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More Password Cracking Decrypted

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More Password Cracking Decrypted

Post by HELLU on Sun Aug 26, 2012 4:16 am

Welcome to another edition of Password Cracking Decrypted. In this manual we will learn, you guessed it, how to crack passwords. In this edition we have explanations to how to break more kinds of passwords.

Although this manual is quite easy to understand, I would definitely like to make one suggestion. To truly enjoy reading this manual, you need to know C relatively well. However, even if you have no idea what C is, I assure you that this manual will definitely be of use to you.

Cracking the Netzero (Free ISP) Dial Up Password

Today, the number of Internet Service Providers (both free and the not so free ones) has really reached a very high figure. All of them aim at providing better services and making the process of connecting to the Internet easier for the user. One common practice amongst both Internet Service Providers and popular browsers like Internet Explorer, have this option called ‘Save Password’, which makes life easier for the user, as it allows the user to not type in the password each time he has to connect to the Internet.

Although, like all other software, as soon as the developer tries to add a user friendly feature or make the software easier to use or more efficient, he has to make at least some compromise in the security or safety field. One popular example would be Outlook Express, ever since the Preview Pane has been introduced within the email client, Outlook Express users have become prone to Email-Borne Viruses.

Anyway, getting back to the subject of this tutorial, even including the ‘Save Password’ feature has made the User’s Password unsafe. Now, what happens is that, when you check on this option or enable it, then the concerned software (Browser or Internet Service Provider Software) takes it passes it through an algorithm to encrypt it. Once, the Password is encrypted, it is then stored in the Windows Registry or in some .ini or .dat or a similar file. Now, this system sounds quite safe, however, if you look deeper, then you find that it is trouble waiting to happen.

The very fact that the encrypted password has to be stored somewhere, makes this feature vulnerable. Also, almost all software providing this feature does not use a strong algorithm. This makes the work of a hacker really easy. Some software even stores the password as plaintext in the registry!!! So, basically the weakest chain in this feature is that most software developers are weary of the fact that the encrypted password can be easily decrypted, once we study the software inside out. So, what I mean to say is that using this feature although surely makes life easy, for those of you who cannot remember passwords, but it does leave your Internet Account vulnerable. However, if you are one of those people who needs to write down your password on a piece of paper and stick it to the front of your monitor, then this feature is definitely for you.

So how do I crack the Netzero Dial Up Password?

Anyway, Netzero is a free ISP, which asks only for a advertising bar in return for Internet Access. It too provides this ‘Save Password’ feature, however, it too like most services, uses an extremely weak algorithm to encrypt the password. The following process of decryption works on Netzero version 3.0 and earlier and requires Win 9x, NT or Win 2K to be running.

For this exploit, you need to have local access to the machine, which has the Netzero software installed.

This vulnerability cannot be exploited unless and until you get the required file, for that you either have to have local access or need to devise a method of getting the file, which contains the password.

The Netzero Username and Password are stored in an ASCII file named, id.dat, which is located in the Netzero directory. If the user has enabled the ‘Save Password’ option, then the Username and Password are also stored in the jnetz.prop file. The passwords stored in both these files are encrypted using a very simply easy to crack algorithm. Although the algorithms used to get the encrypted information (to be stored in the two files), are not same, however they are derived from the same main algorithm. Both the algorithms differ very slightly. In this manual we will learn as to how this weak algorithm can be exploited.

The Netzero Password is encrypted using a substitution cipher system. The cipher system used is a typical example of a 1 to 1 mapping between characters where each single plaintext character is replaced by a single encrypted character.

Are you lost? Well, to understand better read on.

Say, the Netzero application is running, and the user clicks on the ‘Save Password’ option and types his password in the required field. Now, then what happens is that, the Netzero Application loads the encrypting file, which contains the plaintext to cipher-text database into memory. Now, for example your password is xyz and it is stored in location ‘m’ of the memory and the corresponding encrypted password abc is stored in the location ‘n’ of the memory, then the password xyz actually is stored as abc.

Well it is quite simple, right? Well, almost. The part of the encryption algorithm used by Netzero which is difficult to understand, is that two encrypted characters replace each character of the plaintext password. These two encrypted characters replacing a single plaintext character, are however not stored together.

When substituting character x stored in i of a password ‘n’ characters long, the first encrypted character would be stored in ‘i’ and the next in ‘n+i.’

