Sorting

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# Sorting - PowerPoint PPT Presentation

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 &gt; k j , k i &lt; k j , or k i = k j. Sorting Terminology. The Sorting Problem

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## PowerPoint Slideshow about 'Sorting' - quang

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### 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 ki and kj,

ki > kj , ki < kj ,or ki = kj

Sorting Terminology
• The Sorting Problem

Arrange a set of records so that the values of their key fields are in non-decreasing order.

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.

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
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] )

}

}

}

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 )
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] )

}

}

}

}

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

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 ] )

}

}

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

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] );

}

}

}

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

}

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