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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
This puzzle is very strange. If it's for measuring the world's brute forcing capacity, 161-256 are just a waste (RIPEMD160 entropy is filled by 160, and by all of P2PKH Bitcoin). The puzzle creator could improve the puzzle's utility without bringing in any extra funds from outside - just spend 161-256 across to the unsolved portion 51-160, and roughly treble the puzzle's content density.
If on the other hand there's a pattern to find... well... that's awfully open-ended... can we have a hint or two?
If on the other hand there's a pattern to find... well... that's awfully open-ended... can we have a hint or two?
I am the creator.
You are quite right, 161-256 are silly. I honestly just did not think of this. What is especially embarrassing, is this did not occur to me once, in two years. By way of excuse, I was not really thinking much about the puzzle at all.
I will make up for two years of stupidity. I will spend from 161-256 to the unsolved parts, as you suggest. In addition, I intend to add further funds. My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key). Probably in the next few weeks. At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.
A few words about the puzzle. There is no pattern. It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty). It is simply a crude measuring instrument, of the cracking strength of the community.
Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology. The "large bitcoin collider" is especially innovative and interesting!
How to do the masked with leading 000...0001 to set difficulty in python code?
Re: Biden proposed 30% mining tax. what would impact on Bitcoin mining?
⭐ Merited by larry_vw_1955 (1)
USA sucks to begin with in the first place.
nope brother
the private key starts with L3
the key ranges you posted private keys start with K
3803496f04b8c0f3e8335eb24b07a4a15335d37bff03bafbb19640a287e0609a
Ky6bKVgbJPFRiAPxvBo4ftiAMkKCBjAbWfVRWEevsHxgULTthBFB
Sorry, I made two mistakes. the private key starts with L3
the key ranges you posted private keys start with K
3803496f04b8c0f3e8335eb24b07a4a15335d37bff03bafbb19640a287e0609a
Ky6bKVgbJPFRiAPxvBo4ftiAMkKCBjAbWfVRWEevsHxgULTthBFB
First is adding 1 less character (total 51 instead of 52). The correct values are:
Minimum
Code:
L39HAHFF1p6vxjBEe4YMxno11111111111111111111111111111
b0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f
Maximumb0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f
Code:
L39HAHFF1p6vxjBEe4YMxnozzzzzzzzzzzzzzzzzzzzzzzzzzzzz
b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459
Diff: b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459
1253753004473247624297767682974128968010 = 1.25E+39 (which is close to your initial value).
Second is that the number I posted is all permutations of the base58 characters if they are placed in their missing positions (ie. 5829) instead of using the method above since all missing characters are from one place at the end without anything in between.
Quote
i can able to recover last or even mid where missing seven to eight or upto 10 in matter of minutes thats not an issue
its more than 10 that whats the issue is i am expermienting on it anyway thanks for your suggestions brother
Up to 12 characters from the end can be recovered very easily since you'd essentially be checking less than a million keys. The problem is that the numbers grow very fast as the number of missing chars increase.its more than 10 that whats the issue is i am expermienting on it anyway thanks for your suggestions brother
This is wrong:
Code:
L39HAHFF1p6vxjBEe4YMxno11111111111111111111111111111
b0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f
Maximumb0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f
Code:
L39HAHFF1p6vxjBEe4YMxnozzzzzzzzzzzzzzzzzzzzzzzzzzzzz
b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459
b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459
The correct is:
b0bea32711dbb7429ba37464ffbb4c8cda31ea17c6d85d063c0aa4d2c8d5e30f=L39HAHFF1p6vxjBEe4YMxnnzzzzzzzzzzzzzzzzzzzzzzfskX9dq
b0bea32711dbb7429ba37464ffbb4c908969f5b84b442e0734743c599dec7459=L39HAHFF1p6vxjBEe4YMxnozzzzzzzzzzzzzzzzzzzzzzv62YdrY
Find the WIF challenge
The known part of the key is 40 characters from 52
The address 1PfNh5fRcE9JKDmicD2Rh3pexGwce1LqyU
How to participate in the challenge:
If you have a GPUs:
RTX 2070, 2080, 2090, 3060, 3070, 3080, 3090, A5000, A6000 and are ready to search for a key 24/7
Challenge is a collective search for a key
A large range of 12 characters is divided into 3364 small ranges.
The program is configured correctly, it takes into account many technical aspects of searching for the initial part of the key.
read more : https://github.com/phrutis/wif500
The known part of the key is 40 characters from 52
The address 1PfNh5fRcE9JKDmicD2Rh3pexGwce1LqyU
How to participate in the challenge:
If you have a GPUs:
RTX 2070, 2080, 2090, 3060, 3070, 3080, 3090, A5000, A6000 and are ready to search for a key 24/7
Challenge is a collective search for a key
A large range of 12 characters is divided into 3364 small ranges.