The two encrypted characters are derived from the following table:

| 1 a M Q f 7 g T 9 4 L W e 6 y C
g | ` a b c d e f g h i j k l m n o
T | p q r s t u v w x y z { | } ~
f | @ A B C D E F G H I J K L M N O
7 | P Q R S T U V W X Y Z [ \ ] ^ _
Q | 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
M | SP ! " # $ % & ' ( ) * + , - . /

NOTE: SP represents a single space and the above chart represents ASCII characters.

To encrypt a string of length ‘n’, we need to find each character in the above table and place the column header into i and place the row header into n+i.

For example:
E(a) = ag
E(aa) = aagg
E(aqAQ1!) = aaaaaagTf7QM
E(`abcdefghijklmno) = 1aMQf7gT94LWe6yCgggggggggggggggg

On the other hand, while decrypting the password of length 2n, then I will be become the element in the element in the above table where the column is headed by i and the row headed by n+i intersect.

For example:
D(af) = A
D(aaff) = AA
D(aaMMQQfgfgfg) = AaBbCc

Decrypting the password manually would be quite fun, but would definitely be a very time consuming process. Anyhow, I do suggest you try to decrypt the Netzero Password manually atleast once. For those of you, who do not enjoy decrypting passwords manually, I also have a C program, which will do it for you.

The following C program demonstrates how the Netzero Password is decrypted. Simply compile and execute in the directory in which the jnetz.prop exists.


#include <stdio.h>

#include <string.h>

#define UID_SIZE 64

#define PASS_CIPHER_SIZE 128

#define PASS_PLAIN_SIZE 64

#define BUF_SIZE 256

const char decTable[6][16] = {






{' ','!','"','#','$','%','&','\'','(',')','*','+',',','-','.','/'}


int nz_decrypt(char cCipherPass[PASS_CIPHER_SIZE],

char cPlainPass[PASS_PLAIN_SIZE])


int passLen, i, idx1, idx2;

passLen = strlen(cCipherPass)/2;

if (passLen > PASS_PLAIN_SIZE)


printf("Error: Plain text array too small\n");

return 1;


for (i = 0; i < passLen; i++)




case '1':

idx2 = 0; break;

case 'a':

idx2 = 1; break;

case 'M':

idx2 = 2; break;

case 'Q':

idx2 = 3; break;

case 'f':

idx2 = 4; break;

case '7':

idx2 = 5; break;

case 'g':

idx2 = 6; break;

case 'T':

idx2 = 7; break;

case '9':

idx2 = 8; break;

case '4':

idx2 = 9; break;

case 'L':

idx2 = 10; break;

case 'W':

idx2 = 11; break;

case 'e':

idx2 = 12; break;

case '6':

idx2 = 13; break;

case 'y':

idx2 = 14; break;

case 'C':

idx2 = 15; break;


printf("Error: Unknown Cipher Text index: %c\n", cCipherPass[i]);

return 1;





case 'g':

idx1 = 0; break;

case 'T':

idx1 = 1; break;

case 'f':

idx1 = 2; break;

case '7':

idx1 = 3; break;

case 'Q':

idx1 = 4; break;

case 'M':

idx1 = 5; break;


printf("Error: Unknown Cipher Text Set: %c\n",


return 1;



cPlainPass[i] = decTable[idx1][idx2];


cPlainPass[i] = 0;

return 0;


int main(void)


FILE *hParams;

char cBuffer[BUF_SIZE], cUID[UID_SIZE];

char cCipherPass[PASS_CIPHER_SIZE], cPlainPass[PASS_PLAIN_SIZE];

int done = 2;

printf("\nNet Zero Password Decryptor\n");

printf("Brian Carrier [bcarrier@atstake.com]\n");

printf("@Stake L0pht Research Labs\n");


if ((hParams = fopen("jnetz.prop","r")) == NULL)


printf("Unable to find jnetz.prop file\n");

return 1;


while ((fgets(cBuffer, BUF_SIZE, hParams) != NULL) && (done > 0))


if (strncmp(cBuffer, "ProfUID=", 8) == 0)



strncpy(cUID, cBuffer + 8, UID_SIZE);

printf("UserID: %s", cUID);


if (strncmp(cBuffer, "ProfPWD=", 8) == 0)



strncpy(cCipherPass, cBuffer + 8, PASS_CIPHER_SIZE);

printf("Encrypted Password: %s", cCipherPass);

if (nz_decrypt(cCipherPass, cPlainPass) != 0)

return 1;


printf("Plain Text Password: %s\n", cPlainPass);




if (done > 0)


printf("Invalid jnetz.prop file\n");

return 1;

} else {

return 0;




More NetZero Fun

Reinaldo Trujilo Adds:

Today we're going to tear apart the NetZero logon password.
Things you must keep in mind.

