Sorting numbers in an array without sort function

Hello,

I want to sort numbers in an arrey without using sort function, but with bubblesort (swapping number next each other until all numbers are sorted)

Here is what I have...

X=[10 1 29 89 5 56 76 8];
finished=0;
while finished==0
    if finished==0
       for i=1:size(X,2)-1
          if X(i)>X(i+1)
              c=X(i);
              X(i)=X(i+1);
              X(i+1)=c;
          end
       end
   end
   if X(i)<X(i+1) || X(i)==X(i+1)
      finished=1; 
    end
end
disp(X)

I understand that this loop works by comparing numbers next each other (like comparing 10 with 1). Then, if the front number (10) is bigger than the next number (1), the numbers are swaped. However, I want to break the loop once all numbers are sorted from the minimum number to the maximum number.

The code above only sorts the numbers until [1,10,29,5,56,76,8,89], so I want to loop several more to complete the sorting, but I want the code stops when the sorting is done. Otherwise, the code keeps on looping again and again.

I really appreciate if I get some help with this problem.

Answer 1

[James Tursa]

Simply set your "finished" flag depending on whether a swap was done or not. E.g.,

while true
    finished = 1;
    for i=1:size(X,2)-1
        if X(i)>X(i+1)
            c=X(i);
            X(i)=X(i+1);
            X(i+1)=c;
            finished = 0;
        end
    end
    if finished
        break
    end
end

The for-loop can also be shortened by realizing that after each pass you are guaranteed to have one more element sorted at the bottom. I.e., the for-loop doesn't have to examine the entire array each time ... at each iteration you only have to examine one fewer elements.

[Takashi Fukushima]

Thank you so much! It worked perfectly. Could you explain (or refer websites) how the flag works because I do not understand why "if finished" works.

[James Tursa]

If you do the entire for-loop and don't do any swaps, then the array is already sorted and you are done. In this case the "finished" flag will be 1 because it was never set to 0 inside the swapping code. So you simply test for this condition right after the for-loop is done.

[the cyclist]

In case you were asking something different ... the syntax

if finished

works because when the variable finished takes the value "1", that value will be converted to the boolean "true", and the if statement will be entered. In general, the condition always has to be evaluated to true/false.

I would argue that slightly better programming practice would have been to define the variable finished as a boolean false in the first place. But that's fairly precious. It's more important to get the code to work correctly first, before worrying about such nuances.

[Takashi Fukushima]

Thank you so much for your answer James and the cyclist !

I figured it out and think I understand it!

Answer 2

[the cyclist]

if all(diff(X)) >= 0

Also, rather than defining the finished variable, you could have just made that the condition for the while loop:

while not(all(diff(X)) >= 0)

or, mathematically equivalently but a bit more elegantly

while any(diff(X)) < 0

[Takashi Fukushima]

Thank you so much! I did not know diff function would be useful here!