The program is configured correctly, it takes into account many technical aspects of searching for the initial part of the key.
read more : https://github.com/phrutis/wif500
In the photo of the diary, it only shows 38 characters. What happened to the other two characters?
It shows 40 characters:
Code:
5bCRZhiS5sEGMpmcRZdpAhmWLRfMnutGnPHtjVob
You are right. It does. I apologize.
Find the WIF challenge
The known part of the key is 40 characters from 52
The address 1PfNh5fRcE9JKDmicD2Rh3pexGwce1LqyU
How to participate in the challenge:
If you have a GPUs:
RTX 2070, 2080, 2090, 3060, 3070, 3080, 3090, A5000, A6000 and are ready to search for a key 24/7
Challenge is a collective search for a key
A large range of 12 characters is divided into 3364 small ranges.
The program is configured correctly, it takes into account many technical aspects of searching for the initial part of the key.
read more : https://github.com/phrutis/wif500
The known part of the key is 40 characters from 52
The address 1PfNh5fRcE9JKDmicD2Rh3pexGwce1LqyU
How to participate in the challenge:
If you have a GPUs:
RTX 2070, 2080, 2090, 3060, 3070, 3080, 3090, A5000, A6000 and are ready to search for a key 24/7
Challenge is a collective search for a key
A large range of 12 characters is divided into 3364 small ranges.
The program is configured correctly, it takes into account many technical aspects of searching for the initial part of the key.
read more : https://github.com/phrutis/wif500
In the photo of the diary, it only shows 38 characters. What happened to the other two characters?
I want to say that this so called challenge is a scam. If the owner of the address is only missing the first 5 characters of the BTC WIF, then he won't be asking for help because 58^5 is 656,356,768 and WIFSolverCuda can find it in no time. I wouldn't waste my time on the challenge.
Re: VanBitCracken - a program to use for 32 BTC challenge (supports RTX 30xx cards)
I just dont fucking understand. You created the damn program. Why didn't and haven't you upload the source code? It is common knowledge that when you upload a program to github, you also upload the source code as well.
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
Hello guys, a newbie question here...
Can anyone do the math for me? because i'd really like to finally crack that puzzle 64.
How many 2080 TI's GPUS do i need to be able to reach a cracking speed of a 1 Trillion/s? (i think it's also called Tera or 1000 gigas)
Also, can someone calculate the time needed to crack puzzle 64 at speed of 1 Trillion key/s ?
Much Thanks for your time guys.
We're here together to crack puzzle 64.
Can anyone do the math for me? because i'd really like to finally crack that puzzle 64.
How many 2080 TI's GPUS do i need to be able to reach a cracking speed of a 1 Trillion/s? (i think it's also called Tera or 1000 gigas)
Also, can someone calculate the time needed to crack puzzle 64 at speed of 1 Trillion key/s ?
Much Thanks for your time guys.
We're here together to crack puzzle 64.Please tell me that you are joking. You will be spending more money on GPUS than the reward you could potential get.
Re: Vanitygen: Vanity bitcoin address generator/miner [v0.22]
When i try to build to build the program in Win32, I get this error:
vanitygen.c(66): error C2057: expected constant expression
vanitygen.c(66): error c2466: cannot allocate an array of constant size 0
vanitygen.c(66): error c2133: 'ppnt': unknown size
The code is from:
vg_thread_loop(void *arg)
{
unsigned char hash_buf[128];
unsigned char *eckey_buf;
unsigned char hash1[32];
int i, c, len, output_interval;
int hash_len;
const BN_ULONG rekey_max = 10000000;
BN_ULONG npoints, rekey_at, nbatch;
vg_context_t *vcp = (vg_context_t *) arg;
EC_KEY *pkey = NULL;
const EC_GROUP *pgroup;
const EC_POINT *pgen;
const int ptarraysize = 256;
EC_POINT *ppnt[ptarraysize];
I haven't change anything in the code. How do I fix the errors?
vanitygen.c(66): error C2057: expected constant expression
vanitygen.c(66): error c2466: cannot allocate an array of constant size 0
vanitygen.c(66): error c2133: 'ppnt': unknown size
The code is from:
vg_thread_loop(void *arg)
{
unsigned char hash_buf[128];
unsigned char *eckey_buf;
unsigned char hash1[32];
int i, c, len, output_interval;
int hash_len;
const BN_ULONG rekey_max = 10000000;
BN_ULONG npoints, rekey_at, nbatch;
vg_context_t *vcp = (vg_context_t *) arg;
EC_KEY *pkey = NULL;
const EC_GROUP *pgroup;
const EC_POINT *pgen;
const int ptarraysize = 256;
EC_POINT *ppnt[ptarraysize];
I haven't change anything in the code. How do I fix the errors?
secp256k1::uint256 CLKeySearchDevice::getNextKey()
{
uint64_t totalPoints = (uint64_t)_points * _threads * _blocks;
return _start + secp256k1::uint256(totalPoints) * _iterations * _stride;
}
Can I add some creative counting to the function? For example, ×2 private key, ×3 private key, or using fibonacci sequence.
