I think it should be pretty hard, but don't know exactly how hard it is.
Maybe its a weird question. I'm just very curious... Thought maybe some tech guru here knows the answer.
Thanks in advance!
An alternative approach. It is believed that there are roughly 2^160 different possible Bitcoin addresses (RIPEMD160). If you want to specify half of the address, you must generate roughly 2^80 different random addresses for a reasonable chance of finding one (probability about e ^{1}). Assuming for simplicity that all hash types are the same, this is approximately the total work done by the Bitcoin network to date (the total work in the main chain up to block 326696 is 2 ^{81.2} hashes to 3.s.f). So to make such an address, you'd probably have to compute as many hashes as all Bitcoin miners have ever computed. That's about one hash for every atom in a 1kg gold bar. Wow, that's an great explanation! So, the difficulty level is even beyond what my expected. Thanks teukon!




amaclin
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October 24, 2014, 06:52:18 AM 

Each position has 1/58 probability to be the desired symbol. Note: this is not exact correct. This slightly changes the answer.




deepceleron
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October 24, 2014, 07:38:23 PM 

Elementary probability theory.
Each position has 1/58 probability to be the desired symbol.
Each position has 1/58 probability to be the desired symbol. Note: this is not exact correct. This slightly changes the answer. If by "not exact correct", you mean potentially orders of magnitudes off (by not understanding how Bitcoin addresses are encoded), than you are correct. The chance the second digit is "A" is about 1/22. The probability the same position is "1" is 1/256; for it to be "z", 1/1353. The probability that the 34th digit doesn't exist is about 1/41. The probability that positions 120 are all "1"s is 0.




amaclin
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October 25, 2014, 06:43:25 AM 

deepceleron, thanks for comment




polarhei
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October 25, 2014, 07:39:43 AM 

During the power form of the basic curve function, also must fulfill conditions defined. It will be very difficult to get the same.
I think, you will need to test the exponential to try.




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October 25, 2014, 08:03:42 AM 

I have a feeling that OP believes that when you generate an address that half of it is similar to another address that numerically(private keys) you are getting closer to the an existing address. But this is not the case.

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novacn


October 25, 2014, 11:32:30 AM 

I have a feeling that OP believes that when you generate an address that half of it is similar to another address that numerically(private keys) you are getting closer to the an existing address. But this is not the case.
Sorry but I don't quite understand what you mean by numerically(private keys). We never show our private keys to anyone right? I mean that "half same" only in public key, and if it's darn impossible for a today's normal pc to run years to acheive the result.




Jungian
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October 25, 2014, 02:31:42 PM 





