1. ## Re: How to generate 8 Digit Possible passwords?

The y+1 is the length of the string plus the LF in Linux. In Linux you use a Line Feed LF character to end a line. In Windows you use a Carriage Return and Line Feed CR-LF. This all dates back to typewriters. Anyhow the formula is number of permutations times string length +1. If you are calculating the filesize for windows the formula is:
(x^y)*(y+2)

2. ## Re: How to generate 8 Digit Possible passwords?

So shouldn't it be X^Y(All chars in all places) * 2 (for each permutation add the LF char.) then -1 because the last permutation wouldn't need the LF. It seems you're adding more than necessary with y + 1 (||2) .
Shouldn't it be ((36^8)*2)-1
(a-z+0-9) ^ (length of strings) + number of LF's needed (total permutations(36^8)) - 1 for the last one? This should give the size of a linux dictionary file of all 8 character A-Z 0-9 passwords.
Looking like this: (36 possible chars) * (36 possible chars) * (36 possible chars) * (36 possible chars) * (36 possible chars) * (36 possible chars) * (36 possible chars) * (36 possible chars) = 36^8
Plus the LF for one: (For each permutation add one LF char) = (36^8) *2
Minus the extra char at the end ((36^8)*2)-1
Still not getting your formula. Sorry....

Edit: Did some research and got it. But a really clear explanation would still be cool.

3. ## Re: How to generate 8 Digit Possible passwords?

Thats because the formula is wrong.

It should be:

x = possible characters
y = length of string

Unix
(x^y)+((x^y)/y)

Dos
(x^y)+(((x^y)/y)*2)

4. ## Re: How to generate 8 Digit Possible passwords?

Sounds cool. But I found his formula online, and this formula is new.... Would you mind explaining please?

5. ## Re: How to generate 8 Digit Possible passwords?

Sounds cool. But I found his formula online, and this formula is new.... Would you mind explaining please?
Ok.

x^y calculates how many characters will be in the file. Each character should equal 1 byte.

(x^y)/y calculates the number of lines in the file. The concept is to calculate all possible characters and then divide by the length of each string to get your string count. This is simply added on to your character bytes. If it's Dos you just multiply the string count by 2 since it uses 2 bytes for CRLF.

If I'm missing something let me know but I believe this is how you want to calculate it. Maybe the other formula accounts for something I'm missing could you please post the link for it?

6. ## Re: How to generate 8 Digit Possible passwords?

http://projects.z-dev.org/hashing/

That formula is a summation for a string lengths 0-i plus hash. But if you edit it for his case, 8 char only without hash, you get this:
36^8 = charset ^ length = x^y = amount of possible permutations. This number ONLY describes the possible AMOUNT of permutations, and nothing else. If it was a 10 character charset with a 3 length, there would be 1000 permutations, but each permutation would be 3 bytes long, which moves us on to the next part of the formula
* (y+format char(1||2))
This is: for each permutation possible, the string will be y+1||2 chars long. Therefore, the correct formula, his, would be:
(Charset^Length) * (Length+FormatChars) = (x^y)*(y+1)
However, this still gives us the extra format char at the end, since the last line has no need of a newline char, so my addendum -1 will take care of that. Leaving the final formula for calculating the size of all the passwords of a given length and charset:
((x^y)*(y+fc))-fc in this case ((x^y)*(y+1))-1 = ((36^8)*(9))-1= 25389989167103 Bytes = ~23 Terabytes
To edit for multiple lengths, just use a summation, or, using the same formula, add that sepcific length to the final total.
Any questions? I'm curious if this checks out like I said...

7. ## Re: How to generate 8 Digit Possible passwords?

I dug up my formula that I made up a while ago and I was wrong, my apologies.
This is what I had. Now using bofh's. Would be trivial to subtract one or two for the last LF.
((x^y)*y)+(x^y)

I get the same results as bofh's posted formula.

25,389,989,167,104 or 25.39TB

8. ## Re: How to generate 8 Digit Possible passwords?

Yours is equivalent to his with the only difference being where you add the final x^y. You require four complex operations to his two complex, one simple, so it's a good thing you now use his. By the way, since you divide by 1024 to get the next level byte count, it's 23 terabytes. I know I'm being an annoying nitpicker here, but since this guy might actually be considering doing this, then those two terabytes+ make a large difference. Thanks for your patience, and help.

9. ## Re: How to generate 8 Digit Possible passwords?

Haha, yes. Your result will be in a more accurate KB once you divide by 1024. But an OS has to go somewhere and you will lose space due to formatting.

The time it would take to generate 23TB of data is going to take years and nearly \$800 just for your hard drives. Not to mention all your files will have to be 2GB in length and would take an even more tremendous amount of time to crack a wireless AP. If you want to get more speed you would require even more space for hashes and time to generate. It's simply unefficient. If anyone's planning on doing this I hope you have a lot of time to wait and money to burn.

10. ## Re: How to generate 8 Digit Possible passwords?

Hello everyone,

Do you think we need 25 Terabytes ?

Not in my ...

Returning to the initial state:

To generate 8 Digit Possible passwords = 0k (it is clear)

I think there are ways to dramatically speed up the processus by using some tables arc-en !

But there are conditions (Quality, Time, Speed​​, which depends on the RAM, processor, memory cache...)

What really interests me is Quality !

what did you say?

Big thanks to bofh28, Shadowmaster, hhmatt

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