2. Merge Sort
7 29 4 → 2 4 7 9
7 2 → 2 7
7→ 7
2→ 2
9 4 → 4 9
9→ 9
4→ 4

3. Merge Sort
Merge sort is based on
the divide-and-conquer
paradigm. It consists of
three steps:
 Divide: partition input
sequence S into two
sequences S1 and S2 of
about n/2 elements each
 Recur: recursively sort S1
and S2
 Conquer: merge S1 and S2
into a unique sorted
sequence
Algorithm mergeSort(S, C)
Input sequence S, comparator C
Output sequence S sorted
according to C
if S.size() > 1 {
(S1, S2) := partition(S, S.size()/2)
S1 := mergeSort(S1, C)
S2 := mergeSort(S2, C)
S := merge(S1, S2)
}
return(S)

4. Merge Sort Execution Tree (recursive calls)
An execution of merge-sort is depicted by a binary
tree
 each node represents a recursive call of merge-sort and stores
unsorted sequence before the execution and its partition
 sorted sequence at the end of the execution
 the root is the initial call
 the leaves are calls on subsequences of size 0 or 1

7 2 9 4 → 2 4 7 9
7 2 → 2 7
7→ 7
2→ 2
Divide-and-Conquer
9 4 → 4 9
9→ 9
4→ 4

5. Execution Example

Partition
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 2 9 4 → 2 4 7 9
7 2 → 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

6. Execution Example (cont.)

Recursive call, partition
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2 → 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

7. Execution Example (cont.)

Recursive call, partition
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

8. Execution Example (cont.)

Recursive call, base case
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

9. Execution Example (cont.)
Recursive call, base case
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

10. Execution Example (cont.)

Merge
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

11. Execution Example (cont.)

Recursive call, …, base case, merge
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
72→2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

12. Execution Example (cont.)

Merge
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 8 6
9 4 → 4 9
9→ 9
4→ 4
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

13. Execution Example (cont.)
Recursive call, …, merge, merge
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 6 8
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
8→ 8
6 1 → 1 6
6→ 6
1→ 1

14. Execution Example (cont.)

Merge
7 2 9 43 8 6 1 → 1 2 3 4 6 7 8 9
7 29 4→ 2 4 7 9
7 2→ 2 7
7→ 7
2→ 2
3 8 6 1 → 1 3 6 8
9 4 → 4 9
9→ 9
4→ 4
Divide-and-Conquer
3 8 → 3 8
3→ 3
6 1 → 1 6
8→ 8
14
6→ 6
1→ 1

15. Static Method mergeSort()
Public static void mergeSort(a, int left, int right)
{
// sort a[left:right]
if (left < right)
{// at least two elements
int mid = (left+right)/2;
//midpoint
mergeSort(a, left, mid);
mergeSort(a, mid + 1, right);
merge(a, b, left, mid, right); //merge from a to b
copy(b, a, left, right); //copy result back to a
}
}

16. MERGE-SORT(A,p,r)
1. if lo < hi
2. then mid ← (lo+hi)/2
3.
4.
5.
MERGE-SORT(A,lo,mid)
MERGE-SORT(A,mid+1,hi)
MERGE(A,lo,mid,hi)
Call MERGE-SORT(A,1,n) (assume n=length of list A)
A = {10, 5, 7, 6, 1, 4, 8, 3, 2, 9}

17. Another Analysis of Merge-Sort
 The height h of the merge-sort tree is O(log n)
 at each recursive call we divide in half the sequence,
 The work done at each level is O(n)
 At level i, we partition and merge 2i sequences of size n/2i
 Thus, the total running time of merge-sort is O(n log n)
depth
#seqs
size
0
1
n
Cost for
level
n
1
2
n/2
n
i
2i
n/2i
n
…
…
…
…
logn
2logn = n n/2logn = 1
n
Divide-and-Conquer

18. Summary of Sorting Algorithms
(so far)
Vectors
Algorithm
Time
Notes
Selection Sort
O(n2)
Slow, in-place
For small data sets
Insertion Sort
O(n2) WC, AC
O(n) BC
Slow, in-place
For small data sets
Heap Sort
O(nlog n)
Fast, in-place
For large data sets
Merge Sort
O(nlogn)
Fast, sequential data access
For huge data sets

