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Discrete Mathematics and its Applications
Found in: Page 203
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Compare the number of comparisons used by the insertion sort and the binary insertion sort to sort the list 7, 4, 3, 8, 1, 5, 4, 2.

Hence, The number of comparisons in insertion sort is whereas the number of comparisons in binary insertion sort is and the given array 7,4,3,8,15,4,2 became 1,2,3,4,4,5,7,8 .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1:

  • first find the number of comparisons done by the insertion sort.
  • Then find the number of comparisons done by the binary sort.
  • After completion compare the number of comparisons done by both of these algorithms.
  • The list to be short is 7,4,3,8,1,5,4,2

Step 2:

In the case of insertion sort:

Case 1:

Compare the first two elements in this case.

Since 7 > 4 so 4 is inserted before the 7 .

Number of comparisons done: 1

4,7,3,8,1,5,4,2

Case 2:

Compare the third element with the elements before it.

Since 3 < 4 so inserted 3 before 4 .

Number of comparisons made: 1

3,4,7,8,1,5,4,2

Case 3:

Now, compare the fourth element in this case with the sorted arrays before it.

Since , 8>3 8>4and 8>7

so, insert 8 after 7 .

Number of comparisons made: 3

3,4,7,8,1,5,4,2

Case 4:

Now, compare the fifth element in this case with the sorted arrays before it.

Since 1 < 3 so insert before 3 .

Number of comparisons made: 1

1,3,4,7,8,5,4,2

Case 5:

Now, compare the sixth element in this case with the sorted arrays before it.

Since 5>1 5>3 5>4and 5<7

so, insert 5 between 4 and 7 .

Number of comparisons made: 4

1,3,4,5,7,8,4,2

Case 6

Now, compare the seventh element in this case with the sorted arrays before it.

Since 4>1 4>3and 4=4

So, inserted 4 after 3 and before the previous 4 .

Number of comparisons made: 3

1,3,4,4,5,7,8,2

Case 7:

Now, compare the eighth element in this case with the sorted arrays before it.

Since 2>1and 2<3

So, insert 2 between 1 and 3 .

Number of comparisons made: 2

1,2,3,4,4,5,7,8

Thus, the total number of comparisons in Insertion Sort is:

1+1+3+1+4+3+2 = 15

Step 3:

In the case of binary insertion sort:

Case 1:

First, compare the first two elements in this case.

Since 7 > 4 so is inserted before the 7 .

Number of comparisons done: 1

4,7,3,8,1,5,4,2

Case 2:

Compare the third element with the elements before it.

Since 3 < 4 so inserted 3 before 4 .

Number of comparisons made: 1

3,4,7,8,1,5,4,2

Case 3:

Now, compare the fourth element in this case with the sorted arrays before it.

Since 8 > 4

and 8 > 7

so, insert 8 after 7 .

Number of comparisons made: 2

3,4,7,8,1,5,4,2

Case 4:

Now, compare the fifth element in this case with the sorted arrays before it.

Since 1 < 4 so insert 1 before 4 .

Then compare 1 with the first element 3 .

Since 1 < 3 so insert 1 before 3 .

Number of comparisons made: 2

1,3,4,7,8,5,4,2

Case 5:

Now, compare the sixth element in this case with the sorted arrays before it.

Since 5 > 4

and 5 < 7

so, insert between 4 and 7 .

Number of comparisons made: 2

1,3,4,5,7,8,4,2

Case 6

Now, compare the seventh element in this case with the sorted arrays before it.

Since 4>1 4>3and 4=4

So, inserted 4 after 3 and before the previous 4 .

Number of comparisons made:

1,3,4,4,5,7,8,2

Case 7:

Now, compare the eighth element in this case with the sorted arrays before it.

Since 2>1 2<3and 2<4

So, insert between 1 and 3 .

Number of comparisons made: 2

1,2,3,4,4,5,7,8

Thus, the total number of comparisons in Insertion Sort is:

1+1+2+2+2+3+2 = 13

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