Exchange sort (Bubblesort) compares two consecutive items in the list, and swap them if they are out of order. for (i=1;i<n;i++) for(j=n-1;j>=i;j--) if.

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Exchange sort (Bubblesort) compares two consecutive items in the list, and swap them if they are out of order. for (i=1;i<n;i++) for(j=n-1;j>=i;j--) if (a[j]<a[j-1]) {temp=a[j]; a[j]=a[j-1]; a[j-1]=temp;}

Bubblesort

int ilast; i=n-1; while (i>0) { ilast=0; for(j=0;j<i;j++) if (a[j]>a[j+1]) {temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; ilast=j; } i=ilast; }

Pass 0 ilast=3 Pass1 ilast=2

Pass 2 ilast=0 O(n 2 )

Shaker sort O(n 2 )

Partition sort (Quicksort) Pick a middle item (and call it mid); scan the array from the left until an item a[i] > mid is found scan from the right until an item a[j] < mid is found. Exchange the two items and continue this scan and swap process until the two scans meet somewhere in the middle of the array O(n*log(n))

void sort(int a[], int low, int high) { … scanUp=low; scanDown=high; mid=(low+high)/2; item=a[mid]; do { while (a[scanUp]<item) scanUp++; while (item<a[scanDown]) scanDown--; if (scanUp<=scanDown) { temp=a[scanUp]; a[scanUp]=a[scanDown]; a[scanDown]=temp; scanUp++; scanDown --; } while (scanUp < scanDown); if (low<scanDown) sort(a,low, scanDown); if (scanUp <high) sort(a, scanUp,high); }

Finding the median The median of n items is defined as that item which is less or equal to half of the n items and which is larger or equal to the other half of the n items. The median of is 18. The main idea is: finding the k-th smallest of n items. k = n/2

int find(int a[],int first, int last, int k) {while (first<last) { val=a[k]; i=first; j=last; do { while (a[i]<val) i++; while (a[j]>val) j--; if (i<=j) { temp=a[i]; a[i]=a[j]; a[j]=temp; i++; j--; } } while (j>=i); if (j<k) first=i; if (k<i) last=j; } return last; } O(n*log(n))

Straight merging 1. Split the sequence into two halves, called A and B. 2. Merge A and B by combining single items into ordered pairs. 3. Repeat steps 1 and 2, merging ordered pairs into ordered quadruples. 4. Repeat the previous steps, merging quadruples into octets, and continue, until the entire sequence is ordered.

fC 6, 34, 1, 64, 89, -2, 12, 46, -78, 3, 11, 49, 80, 20 fA 6, 1, 89, 12, -78, 11, 80 fB 34, 64, -2, 46, 3, 49, 20 fC (6,34),(1,64),(-2,89),(12,46),(-8,3),(11,49),(20,80) fA (6,34),(-2,89),(-78,3),(20,80) fB (1,64),(12,46),(11,49) fC (1,6,34,64),(-2,12,46,89),(-78,3,11,49),(20,80) O(n*log(n))