Status

• Exam 2 ... sorry, not yet. :( Wed!

Last time

• Insertion sort, Selection sort
  • both fairly useful: O(n^2)
• Other approaches:
  • What's the simplest sort we could write?
  • Which sorts would be more efficient?

Bubble sort

public static <T extends Comparable<T>> void bubbleSort(T[] array) {
  boolean swapped;
  do {
    swapped = false;
    for (int i = 0; i < array.length - 1; i++) {
      if (array[i].compareTo(array[i+1]) > 0) {
        //swap the two
        E temp = array[i];
        array[i] = array[i + 1];
        array[i + 1] = temp;
        swapped = true;
      }
    }
  }while (swapped);
}

• Trace: [5, 3, 1, 4]
• show i, swapped, and array contents every time you reach the end of the for loop

Bubble sort performance

• How would it perform (big-O) on:
  • Already sorted?
    • Sorted but smallest element at end?
  • Reverse sorted?
  • Average case? O(n^2)
• stable? yes
• How could we improve it?
  • Go through 1 fewer elements each time through array
  • Even better: track the last swap made on trip through list and only go that far next time
  • (doesn't change big-O, but improves practical overhead)

Merge sort

• Idea: lets use recursion to break the problem in half
• Base case: a single element is already sorted
• Tricky part: merging the sorted sub problems on the way back up
• (example)

Merge sort code (1 + 2 of 3)

public static <T extends Comparable<T>> void mergeSort(T[] array) {
  @SuppressWarnings("unchecked")
  T[] aux = (T[]) new Comparable[array.length];
  mergeSort(array, aux, 0, array.length - 1);
}

private static <E extends Comparable<E>> void mergeSort(T[] array, 
                                                T[] aux, int left, int right) {
  if(left < right) {
    int mid = (left + right) / 2;
    mergeSort(array, aux, left, mid);
    mergeSort(array, aux, mid + 1, right);
    merge(array, aux, left, mid, right);
  }
}

Merge sort code (3/3)

private static <T extends Comparable<T>> void merge(T[] array, T[] aux, 
                                               int startLeft, int mid, int endRight) {
  int endLeft = mid;
  int startRight = mid + 1;

  //cursor indexes
  int leftPos = startLeft;
  int rightPos = startRight;
  int auxPos = startLeft;

  //merge from left or right, whichever is smaller, into aux
  while (leftPos <= endLeft && rightPos <= endRight) {
    if (array[leftPos].compareTo(array[rightPos]) <= 0) {
      aux[auxPos++] = array[leftPos++];
    }else {
      aux[auxPos++] = array[rightPos++];
    }
  }
  //at this point, all of left or all of right copied over
  //so copy the remainder of other
  while(leftPos <= endLeft) {
    aux[auxPos++] = array[leftPos++];
  }
  while(rightPos <= endRight) {
    aux[auxPos++] = array[rightPos++];
  }

  //now copy sorted aux section back into original array
  for(int i = startLeft; i <= endRight; i++)
    array[i] = aux[i];
  }
}

Merge sort analysis

• Big-O? O(n lg n)
  • How do we compute this based on the tree-like structure of sub-problem recursive calls?
• Best case or worst case? No. Always O(n lg n)
• Stable? yes, if implemented correctly
• So what's the drawback? O(n) extra space: the auxillary array
  • There are in-place versions, though slower in practice
  • Even normal version slow in practice: lots of copying overhead here
• Other versions:
  • Natural merge: skip merge if already in correct order (used by Java 6 and earlier)
    • Gives you O(n) if already sorted
  • Bottom-up iterative version
  • On linked lists (actually better than for arrays)

Quicksort

• Named so because fast in practice
  • usually O(n lg n) with lower constant overhead than mergesort
  • Interestingly, more comparisons usually made than with mergesort
  • And O(n²) worst case
  • Not usually stable (though can be made to be)
  • But no extra array needed, nor all of that extra copying back and forth
• Idea:
  • Divide in half and recurse, like merge sort
  • But sort on the way down based on a central pivot
  • (examples)

Quicksort code (for reference)

  public static <T extends Comparable<? super T>> void quickSort(T[] array) {
    quickSort(array, 0, array.length - 1);
  }
  
  private static <T extends Comparable<? super T>> void quickSort(T[] array, 
      int start, int end) {
    if (start < end) {
      int pivot = partition(array, start, end);
      quickSort(array, start, pivot - 1);
      quickSort(array, pivot + 1, end);
    }
  }
  
  /**
   * Picks a random pivot value from the given range and then
   * sorts the remaining values into two groups: those larger 
   * and those smaller than the pivot.  Places the pivot value
   * between these groups and returns its index.
   */
  private static <T extends Comparable<? super T>> int partition(T[] array, 
      int start, int end) {
    
    //find pivot and swap into end position
    int pivot = (int) (Math.random() * (end - start + 1)) + start;
    T temp = array[pivot];
    array[pivot] = array[end];
    array[end] = temp;
    pivot = end;
    
    int less = start - 1;
    for (int greater = start; greater < end; greater++) {
      if (array[greater].compareTo(array[pivot]) < 0) {
        //actually expand lesser group instead
        less++;
        temp = array[less];
        array[less] = array[greater];
        array[greater] = temp;
      }
      //else: just leave where it is in the greater range
    }
    //swap pivot into end of lesser group
    less++;  //actually first of greater set
    temp = array[less];
    array[less] = array[pivot];
    array[pivot] = temp;
    pivot = less;
    
    return pivot;
  }

Quicksort analysis

• Not always divided in half each time
• Pivot selection is vital
  • Easiest: just pick last each time so no swap needed
    • But then already sorted is worst case!
  • Almost as easy: random pick
  • Pick 3 and then take median
• Big-O: usually O(n lg n), can be O(n²)

Summary

• Bubble sort: Easy to write
• Merge and Quick sort: More efficient
• Note: Often the case that shorter code != more efficient runtime
  • Do you want to save your time or the computer's?

For next time...

• Bucket/radix sorts and sorting summary
• Start trees
• A08 to be posted...
• Quiz 09 to be posted...