1. Binary Search
A binary search algorithm finds the position of a specified value within a sorted array. Binary search is a technique for searching an ordered list in which middle item is checked first and - based on that comparison – half of the data is discarded. The same procedure is then applied to the remaining half until a match is found or there are no more items left.
Course Name: Design and Analysis of Algorithms Level(UG/PG): UG
Author : Phani Swathi Chitta
Mentor : Aruna Adil
*The contents in this ppt are licensed under Creative Commons Attribution-NonCommercial-ShareAlike 2.5 India license

2. Learning Objectives
After interacting with this Learning Object, the learner will be able to:
Explain how to find an element in an array using “Binary Search”

3. 1 2 3 4 5 Definitions of the components/Keywords:
Binary Search is also called half-interval search algorithm.
The binary search is best suitable search algorithm for searching a particular value in sorted arrays.
Every iteration eliminates half of the remaining possibilities because the array is sorted. This makes binary searches very efficient - even for large arrays.
Desired element / Item / Search value: An element that is being searched for in the array. 2 3 4 5 3

4. 1 2 3 4 5 Definitions of the components/Keywords:
Process to find the desired element using Binary search:
Length of the array is defined.
Get the middle element -----> floor[(start + end)/2]
start: Index of the first element
end: Index of the last element
If the middle element equals to the searched value, the algorithm stops;
otherwise, two cases are possible:
If searched value is less than the middle element, restrict the search to the first half (start to mid -1) of the list
If searched value is greater than the middle element, restrict the search to the second half (mid + 1 to end) of the list
Repeat steps 2 and 3 until
searched element is found or
sub array has no elements. This implies element is not found in the array 2 3 4 5

5. 1 2 3 4 5 Definitions of the components/Keywords:
Algorithm to implement binary search:
int binarySearch(int arr[ ], int value, int start, int end) {
      while (start <= end)
            int middle = floor(start + end) / 2;
            if (arr[middle] == value)
                  return middle;
            else if (arr[middle] > value)
                  end = middle - 1;
            else
                  start = middle + 1;
      }
      return key_not_found; } 1 2 3 4 5

6. 1 2 3 4 5 Definitions of the components/Keywords:
The advantage of binary search is its speed of finding the desired element in large arrays when compared to sequential search.
The main limitation of binary search is that it can be done only on sorted arrays.
The complexity of any searching method is determined from the number of comparisons performed among the elements of the list in order to find the element.
The time required for a search operation depends on the complexity of the searching algorithm.
The complexity of the binary search algorithm is O(log n). 2 3 4 5

7. Master Layout 1
Give START, PAUSE and STOP buttons
Give 2 radio buttons
Element found
Element not found
Give a STEPPER button that allows the user to follow the simulation procedure step by step. After every step the simulation pauses until the STEPPER button is pressed
Give a text area to display the status of the simulation
Give a slider bar to control the speed of animation
Simulation Area
Index 2 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 3
Array or list
Fig. A 4
Current Indices: Give the value of the Indices i, j and mid at that particular time
Legend: Array Desired element Desired element Match Current Index 5
** For animator: All digits written in white in the array are “elements” or “values at index n “(n refers to index number ex. value at index 4 = 14).

8. 3 Step 1: 1 2 4 5 Case – 1 (Element Found) i=0 j=7 1 2 3 4 5 6 7 8 141 2 3 4 5 6 7
start 8 14 21 25 48 52 60 76
end 2 3
start: i=0
end: j=7
mid: FLOOR ([0+7]/2) = 3
Legend: Array
Desired element
Desired element Match
Current Index 21
Search Value: 4
Instruction for the animator
Text to be displayed in the working area (DT)
When the user clicks start, show the figure in master layout without labeling.
Display array with its values inside and numbers above it.
Also show the search value till the animation is over.
Then show the text and arrows in red as shown in figure.
Show the violet box with values of start, end and calculation of mid value
The text in DT should appear in parallel to the figure
Array of 8 elements with indices
The search value: 21
Initialize start: i=0 and end : j=7
Compare i and j -- i<= j - 0<=7 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

9. Step 2: 1
i=0
mid=3
j=7 1 2 3 4 5 6 7
start 8 14 21 25 48 52 60 76
end 2 21
Comparing 21 and 25 3
start: i=0
end: j=7
mid: 3
Legend: Array
Desired element
Desired element Match
Current Index 21
Search Value: 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 21 and 25” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If less then display the only blue boxes before mid
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([0+7]/2) = 3
The desired element to search for in the array is 21
21 is compared with the element at mid
21 ≠ 25 and 21 < 25
Therefore the search for element 21 is done only in the first half . 5

