[Moved] Wordlist creation question

[im-a-skier]

ok im fairly new at linux, although i did take a few a+, cicso, microsoft server 2003 classes and they touched on linux. i would like to know if there is any way that i could make my own wordlist file with all characters (ex. a,A,1,!) in length from 3 to 16.

ive googled for about 2 hours now and no result. any help would be awesome. step by step would be phenomenal!

Thanks Dylan

[lupin]

ok im fairly new at linux, although i did take a few a+, cicso, microsoft server 2003 classes and they touched on linux. i would like to know if there is any way that i could make my own wordlist file with all characters (ex. a,A,1,!) in length from 3 to 16.

ive googled for about 2 hours now and no result. any help would be awesome. step by step would be phenomenal!

Thanks Dylan

I have moved this to a new thread for you. Dont hijack threads like this again please.

Anyway do you mean you want all possible combinations of all possible characters for all lengths from 3 to 16? That is going to be gigantic. Your hard disk won't be big enough to hold it. Crunch or a custom script will generate it for you though, should you have the desire to try.

Theres examples of using crunch all over the forum and in the wiki.

[im-a-skier]

thanks a bunch!!
sorry for the wrong thread :P

[thorin]

Considering a wordlist that contains only length 8 "words" using 62 chars (a-zA-Z0-9) is 218.3TB you aren't going to be storing everything length 3-16 using 62 chars anytime soon (let alone using a 95 char character set).

[streaker69]

Considering a wordlist that contains only length 8 "words" using 62 chars (a-zA-Z0-9) is 218.3TB you aren't going to be storing everything length 3-16 using 62 chars anytime soon (let alone using a 95 char character set).

4 of these should do it: The biggest flash drive in the world • The Register

[thorin]

They are definitely interesting.

But even 400TB isn't enough to hold 95^16Bytes (well actually slightly less since we're ignoring 95^3Bytes).

[wif1bust3r]

Considering a wordlist that contains only length 8 "words" using 62 chars (a-zA-Z0-9) is 218.3TB you aren't going to be storing everything length 3-16 using 62 chars anytime soon (let alone using a 95 char character set).

218.3 T words you mean... and multiplying by 9 (for 8-character word + \n), google gives me, 1 787.21252 terabytes.

i would like to know if there is any way that i could make my own wordlist file with all characters (ex. a,A,1,!) in length from 3 to 16.

Seriously, it's not really that impossible. If you are immortal, you can start creating your wordlist now and wait for 600 trillion years. After that, it's probably half-completed already. Then all you have to do is wait for another 600 trillion years and it will all be over. But do remember to split your wordlist into smaller chunks so you could store it into file systems (e.g., FAT32,etc.). Also, some cracking programs do not support more than 2gb file size.

On the other hand, you can start reading about various researches made by people in order to crack human-generated passwords easier without brute forcing, also known as smart dictionary attacks. Letter frequency analysis and Markov chains to name a few.

[MikeCa]

Technically your math is not correct because you are not accounting for lengths outside 16 and 3.

It is not important for this thread at all, but this thread is dead.

We care about all combinations 4 to 16:
95^4 + 95^5 + ... + 95^16 = 4.45e31

You were only accounting for combinations that are 16 characters, and you wanted to subtract the combinations that are 3 characters. Your mistake is that you didn't count combinations from 4 to 15, and didn't want to subtract combinations from 1 to 2 (which you wouldn't have to do if you only counted from 4 to 16).

EDIT: I am responding to Thorin's last post, wif1bust3r is correct about counting bytes correctly.

[thorin]

218.3 T words you mean... and multiplying by 9 (for 8-character word + \n), google gives me, 1 787.21252 terabytes.

Yes you're right.

If I was doing length 2 words with a binary character set, the list would be 8 bytes (assuming you ignore CR/LF).

2^2=4 Words (each word is 2 bytes) [Well actually since I chose a binary character set you could argue it's only 2 bits but for illustration purposes stick with me].
00
11
10
01
8 Bytes for the list.

Sadly I didn't fully think through my original math. Either way given the scope of the data we're talking about it only gets worse.