{
uint64_t totalPoints = (uint64_t)_points * _threads * _blocks;
return _start + secp256k1::uint256(totalPoints) * _iterations * _stride;
}
Can I add some creative counting to the function? For example, ×2 private key, ×3 private key, or using fibonacci sequence.
Why do I get two different private keys for the same BTC address?
C:\bitcrack>test -u -i taddress.txt
[2019-05-18.18:30:32] [Info] Compression: uncompressed
[2019-05-18.18:30:32] [Info] Starting at: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:30:32] [Info] Ending at: FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
[2019-05-18.18:30:32] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:30:32] [Info] Initializing GeForce GTX 1050
[2019-05-18.18:30:32] [Info] Generating 262,144 starting points (10.0MB)
[2019-05-18.18:30:33] [Info] 10.0%
[2019-05-18.18:30:33] [Info] 20.0%
[2019-05-18.18:30:33] [Info] 30.0%
[2019-05-18.18:30:33] [Info] 40.0%
[2019-05-18.18:30:33] [Info] 50.0%
[2019-05-18.18:30:33] [Info] 60.0%
[2019-05-18.18:30:33] [Info] 70.0%
[2019-05-18.18:30:33] [Info] 80.0%
[2019-05-18.18:30:34] [Info] 90.0%
[2019-05-18.18:30:34] [Info] 100.0%
[2019-05-18.18:30:34] [Info] Done
[2019-05-18.18:30:34] [Info] Loading addresses from 'taddress.txt'
[2019-05-18.18:30:34] [Info] 1 addresses loaded (0.0MB)
[2019-05-18.18:30:34] [Info] Address: 1KWj99Jwd9LGGC2Y1c9c4cmvWvYTQrLFVc
Private key: 0000000000000000000000000000000000000000000000000000000220082201
Compressed: no
Public key:
6A245BF6DC698504C89A20CFDED60853152B695336C28063B61C65CBD269E6B4
E022CF42C2BD4A708B3F5126F16A24AD8B33BA48D0423B6EFD5E6348100D8A82
C:\bitcrack>cubitcrack -u --keyspace 1e -i taddress.txt
[2019-05-18.18:16:36] [Info] Compression: uncompressed
[2019-05-18.18:16:36] [Info] Starting at: 000000000000000000000000000000000000000000000000000000000000001E
[2019-05-18.18:16:36] [Info] Ending at: FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
[2019-05-18.18:16:36] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:16:36] [Info] Initializing
[2019-05-18.18:16:36] [Info] Generating 262,144 starting points (10.0MB)
[2019-05-18.18:16:36] [Info] 10.0%
[2019-05-18.18:16:36] [Info] 20.0%
[2019-05-18.18:16:36] [Info] 30.0%
[2019-05-18.18:16:36] [Info] 40.0%
[2019-05-18.18:16:36] [Info] 50.0%
[2019-05-18.18:16:37] [Info] 60.0%
[2019-05-18.18:16:37] [Info] 70.0%
[2019-05-18.18:16:37] [Info] 80.0%
[2019-05-18.18:16:37] [Info] 90.0%
[2019-05-18.18:16:37] [Info] 100.0%
[2019-05-18.18:16:37] [Info] Done
[2019-05-18.18:16:37] [Info] Loading addresses from 'taddress.txt'
[2019-05-18.18:16:37] [Info] 1 addresses loaded (0.0MB)
[2019-05-18.18:16:37] [Info] Address: 1KWj99Jwd9LGGC2Y1c9c4cmvWvYTQrLFVc
Private key: 000000000000000000000000000000000000000000000000000000000000001F
Compressed: no
Public key:
6A245BF6DC698504C89A20CFDED60853152B695336C28063B61C65CBD269E6B4
E022CF42C2BD4A708B3F5126F16A24AD8B33BA48D0423B6EFD5E6348100D8A82
C:\bitcrack>test -u -i taddress.txt
[2019-05-18.18:30:32] [Info] Compression: uncompressed
[2019-05-18.18:30:32] [Info] Starting at: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:30:32] [Info] Ending at: FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
[2019-05-18.18:30:32] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:30:32] [Info] Initializing GeForce GTX 1050
[2019-05-18.18:30:32] [Info] Generating 262,144 starting points (10.0MB)
[2019-05-18.18:30:33] [Info] 10.0%
[2019-05-18.18:30:33] [Info] 20.0%
[2019-05-18.18:30:33] [Info] 30.0%
[2019-05-18.18:30:33] [Info] 40.0%
[2019-05-18.18:30:33] [Info] 50.0%
[2019-05-18.18:30:33] [Info] 60.0%
[2019-05-18.18:30:33] [Info] 70.0%
[2019-05-18.18:30:33] [Info] 80.0%
[2019-05-18.18:30:34] [Info] 90.0%
[2019-05-18.18:30:34] [Info] 100.0%
[2019-05-18.18:30:34] [Info] Done
[2019-05-18.18:30:34] [Info] Loading addresses from 'taddress.txt'
[2019-05-18.18:30:34] [Info] 1 addresses loaded (0.0MB)
[2019-05-18.18:30:34] [Info] Address: 1KWj99Jwd9LGGC2Y1c9c4cmvWvYTQrLFVc