19. 7 4 9 6 2 → 2 4 6 7 9
4 2 → 2 4
2→ 2
Divide-and-Conquer
7 9 → 7 9
9→ 9

20. Quick-Sortrandomized
Quick-sort is a
sorting algorithm based on
the divide-and-conquer
paradigm:
x
 Divide: pick a random
element x (called pivot) and
partition S into



L elements less than x
E elements equal x
G elements greater than x
x
L
E
 Recur: sort L and G
 Conquer: join L, E and G
Divide-and-Conquer
x
G

21. Quicksort
Efficient sorting algorithm
Discovered by C.A.R. Hoare
Example of Divide and Conquer algorithm
Two phases
Partition phase

Divides the work into half
Sort phase

Conquers the halves!

22. Quicksort
Partition
Choose a pivot
Find the position for the pivot so that
all elements to the left are less
 all elements to the right are greater

pivot
< pivot
> pivot

23. Quicksort
Conquer
Apply the same algorithm to each half
< pivot
< p’
p’
> pivot
> p’
pivot
< p”
p”
> p”

24. 78 70 65 84 98 78 100 93 55 61 81 68
78 70 65 84 98 78 100 93 55 61 81 68
78 70 65 84 98 78 100 93 55 61 81 68
78 70 65 68 98 78 100 93 55 61 81 84

25. 78 70 65 68 98 78 100 93 55 61 81 84
78 70 65 68 61 78 100 93 55 98 81 84
78 70 65 68 61 78 100 93 55 98 81 84
78 70 65 68 61 55 100 93 78 98 81 84
55 70 65 68 61 78 100 93 78 98 81 84

26. 78
55 70 65 68 61
100 93 78 98 81 84

27. Quicksort
Implementation
quicksort( void *a, int low, int high )
{
int pivot;
/* Termination condition! */
if ( high > low )
{
pivot = partition( a, low, high );
quicksort( a, low, pivot-1 );
quicksort( a, pivot+1, high );
}
}
Divide
Conquer

28. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
}
/* right is final position for the pivot */
a[low] = a[right];
a[right] = pivot_item;
return right;
}

29. Quicksort - Partition
This example
uses int’s
to keep things
simple!
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
Any item will do as the pivot,
while ( left < right ) { choose the leftmost one!
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
}
23 12 15 38 42 18 36 29 27
/* right is final position for the pivot */
a[low] = a[right];
a[right] = pivot_item;
return right;
low
high
}

30. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
Set left and right markers
right = high;
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++; right
left
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if (23 12 right 38 SWAP(a,left,right);27
left < 15
) 42 18 36 29
}
/* right is final position for the pivot */
a[low]low a[right];
=
high
pivot: 23
a[right] = pivot_item;
return right;
}

31. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
Move the markers
until they cross over
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
left
}
/* right is final position for the pivot */
a[low] = a[right]; 15 38 42 18 36 29
23 12
a[right] = pivot_item;
return right;
low
pivot: 23
}
right
27
high

32. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
Move the left pointer while
it points to items <= pivot
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
left
right
}
Move right
/* right is final position for the pivot */ similarly
a[low] = a[right];
23 12 15 38 42
a[right] = pivot_item; 18 36 29 27
return right;
}
low
high
pivot: 23

33. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
Swap the two items
on the wrong side of the pivot
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
}
left
right
/* right is final position for the pivot */
a[low] = a[right];
a[right] = pivot_item; 18 36 29 27
23 12 15 38 42
return right;
pivot: 23
}
low
high

34. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
left and right
have swapped over,
so stop
while ( left < right ) {
/* Move left while item < pivot */
while( a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
}
right
left
/* right is final position for the pivot */
a[low] = a[right];
a[right] = pivot_item; 38 36 29 27
23 12 15 18 42
return right;
}
low
high
pivot: 23

35. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
right
left
while ( left < right ) {
/* Move left while item < pivot */
23
while( 18 42 38 36 29 27
12 15 a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
low
high
pivot: 23
}
/* right is final position for the pivot */
a[low] = a[right];
Finally, swap the pivot
a[right] = pivot_item;
and right
return right;
}

