This issue is now resolved thank you all.
And who won the price?
Topic: Prize award to whoever can decipher the most characters in the attached image (Read 194 times)
July 13, 2022, 06:10:56 PM
And who won the price?
July 13, 2022, 06:04:39 PM
This issue is now resolved thank you all.
March 09, 2022, 03:56:29 PM
Hi I desperately need help deciphering characters of a lost private key, the only remnant is a blurred out image from the past.
I've got some variants, but far too many for a person (110,592,000); maybe your tool can get use of subsets?
Some notations:
- [x,y,z] means a set: it can be one of those symbols on that one position
- * means it can be anything on that position and it's equivalent with [0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f]; it's your #, just I've seen late it should be # and not *.
- on the image (sorry, no apple, no Helvetica) black means I'm pretty much sure about the value, red means that I have an set of possible values for that position (including *)
Code:
2[6,8]b6[0,9]dfb[3,6,8,9][a,c,e]611*23435d[3,6,8,9]f[c,e]2e2[1,2]f[4,f][c,e]2[a,c,e]145[0,3,6,8,9][0,3,6,8,9]7[a,c,e]95e4[0,3,6,8,9][c,e]0
Good luck.
March 09, 2022, 08:44:41 AM
here's a code which will bruteforce this stuff in half an hour
https://github.com/alexeyneu/secp256k1-cxx
but this paranoid bastard doesn't wanna run it.
upfront paid so i have np
Code:
#include "src/secp256k1-cxx.hpp"
#include "src/sha/sha2.hpp"
#include
#include
#include
#include "libbase58.h"
#include
#include
#include
#include "keccak.h"
#include
#include
#include
#include
#include
const std::string bin2hex(const unsigned char *p, size_t length) {
std::stringstream f;
f << std::hex << std::setfill('0');
for (int i = 0; i < length; i++) f << std::setw(2) << (int)p[i];
return f.str();
}
size_t hex2bin(unsigned char *p , const char *hexstr, const size_t length) {
size_t wcount = 0;
while ( wcount++ < length && *hexstr && *(hexstr + 1)) { //last condition cause np if check fails on middle one.thats coz of short-circuit evaluation
sscanf(hexstr, "%2hhx",p++); //7x slower than tables but doesnt metter
hexstr = hexstr+2;
}
return --wcount; // error check here is a waste
}
void brough(unsigned long long f);
int main()
{
std::vector a{2, 4, 6, 8, 10, 12, 16};
std::cout << std::endl;
std::thread h[7];
for(auto f : a)
{
h[f == 16 ? 6 : f/2 - 1] = std::thread(&brough, f);
}
h[0].join();
h[1].join();
h[2].join();
h[3].join();
h[4].join();
h[5].join();
h[6].join();
return 0;
}
void brough(unsigned long long f)
{
unsigned long long t = f == 16 ? 12 : f - 2;
std::string key(32, ' ');
std::vector keyb;
std::pair , unsigned short> mulligan{{0x6, 0x0}, 0};
std::pair , unsigned short> derby{{0x9, 0x6, 0x0, 0x8}, 0};
std::pair , unsigned short> dendra{{0xb, 0xa, 0xc/*remove*/, 0xe}, 0};
std::pair , unsigned short> remm{{0x2, 0x7}, 0};
std::pair , unsigned short> alfa{{0x2, 0x1}, 0};
std::pair , unsigned short> tetra{{0x7, 0x5}, 0};
std::pair , unsigned short> epsilon{{0x0, 0x9, 0x6}, 0};
std::pair , unsigned short> omega{{0xb, 0x2, 0x8, 0xe}, 0};
std::pair , unsigned short> beta{{0xa, 0xe, 0xc}, 0};
std::pair , unsigned short> zeta{{0x9, 0x0, 0x8}, 0};
std::pair , unsigned short> magna{{0x6, 0x3, 0x8, 0x0}, 0};
std::pair , unsigned short> medley{{0xe, 0xc, 0xa}, 0};
hex2bin((unsigned char *)&key[0], "eeeeeeeeeeeeeeeeee28b69dfb9d611d23435d6fc2e2277c2a145087d95e46e0", 32);
std::string z = " m dd {} r teo bz mm ";
for(int k = 0; k < 64; k++)
{
keyb.insert(keyb.end(), k % 2 == 0 ? key[k / 2] >> 4 : key[k / 2] & 0x0f);
}
do
{
keyb[31] = *((unsigned char *)&t);
keyb[24] = mulligan.first[mulligan.second]; // mulligan
keyb[26] = derby.first[derby.second]; // derby
keyb[27] = dendra.first[dendra.second]; // dendra
keyb[43] = remm.first[remm.second]; // remm
keyb[44] = alfa.first[alfa.second]; // alfa
keyb[52] = tetra.first[tetra.second]; // tetra
keyb[53] = epsilon.first[epsilon.second]; //epsilon
keyb[54] = omega.first[omega.second]; // omega
keyb[56] = beta.first[beta.second]; // beta
keyb[57] = zeta.first[zeta.second]; // zeta
keyb[61] = magna.first[zeta.second]; // magna
keyb[62] = medley.first[zeta.second]; // magna
for(int b = 0; b < 32; b++)
{
key[b] = (keyb[b * 2] << 4) + keyb[b * 2 + 1];
}
medley.second++;
if(medley.second > medley.first.size() - 1) { magna.second++; medley.second = 0; }
if(magna.second > magna.first.size() - 1) { zeta.second++; magna.second = 0; }
if(zeta.second > zeta.first.size() - 1) { beta.second++; zeta.second = 0; }
if(beta.second > beta.first.size() - 1) { omega.second++; beta.second = 0; }