1. password format:0,n,i-n,1
2. based on the 0 counting system..ie. 0,1,2,3,4,5,etc
3. all passwords begin with a 0 and end with a 1


a=a A=#
b=? B=@
c=> C=!
d=< D=~
e=/ E==
f=. F=-
g=, G=`
h=" H=9
i=: I=8
j=' J=7
k=; K=6
l=| L=5
m=} M=4
n={ N=3
o=\ O=2
p=] P=1
q=[ Q=0
r=+ R=Z
s=_ S=Y
t=) T=X
u=( U=W
v=* V=V
w=& W=U
x=^ X=T
y=% Y=S
z=$ Z=R

SPECIAL NOTE MUST READ: the letter "a" can be equal to "a" if its the first
letter of the password,but anywhere else in the password it will take its N
value's numeric value in the alphabet. I.e if "a" is the 5th letter in the
password, its value would be the LETTER "e".(not e's encrypted value).

Alright..so lets decrypt our first password, lets keep it simple. Our
plaintext password is going to be the word "amore".

1. Count the number of characters contained in the word.
*in this case we have 5 characters
Note: we are counting on a 0 based system so the the word amore would look
something like this:
a m o r e
0 1 2 3 4
Just keep that in mind

Note: each character in the word is assigned a numeric N value.(as
shown above)

3. now write down the password like this:

a m o r e
0 1 2 3 4 <-N values
0 1 <--encrypted password

4.Now the chart says the encrypted character for "a"=a so we place that "a"
under our plaintext "a"
a m o r e
0 1 2 3 4 <--N values
0 a 1 <--encrypted password
Note: every single first character of any password will equal
its value in the chart

5.now to get the second character we have a special equation"i-n=V" where
"i" is the plaintext character, "n" is that characters numerical value, and
V is the new encrypted value.

a m o r e
0 1 2 3 4 <--the N values.(ie.N value for "m" is 1)
0 a 1 <--encrypted password

now we go to the chart and find "m". You'll find that m=} but since our
equation tells us that i-n=V we get our answer like this. m-1=l so our
encrypted value for "m" now equals "|".
so our encrypted password now looks like this

a m o r e
0 1 2 3 4 <--N values
0 a | 1 <--Encrypted password

6.now we do the same thing for letter "o".
o-2=m. "o" now equals "}"

a m o r e
0 1 2 3 4 <-- N values
0 a | } 1 <--encrypted password

7. do the same for "r".
r-3=o. "r" now equals "\"

a m o r e
0 1 2 3 4 <--N values
0 a | } \ 1 <--encrypted password

8. an now our last value "e".
e-4=a. "e" now equals "a"

a m o r e
0 1 2 3 4 <--N values
0 a | } \ a 1 <--Our full encrypted password

Result amore= 0a|}\a1

I hope this has been useful. Enjoy.

Now you can type your encrypted password into the password field of your
dialup program instead of having to use netzero's software.

P.S. one more thing.. the user name also has a special format.
it goes:


example if your user name is BigDaddy you'd put this in the user field of
your dialup program


P.S.S. very important..i almost forgot, say you get the letter "b" as the
5th letter of your password, according to the chart(above) there is no more
spaces to move to. so what you do is follow the chart below like this

a a a a a b a a a a a a a a a a a a a a a <--original plaintext
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 <--N values
a b c d e f g h i j k l m n o p q r s t u <--password format

so "b" is the 5th letter. now its N value should be equalto "f" but since
there is a "b" is the second letter in the password format alphabet its N
value gets shifted back one space (remember we're on the 0 counting system
b=1 not 2). If 5th letter happened to be a "c" its N value would have been
shifted back two spaces. etc etc. Then you would just go on encrypting the
password as normal.

*in short, you take the the original plaintext letter's N value (5 in this
case) and you subract that letter's N (1 in this case) value in the password
format to get the new N value (4 in this case, which would make the new
letter "e").

Hopefully this cleared up what i meant, keep in mind that you only reference
this second chart below when you get a letter in the original plaintext
format who's plaintext N value is greater than its password format N value


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