forumaster


August 28, 2016, 02:51:01 PM 

How to choose what pattern to generate? I don't have a very good GPU and with the autochosen pattern it says 50% in 25 years xD I want to generate the one with the lowest difficult





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bitcoiner23


September 10, 2016, 02:36:17 PM 

If you give a damn a GTX 1080 hashes with 69 Mkey/s. Still not profitable.




Timelord2067


September 10, 2016, 09:55:11 PM 

If you give a damn a GTX 1080 hashes with 69 Mkey/s. Still not profitable.
Nice. Was thinking of upgrading at least one of my cards  that'd give me a cool 100 MKeys/sec ... HHmmm...




okbaby123
Jr. Member
Offline
Activity: 34


October 21, 2016, 11:02:23 AM 

An example (available from gobittest website): We have a private key: 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 which maps to public key: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
and say we want to find a pattern "166". One of the solutions takes a form of a private key B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F mapping to public key: 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h
If we add the two private keys (like the person requesting the address would), we get: CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 which maps to the same public key as the sum of the public keys, and thus  to the same address.
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h how?0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA 6+ 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C840 4= 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A2748 7how to calculate it ? the last number 6+4=7? not 6+4=a?




okbaby123
Jr. Member
Offline
Activity: 34


October 22, 2016, 01:49:16 AM 

An example (available from gobittest website): We have a private key: 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 which maps to public key: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
and say we want to find a pattern "166". One of the solutions takes a form of a private key B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F mapping to public key: 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h
If we add the two private keys (like the person requesting the address would), we get: CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 which maps to the same public key as the sum of the public keys, and thus  to the same address.
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h how?0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA 6+ 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C840 4= 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A2748 7how to calculate it ? the last number 6+4=7? not 6+4=a?No one helping?




Timelord2067


October 22, 2016, 09:57:08 AM 

how to calculate it ? the last number 6+4=7? not 6+4=a?
No one helping? Might have something to do with the fact we have lives and don't monitor ever thread all day. Did you try multiplying the two values together?




oinkoink
Jr. Member
Offline
Activity: 45


November 18, 2016, 09:02:37 PM 

An example (available from gobittest website): We have a private key: 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 which maps to public key: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
and say we want to find a pattern "166". One of the solutions takes a form of a private key B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F mapping to public key: 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h
If we add the two private keys (like the person requesting the address would), we get: CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 which maps to the same public key as the sum of the public keys, and thus  to the same address.
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h how?No one helping? interested too

piggybank: 1oinkBob5MDsZukoY8Vmy8TxDH7VovPx



alfaboy23


November 20, 2016, 03:31:52 AM 

Good day!
I've seen this wallet address from the list of my referrals in a faucet, look at the last 3 character: 3GZLKxx7SgYkUXarSRNULxjxRQf4P3DAY3
I thought the addresses only starting in 1 can be generated for vanity address, isn't that a multisig address? If that is possible, how?




oinkoink
Jr. Member
Offline
Activity: 45


November 20, 2016, 09:01:06 AM 

Good day!
I've seen this wallet address from the list of my referrals in a faucet, look at the last 3 character: 3GZLKxx7SgYkUXarSRNULxjxRQf4P3DAY3
I thought the addresses only starting in 1 can be generated for vanity address, isn't that a multisig address? If that is possible, how? afaik before you base58encode it, instead of 0x00 at the beginning, you start with a 0x05 (prefix/version number) this results in a leading 3 address e.g. 05.payload.checksum > base58encode > 3..... 00.payload.checksum > base58encode > 1.....

