The trick and the solution is in:

VerticalAlignment="";
HorizontalAlignment="";
Width="";
Heigth="";
Grid.Column="";
Grid.Row="";

Play around with it and you will see the solution.

You got the length of your array so u just say:

int k;
int []room = {.......};

for (int i = 0; i < room.length; i++) {
      k = room[i]  
      system.out.println("Location in Array: " +i+ ". Value at Location: " +k \n);
}

There you go. Have fun.

 int k = 0; 

of course.

Hello there MatrimC.

You dont take in consideration that if num1 = 1 and num2 = 1, they are both low and mid value.
You have to write specific what value should be low / mid.
Something like this:

if ((num1 = num2) && (num3 >= num2)) {
    low = num1;
    mid = num2;
    hi = num3;
}

And the other two ways around:

if ((num3 = num2) && (num1>= num3)) {
       low = num2;
       mid = num3;
       hi = num1;
}else if ((num3 = num1 && (num2 >= num1)) {
       low = num3;
       mid = num1;
       hi = num2;
}

Define these three if-statements before the other onces and u should succed!
And look forward to use loops (for loops) its way easier :-)!

Close the thread I figured it out by myself.

Like this ofcourse

member2a(X,[X,X|_]).
member2a(X,[_|T]) :- append(_,[X,X|_],X), append(_,[T|_],X).