Originally Posted by
modacompany
he gets 58 k password per secont with GTX ,TESLA must be more than 4 times faster than gtx
First of all, be specific with your figures, "more than 4 times faster" could be anywhere from 5 times faster to 20 times faster.
I'm gonna assume the figure you mean is about 5. So now use a calculator:
5 times 58000 is 290000 keys per second.
and each rack s1070 got 2 x960 core gpu mean is speed will be double.
Mr Calculator gives me 580000 keys per second.
and if we get multiple computers wpa web cracking will be so easy
OK, let's say 10 computers. Now we're at 5 800 000 keys per second.
I'm gonna be extremely generous to you here, and say that the password is exactly 8 characters long, and it only uses the 95 printable characters of ASCII.
Here's the C code for it:
Code:
#include <stdio.h>
#include <gmp.h>
void CalcAmountPasswords(mpz_t total,unsigned min,
unsigned const max,
unsigned const radix)
{
mpz_t temp; mpz_init(temp);
mpz_set_ui(total,0);
for ( ; min <= max; ++min)
{
mpz_ui_pow_ui(temp,radix,min);
mpz_add(total,total,temp);
}
mpz_clear(temp);
}
int main(void)
{
char buf[1024];
mpz_t total; mpz_init(total);
CalcAmountPasswords(total,8,8,95);
mpz_get_str(buf,10,total);
printf("Total amount of keys = %s\n\n",buf);
mpz_cdiv_q_ui(total,total,5800000lu);
mpz_get_str(buf,10,total);
printf("Cracking at 1000000 k/s, so that's %s seconds\n\n",buf);
mpz_cdiv_q_ui(total,total,60ul * 60 * 24 * 365);
mpz_get_str(buf,10,total);
printf("In years, that's %s years\n\n",buf);
mpz_clear(total);
return 0;
}
And here's the output:
Code:
Total amount of keys = 6634204312890625
Cracking at 1000000 k/s, so that's 1143828330 seconds
In years, that's 37 years
There you go: 37 years to crack an 8-character-long password that only uses the 95 printable characters of ASCII. I dunno if I'm gonna be alive in 37 years.
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