[Question] How do I do this?

So I was planing to make a "password brute forcer" that will try all the possible combinations to crack a password but I would like to know how would I make a string change like this? Example 1,2,3,4,5,6,7,8,9,0,-,=,q ect…

I think you would need to make tables with all possible values o.O no clue the rest lol

I'm not sure how you are going to use the generated string but you would do so by using. string.char(n) . I'm not sure which character code will generate which character on computercraft, but here is an example.

for i = 1 , 100 do
	somestring = string.char(i)
	dosomething(somestring)
end

This would go through every character from character code 1 to 100

If you wish to combine 2 characters you can by using a for loop inside a for loop

for i = 1 , 100 do
	for j = 1, 100 do
		somestring = string.char(i,j)
		dosomething(somestring)
	end
end

This would create every combination of the 100 characters combined when used twice in one string.

But remember, if you start generating strings this way the time it takes increases exponentially for every character in the finished string, if you are trying to get 26 characters, every time you add another character to the string the time necessary to calculate it is the old time * 26

XoX, on 02 November 2012 - 01:33 AM said:

I'm not sure how you are going to use the generated string but you would do so by using. string.char(n) . I'm not sure which character code will generate which character on computercraft, but here is an example.

for i = 1 , 100 do
	somestring = string.char(i)
	dosomething(somestring)
end

This would go through every character from character code 1 to 100

If you wish to combine 2 characters you can by using a for loop inside a for loop

for i = 1 , 100 do
	for j = 1, 100 do
		somestring = string.char(i,j)
		dosomething(somestring)
	end
end

This would create every combination of the 100 characters combined when used twice in one string.

But remember, if you start generating strings this way the time it takes increases exponentially for every character in the finished string, if you are trying to get 26 characters, every time you add another character to the string the time necessary to calculate it is the old time * 26

So how many characters are there in total? Or is 100 the max? You helped a lot thanks. :D/>/>

I'm not sure, I would simple figure out by trial and error which characters you want to use, 97 - 120 or so is lowercase letters and symbols, somewhere around 70 is uppercase letters and different symbols.
One sec I have an idea :D/>/>

a = fs.open("chars", "w") ; for i = 1 , 150 do a.writeLine( i .. " : " .. string.char(i) ) end ; a.close()

Now you just open the file chars and you got all the chars you need with their charcode.
I already checked it out, it's 33 - 126 for pretty much all letters and symbols.

So

for i = 33 , 126 do
        somestring = string.char(i)
        dosomething(somestring)
end

will do?

This would make the string somestring the value of the character and then pass it to the function dosomething(s)

But this would only make it 1 character long, so it would make it a, then b, then c, then d, etc.

here is my take on it. replace the print with the submit of password of course

tChoices={}
for i=33,126 do
  tChoices[#tChoices+1]=string.char(i)
end


local function cycleChar(char)
  for num=1,#tChoices do
   if char==tChoices[num] then
	return tChoices[num+1]
   end
  end
  return false
end

local function cycleStr(str)
  for dist=#str,1,-1 do
   local cycled=cycleChar(string.sub(str,dist,dist))
   if cycled then
	return (dist>1 and string.sub(str,1,dist-1) or '')..cycled..string.rep(tChoices[1],#str-dist)
   end
  end
  return string.rep(tChoices[1],#str+1)
end

local s=tChoices[1]
while true do
  term.clear()
  term.setCursorPos(1,1)
  print(s)
  s=cycleStr(s)
  sleep(0)
end

Thanks Kaos, but is it possible to start with three digits instead of one?

Sure, but why bother? You shouldn't make a brute-forcer that can be defeated by a two character password. And, as mentioned above, doing the one and two character combos takes no time at all compared to even doing all the three character strings, let alone the rest.

sure. just replace

local s=tChoices[1]

with

local s=string.rep(tChoices[1],3)