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Bubble-sort algorithm

  1. Jul 13, 2012 #1
    Q: Find the maximum number of interchanges needed to sort a list of six pieces of data using the bubble-sort algorithm

    working:

    for the first past, maximum needed is n-1 swaps where n is the amount of the pieces of data
    2nd - n-2
    3rd - n-3
    4th - n-4
    5th - n-5

    so generally it takes

    (n-1) + (n-2) +... (n-(n-1))

    is this correct? It works for six as the it will be a maximum of 15 swaps, also how can I generalise this?
     
  2. jcsd
  3. Jul 13, 2012 #2

    Zondrina

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    This would be more suitable for the computer science section, but hey ill help.

    Consider an array of natural numbers ( I'll be using some java syntax here to provide a practical example ) :

    int[] nums = {0,1,2,3,4,5,...n};

    and presume you want to sort it from highest to lowest ( So we have a worst case scenario here ).

    There are n elements in the array, so to move 0 to the nth position would require n-1 exchanges.

    The next exchange for 1 would require n-2 exchanges.

    This continues until you have n-(n-1) exchanges since n-n would be redundant because the list is sorted at that point.

    I suppose a way you could generalize this would be to say :

    [itex]\sum_{k=1}^{n-1}[/itex] (n-k)
     
  4. Jul 13, 2012 #3
    Cheers,

    btw I had no idea it was the wrong section, in the UK, this is under our maths syllabus and we're not programming anything, but I can see how it leads to things :)

    thanks again.
     
  5. Jul 13, 2012 #4

    rcgldr

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    There's a trick to adding up an incrementing series of numbers, arrange the numbers forwards and backards and add up the sum:

    Code (Text):

    n-1 + n-2 + n-3 + ... +   1
      1 +   2 +   3 _ ... + n-1
    ---------------------------
      n +   n +   n + ... +   n
     
    This sum equals n (n-1), but since the array was added twice divide by 2:

    (n-1) + (n-2) + (n-3) + ... + 1 = n (n-1) / 2
     
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