if(omega.second > omega.first.size() - 1) { epsilon.second++; omega.second = 0; }
if(epsilon.second > epsilon.first.size() - 1) { tetra.second++; epsilon.second = 0; }
if(tetra.second > tetra.first.size() - 1) { alfa.second++; tetra.second = 0; }
if(alfa.second > alfa.first.size() - 1) { remm.second++; alfa.second = 0; }
if(remm.second > remm.first.size() - 1) { dendra.second++; remm.second = 0; }
if(dendra.second > dendra.first.size() - 1) { derby.second++; dendra.second = 0; }
if(derby.second > derby.first.size() - 1) { mulligan.second++; derby.second = 0; }
if(mulligan.second > mulligan.first.size() - 1) { t++; mulligan.second = 0; }
Secp256K1 p { key };
if(f == 4) std::cout << "Private key: " << bin2hex((const unsigned char*)key.c_str(), 32) << '\r';
unsigned char ethashtag[32] = {};
unsigned char etaddr[20] = {};
Keccak keccak256(Keccak::Keccak256);
hex2bin(ethashtag, keccak256((char *)p.pubkey_ + 1, 64).c_str(), 32);
memcpy(etaddr, ethashtag + 12, 20);
std::string etaddrstring = "0x" + bin2hex(etaddr, 20);
if (etaddrstring.contains("address")) { std::cout << std::endl << "yeah" << std::endl << "Private key: " << bin2hex((const unsigned char*)key.c_str(), 32) << std::endl; break; }
}while(t < f);
}
#include "src/sha/sha2.hpp"
#include
#include
#include
#include "libbase58.h"
#include
#include
#include
#include "keccak.h"
#include
#include
#include
#include
#include
const std::string bin2hex(const unsigned char *p, size_t length) {
std::stringstream f;
f << std::hex << std::setfill('0');
for (int i = 0; i < length; i++) f << std::setw(2) << (int)p[i];
return f.str();
}
size_t hex2bin(unsigned char *p , const char *hexstr, const size_t length) {
size_t wcount = 0;
while ( wcount++ < length && *hexstr && *(hexstr + 1)) { //last condition cause np if check fails on middle one.thats coz of short-circuit evaluation
sscanf(hexstr, "%2hhx",p++); //7x slower than tables but doesnt metter
hexstr = hexstr+2;
}
return --wcount; // error check here is a waste
}
void brough(unsigned long long f);
int main()
{
std::vector
std::cout << std::endl;
std::thread h[7];
for(auto f : a)
{
h[f == 16 ? 6 : f/2 - 1] = std::thread(&brough, f);
}
h[0].join();
h[1].join();
h[2].join();
h[3].join();
h[4].join();
h[5].join();
h[6].join();
return 0;
}
void brough(unsigned long long f)
{
unsigned long long t = f == 16 ? 12 : f - 2;
std::string key(32, ' ');
std::vector
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
std::pair
hex2bin((unsigned char *)&key[0], "eeeeeeeeeeeeeeeeee28b69dfb9d611d23435d6fc2e2277c2a145087d95e46e0", 32);
std::string z = " m dd {} r teo bz mm ";
for(int k = 0; k < 64; k++)
{
keyb.insert(keyb.end(), k % 2 == 0 ? key[k / 2] >> 4 : key[k / 2] & 0x0f);
}
do
{
keyb[31] = *((unsigned char *)&t);
keyb[24] = mulligan.first[mulligan.second]; // mulligan
keyb[26] = derby.first[derby.second]; // derby
keyb[27] = dendra.first[dendra.second]; // dendra
keyb[43] = remm.first[remm.second]; // remm
keyb[44] = alfa.first[alfa.second]; // alfa
keyb[52] = tetra.first[tetra.second]; // tetra
keyb[53] = epsilon.first[epsilon.second]; //epsilon
keyb[54] = omega.first[omega.second]; // omega
keyb[56] = beta.first[beta.second]; // beta
keyb[57] = zeta.first[zeta.second]; // zeta
keyb[61] = magna.first[zeta.second]; // magna
keyb[62] = medley.first[zeta.second]; // magna
for(int b = 0; b < 32; b++)
{
key[b] = (keyb[b * 2] << 4) + keyb[b * 2 + 1];
}
medley.second++;
if(medley.second > medley.first.size() - 1) { magna.second++; medley.second = 0; }
if(magna.second > magna.first.size() - 1) { zeta.second++; magna.second = 0; }
if(zeta.second > zeta.first.size() - 1) { beta.second++; zeta.second = 0; }
if(beta.second > beta.first.size() - 1) { omega.second++; beta.second = 0; }
if(omega.second > omega.first.size() - 1) { epsilon.second++; omega.second = 0; }
if(epsilon.second > epsilon.first.size() - 1) { tetra.second++; epsilon.second = 0; }
if(tetra.second > tetra.first.size() - 1) { alfa.second++; tetra.second = 0; }
if(alfa.second > alfa.first.size() - 1) { remm.second++; alfa.second = 0; }
if(remm.second > remm.first.size() - 1) { dendra.second++; remm.second = 0; }
if(dendra.second > dendra.first.size() - 1) { derby.second++; dendra.second = 0; }
if(derby.second > derby.first.size() - 1) { mulligan.second++; derby.second = 0; }
if(mulligan.second > mulligan.first.size() - 1) { t++; mulligan.second = 0; }
Secp256K1 p { key };
if(f == 4) std::cout << "Private key: " << bin2hex((const unsigned char*)key.c_str(), 32) << '\r';
unsigned char ethashtag[32] = {};
unsigned char etaddr[20] = {};
