Module 2_ Divide and Conquer Approach.pptx

Module 2:
Divide and Conquer Approach
CSC402.2: Describe, apply and analyze the complexity of divide and conquer strategy.
Approx Weightage: 20-25 marks

Contents
● General method
● Analysis of Binary search.
● Analysis of Merge sort
● Analysis of Quick sort
● Analysis of Finding minimum and maximum algorithms

Divide and Conquer
If we can break a single big problem into smaller sub-problems, solve the smaller
subproblems and combine their solutions to find the solution for the original big
problem, it becomes easier to solve the whole problem.
The concept of Divide and Conquer involves three steps:
1. Divide the problem into multiple small problems.
2. Conquer the subproblems by solving them. The idea is to break down the
problem into atomic subproblems, where they are actually solved.
3. Combine the solutions of the subproblems to find the solution of the actual
problem.

Divide and Conquer
The concept of Divide and Conquer involves three steps:
1. Divide the problem into multiple subproblems.
2. Solve the Sub Problems. The idea is to break down the
problem into atomic subproblems, where they are actually
solved.
3. Combine the solutions of the subproblems to find the
solution of the actual problem.

Divide and Conquer
Divide/Break
This step involves breaking the problem into smaller sub-problems. Sub-problems should represent a part
of the original problem. This step generally takes a recursive approach to divide the problem until no sub-
problem is further divisible. At this stage, sub-problems become atomic in size but still represent some
part of the actual problem.
Conquer/Solve
This step receives a lot of smaller subproblems to be solved. Generally, at this level, the problems are
considered 'solved' on their own.
Merge/Combine
When the smaller subproblems are solved, this stage recursively combines them until they formulate a
solution of the original problem. This algorithmic approach works recursively and conquer & merge steps
works so close that they appear as one.

Divide And Conquer algorithm
DAC(a, i, j)
{
if(small(a, i, j))
return(Solution(a, i, j))
else
mid = divide(a, i, j) // f1(n)
b = DAC(a, i, mid) // T(n/2)
c = DAC(a, mid+1, j) // T(n/2)
d = combine(b, c) // f2(n)
return(d)
}

Recurrence Relation for DAC algorithm
This is a recurrence relation for the above program.
O(1) if n is small
T(n) =
f1(n) + 2T(n/2) + f2(n)

Time Complexity of Divide and Conquer Algorithm
T(n) = aT(n/b) + f(n),
where,
n = size of input
a = number of subproblems in the recursion
n/b = size of each subproblem. All subproblems are assumed to have the
same size.
f(n) = cost of the work done outside the recursive call, which includes
the cost of dividing the problem and cost of merging the solutions

Pros and cons of Divide and Conquer Approach
● Divide and conquer approach supports parallelism as sub-
problems are independent.
● Hence, an algorithm, which is designed using this technique,
can run on the multiprocessor system or in different machines
simultaneously.
● In this approach, most of the algorithms are designed using
recursion, hence memory management is very high.
● For recursive function stack is used, where function state needs
to be stored.

Binary Search
https://binary-search-visualization.netlify.app/

Binary Search
Binary Search algorithm is used extensively in computer
science and mathematics that locates a specific element in
a sorted dataset. It works by repeatedly dividing the dataset
in half and comparing the target value with the middle
value until the target value is discovered or determined to
be absent.

Binary_Search(a, lower_bound, upper_bound, val) // 'a' is the given array, 'lower_bound' is the index of the first array
element, 'upper_bound' is the index of the last array element, 'val' is the value to search
1. Step 1: set beg = lower_bound, end = upper_bound, pos = - 1
2. Step 2: repeat steps 3 and 4 while beg <=end
3. Step 3: set mid = (beg + end)/2
4. Step 4: if a[mid] = val
5. set pos = mid
6. print pos
7. go to step 6
8. else if a[mid] > val
9. set end = mid - 1
10. else
11. set beg = mid + 1
12. [end of if]
13. [end of loop]
14. Step 5: if pos = -1
15. print "value is not present in the array"
16. [end of if]
17. Step 6: exit

Complexity Analysis of Binary Search
● Since analysis starts with the middle element, in the best
case, it’s possible that the middle element of the array
itself is the target element.
● Otherwise, the length of the array is halved.
● The equation T(n)= T(n/2)+1 is known as the recurrence
relation for binary search.