piggybank: 1oinkBob5MDsZukoY8Vmy8TxDH7VovPx



alfaboy23


December 02, 2016, 03:40:28 AM 

Good day!
I've seen this wallet address from the list of my referrals in a faucet, look at the last 3 character: 3GZLKxx7SgYkUXarSRNULxjxRQf4P3DAY3
I thought the addresses only starting in 1 can be generated for vanity address, isn't that a multisig address? If that is possible, how? afaik before you base58encode it, instead of 0x00 at the beginning, you start with a 0x05 (prefix/version number) this results in a leading 3 address e.g. 05.payload.checksum > base58encode > 3..... 00.payload.checksum > base58encode > 1..... Thanks for the explanation. But I'm still having trouble understanding codes regarding addresses. Can I just do that using the vanity miner?




kolloh
Legendary
Offline
Activity: 1148


December 06, 2016, 01:59:58 AM 

ThePiachu,
I'm not sure how much you are actively maintaining this project, but any chance you could update the "Solved work" page to include a "Solved on Date" column or similar? I'd also like to see a "Requested on Date" column on the available work page if possible.
It would be nice to be able to look at dates to see if there are recent requests or solved work to monitor the recent activity of the pool.




arulbero


December 25, 2016, 09:50:33 PM 

An example (available from gobittest website): We have a private key: 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 which maps to public key: 0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
and say we want to find a pattern "166". One of the solutions takes a form of a private key B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F mapping to public key: 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h
If we add the two private keys (like the person requesting the address would), we get: CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 which maps to the same public key as the sum of the public keys, and thus  to the same address.
If we add the two public keys (like the person looking for the solution would do), we get a public key: 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487 which is equivalent to this address: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h how?0450863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA 6+ 049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C840 4= 0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A2748 7how to calculate it ? the last number 6+4=7? not 6+4=a?Curve secp256k1: base point : G = (79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B 16F81798, 483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8) order curve: n = FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFBAAEDCE6AF48A03BBFD25E8CD0364141 First private key : a = 18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 First public key : A = a*G = (50863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B23522CD470243453A, 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6) see for yourself here > https://gobittest.appspot.com/AddressSecond private key : b = B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F Second public key : B =b*G=(36970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98, 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487) If we add the two private keys : a+b = CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54 ~$ python Python 2.7.12+ (default, Sep 17 2016, 12:08:02) [GCC 6.2.0 20160914] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> a=0x18E14A7B6A307F426A94F8114701E7C8E774E7F9A47E2C2035DB29A206321725 >>> b=0xB18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F >>> n=0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 #order curve >>> print hex((a+b) % n) # add mod 'n' 0xca65722cd418ed28ec369e36cfe3b7f3cc1cd035bfbf6469ce759fca30ad6d54
and if we add the two public keys: A+B = a*G + b*G = (a+b)*G see for yourself here > https://gobittest.appspot.com/Address(a+b)*G = (36970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461, E100E705CCA9854436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487) and the address is: 166ev9JXn2rFqiPSQAwM7qJYpNL1JrNf3h




adaseb
Legendary
Offline
Activity: 1428


January 12, 2017, 12:20:23 AM 

I got a bunch of old GPUs that can't mine any of the current profitable coins such as ZEC or ETH. Wondering if any of the work available at: https://vanitypool.appspot.com/availableWorkStill will pay out if a solution is found. For example, 1qwertyuiop Pays 1.2 BTC however the guy was looking for this vanity key since 2013. EDIT: Nevermind, didn't know the results had to be case sentitive, in that case it would take millions of years to find a result.

FORTUNEJACK.COM  [  9 BTC WELCOME PACK FOR 1ST 5 DEPOSITS FREE 1,000 mBTC daily for LuckyJack winners  [   ] 



