ken3go
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March 25, 2018, 11:51:58 AM 

My searches usually error out after 7 days or so of running at 16Mkeys/sec (~12 quadrillion tries). If you just restart it, nothing is lost. It just makes a clean random start at another point than where you started before. What does "nothing is lost" mean? It went through 12 quadrillion tries before crashing. Is every try completely random (it doesn't "save" a list of previous attempts or go in some methodical order)? The probability of repeating one of those 12q tries is the same as trying an untried one? If so, then it wouldn't make any difference if I ran the program for 200 hours straight or for 10 hours on each of 20 days or for 2 mins on each of 6000 days. Right? Sorry, I'm sure that question has been asked 12q times.






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LoyceV
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March 25, 2018, 12:09:12 PM 

Is every try completely random (it doesn't "save" a list of previous attempts or go in some methodical order)? Yes. The probability of repeating one of those 12q tries is the same as trying an untried one? Yes. If so, then it wouldn't make any difference if I ran the program for 200 hours straight or for 10 hours on each of 20 days or for 2 mins on each of 6000 days. Right? Correct. Sorry, I'm sure that question has been asked 12q times. More or less, yes Think of it this way: you have 3 dice, and you're trying to throw them all at 6 in one throw. The odds of doing this are 1 in 216. After trying 10, 100 or 1000 times, the odds of throwing it at your next try are still exactly the same. Vanitygen works the same: say the odds of finding it are 50% in 20 minutes. It doesn't matter if it been running for 1 minute or 1 hour, the odds are still exactly the same: 50% for the next 20 minutes.




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March 25, 2018, 12:47:03 PM 

Think of it this way: you have 3 dice, and you're trying to throw them all at 6 in one throw. The odds of doing this are 1 in 216. After trying 10, 100 or 1000 times, the odds of throwing it at your next try are still exactly the same.
I can see where the confusion can stem from, people seem to forget that individual rolls have no effect on future rolls as they are mutually exclusive. To calculate the true probability you can't add the probability. A common logical fallacy is people believe if the chance of getting a 6 when rolling a single dice is 1/6 then the probability of getting a 6 in six rolls is = 1/6 +1/6 +1/6 +1/6 +1/6 +1/6 = 1. You need keep in mind the probability is always fixed to exactly 1/6 every single roll. Doesn't matter you roll non stop or if you roll once a year.




nullius
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March 25, 2018, 12:54:51 PM 

Each extra character makes it 58 times more difficult to find. Note that starting with a Capital can be 58 times faster (depending on which character you use): 1Abcdef or 1ABCDEF are much faster than 1abcdef.
Just two questions about the two points:  Why is it 58 exactly? My guess would be: is it something like 26 +26+ 10 = 62 (alphabet sets in caps and regular making the 26 each and the 10 being the number of numbers zero to nice) minus four illegal characters?
Yes, 62 minus four illegal characters. That equals “58 exactly”. Oldstyle (preBech32) Bitcoin addresses use base58, not base62. Each character is a radix58 digit, in the range of [0, 57]. Following are the “digits” used by Bitcoin, from an old code snippet of mine. Observe that “I” (uppercase i), “O” (uppercase o), “0” (numeral zero), and “l” (lowercase L) are excluded. const char base58[59] = "123456789" /* 9, [0..8] */ "ABCDEFGHJKLMNPQRSTUVWXYZ" /* 24, [9..32] */ "abcdefghijkmnopqrstuvwxyz"; /* 25, [33..57] */
 Why are capital letters easier to find as compared to regular numbers and what's like the "math" behind it?
Capital letters are not generally easier to find. However, at the beginning, they represent a lower number. Since the large integer being represented is in a range which is not a power of 58, higher digits at the beginning may be rare, or even impossible. For an analogy: Imagine that you are searching for a pattern of base10 digits in a 30bit base2 (binary) number. The number you seek has a range of [0, 1073741823]. Digits [29] are impossible in the first position; and digit 1 is only in the first position for 73741824/1073741823 ≈ 6.9% of randomly selected 30bit numbers. Here, you are searching for a 192bit base2 (binary) number, where the upper 160 bits are uniformly distributed and the lower 32 bits are also uniformly distributed (but dependent on the upper 160 bits). You are representing that number as a base58 number. Probability of hitting various base58 digits in the first position is left as an exercise to the reader. <g>
Hmm, I'm guessing this more of a practical real life data rather than actual theoretical analysis?
Yes. The theoretical answer must be somewhere within the hashing algorithm, but that's beyond my understanding. The theoretical answer is actually not in the hashing algorithm at all, but rather, in how a pseudorandom number uniformly distributed across a binary search space is represented in radix58 (base58).
If you just restart it, nothing is lost. It just makes a clean random start at another point than where you started before. What does "nothing is lost" mean? It went through 12 quadrillion tries before crashing. Is every try completely random (it doesn't "save" a list of previous attempts or go in some methodical order)? LoyceV provided a good explanation by analogy to dice throws. I have only to add: This is a probabilistic search. You could hit your lucky address on the very first try (like winning a lottery). Considering your previous 12 quadrillion “losses” is actually an instance of classic Gambler’s Fallacy.
The probability of repeating one of those 12q tries is the same as trying an untried one?
In both cases, the probability is negligible = practically impossible. 12 quadrillion (1.2 x 10^{16}) is a drop in the ocean of a 2^{160} search space (>10^{48}, more than a thousand quadrillion quadrillion quadrillion).
(N.b. that the search space is of size 2^{160} although its input is 33 octets for compressed keys and 65 octets for uncompressed keys, and the output is a 192bit number due to the 32bit checksum.)




