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Sorting Algorithms
Selection, Bubble, Insertion and
Radix Sort
Worst-casevs. Average-case Analyses
- An algorithm can require different times tosolve different problems of the same size.
- Worst-case analysis (find the maximumnumber of operations an algorithm can
execute in all situations)– is easier to calculate and is more common
- Average-case (enumerate all possiblesituations, find the time of each of the m
possible cases, total and dive by m)
- is harder to compute but yields a more realisticexpected behavior
Selection Sort
Divides the array into two parts: already sorted, andnot yet sorted. On each pass, finds thesmallest of the unsorted elements, and swap it intoits correct place, thereby increasing the number ofsorted elements by one.
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
Selection Sort: Pass One
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
U N S O R T E D
Selection Sort: End Pass One
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
U
NS
OR
TE
D
SORTED
Selection Sort: Pass Two
values [ 0 ] SORTED [ 1 ] [ 2 ] [ 3 ] [ 4 ]
U
NS
OR
T
ED
Selection Sort: Pass Four
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
SO
RT
ED
UNSORTED
Selection Sort: End Pass Four
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
S O R T E D
Selection Sort:
How many comparisons?
values [ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ]
4 compares for values[0] 3 compares for values[1] 2 compares for values[2] 1 compare for values[3] = 4 + 3 + 2 + 1
For selection sort in general
• The number of comparisons when the
array contains N elements is
Sum = (N-1)+(N-2)+ …+2+
Notice that...
Sum = (N-1) + (N-2) +... + 2 + 1
+ Sum = 1 + 2 +... + (N-2) + (N-1)
2* Sum = N + N +... + N + N
2 * Sum = N * (N-1)
Sum = N * (N-1)/2= .5 N 2 - .5 N
• How many number of swaps?
• O(N)
• Therefore, the complexity is O(N^2 )
Code for Selection Sort
void SelectionSort (int values[], int numValues) //// Post:into ascendingSorts array order values[0 by key.. numValues-1 ] {int endIndex = numValues - 1 ; forSwap (int (values, current=0;current<endIndex;current++) current, MinIndex(values,current,endIndex)); }
Bubble Sort: Pass One
Bubble Sort: Pass Two
Snapshot of BubbleSort
Code for Bubble Sort
void{ BubbleSort(int values[], int numValues) intwhile current (current = 0; < numValues - 1) {BubbleUp(values, current, numValues-1); }current++; }
Bubble Sort code (contd.)
void BubbleUp(intint values[],startIndex, int endIndex) //// Post:order Adjacent have been pairs switched that arebetween out of //// values[startIndex]..values[endIndex]beginning at values[endIndex]. {for (int index = endIndex; indexif (values[index] > startIndex; < index--)values[index-1]) } Swap(values,^ index,^ index-1);
Observations on Bubble Sort
• There can be a large number of
intermediate swaps.
• Can this algorithm be improved?
- What are the best/worst cases?
• This algorithm is O(N^2 ) in the worst-case
