Combinations of letters

[Cbomb]

I'm pretty sure this is called permutation. WHat I need to do is take a variable number of letters/numbers an compute all the possible combinations. Example:
Given string: abc
Results:
abc
acb
bca
bac
cab
cba

And I have no clue as to how to do this. I need to be able to do this with any number of letters up to 10 at least. Code would be great. But if you could explain the process that would be good too.

Thanks in advance.

[dimava]

I dont have the time to write it but this is how it works... you break up every letter in the word into a seperate string...then you randmoize putting the string togehter in diffrent ways and then putting all of that into a list box, or what ever you need... If I had lots of time on my hands I would give you the code. and If i did then it will be like 100 pages long becuase II dont know how to make it short

[HarryW]

This is like the nPr thing you see on your scientific calculator. I'm not sure what the formula is off the top of my head, I'll look it up a sec (assuming someone doesn't answer it already).

[HarryW]

P(n,k) = n·(n-1) ···(n-k+1) = n!/k!

wheere P(n,k) is the number of permutations (a function of n & k), n is the number of different possible values (3 if you're using the set {a, b, c} for example), and k is the length of your string.

[HarryW]

Oh yeah I guess I'll give you some helping using the formula too Basically I'd say write yourself a recursive function that will calculate factorials for you, then just use P = n!/k!.

Example of code to go in a recursive function to find a factorial:

Public Function Factorial(x as integer)

If x = 1 then
	Factorial = 1
Else
	Factorial = x * Factorial(x-1)
End If

End Function

Hmm, it looks simple... too simple... think I may have fogotten something See if it works anyway. And try not to use it with too high an x value, because the function headers will really add up. Well obviously

Try this:

'Code from Megatron
For I = 0 To 25
    fChar = 97 + I
    For x = 0 To 25
        sChar = 97 + x
        For y = 0 To 25
            List1.AddItem Chr(fChar) & Chr(sChar) & Chr(97 + y)
        Next y
    Next x
Next I

[HarryW]

D'oh! I'm not just racking up my post count here, honest Looking at that formula something's wrong with it, here's the right one:

nPr = n!/(n-r)!

If n = r then nPr = n!.

Sorry, wrong letters so I edited it.

[Edited by HarryW on 09-07-2000 at 10:10 PM]

[Cbomb]

Ok I now can get the number of combinations. But how do I get the combinations themselves? If you told me in that math stuff do forgive me I'm not great at math.

[Guv]

The number of permutations goes up fast: 8 Letters (40,320), 10 (3,628,800), 12 (479,001,600).

The best way to generate them can be done recursively. The general idea is to move the first letter in the string right and left (ABCDEFG, BACDEFG. BCADEFG, BCDAEFG). When first letter gets to one end or the other, advance second letter once before reversing direction of first letter (BCDEFGA, CBDEFGA, CBDEFAG, CBDEAFG). Similarly, when second letter gets to an extreme position, advance third letter.

The advantage of this method, is that the first letter is moved a huge number of times. It is kept in a temporary location and inserted into the string as required. Thus, most of the exchanges are "half-exchanges" instead of "full exchanges."

I created a simple VB application which does the above to illustrate the algorithm for a few bright teenage friends. I can Zip the VB files I used and send them to you if you want.

[Cbomb]

Oh ok I get it now. If you could send that it would be great! [email protected]

Thanks!