Merge Sort
The merge sort working rule involves the following steps:
1. Divide the unsorted array into subarray, each containing a single element.
2. Take adjacent pairs of two single-element array and merge them to form an
array of 2 elements.
3. Repeat the process till a single sorted array is obtained.

Merge Sort
● An array of Size ‘N’ is divided into two parts ‘N/2’ size of each.
● Then the arrays are further divided till we reach a single
element.
● The base case here is reaching one single element.
● When the base case is hit, we start merging the left part and the
right part and we get a sorted array at the end.
● Merge sort repeatedly breaks down an array into several
subarrays until each subarray consists of a single element and
merging those subarrays in a manner that results in a sorted
array.

Complexity Analysis of Merge Sort
Merge sort repeatedly divides the array into two equally sized parts.
Thus merge sort time complexity depends on the number of division stages.
The number of division stages is log2 n.
On each merge stage, n elements are merged.
● Step 1 - n×1
● Step 2 - n/2×2
● Step 3 - n/4×4
Merge Sort time complexity is calculated using time per division stage. Since the merge process has
linear time complexity, for n elements there will be n∗log 2 n division and merge stages.
Hence, regardless of the arrangement, the time complexity of Merge Sort is O(nlogn)

Analysis of Merge Sort
Let T (n) be the total time taken by the Merge Sort algorithm.

Analysis of Merge Sort Time Complexity
Best Case Time Complexity of Merge Sort
The best case scenario occurs when the elements are already sorted in ascending order. If two
sorted arrays of size n need to be merged, the minimum number of comparisons will be n. This
happens when all elements of the first array are less than the elements of the second array. The
best case time complexity of merge sort is O(n∗logn).
Average Case Time Complexity of Merge Sort
The average case scenario occurs when the elements are jumbled (neither in ascending or
descending order). This depends on the number of comparisons. The average case time complexity
of merge sort is O(n∗logn).
Worst Case Time Complexity of Merge Sort
The worst-case scenario occurs when the given array is sorted in descending order leading to the
maximum number of comparisons. In this case, for two sorted arrays of size n, the minimum
number of comparisons will be 2n. The worst-case time complexity of merge sort is O(n∗logn).

• Quick sort algorithm:
• Is one of the most efficient sorting algorithms
• Is based on the divide and conquer approach
• Successively divides the problem into smaller parts until the
problems become so small that they can be directly solved
Sorting Data by Using Quick Sort

• In quick sort algorithm, you:
• Select an element from the list called as pivot.
• Partition the list into two parts such that:
• All the elements towards the left end of the list are smaller than
the pivot.
• All the elements towards the right end of the list are greater than
the pivot.
• Store the pivot at its correct position between the two parts of
the list.
• You repeat this process for each of the two sublists created
after partitioning.
• This process continues until one element is left in each
sublist.
Implementing Quick Sort Algorithm

• To understand the implementation of quick sort algorithm,
consider an unsorted list of numbers stored in an array.
Implementing Quick Sort Algorithm (Contd.)
arr
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7

• Let us sort this unsorted list.
arr
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7

• Select a Pivot.
arr
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7
Pivot

• Start from the left end of the list (at index 1).
• Move in the left to right direction.
• Search for the first element that is greater than the pivot
value.
arr
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7
Pivot
Greater element
Greater Value

• Start from the right end of the list.
• Move in the right to left direction.
• Search for the first element that is smaller than or equal to
the pivot value.
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7
Smaller element
Smaller Value
arr
Pivot
Greater Value

• Interchange the greater value with smaller value.
arr
2
1
0 4
3
55 46 38
28 16 89 83 30
5 6 7
Pivot
Greater Value Smaller Value
Swap
55
16

• Continue the search for an element greater than the pivot.
• Start from arr[2] and move in the left to right direction.
value.
arr
2
1
0 4
3
16 46 38
28 55 89 83 30
5 6 7
Pivot
Greater element
Greater Value