ken3go
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March 25, 2018, 01:47:42 PM 

Think of it this way: you have 3 dice, and you're trying to throw them all at 6 in one throw. The odds of doing this are 1 in 216. After trying 10, 100 or 1000 times, the odds of throwing it at your next try are still exactly the same.
I can see where the confusion can stem from, people seem to forget that individual rolls have no effect on future rolls as they are mutually exclusive. To calculate the true probability you can't add the probability. A common logical fallacy is people believe if the chance of getting a 6 when rolling a single dice is 1/6 then the probability of getting a 6 in six rolls is = 1/6 +1/6 +1/6 +1/6 +1/6 +1/6 = 1. You need keep in mind the probability is always fixed to exactly 1/6 every single roll. Doesn't matter you roll non stop or if you roll once a year. Yup, I got it now. What I was imagining was that there could be a simple loop in the program. Start with the "first" private key (...001), go through the various hashing steps and see if you get a public address with the desired pattern. If not, then increment the private key by 1 (...002) and do the hashes again. That way, the attempted private keys would effectively get "burned" and not be reused. It's like buying millions of lottery tickets in the same draw to try to cover as many numbers as possible. You might as well start with 123456 and then 123457 and so on methodically than to choose a bunch of random "pick 6" numbers. The chance to win is the same for any set of numbers, but there is a slight chance that a "pick 6" could be generated twice, thereby wasting the ticket (i.e. if you win, you would be splitting the jackpot with yourself). I suppose the "slight" chance is so slight that maybe it doesn't matter.




LoyceV
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March 25, 2018, 02:10:42 PM 

What I was imagining was that there could be a simple loop in the program. Start with the "first" private key (...001), go through the various hashing steps and see if you get a public address with the desired pattern. If not, then increment the private key by 1 (...002) and do the hashes again. A fixed instead of truely random starting point would mean your private key isn't secure. It would mean anyone could reproduce your search and steal your coins.




nullius
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March 25, 2018, 03:07:35 PM 

