How to interchange the position of each of my four ui-elements randomly? - algorithm for the 24 possibilities - objective-c

Question

I have a program with four different buttons. I want to interchange the position of the buttons randomly. For example: 1 2 3 4 Later: 3 4 1 2 Later: 1 3 2 4
Is there a algorithms for that? The only way I can think is to make a random number from 1 to 24 (24 possibilities) and then code all the possible button postitions.
int foo = arcrandom() % 23;
switch(foo){
case 0:
button1postiton = 100; //just an example
button2position = 200;
button3position = 300;
button4position = 400;
break;
case 2:
button1postiton = 200;
//blablabla and so on and so on
}
But is there a more efficient way?
Thanks!

Answer 1

You could shuffle the buttons or their positions, e.g. with a Fisher-Yates shuffle.

Answer 2

There is code in this website to get a list of all permutations of an array (see method perm2), it is coded for char arrays, but can be modified to do int arrays as well and to other languages as well, then you can use mjv's idea.
http://www.cs.princeton.edu/introcs/23recursion/Permutations.java.html
If in Java, this is what I would try....
Once you get all the possible permutations maybe in a vector, I think you can use a grid bag layout and change the grid constraints, picking one of the elements of the vector randomly. I have not tried this out, but I am thinking along the lines of
Vector permutations = ... //get the permutation using a class similar to the one in the website for an array of ints {0,1,2,3}
//The panel
JPanel pane;
JButton button;
pane.setLayout(new GridBagLayout());
GridBagConstraints c = new GridBagConstraints();
//Choose one permutation at random
int foo = arcrandom() % 23;
int current[] = permutations.get(foo);
//Add the buttons in the chosen order
button = new JButton("Button 1");
c.gridx = current[0];
c.gridy = 0;
pane.add(button, c);
button = new JButton("Button 2");
c.gridx = current[1];
c.gridy = 0;
pane.add(button, c);
button = new JButton("Button 3");
c.gridx = current[2];
c.gridy = 0;
pane.add(button, c);
button = new JButton("Button 4");
c.gridx = current[3];
c.gridy = 0;
pane.add(button, c);
Let me know if this works!

Answer 3

Start with a random number 0 <= r < 24
Start with your first position. Derive rr = r % 4 and r = r / 4. Those are the remainder and quotient respectively after division by 4.
The remainder specifies a position. Swap position 0 with the specified position.
For the next position, derive rr = r % 3 and r = r / 3. Again the remainder specifies a position, this time 0, 1 or 2, but relative to your current position (1).
Swap position 1 with position rr+1.
For the next position, derive rr = r % 2 and r = r / 2. Again the remainder specifies a position, this time 0 or 1, and relative to your current position again (2).
Swap position 2 with position rr+2.
For position 3, there is nothing to do.
Note - for each swap, one possibility is to swap a position with itself. Obviously no swap is needed for that.
This is probably the Fisher-Yates shuffle - I had no idea it had a name until today.

Answer 4

Thanks for all your answers! I used the Fisher-Yates shuffle! I found here a nice tutorial, how to use the algorithm in Objective-C: gorbster.net