36. Quicksort - Partition
int partition( int *a, int low, int high ) {
int left, right;
int pivot_item;
pivot_item = a[low];
pivot = left = low;
right = high;
right
while ( left < right ) {
/* Move left while item < pivot */
18
pivot: 23
while( 23 42 38 36 29 27
12 15 a[left] <= pivot_item ) left++;
/* Move right while item > pivot */
while( a[right] >= pivot_item ) right--;
if ( left < right ) SWAP(a,left,right);
low
high
}
/* right is final position for the pivot */
a[low] = a[right];
Return the position
a[right] = pivot_item; of the pivot
return right;
}

37. Quicksort - Conquer
pivot
pivot: 23
18
12
15
Recursively
sort left half
23
42
38
36
29
27
Recursively
sort right half

39. Why study Heapsort?
It is a well-known, traditional sorting
algorithm you will be expected to know
Heapsort is always O(n log n)
Quicksort is usually O(n log n) but in the worst
case slows to O(n2)
Quicksort is generally faster, but Heapsort is
better in time-critical applications
Heapsort is a really cool algorithm!

40. What is a “heap”?
 Definitions of heap:
1. A large area of memory from which the
programmer can allocate blocks as needed,
and deallocate them (or allow them to be
garbage collected) when no longer needed
2. A balanced, left-justified binary tree in
which no node has a value greater than the
value in its parent
 These two definitions have little in
common
 Heapsort uses the second definition

41. Balanced binary trees
Recall:
The depth of a node is its distance from the root
The depth of a tree is the depth of the deepest node
A binary tree of depth n is balanced if all the
nodes at depths 0 through n have two children
n-2
n-1
n
Balanced
Balanced
Not balanced

42. Left-justified binary trees
A balanced binary tree is left-justified if:
all the leaves are at the same depth, or
all the leaves at depth n+1 are to the left of
all the nodes at depth n
Left-justified
Not left-justified

43. Plan of attack
First, we will learn how to turn a binary tree into a
heap
Next, we will learn how to turn a binary tree back into
a heap after it has been changed in a certain way
Finally (this is the cool part) we will see how to use
these ideas to sort an array

44. The heap property
A node has the heap property if the value in the
node is as large as or larger than the values in its
children
12
8
12
3
Blue node has
heap property
8
12
12
Blue node has
heap property
8
14
Blue node does not
have heap property
All leaf nodes automatically have the heap property
A binary tree is a heap if all nodes in it have the
heap property

45. shiftUp
Given a node that does not have the heap property,
you can give it the heap property by exchanging its
value with the value of the larger child
12
8
14
14
Blue node does not
have heap property
8
12
Blue node has
heap property
This is sometimes called shifting up
Notice that the child may have lost the heap property

46. Constructing a heap I
A tree consisting of a single node is automatically a
heap
We construct a heap by adding nodes one at a
time:
Add the node just to the right of the rightmost node
in the deepest level
If the deepest level is full, start a new level
Examples:
Add a new
node here
Add a new
node here

47. Constructing a heap II
Each time we add a node, we may destroy the heap
property of its parent node
To fix this, we sift up
But each time we sift up, the value of the topmost
node in the shift may increase, and this may destroy
the heap property of its parent node
We repeat the shifting up process, moving up in the
tree, until either
We reach nodes whose values don’t need to be swapped
(because the parent is still larger than both children), or
We reach the root

48. 8
8
10
10
8
1
12
8
5
2
10
8
10
3
10
5
12
8
12
5
10
8
5
4

49. Other children are not affected
12
10
8
12
5
14
14
8
14
5
10
12
8
5
10
 The node containing 8 is not affected because its parent gets
larger, not smaller
 The node containing 5 is not affected because its parent gets
larger, not smaller
 The node containing 8 is still not affected because, although its
parent got smaller, its parent is still greater than it was originally

50. A sample heap

Here’s a sample binary tree after it has been heapified
25
22
19
18
17
22
14
21
14
3
9
15
11
Notice that heapified does not mean sorted
Heapifying does not change the shape of the binary
tree; this binary tree is balanced and left-justified
because it started out that way

51. Removing the root
Notice that the largest number is now in the root
Suppose we discard the root:
11
22
19
18
17
22
14
21
14
3
9
15
11
How can we fix the binary tree so it is once again
balanced and left-justified?
Solution: remove the rightmost leaf at the deepest
level and use it for the new root