What I was imagining was that there could be a simple loop in the program. Start with the "first" private key (...001), go through the various hashing steps and see if you get a public address with the desired pattern. If not, then increment the private key by 1 (...002) and do the hashes again. That way, the attempted private keys would effectively get "burned" and not be reused.
It's like buying millions of lottery tickets in the same draw to try to cover as many numbers as possible. You might as well start with 123456 and then 123457 and so on methodically than to choose a bunch of random "pick 6" numbers. The chance to win is the same for any set of numbers, but there is a slight chance that a "pick 6" could be generated twice, thereby wasting the ticket (i.e. if you win, you would be splitting the jackpot with yourself). I suppose the "slight" chance is so slight that maybe it doesn't matter.
The entire security of Bitcoin, PGP, TLS/SSL, Tor, disk encryption, and all other crypto using fixedlength keys rests on the premise that the “slight” chance of a collision is impossible as a practical matter. Think: The probability of you picking the same key twice is equal to the probability of an attacker randomly picking your key in a bruteforce attack. Theoreticians use terms such as “negligible probability” because such a thing is possible in theory. But it will never actually happen that you generate the same key twice, unless your random number generator is so badly broken as to be worse than useless. Conceptually, think of randomly picking one drop of water from the ocean, then another, and getting the same drop; or randomly picking one grain of sand from anywhere on Earth, then another, and getting the same grain of sand. 2^{160} is much bigger than that.Whereas LoyceV speaks truly: What I was imagining was that there could be a simple loop in the program. Start with the "first" private key (...001), go through the various hashing steps and see if you get a public address with the desired pattern. If not, then increment the private key by 1 (...002) and do the hashes again. A fixed instead of truely random starting point would mean your private key isn't secure. It would mean anyone could reproduce your search and steal your coins. Note, however, that Vanitygen does try sequential points from a randomly chosen starting point. (“Sequential” here does not mean linear “1, 2, 3”; rather, it uses elliptic curve point addition.) It does this for reason of efficiency. sipa’s keygrinder used in the current development branch of segvan uses similar methods to rapidly generate a great quantity of keys (or optionally, tweaks) from a single random seed. This can be secure if and only if all seed and key material other than the “winning” key is destroyed and never reused.




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March 25, 2018, 05:32:24 PM 

Darkstar has if I'm not wrong 1darkstr or something similar. People recongise those addresses quickly as compared to random ones which are hard to remember
1 DarkStrRagcDjWtsPGxkav4WG3poLXz DSI'd get 1DarkStar, but that's too long for me to feasibly make or pay someone to get at a reasonable price.




nullius
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March 25, 2018, 05:44:50 PM 

Darkstar has if I'm not wrong 1darkstr or something similar. People recongise those addresses quickly as compared to random ones which are hard to remember
1 DarkStrRagcDjWtsPGxkav4WG3poLXz DSI'd get 1DarkStar, but that's too long for me to feasibly make or pay someone to get at a reasonable price. I just want to note, this is NOT a good means to recognize an address. There are at least 2100254120907352485526230505830591911428096 (58 ^{24}) addresses which match the pattern ^1DarkStr.+DS$. Somebody else could easily find a different one to spoof DarkStar_’s address. I know that this is a real problem with Tor .onion vanity addresses; and I suspect it may be with Bitcoin vanity addresses, too. A vanity address is good for showing off, and/or making a statement such as with my 35 segwitgLKnDi2kn7unNdETrZzHD2c5xh address. But it is highly insecure as a user interface feature.




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March 25, 2018, 05:52:05 PM 

I'd get 1DarkStar, but that's too long for me to feasibly make or pay someone to get at a reasonable price.
I've added these to my search: 1DarkStar 1Darkstar 1DARKSTAR 1darkstar 1darkStar I don't run it often anymore, but eventually I'll find one. Just to show off of course, as you shouldn't trust anyone else's keys. @Spammers here: please go back to the spamboards! I've reported one for removal, the other for a ban. Leave the serious boards alone!




DaveF


March 31, 2018, 03:27:22 PM 

I know I asked a few pages back in the thread, but does anyone have and performance numbers for the GTX 1050 or 1050Ti?
I can pick one up for a good price, but I can't return it if I do. They are mostly useless for mining now so I would dedicate it to this. But, if it does not have the speed I'll stick with the AMD that I have now.
Thanks, Dave




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April 02, 2018, 01:31:38 PM 

I don't understand how to run oclvanitygen.
If I type in "oclvanitygen v d 0 1a" I keep getting GPU and CPU hashes.
EDIT: nvm I got my address with vanitygen64




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April 05, 2018, 10:10:10 AM 

what is the most interesting vanity address created so far? anybody care to share?