• Continue the search for an element smaller than the pivot.
• Start from arr[3] and move in the right to left direction.
the pivot value.
2
1
0 4
3
16 46 38
28 55 89 83 30
5 6 7
Pivot
Smaller element
Smaller Value
Greater Value
arr

• The smaller value is on the left hand side of the greater
value.
• Values remain same.
2
1
0 4
3
16 46 38
28 55 89 83 30
5 6 7
Pivot
Smaller Value
Greater Value
arr

• List is now partitioned into two sublists.
• List 1 contains all values less than or equal to the pivot.
• List 2 contains all the values greater than the pivot.
2
1
0 4
3
16 46 38 55 89 83 30
5 6 7
Pivot
28 16
List 1 List 2
55 89 83 30
46 38
arr
2
1
0 4
3 5 6 7
28

• Replace the pivot value with the last element of List 1.
• The pivot value, 28 is now placed at its correct position in
the list.
arr 16
List 1 List 2
55 89 83 30
38
Swap
28
2
1
0 4
3 5 6 7
46
28
16

• Truncate the last element, that is, pivot from List 1.
List 2
55 89 83 30
38
2 4
3 5 6 7
arr
List 1
1
0
16 28
16
16
16
0
46

• List 1 has only one element.
• Therefore, no sorting required.
arr 16
List 1 List 2
55 89 83 30
38
16
16
0 2 4
3 5 6 7
46

• Sort the second list, List 2.
16 55 89 83 30
38
16
16
0 2 4
3 5 6 7
46
List 1 List 2
arr

• Select a pivot.
• The pivot in this case will be arr[2], that is, 46.
16 55 89 83 30
38
16
16
0
Pivot
2 4
3 5 6 7
46
List 1 List 2
arr

16 55 89 83 30
38
16
16
0
Pivot
• Start from the left end of the list (at index 3).
• Move in the left to right direction.
value. Greater element
Greater Value
2 4
3 5 6 7
46
List 1 List 2
arr

16 55 89 83 30
38
16
16
0
Pivot
Greater Value
• Start from the right end of the list (at index 7).
• Move in the right to left direction.
the pivot value. Smaller element
Smaller Value
2 4
3 5 6 7
46
List 1 List 2
arr

16 55 89 83 30
38
16
16
0
Pivot
Greater Value Smaller Value
• Interchange the greater value with smaller value.
Swap
30 55
2 4
3 5 6 7
46
List 1 List 2
arr

16 89 83
38
16
16
0
Pivot
• Continue the search for an element greater than the pivot.
• Start from arr[5] and move in the left to right direction.
value. Greater element
Greater Value
2 4
3 5 6 7
46
List 1 List 2
arr 30 55

16 89 83
38
16
16
0
Pivot
• Continue the search for an element smaller than the pivot.
• Start from arr[6] and move in the right to left direction.
• Search for the first element that is smaller than the pivot
value.
Greater Value
Smaller element
Smaller Value
2 4
3 5 6 7
46
List 1 List 2
arr 30 55

16 89 83
38
16
16
0
Pivot
Greater Value
Smaller Value
• The smaller value is on the left hand side of the greater
value.
• Values remain same.
2 4
3 5 6 7
46
List 1 List 2
arr 30 55

• Divide the list into two sublists.
• Sublist 1 contains all values less than or equal to the pivot.
• Sublist 2 contains all the values greater than the pivot.
2 4
3 5 6 7
83
89 55
38
46 30
16 28
1
0
arr

• Replace the pivot value with the last element of Sublist 1.
• The pivot value, 46 is now placed at its correct position in
the list.
• This process is repeated until all elements reach their
correct position.
arr
arr
2 4
3 5 6 7
83
89 55
38
46 30
16 28
1
0
Sublist 1 Sublist 2
Swap
46
30