Private key: 0000000000000000000000000000000000000000000000000000000220082201
Compressed: no
Public key:
6A245BF6DC698504C89A20CFDED60853152B695336C28063B61C65CBD269E6B4
E022CF42C2BD4A708B3F5126F16A24AD8B33BA48D0423B6EFD5E6348100D8A82
C:\bitcrack>cubitcrack -u --keyspace 1e -i taddress.txt
[2019-05-18.18:16:36] [Info] Compression: uncompressed
[2019-05-18.18:16:36] [Info] Starting at: 000000000000000000000000000000000000000000000000000000000000001E
[2019-05-18.18:16:36] [Info] Ending at: FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364140
[2019-05-18.18:16:36] [Info] Counting by: 0000000000000000000000000000000000000000000000000000000000000001
[2019-05-18.18:16:36] [Info] Initializing
[2019-05-18.18:16:36] [Info] Generating 262,144 starting points (10.0MB)
[2019-05-18.18:16:36] [Info] 10.0%
[2019-05-18.18:16:36] [Info] 20.0%
[2019-05-18.18:16:36] [Info] 30.0%
[2019-05-18.18:16:36] [Info] 40.0%
[2019-05-18.18:16:36] [Info] 50.0%
[2019-05-18.18:16:37] [Info] 60.0%
[2019-05-18.18:16:37] [Info] 70.0%
[2019-05-18.18:16:37] [Info] 80.0%
[2019-05-18.18:16:37] [Info] 90.0%
[2019-05-18.18:16:37] [Info] 100.0%
[2019-05-18.18:16:37] [Info] Done
[2019-05-18.18:16:37] [Info] Loading addresses from 'taddress.txt'
[2019-05-18.18:16:37] [Info] 1 addresses loaded (0.0MB)
[2019-05-18.18:16:37] [Info] Address: 1KWj99Jwd9LGGC2Y1c9c4cmvWvYTQrLFVc
Private key: 000000000000000000000000000000000000000000000000000000000000001F
Compressed: no
Public key:
6A245BF6DC698504C89A20CFDED60853152B695336C28063B61C65CBD269E6B4
E022CF42C2BD4A708B3F5126F16A24AD8B33BA48D0423B6EFD5E6348100D8A82
Re: Half of any bitcoin (crypto) public key - (public key half) in Python
2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 = 1 (mod n)
Multiplying a point by this number will "half" the point.
How do I multiply a point by the number?
private key 3
x = f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y = 388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
half of the above public key is 1 given below
x = 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
y = 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
How to half to multiply f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 to get to 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798?
Multiplying a point by this number will "half" the point.
How do I multiply a point by the number?
private key 3
x = f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y = 388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
half of the above public key is 1 given below
x = 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
y = 483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
How to half to multiply f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 to get to 79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798?
Re: Half of any bitcoin (crypto) public key - (public key half) in Python
The forum I got the information on is https://bitcointalk.org/index.php?topic=4455904.0
Half of any bitcoin (crypto) public key - (public key half) in Python
⭐ Merited by ETFbitcoin (1)
I know from reading another forum that a public key in bitcoin is a point P. To do P2, you multiply 12 to P where 12 is the multiplicative inverse of 2 in Zn. It is an integer that can be found using the extended Euclidean algorithm and is 57896044618658097711785492504343953926418782139537452191302581570759080747169 in the case of secp256k1.
2^-1 (mod n) = 57896044618658097711785492504343953926418782139537452191302581570759080747169
2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 = 1 (mod n)
Multiplying a point by this number will "half" the point.
My question is how would I code this in python?
2^-1 (mod n) = 57896044618658097711785492504343953926418782139537452191302581570759080747169
2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 = 1 (mod n)
Multiplying a point by this number will "half" the point.
My question is how would I code this in python?