Question

11) a) Using RSA to encode with p-5,q-7 and c-7, what information do we make publicly available? b) What is d, the private key? 12) Describe how, using RSA, Bob knows that, as claimed, it's really Alice sending the message. 13) Using non-randomized Select(A, n, i) to find the 4th largest element in array A, with A as I first pivot 8 11 14 09 4 25 33 98 5 22 63 54 2 111 45 23 19 728 16 116 10 I put the first pivot in the right place for you ow the array in sequence, one in each row, after each recursive call of the algorithm. You must use blanks (or s out) for the parts of the array will get ignored in th e next recursive call. You should not need as many rows as provided. 10 14) For the Blum-Floyd-Pratt-Rivest-Tarjan selection algorithm, BFPRT(A, n, i), in which we first split the array columns of 5, let's say we want the median of an array of 115 integers, so we want BFPRT(A, 115, 53). Array A. rearranged according to the algorithm

Answer 1

(Note: The figure contains 4 questions: 11, 12, 13, and 14. Question number 11 and 12 require information from some previous context, while Question 14 is incomplete. So, I'm assuming that the student intends to ask only Question 13. So, the following answer belongs to Question 13.)

The question asks to find the 4th largest element of the given array. The length of the array is 22. This means that we have to stop when we find the element with index 18 (considering the array starts from 0), as it will be the 4th largest element.

We are assuming that the pivot is always the last element.

The array is:

8

11

14

0

9

4

25

33

98

5

22

63

54

2

111

45

23

19

728

16

116

10

Now, 10 is the pivot. So, all the elements less than 10 will come to the left side of 10 while all elements greater than 10 will come to its right side. Thus, the array will be:

8

0

9

4

5

2

10

11

14

25

33

98

22

63

54

111

45

23

19

728

16

116

Now, the index of 10 is 6, which is less than 18. Therefore, we will ignore the part of the array that is on the left side of 10 and apply the algorithm to the elements on the right side of 10. In this case, 116 will be the pivot. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

25

33

98

22

63

54

111

45

23

19

16

116

728

The index of 116 is 20, which is greater than 18. Therefore, we will now apply the algorithm to the elements on the left side of 116. The pivot will be 16. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

25

33

98

22

63

54

111

45

23

19

116

728

The index of 16 is 9, which is less than 18. Therefore, we will now apply the algorithm to the elements on the right side of 16. The pivot will be 19. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

25

33

98

22

63

54

111

45

23

116

728

The index of 19 is 10, which is again less than 18. Therefore, we will now apply the algorithm to the elements on the right side of 19. The pivot will be 23. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

22

23

25

33

98

63

54

111

45

116

728

The index of 23 is 12, which is again less than 18. Therefore, we will now apply the algorithm to the elements on the right side of 23. The pivot will be 45. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

22

23

25

33

45

98

63

54

111

116

728

The index of 45 is 15, which is again less than 18. Therefore, we will now apply the algorithm to the elements on the right side of 45. The pivot will be 111. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

22

23

25

33

45

98

63

54

111

116

728

The index of 111 is 19, which is greater than 18. Therefore, we will now apply the algorithm to the elements on the left side of 111. The pivot will be 54. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

22

23

25

33

45

54

63

98

111

116

728

The index of 54 is 16, which is less than 18. Therefore, we will now apply the algorithm to the elements on the right side of 54. The pivot will be 98. The array, after applying the algorithm, would look like:

8

0

9

4

5

2

10

11

14

16

19

22

23

25

33

45

54

63

98

111

116

728

The index of 98 is 18. This means that we have found our element. We will stop the algorithm here.

This means that the 4th largest element in array A is 98.

This is how the given question will be solved.

Hope it helped. I tried my best to explain the solution in the best manner possible. If you have any doubts or queries, please feel free to ask in the comments section. If it helped in any way, please consider giving positive ratings.