10. Step 3: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76
i=0
j=2 2 1 2
start: i=0
end: j= mid -1 = 3-1= 2
mid: FLOOR ([0+2]/2) = 1
start 8 14 21
end 3 21
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
Compare i and j -- i<= j - 0<=2 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

11. Step 4: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76
i=0
mid=1
j=2
start: i=0
end: j=2
mid: 1 2 1 2 8 14 21
start
end 21 21
Search Value: 3
Legend: Array
Desired element
Desired element Match
Current Index
Comparing 21 and 14 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 21 and 14” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If greater then display the only blue boxes after mid
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([0+2]/2) = 1
The desired element to search for in the array is 21
21 is compared with the element at mid
21 ≠ 14 and 21 > 14
Therefore the search for element 21 is done only in the second half . 5

12. Step 5: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 1 2 8 14 21
start: i=mid +1 = 1+1 = 2
end: j=2
mid: FLOOR ([2+2]/2) = 2 2
i=2
j=2 2
start 21
end 3 21
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
Compare i and j -- i<= j - 0<=2 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

13. Step 6: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 1 2
start: i=2
end: j=2
mid: 2 8 14 21 2
mid=2
i=2
j=2 2 21
Search Value: 3
start 21
end
Legend: Array
Desired element
Desired element Match
Current Index 21
Comparing 21 and 21
Element found at location 2 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 21 and 21” must appear.
If the values in those 2 boxes match, change the color of both the boxes to green.
Display the location of desired element. With the text “ Element found at location 2”
Always update the value in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([2+2]/2) = 2
The desired element to search for in the array is 21
21 is compared with the element at mid
21 = 21
Therefore the search for element 21 is done
Element found at location 2 . 5

14. 3 Step 7: 1 2 4 5 Case – 2 (Element not found) i=0 j=7 1 2 3 4 5 6 7 81 2 3 4 5 6 7
start 8 14 21 25 48 52 60 76
end 2
start: i=0
end: j=7
mid: FLOOR ([0+7]/2) = 3 3 65
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
When the user clicks start, show the figure in master layout without labeling.
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Show the violet box with values of start, end and calculation of mid value
The text in DT should appear in parallel to the figure
Array of 8 elements with indices
The search value: 65
Initialize start: i=0 and end : j=7
Compare i and j -- i<= j - 0<=7 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

15. Step 8: 1
i=0
mid=3
j=7 1 2 3 4 5 6 7
start 8 14 21 25 48 52 60 76
end 2
start: i=0
end: j=7
mid: 3 3 65
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Always update the value in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([0+7/2) = 3 5

16. Step 9: 10: 1
i=0
mid=3
j=7 1 2 3 4 5 6 7
start 8 14 21 25 48 52 60 76
end 2 65
start: i=0
end: j=7
mid: 3
Comparing 65 and 25 3 65
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 65 and 25” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If greater then display the only blue boxes after mid
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
The desired element to search for in the array is 65
65 is compared with the element at mid
65 ≠ 25 and 65 > 25
Therefore the search for element 65 is done only in the second half . 5

17. Step 10: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76
i=4
j=7 2 4 5 6 7
start 48 52 60 76
end
start: i=mid +1 = 3+1 = 4
end: j= 7
mid: FLOOR ([4+7]/2) = 5 3 65
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
Compare i and j -- i<= j - 4<=7 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

18. Step 11: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76
i=4
mid=5
j=7 2 4 5 6 7
start 48 52 60 76
end
start: i=4
end: j= 7
mid: 5 3 65
Search Value:
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Always update the value in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([4+7/2) = 5 5

19. Step 12: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76
i=4
mid=5
j=7 2
start: i=4
end: j= 7
mid: 5 4 5 6 7
start 48 52 60 76
end 65 3 65
Search Value:
Comparing 65 and 52
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 65 and 52” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If greater then display the only blue boxes after mid
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
The desired element to search for in the array is 65
65 is compared with the element at mid
65 ≠ 52 and 65 > 52
Therefore the search for element 65 is done only in the second half . 5

20. Step 13: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76
start: i=mid+1 = 5+1 = 6
end: j= 7
mid: FLOOR ([6+7]/2) = 6 2
i=6
j=7 6 7 3 65
Search Value:
start 60 76
end
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
Compare i and j -- i<= j - 6<=7 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

21. Step 14: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76 2
start: i=6
end: j= 7
mid: 6
mid=6
j=7
i=6 6 7 3 65
Search Value: 60 76
end
start
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Always update the value in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([6+7/2) = 6 5

