When will this be live? Would like to get a vanity address asap.
Topic: Vanity Pool - vanity address generator pool - page 5. (Read 147183 times)
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.
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?
Any possibility in bumping up the fee a bit to make confirmations faster for these transactions?
Is it possible to merge-mine 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?
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?
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/availableWork
Still 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.
Wondering if any of the work available at:
https://vanitypool.appspot.com/availableWork
Still 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.
Quote
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.
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 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
+
049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
=
0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487
how 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/Address
Second private key : b = B18427B169E86DE681A1A62588E1D02AE4A7E83C1B413849989A76282A7B562F
Second public key :
B =b*G=(36970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98, 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487)
If we add the two private keys : a+b = CA65722CD418ED28EC369E36CFE3B7F3CC1CD035BFBF6469CE759FCA30AD6D54
Code:
~$ 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
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.
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.
Quote
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 multi-sig address? If that is possible, how?
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 multi-sig 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.....
Can I just do that using the vanity miner?
Quote
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 multi-sig address? If that is possible, how?
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 multi-sig 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.....
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 multi-sig address? If that is possible, how?
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 multi-sig address? If that is possible, how?
Quote
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.
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?
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?
Quote
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.
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 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
+
049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
=
0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487
how to calculate it ? the last number 6+4=7? not 6+4=a?
No one helping?
Quote
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.
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 299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6
+
049C95E0949E397FACCECF0FE8EAD247E6FD082717E4A4A876049FB34A9ADED110DFEA2EF691CC4 A1410498F4C312F3A94318CD5B6F0E8E92051064876751C8404
=
0436970CE32E14DC06AC50217CDCF53E628B32810707080D6848D9C8D4BE9FE461E100E705CCA98 54436A1283210CCEFBB6B16CB9A86B009488922A8F302A27487
how to calculate it ? the last number 6+4=7? not 6+4=a?
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...
If you give a damn a GTX 1080 hashes with 69 Mkey/s. Still not profitable.
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
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
VanityPool still lists 1Private but refuses to accept solutions (one of my rig found the solution and all of the ones I pointed at VanityPool tried to submit new ones several times).
Unfortunately OgNasty's Life Boat patch of VanityGen didn't extend to the VanityMiner and after about two weeks of ~42MKeys/sec "mining" I get this:
Oh well...
Code:
Searching for pattern: "1EXAmoney" Reward: 0.008000 Value: 0.000001 BTC/MkeyHr
Difficulty: 50656515217834
[42.28 Mkey/s][total 41252892442624][Prob 55.7%][75% in 7.9d] M
atch idx: 1
CPU hash: 900954e1a558005d304c31b07bfa34eab9205fda
GPU hash: 944e4590b1334a04131c2a5c3e2ebc1170ebf90c
Found delta: 4487231 Start delta: 50331649
[43.30 Mkey/s][total 41475182166016][Prob 55.9%][75% in 7.7d]
Oh well...
Hey guys, im trying to put up some new work on the site, but having trouble with public key?
keeps telling me "public key too short" ;\
not quite sure what it's looking for as i am admittedly pretty noob with ECDSA keypairs ;\
i am using puttygen.
Any help appreciated, thx!
keeps telling me "public key too short" ;\
not quite sure what it's looking for as i am admittedly pretty noob with ECDSA keypairs ;\
i am using puttygen.
Any help appreciated, thx!
Paste it here so we can see what did you copy, probably the wrong thing.
this is the public key i generated...
You did something wrong, use this site to generate a public key, just move the mouse around to get 100% and then click "Vanity wallet", you'll get a public key, and a first private key, then when somebody solves you problem you'll get from them the second part private key and that is it.
https://www.bitaddress.org
awesome. well i see you yourself provide a service so have PM'd you with some work
ty!I've sent you a PM, I have to check it first.
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