QuickSort(low,high)
1. If (low > high):
a. Return
2. Set pivot = arr[low]
3. Set i = low + 1
4. Set j = high
5. Repeat step 6 until i > high or arr[i] > pivot // Search for an
element greater than pivot
6. Increment i by 1
7. Repeat step 8 until j < low or arr[j] < pivot // Search for an
element smaller than pivot
8. Decrement j by 1
9. If i < j: // If greater element is on the left of smaller element
a. Swap arr[i] with arr[j]

10. If i <= j:
a. Go to step 5 // Continue the search
11. If low < j:
a. Swap arr[low] with arr[j] // Swap pivot with last element in
// first part of the list
12. QuickSort(low, j – 1) // Apply quicksort on list left to
pivot
13. QuickSort(j + 1, high) // Apply quicksort on list right to
pivot

• The total time taken by this sorting algorithm depends on
the position of the pivot value.
• The worst case occurs when the list is already sorted.
• If the first element is chosen as the pivot, it leads to a worst
case efficiency of O(n2).
• If you select the median of all values as the pivot, the
efficiency would be O(n log n).
Determining the Efficiency of Quick Sort Algorithm

• Merge sort algorithm:
• Is based on the divide and conquer approach
• Divides the list into two sublists of sizes as nearly equal as
possible
• Sorts the two sublists separately by using merge sort
• Merges the sorted sublists into one single list
Sorting Data by Using Merge Sort

• To understand the implementation of merge sort algorithm,
consider an unsorted list of numbers stored in an array.
Implementing Merge Sort Algorithm
arr
2
1
0 4
3
10 30 76
53 3 57 24
5 6

• Let us sort this unsorted list.
Implementing Merge Sort Algorithm (Contd.)
arr
2
1
0 4
3
10 30 76
53 3 57 24
5 6

• The first step to sort data by using merge sort is to split the
list into two parts.
arr
2
1
0 4
3
10 30 76
53 3 57 24
5 6

• The first step to sort data by using merge sort is to split the
list into two parts.
arr
2
1
0 4
3 5 6
53 10 30 76 3 57 24

• The list has odd number of elements, therefore, the left
sublist is longer than the right sublist by one entry.
arr
2
1
0 4
3 5 6
53 10 30 76 3 57 24

• Further divide the two sublists into nearly equal parts.
arr
2
1
0 4
3 5 6
53 10 30 76 3 57 24
1
0
53 10 30 76
2
1
0 4
3 5 6
53 10 30 76 3 57 24

• Further divide the sublists.
arr
1
0
53 10 30 76
2
1
0 4
3 5 6
53 10 30 76 3 57 24
2
1
0 4
3 5 6
24
57
3
76
30
10
53

• There is a single element left in each sublist.
• Sublists with one element require no sorting.
arr
arr
2
1
0 4
3 5 6
24
57
3
76
30
10
53

• Start merging the sublists to obtain a sorted list.
arr
2
1
0 4
3 5 6
24
57
3
76
30
10
53
0 5 6
57
2
1 4
3
10 53 30 76 3 57 24

• Further merge the sublists.
arr
0 5 6
57
2
1 4
3
10 53 30 76 3 57 24
2
1
0 4
3 5 6
10 30 53 76 3 24 57

• Again, merge the sublists.
arr
2
1
0 4
3 5 6
10 30 53 76 3 24 57
2
0 3 5
1 4
10 24 30
3 53 57 76
6

• The list is now sorted.
arr
2
0 3 5
1 4
10 24 30
3 53 57 76
6

• Write an algorithm to implement merge sort:
MergeSort(low,high)
1. If (low >= high):
a. Return
2. Set mid = (low + high)/2
3. Divide the list into two sublists of nearly equal lengths, and
sort each sublist by using merge sort.
The steps to do this are as
follows:
a. MergeSort(low, mid
b. MergeSort(mid + 1, high)
4. Merge the two sorted sublists:
a. Set i = low
b. Set j = mid + 1

i. If (arr[i] <= arr[j])
Store arr[i] at index k
in array B
Increment i by 1
Else
Store arr[j] at index k in array B
Increment j by 1
ii. Increment k by 1
e. Repeat until j > high: // If there are still some
elements in the
// second sublist append them to the new list
i. Store arr[j] at index k in array B
ii. Increment j by 1
iii. Increment k by 1