22. Step 15: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76 2
start: i=6
end: j= 7
mid: 6
mid=6
j=7
i=6 6 7 3 60 76
end 65
start
Search Value: 65
Legend: Array
Desired element
Desired element Match
Current Index
Comparing 65 and 60 4
Instruction for the animator
Text to be displayed in the working area (DT)
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 65 and 60” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If greater then display the only blue boxes after mid
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
The desired element to search for in the array is 65
65 is compared with the element at mid
65 ≠ 60 and 65 > 60
Therefore the search for element 65 is done only in the second half . 5

23. Step 16: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76 2
start: i=mid+1 = 6+1 = 7
end: j= 7
mid: FLOOR ([7+7]/2) = 7 6 7 60 76 3
i=7
j=7 65
Search Value: 7
start 76
end
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Display array with its values inside and numbers above it.
Then show the text and arrows in red as shown in figure.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
Compare i and j -- i<= j - 7<=7 TRUE
Finding middle element
mid = FLOOR([start +end]/2) 5

24. Step 17: 1 1 2 3 4 5 6 7 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76 2 6 7 60 76
start: i=7
end: j= 7
mid: 7 3
mid=7
i=7
j=7 65 7
Search Value:
start 76
end
Legend: Array
Desired element
Desired element Match
Current Index 4
Instruction for the animator
Text to be displayed in the working area (DT)
Show the text and arrow for mid as shown in figure.
Always update the value in violet box accordingly.
The text in DT should appear in parallel to the figure
mid = FLOOR ([7+7/2) = 7 5

25. Step 18: 1 2 3 4 5 6 7 1 8 14 21 25 48 52 60 76 4 5 6 7 48 52 60 76
start: i=0
end: j= 7
mid: 7 2 6 7 60 76
mid=7 65
Search Value:
i=7
j=7 3
Legend: Array
Desired element
Desired element Match
Current Index 7
start 76
end 65
Comparing 65 and 76 4
Instruction for the animator
Text to be displayed in the working area (DT)
Then blink the orange box and blue box to which mid arrow is pointing to.
While blinking, a text box saying “ Comparing 65 and 76” must appear.
If the values in those 2 boxes doesn’t match, then see whether the number in orange box is greater or less than the number in blue box
If greater then display the only blue boxes after mid
Here this is the end of the array.
Always update the values in violet box accordingly.
The text in DT should appear in parallel to the figure
The desired element to search for in the array is 65
65 is compared with the element at mid
65 ≠ 76 and 65 > 76
Therefore the search for element 65 is done only in the second half .
So i=mid +1 =7+1 = 8
Compare i and j -- i<= j - 8<=7 FALSE
This implies array is completed.
Element not found in the array. 5

26. Test your understanding
Electrical Engineering
Slide 1
Slide 3
Slide
27-30
Slide 31
Introduction
Definitions
Analogy
Test your understanding
(questionnaire)‏
Lets Sum up (summary)‏
Want to know more…
(Further Reading)‏
Interactivity:
Try it yourself
Array Size :
Values in the array: :
Give a dropdown to select the array size from
Place an input box and an ENTER button to enter the values in the array.
After ENTER is pressed, show the array with indices on the top in the simulation area.
Give another input box to enter the desired element value
After the desired element value is entered, show the value in the simulation area.
After desired element value is shown, enable START button.
Once the animation is started, then enable PAUSE and RESET buttons.
Desired element Value :
In the input box the user will input all the numerical values that he/she wants to be in the array. 26
Credits

27. Questionnaire 1
Which of the following are TRUE about Binary search?
It is a sequential search
Its complexity is O(log n)
It is also called half- interval search
Answers:
a) I and II
b) Only II
c) II and III
d) I, II and III 2 3 4 5

28. Questionnaire 1
2. For a searching operation to be done using binary search, the array must be
Answers:
a) Sorted
b) Unsorted
c) either sorted or unsorted 2 3 4 5

29. 4 Questionnaire 1 2 3 5 3. The following statement is TRUE/ FALSE
Binary search is very efficient for large arrays.
Answers:
a) TRUE
b) FALSE 2 3 4 5

30. Questionnaire 1 4.
Binary search takes more comparisons than linear search to find an element
Time taken to find an element is more and is the main disadvantage in binary search
which of the above are TRUE?
Answers:
a) I and II
b) Only I
c) Only II
d)None 2 3 4 5

31. Links for further reading
Reference websites:
Books:
“Introduction to Algorithm” Thomas H. Coremen