ArcCsch


January 12, 2017, 01:22:56 AM 

Is it possible to mergemine vanity addresses? Each user enters a pattern and a public key, for example: User Alice sends pattern "1alice" and public key a1 (private key is A1). User Bob sends pattern "1bob" and public key b1 (private key is B1). User Charlie sends pattern "1charlie" and a public key c1 (private key is C1). Now the miners begin hashing, combining all the public keys together, adding arbitrary keys, hashing and comparing each hash to all the provided patterns. Assume they find a solution for Bob first: Hash160(a1+b1+c1+x1)=1bob... Now, Alice and Charlie publish their private keys A1 and C1, and the miner publishes X1 (the private key that generates x1). Now, Bob combines A1, B1, C1, and X1 to create his private key, and no one else can do this because only Bob has B1. Bob now has a vanity address which fits his pattern, and goes offline. Now, Alice and Charlie generate new key pairs, and send the public keys: Alice sends a2 (private key is A2). Charlie sends c2 (private key is C2). Now the pool continues hashing, combining the two keys, hashing, and comparing each hash to both provided patterns, until they find a hash that fits 1alice or 1charlie. When this happens, the same thing happens again, one user (probably Charlie) publishes the private key, and the other gets control over the new address. The main problem I see is that a user can refuse to publish the private key after an address is found for someone else, wasting all the mining effort. Can this problem be overcome?

If you don't have sole and complete control over the private keys, you don't have any bitcoin! Signature campaigns are OK, zero tolorance for spam! 1JGYXhfhPrkiHcpYkiuCoKpdycPhGCuswa



kolloh
Legendary
Offline
Activity: 1148


April 12, 2017, 09:09:22 PM 

The Vanity Pool sends payments with quite low fees due to the network requiring higher fees now. A recent payment used 44.444 sat/byte.
Any possibility in bumping up the fee a bit to make confirmations faster for these transactions?




ciobanuionut


April 17, 2017, 07:55:36 AM 

mean INVEST... like if you've solved this problem, someone needs to be throwing money at you to buy hardware and further develop this proprietary software for a promise in sharing future profits. Maybe I'm wrong, but I see a big demand for this kind of service as Bitcoin awareness spreads.




iMaster


April 20, 2017, 07:33:03 PM 

When will this be live? Would like to get a vanity address asap.

DeepOnion it is SCAM And fuck this coin



kolloh
Legendary
Offline
Activity: 1148


April 21, 2017, 02:18:51 AM 

When will this be live? Would like to get a vanity address asap.
It is already live and has been working for quite some time. You can use this to get a vanity address right now




addrstore.com
Jr. Member
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Activity: 45
1DonationLd73pZbX3fAitt4JZv5kdrZEt


June 09, 2017, 07:56:35 AM 

Good day!
I've seen this wallet address from the list of my referrals in a faucet, look at the last 3 character: 3GZLKxx7SgYkUXarSRNULxjxRQf4P3DAY3
I thought the addresses only starting in 1 can be generated for vanity address, isn't that a multisig address? If that is possible, how? afaik before you base58encode it, instead of 0x00 at the beginning, you start with a 0x05 (prefix/version number) this results in a leading 3 address e.g. 05.payload.checksum > base58encode > 3..... 00.payload.checksum > base58encode > 1..... Thanks for the explanation. But I'm still having trouble understanding codes regarding addresses. Can I just do that using the vanity miner? Yes it is possible. Here's the command to generate a vanity address with the "3Love" prefix: ./vanitygen F script 3Love




mraksoll
Jr. Member
Offline
Activity: 30


June 10, 2017, 07:56:49 AM 

i hv hardware rnd generator i can use it on vanitygen ? if i can how use it ? add rand bit's how seed ? Or what need do for vanity gen use my rnd bit's my rng gen bit's entropy
Entropy = 7.999506 bits per byte.
Optimum compression would reduce the size of this 333348 byte file by 0 percent.
Chi square distribution for 333348 samples is 227.67, and randomly would exceed this value 89.00 percent of the times.
Arithmetic mean value of data bytes is 127.4296 (127.5 = random). Monte Carlo value for Pi is 3.143309694 (error 0.05 percent). Serial correlation coefficient is 0.001518 (totally uncorrelated = 0.0).

14V57LBZCKnb6jCBofiNMJGKSwrddZhBGy



