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Sorting

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Sorting

- 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

- each element to be sorted must be associated with a sort key which can be compared with other keys

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

- Things to measure
- comparisons bet. keys
- swaps

The measure of these things

usually approximate fairly

accurately the running time

of the algorithm.

- Insertion SortO( n2 )
- Bubble SortO( n2 )
- Selection SortO( 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

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] )

}

}

}

- 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 )

- worst case: reverse sorted (each ith element must travel all the way up)

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] )

}

}

}

}

- 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 )

- running time:

n

i = 1

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 ] )

}

}

- 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 )

- running time:

n

i = 1

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] );

}

}

}

- O( n1.5 )

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

}

- 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

- given by the recurrence relation

n-1

k = 0