Keccak keccak256(Keccak::Keccak256);
hex2bin(ethashtag, keccak256((char *)p.pubkey_ + 1, 64).c_str(), 32);
memcpy(etaddr, ethashtag + 12, 20);
std::string etaddrstring = "0x" + bin2hex(etaddr, 20);
if (etaddrstring.contains("address")) { std::cout << std::endl << "yeah" << std::endl << "Private key: " << bin2hex((const unsigned char*)key.c_str(), 32) << std::endl; break; }
}while(t < f);
}
https://github.com/alexeyneu/secp256k1-cxx
but this paranoid bastard doesn't wanna run it.
upfront paid so i have np
March 04, 2022, 09:57:48 AM
it's np to do. they all are really visible . i can supply bruteforce app source (with quicknode api with my key) for that so you can work it out . it'll find pubkey with balance from this.
$50 upfront (and it's final cost). alexneud tg
$50 upfront (and it's final cost). alexneud tg
March 03, 2022, 02:18:17 PM
The answers provided on the other post were two characters short and I was told Id get better responses here as more people are active on the service board. The more answers I get for the image the better, to assist my efforts for private key recovery.
If you get a chance to look and compare I would be forever grateful.
If you get a chance to look and compare I would be forever grateful.
March 03, 2022, 07:24:49 AM
No need to duplicate topics. You have already posted your question:
https://bitcointalksearch.org/topic/m.59369620
And you received some answers- why to ask again?
https://bitcointalksearch.org/topic/m.59374598
https://bitcointalksearch.org/topic/m.59369620
And you received some answers- why to ask again?
https://bitcointalksearch.org/topic/m.59374598
March 03, 2022, 07:20:39 AM
Hi I desperately need help deciphering characters of a lost private key, the only remnant is a blurred out image from the past.
Prize award to whoever can decipher the most characters in the attached image accurately by mapping the font atop the image.
Prize award to whoever can decipher the most characters in the attached image accurately by mapping the font atop the image.
Hi. Here is my best guess at each number. Not all are 100% sure.
28h69dfb6o611#23435d6fc2P#277C2a145387c95e46c0
March 02, 2022, 07:07:18 PM
*SNIP*
It seems that only a few characters are completely unreadable. I am sure that it is enough for you to learn how brute force works, the way is password finding and I am sure that you will be able to find the key. For sure, the first 18 characters will be very helpful in this case when recovering the key using brute force.
https://en.wikipedia.org/wiki/Brute-force_attack
March 02, 2022, 06:38:14 PM
Hi I desperately need help deciphering characters of a lost private key, the only remnant is a blurred out image from the past.
There are 46 characters visible. The first 18 characters I have along with the wallet ID but of course for security reasons I wont share.
If anyone can help decipher the blurred out characters this would be great, if a character is not known please leave a # for me to brute force guess the missing characters using software which I have.
The font used is Helvetica (apple) - this may help you work out what the missing alphanumeric characters are likely to be in the image.
I just need help deciphering the remaining 46 characters from the above image, with this information and the gaps (indicated by #) I should be able to recover the wallet.
Its okay if not all characters are revealed but help guessing as many as possible would be appreciated as my vision is not the best.
Prize award to whoever can decipher the most characters in the attached image accurately by mapping the font atop the image.
There are 46 characters visible. The first 18 characters I have along with the wallet ID but of course for security reasons I wont share.
If anyone can help decipher the blurred out characters this would be great, if a character is not known please leave a # for me to brute force guess the missing characters using software which I have.
The font used is Helvetica (apple) - this may help you work out what the missing alphanumeric characters are likely to be in the image.
I just need help deciphering the remaining 46 characters from the above image, with this information and the gaps (indicated by #) I should be able to recover the wallet.
Its okay if not all characters are revealed but help guessing as many as possible would be appreciated as my vision is not the best.
Prize award to whoever can decipher the most characters in the attached image accurately by mapping the font atop the image.
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