Cracking Passwords: Calculating Outcomes with 3 Characters

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Discussion Overview

The discussion revolves around calculating the number of possible outcomes for passwords of varying lengths and character types, specifically focusing on 3-character passwords composed of digits, alphabets, or a combination of both. Participants explore mathematical approaches to determine these outcomes and consider the implications of different character sets.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant inquires about the total number of outcomes for a 3-digit password, suggesting it is 999 but seeking clarification on the correct calculation.
  • Another participant states that for 3 digits, the total outcomes are 10^3 = 1000, explaining the counting method by considering the range from 000 to 999.
  • A participant explains that for a password with 3 characters that can include both alphabets and numbers, the total outcomes would be (26 + 10)^3.
  • Further elaboration is provided on how to calculate outcomes for different character positions, using examples from number positions and capital letters.
  • One participant expresses excitement about the mathematical principles involved and confirms their understanding of the calculations for a 12-character password using both alphabets and numbers.
  • A later reply challenges the correctness of the previous calculations, suggesting that a different mathematical approach, such as two's-complement, may be necessary.
  • Another participant provides clarification on the outcomes for a 12-character password, indicating that it would be 36^12 for lowercase letters or case-insensitive letters, and 62^12 if both uppercase and lowercase letters are included.

Areas of Agreement / Disagreement

There is some agreement on the methods for calculating outcomes for passwords of specific lengths and character types. However, there is disagreement regarding the correctness of certain calculations, particularly concerning the application of two's-complement and the treatment of character cases in the password length of 12 characters.

Contextual Notes

Participants express varying assumptions about character sets (e.g., lowercase vs. case-insensitive) and the implications of restrictions on order, which may affect the calculations. Some mathematical steps remain unresolved, particularly in the context of the two's-complement claim.

Raza
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Hi,
I have recently cracked a password for a zip file using brute force which got me thinking. Let's say that the password is only 3 digits, how many outcome could it possibly have?
I know it's 999 but is there a equation that I can use to figure this out?
or let's say that it contains only 3 alphabets, how many outcome could it possibly have, or let's say that the password's length is three characters and it could have either alphabets or numbers, what are the possible outcome?

Please give me brief explanation and some equation.

Thanks
 
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in general, if you have a items with b places to but them, the number of outcomes is a^b.

for 3 digits there are actually 10^3= 1000 different choices (which can easily be seen by considering it is 000=choice 1, 001 = choice 2,...999= choice 1000).

if a password is 3 characters and can have alphabets or numbers, it is (26+10)^3.

these questions become harder if you start putting restrictions on order.
 
For each position, you know how many possible options there are, so you just multiply out to get the total number of possible results.

Like, if you have a single number position, and you are working in base 10, then there are 10 possible results, 0-9.

If you have a 2-digit number, then each position (the 1's postion and the 10's position) has 10 possibilities, so you get a total of 10 * 10 = 100 possible results (0-99).

If you have a capital character position, A-Z, that position has 26 possible outcomes. Two capital letter positions have 26 * 26 possible outcomes, etc.

So as a final example, in California, the standard license plates are formatted like this:

<1 number> <3 chars> <3 numbers>

So the total number of possible plates in that format would be 10 * 26^3 * 10^3.

Makes sense?


EDIT -- matticus beat me to it!
 
Thanks, I got it.
It's simply amazing how math works.
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?
 
Raza said:
Thanks, I got it.
It's simply amazing how math works.
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?

So now you're asking for help hacking passwords... :rolleyes:

No, sorry, your answer is all wrong. It should be the two's-complement of that number to be correct. :bugeye:
 
Raza said:
Right now, I am cracking another zip password, which is 12 characters, alphabets and numbers, so it would be (10+26)12,right?

36^12 if all letters are known to be lowercase (or case-insensitive); 62^12 if letters can be of any case.
 

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