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Sorting

Sorting. Sorting Terminology. Sort Key each element to be sorted must be associated with a sort key which can be compared with other keys e.g. for any two keys k i and k j , k i > k j , k i < k j , or k i = k j. Sorting Terminology. The Sorting Problem

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Sorting

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  1. Sorting

  2. Sorting Terminology • Sort Key • each element to be sorted must be associated with a sort key which can be compared with other keys e.g. for any two keys ki and kj, ki > kj , ki < kj ,or ki = kj

  3. Sorting Terminology • The Sorting Problem Arrange a set of records so that the values of their key fields are in non-decreasing order.

  4. Sorting Algorithms(Running Time Analysis) • Things to measure • comparisons bet. keys • swaps The measure of these things usually approximate fairly accurately the running time of the algorithm.

  5. Sorting Algorithms • Insertion Sort O( n2 ) • Bubble Sort O( n2 ) • Selection Sort O( n2 ) • Shellsort O( n1.5 ) • Quicksort O( nlog2n) • Mergesort O( nlog2n) • Heapsort O( nlog2n) • Binsort O( n) w/ qualified input • Radix Sort O( n) w/ qualified input

  6. Insertion Sort: Algorithm void insertionSort( Elem[] a, int n ) { for( int i = 1; i < n; i++ ) { for( int j = i; (j > 0) && ( a[ j].key < a[ j-1].key); j-- ) { swap( a[ j], a[ j-1] ) } } }

  7. Insertion Sort: Illustration

  8. Insertion Sort: Time complexity • outer for loop executed n-1 times • inner for loop depends on how many keys before element i are less than it. • worst case: reverse sorted (each ith element must travel all the way up) • running time: (n-1)(n)/2 => O( n2 ) • best case: already sorted (each ith element does not need to travel at all) • running time: (n-1) => O( n ) • average case: { (n-1)(n)/2 } / 2 => O( n2 )

  9. Bubble Sort: Algorithm void bubbleSort( Elem[] a, int n ) { for( int i = 0; i < n-1; i++ ) { for( int j = n-1; j > i; j-- ) { if( a[ j].key < a[j-1].key ) { swap( a[ j], a[ j-1] ) } } } }

  10. Bubble Sort: Illustration

  11. Bubble Sort: Time complexity • number of comparisons in inner for loop for the ith iteration is always equals to i • running time: i = n(n+1)/2  O( n2 ) n i = 1

  12. Selection Sort: Algorithm void selectionSort( Elem[] a, int n ) { for( int i = 0; i < n-1; i++ ) { int lowindex = i for( int j = n-1; j > i; j-- ) { if( a[ j].key < a[ lowindex ].key ) { lowindex = j; } } swap( a[i], a[ lowindex ] ) } }

  13. Selection Sort: Illustration

  14. Selection Sort: Time complexity • number of comparisons in inner for loop for the ith iteration is always equals to i • running time: i = n(n+1)/2  O( n2 ) n i = 1

  15. Shellsort: Algorithm void shellsort( Element[] a ) { for( int i = a.length/2; i >= 2; i /=2 ) { for( int j = 0; j < i; j++ ) { insertionSort2( a, j, a.length-j, i ); } } insertionSort2( a, 0, a.length, 1 ); } void insertionSort2( Element[] a, int start, int n, int incr ) { for( int i=start+incr; i<n; i+=incr) { for( j=i; (j>=incr) && (a[ j].key < a[ j-incr].key); j-=incr) { swap( a[j], a[j-incr] ); } } }

  16. Shellsort: Illustration

  17. Shellsort: Time complexity • O( n1.5 )

  18. Quicksort: Algorithm void quicksort( Elem[] a, int I, int j ) { int pivotindex = findpivot( a, i, j ) swap( a[pivotindex], array[j] ) // stick pivot at the end int k = partition( a, i-1, j, a[j].key ) // k will be the first position in the right subarray swap( a[k], a[j] ) // put pivot in place if( k-i > 1 ) quicksort( a, i, k-1 ) // sort left partition if( j-k > 1 ) quicksort( a, k+1, j ) // sort right partition } int findpivot( Elem[] A, int i, int j ){ return ( (i+j) / 2 ) } int partition( Elem[] A, int l, int r, Key pivot ) { do // move the bounds inward until they meet { while( a[++l ].key < pivot ) // move left bound right while( r && a[--r].key > pivot ) // move right bound left swap( a[l], a[r] ) // swap out-of-place values }while( l < r ) // stop when they cross swap( a[l], a[r] ) // reverse last, wasted swap return l // return the first position in right position }

  19. Quicksort: Illustration

  20. Quicksort: Illustration

  21. Quicksort: Time complexity • findpivot() takes constant time: 0(1) • partition() depends on the length of the sequence to be partitioned: • O(s) for sequence of length s • Worst-case: when pivot splits the array of size n into partitions of size n-1 and 0. O(n2) • Best case: when pivot always splits the array into two equal halves. • There will be log2n levels (1st level: one n sequence, 2nd level: two n/2 sequences, 3rd level: four n/4 sequences, …): O(nlog2n) • Average case: O( n log2n ) • given by the recurrence relation T(n) = cn + 1/n (T(k) +T(n - 1 - k)), T(0) = c, T(1) = c n-1 k = 0

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