52. The reHeap method I
Our tree is balanced and left-justified, but no longer a
heap
However, only the root lacks the heap property
11
22
19
18
17
22
14
21
14
3
15
9
We can shiftUp() the root
After doing this, one and only one of its children may
have lost the heap property

53. The reHeap method II
Now the left child of the root (still the number 11)
lacks the heap property
22
11
19
18
17
22
14
21
14
3
15
9
We can shiftUp() this node
After doing this, one and only one of its children may
have lost the heap property

54. The reHeap method III
Now the right child of the left child of the root (still
the number 11) lacks the heap property:
22
22
19
18
17
11
14
21
14
3
15
9
We can shiftUp() this node
After doing this, one and only one of its children may
have lost the heap property —but it doesn’t, because
it’s a leaf

55. The reHeap method IV

Our tree is once again a heap, because every node in it
has the heap property
22
22
19
18
17
21
14
11
14
3
15
9
Once again, the largest (or a largest) value is in the root
We can repeat this process until the tree becomes empty
This produces a sequence of values in order largest to
smallest

56. Sorting
What do heaps have to do with sorting an array?
Here’s the neat part:
Because the binary tree is balanced and left justified, it
can be represented as an array
All our operations on binary trees can be represented as
operations on arrays
To sort:
heapify the array;
while the array isn’t empty {
remove and replace the root;
reheap the new root node;
}

57. Mapping into an array
25
22
17
19
18
0
22
14
1
2
14
21
3
4
3
5
6
9
7
8
15
11
9
10
25 22 17 19 22 14 15 18 14 21 3
11
12
9 11
Notice:
The left child of index i is at index 2*i+1
The right child of index i is at index 2*i+2
Example: the children of node 3 (19) are 7 (18) and 8
(14)

58. Removing and replacing the root
The “root” is the first element in the array
The “rightmost node at the deepest level” is the last
element
Swap them...
0
1 2
3
4
5
6
7
8
9
10
25 22 17 19 22 14 15 18 14 21 3
0
1
2
3
4
5
6
7
8
9
10
11 22 17 19 22 14 15 18 14 21 3
11
12
9 11
11
12
9 25
...And pretend that the last element in the array no
longer exists—that is, the “last index” is 11 (9)

59. Reheaproot node repeat
and (index 0, containing
Reheap the
0
1
2
3
4
5
6
7
8
9
10
11 22 17 19 22 14 15 18 14 21 3
0
1
2
3
4
5
6
7
8
9
10
22 22 17 19 21 14 15 18 14 11 3
0
1
2
3
4
5
6
7
8
9
10
11
11)...
12
9 25
11
12
9 25
11
12
9 22 17 19 22 14 15 18 14 21 3 22 25
...And again, remove and replace the root node
Remember, though, that the “last” array index is
changed
Repeat until the last becomes first, and the array is
sorted!

60. Analysis I
Here’s how the algorithm starts:
heapify the array;
Heapifying the array: we add each of n nodes
Each node has to be shifted up, possibly as far as the
root

Since the binary tree is perfectly balanced, sifting up a
single node takes O(log n) time
Since we do this n times, heapifying takes n*O(log n)
time, that is, O(n log n) time

61. Analysis II

Here’s the rest of the algorithm:
while the array isn’t empty {
remove and replace the root;
reheap the new root node;
}
We do the while loop n times (actually, n-1 times),
because we remove one of the n nodes each time
Removing and replacing the root takes O(1) time
Therefore, the total time is n times however long it
takes the reheap method

62. Analysis III
To reheap the root node, we have to follow one path
from the root to a leaf node (and we might stop
before we reach a leaf)
The binary tree is perfectly balanced
Therefore, this path is O(log n) long
And we only do O(1) operations at each node
Therefore, reheaping takes O(log n) times
Since we reheap inside a while loop that we do n
times, the total time for the while loop is n*O(log
n), or O(n log n)

63. Analysis IV
Here’s the algorithm again:
heapify the array;
while the array isn’t empty {
remove and replace the root;
reheap the new root node;
}
We have seen that heapifying takes O(n log n) time
The while loop takes O(n log n) time
The total time is therefore O(n log n) + O(n log n)
This is the same as O(n log n) time