TryNinja


April 05, 2018, 10:40:21 AM 

what is the most interesting vanity address created so far? anybody care to share?
Nobody can know for sure but there is a old thread with a few: Rare address hall of fame I also find this one from Loyce pretty interesting: 166666666LyMNrkpwwNCdUPzvDTh2tNDLu




ken3go
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April 05, 2018, 11:03:21 AM 

what is the most interesting vanity address created so far? anybody care to share?
I have a good one running at 500Mkeys/s but it will take about 630k years, so I'll have to get back to you. etotheipi found the all caps one in a week (he thought it would take 70 days). 1QBDLYTDFHHZAABYSKGKPWKLSXZWCCJQBX Google that and you'll go down a rat hole of "longest" or "most unusual" addresses.




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April 05, 2018, 06:58:12 PM 

I have always wanted to do a vanitygen for my cold storage, but I am too paranoid the program would have some sort of built in preset way to make an address the creator also has keys for. How am I supposed to know for sure? I just wish I could trust do it trustless.




LoyceV
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April 05, 2018, 08:48:46 PM 

I just wish I could trust do it trustless. You can use ue split key for this. See (my thread) Pretty Addy Giveaway  part 2 for instructions. With split key, it doesn't matter if someone else creates the address for you, or you don't trust vanitygen. As long as you keep your private keys offline all the time, it's secure. If you also don't trust bitaddress.org, you'll need to use something else to create a private key.




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April 05, 2018, 08:51:09 PM 

I've been generating Ravencoin addresses but haven't been able to get one of them to import successfully. Anyone had any luck with them?




LoyceV
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April 22, 2018, 08:13:35 PM 

Vanitygen has a command that is not listed in the program options: F script Actually, it is listed: ./vanitygen help F <format> Generate address with the given format (pubkey or script) Amazing I have to test this! ./vanitygen F script 3test Prefix '3test' not possible Hint: valid bitcoin script addresses begin with "3" I'm not sure what the requirements are exactly, but this works: ./vanitygen F script 333333 Difficulty: 264104224 Pattern: 333333 P2SHAddress: 333333dirmCEYVJGZavhjt2fmWL15boJYx Address: 12ZEW67DZX9LP1rNQq6bSnw1mnkemKvo6H Privkey: 5J48B7JDChgvAUGPA5p6yvjwZtyFufJizc6jFSci6Jwu9KzQdFh Only CPU though, and no bc1address. Using nonnative SegWit feels a bit like a 50% solution. It's multisig, not segwit. See Thirdspace's post. Two weeks ago, I sent some coins to the generated P2SH address without knowing about how to access it. I just recently found out a way. I am not a technical guy so please always check if you have an access to the address generated before sending coins. It's all still new to me too. I'll experiment more later.
Why did you delete your post? You cannot merit that post




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April 22, 2018, 08:25:59 PM 

Why did you delete your post?
Loaded it back again. Mistakenly pressed delete. I'm not sure what the requirements are exactly, but this works: Yeah. Wondering about the same. [P2SH address generation Vanitygen]F script The command generates a P2SH address , a P2PKH address and a Private key. Command was first posted by someone here: https://github.com/exploitagency/vanitygenplus/issues/30 How to generate Redeem script?[1] Copy the private key and go to bitaddress.org . Go to Wallet details and paste that private key you copied. Check for the public key (uncompressed) and copy it. [2] Go to https://coinb.in/multisig/ . Click New >> MultiSig Address . Click the "" sign against the uncompressed public key input field until you have only one field to enter your uncompressed public key. Paste your public key that you copied into that only field for entering uncompressed private key. Change the amount of signatures required to release the coins to "1" and click "Submit". It will generate your Redeem script for the P2SH address that you generated. Rest regarding spending your inputs , I would suggest going through mocacinno guide: http://www.mocacinno.com/blog/createsignbroadcasttransactionsusingcoinb/
Two weeks ago,I sent some coins to the generated P2SH address without knowing about how to access it. I just recently found out a way. I am not a technical guy so please always check if you have an access to the address generated before sending coins. [Don't know if its been brought up earlier. Apologies if there is topic already available regarding this ]




