Sorting algos > Data Structures & Algorithums

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

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)

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

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

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


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


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

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


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


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

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


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

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
}
}

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}

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

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

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

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

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!

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

Quicksort
Conquer

Apply the same algorithm to each half
< pivot
< p’

p’

> pivot
> p’

pivot

< p”

p”

> p”

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

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

78

55 70 65 68 61

100 93 78 98 81 84

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

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;
}


This example
uses int’s
to keep things
simple!

int left, right;
int pivot_item;
pivot = left = low;
right = high;
Any item will do as the pivot,
while ( left < right ) { choose the leftmost one!
}
23 12 15 38 42 18 36 29 27
a[low] = a[right];
return right;
low
high
}

int left, right;
int pivot_item;
pivot = left = low;
Set left and right markers
right = high;
while( a[left] <= pivot_item ) left++; right
left
if (23 12 right 38 SWAP(a,left,right);27
left < 15
) 42 18 36 29
}
a[low]low a[right];
=
high
pivot: 23
return right;
}

int left, right;
int pivot_item;
pivot = left = low;
right = high;

Move the markers
until they cross over

left
}
a[low] = a[right]; 15 38 42 18 36 29
23 12
return right;
low
pivot: 23
}

right
27
high

int left, right;
int pivot_item;
pivot = left = low;
right = high;

Move the left pointer while
it points to items <= pivot
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

int left, right;
int pivot_item;
pivot = left = low;
right = high;

Swap the two items
on the wrong side of the pivot



}
left
right
a[low] = a[right];
23 12 15 38 42
return right;
pivot: 23
}
low
high

int left, right;
int pivot_item;
pivot = left = low;
right = high;

left and right
have swapped over,
so stop



}
right
left
a[low] = a[right];
23 12 15 18 42
return right;
}
low
high
pivot: 23

int left, right;
int pivot_item;
pivot = left = low;
right = high;
right

left

23

while( 18 42 38 36 29 27
12 15 a[left] <= pivot_item ) left++;

low
high
pivot: 23
}
a[low] = a[right];
Finally, swap the pivot
and right
return right;
}

int left, right;
int pivot_item;
pivot = left = low;
right = high;
right


18

pivot: 23

while( 23 42 38 36 29 27
12 15 a[left] <= pivot_item ) left++;

low
high
}
a[low] = a[right];
Return the position
a[right] = pivot_item; of the pivot
return right;
}

Quicksort - Conquer
pivot

pivot: 23
18

12

15

Recursively
sort left half

23

42

38

36

29

27

Recursively
sort right half

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!

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

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

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

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

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

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

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

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

8

8

10

10

8

1

12

8

5

2

10
8

10

3

10
5

12
8

12
5

10
8

5

4

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

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

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

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

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

have lost the heap property

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

have lost the heap property —but it doesn’t, because
it’s a leaf

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

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;
}

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)

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)

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!

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

Analysis II


Here’s the rest of the algorithm:
}

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

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)

Analysis IV

Here’s the algorithm again:
heapify the array;
}

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

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