• To sort the list by using merge sort algorithm, you need to
recursively divide the list into two nearly equal sublists until
each sublist contains only one element.
• To divide the list into sublists of size one requires log n
passes.
• In each pass, a maximum of n comparisons are performed.
• Therefore, the total number of comparisons will be a
maximum of n × log n.
• The efficiency of merge sort is equal to O(n log n)
• There is no distinction between best, average, and worst
case efficiencies of merge sort because all of them require
the same amount of time.
Determining the Efficiency of Merge Sort Algorithm

• Which algorithm uses the following procedure to sort a given
list of elements?
1. Select an element from the list called a pivot.
2. Partition the list into two parts such that one part contains
elements lesser than the pivot, and the other part contains
elements greater than the pivot.
3. Place the pivot at its correct position between the two lists.
4. Sort the two parts of the list using the same algorithm.
Just a minute
• Answer:
• Quick sort

• On which algorithm design technique are quick sort and
merge sort based?
Just a minute
• Answer:
• Quick sort and merge sort are based on the divide and
conquer technique.

• Linear Search:
• Is the simplest searching method
• Is also referred to as sequential search
• Involves comparing the items sequentially with the elements in
the list
Performing Linear Search

• The linear search would begin by comparing the required
element with the first element in the list.
• If the values do not match:
• The required element is compared with the second element in
the list.
• If the values still do not match:
• The required element is compared with the third element in the
list.
• This process continues, until:
• The required element is found or the end of the list is reached.
Implementing Linear Search

• Write an algorithm to search for a given employee ID in a list
of employee records by using linear search algorithm:
1. Read the employee ID to be searched
2. Set i = 0
3. Repeat step 4 until i > n or arr[i] = employee ID
4. Increment i by 1
5. If i > n:
Display “Not Found”
Else
Display “Found”
Implementing Linear Search (Contd.)

• The efficiency of a searching algorithm is determined by the
running time of the algorithm.
• In the best case scenario:
• The element is found at the first position in the list.
• The number of comparisons in this case is 1.
• The best case efficiency of linear search is therefore, O(1).
• In the worst case scenario:
• The element is found at the last position of the list or does not
exists in the list.
• The number of comparisons in this case is equal to the number
of elements.
• The worst case efficiency of linear search is therefore, O(n).
Determining the Efficiency of Linear Search

• In the average case scenario:
• The number of comparisons for linear search can be
determined by finding the average of the number of
comparisons in the best and worst case.
• The average case efficiency of linear search is 1/2(n + 1).
Determining the Efficiency of Linear Search (Contd.)

• You have to apply linear search to search for an element in
an array containing 5,000 elements. If, at the end of the
search, you find that the element is not present in the array,
how many comparisons you would have made to search the
required element in the given list?
Just a minute
• Answer:
• 5,000

• Problem Statement:
• Write a program to search a given number in an array that
contains a maximum of 20 numbers by using the linear search
algorithm. If there are more than one occurrences of the
element to be searched, then the program should display the
position of the first occurrence. The program should also
display the total number of comparisons made.
Activity: Performing Linear Search

• Binary search algorithm:
• Is used for searching large lists
• Searches the element in very few comparisons
• Can be used only if the list to be searched is sorted
Performing Binary Search

• Consider an example.
You have to search for the name Steve in a telephone
directory that is sorted alphabetically.
• To search the name Steve by using binary search algorithm:
• You open the telephone directory at the middle to determine which
half contains the name.
• Open that half at the middle to determine which quarter of the
directory contains the name.
• Repeat this process until the name Steve is not found.
• Binary search reduces the number of pages to be searched by
half each time.
Implementing Binary Search

• Consider a list of 9 elements in a sorted array.
Implementing Binary Search (Contd.)
arr
2
1
0 4
3
13 17 19
9 25 29 39
5 6
40 47
7 8

• You have to search an element 13 in the given list.
arr
2
1
0 4
3
13 17 19
9 25 29 39
5 6
40 47
7 8

• Determine the index of the middlemost element in the list:
Mid = (Lower bound + Upper bound)/2
= (0 + 8)/2
= 4
arr
2
1
0 4
3
13 17 19
9 25 29 39
5 6
40 47
7 8
Middle element
Lower bound Upper bound

• 13 is not equal to the middle element, therefore, again
divide the list into two halves:
Mid = (Lower bound + Upper bound)/2
= (0 + 3)/2
= 1
arr
2
1
0 4
3
13 17 19
9 25 29 39
5 6
40 47
7 8
Middle element
Upper bound
Middle element

• 13 is equal to middle element.
• Element found at index 1.
arr
2
1
0 4
3
13 17 19
9 25 29 39
5 6
40 47
7 8
Element found

• Write an algorithm to implement binary search algorithm.
1. Accept the element to be searched
2. Set lowerbound = 0
3. Set upperbound = n – 1
4. Set mid = (lowerbound + upperbound)/2
5. If arr[mid] = desired element:
a. Display “Found”
b. Go to step 10
6. If desired element < arr[mid]:
a. Set upperbound = mid – 1

7. If desired element > arr[mid]:
a. Set lowerbound = mid + 1
8. If lowerbound <= upperbound:
a. Go to step 4
9. Display “Not Found”
10.Exit

• In binary search, with every step, the search area is reduced
to half.
• In the best case scenario, the element to be search is found
at the middlemost position of the list:
• The number of comparisons in this case is 1.
• In the worst case scenario, the element is not found in the
list:
• After the first bisection, the search space is reduced to n/2
elements, where n is the number of elements in the original list.
• After the second bisection, the search space is reduced to n/4
elements, that is, n/22 elements.
• After ith bisections, the number of comparisons would be n/2i
Determining the Efficiency of Binary Search

• In ___________ search algorithm, you begin at one end of
the list and scan the list until the desired item is found or the
end of the list is reached.
Just a minute
• Answer:
• linear

• To implement __________ search algorithm, the list should
be sorted.
Just a minute
• Answer:
• binary

• Problem Statement:
• Write a program to search a number in an array that contains a
maximum of 20 elements by using binary search. Assume that
the array elements are entered in ascending order. If the
number to be searched is present at more than one location in
the array, the search should stop when one match is found.
The program should also display the total number of
comparisons made.
Activity: Performing Binary Search

• In this session, you learned that:
• Quick sort and merge sort algorithms are based on the divide
and conquer technique.
• To sort a list of items by using the quick sort algorithm, you
need to:
• Select a pivot value.
• Partition the list into two sublists such that one sublist contains
all items less than the pivot, and the second sublist contains all
items greater than the pivot.
• Place the pivot at its correct position between the two sublists.
• Sort the two sublists by using quick sort.
Summary

• The total time taken by the quick sort algorithm depends on
the position of the pivot value and the initial ordering of
elements.
• The worst case efficiency of the quick sort algorithm is O(n2).
• The best case efficiency of the quick sort algorithm is O(n log
n).
• To sort a list of items by using merge sort, you need to:
• Divide the list into two sublists.
• Sort each sublist by using merge sort.
• Merge the two sorted sublists.
• The merge sort algorithm has an efficiency of O(n log n).
Summary (Contd.)

• The best case efficiency of linear search is O(1) and the worst
case efficiency of linear search is O(n).
• To apply binary search algorithm, you should ensure that the
list to be searched is sorted.
• The best case efficiency of binary search is O(1) and the
worst case efficiency of binary search is O(log n).
Summary (Contd.)

1. Given the array A = [1, 2, 5, 3, 8, 10], divide it in two subarrays A1 = [1, 2, 5]
and A2 = [3, 8, 10].
2. The min and max of each subarray will be
● A1: min = 1, max = 5
● A2: min = 3, max = 10
3. The min of A will be min of [1, 3] (the min of A1 and A2)
4. The max of A will be max of [5, 10] (the max of A1 and A2)

Module 2_ Divide and Conquer Approach.pptx

Module 2_ Divide and Conquer Approach.pptx

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