To describe how the function reverses that run in θ ( k ) time can runs in O ( n k ) if the length is n = 2 k .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

I need to define and discuss the uses of one monitoring or troubleshooting tool in Windows Server 2019. thank you

I would likr toget help with the following concepts: - Windows Server features - Windows Server versus Windows 10 used as a client-server network

I need to define and discuss the uses of one monitoring or troubleshooting tool in Windows Server 2019. thank you

Answer 2

Question

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Answer 6

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

Answer 7

Chapter 17, Problem 1P

(a)

Program Plan Intro

To describe how the function reverses that run in θ(k) time can runs in O(nk) if the length is n=2k .

Answer 8

(a)

Expert Solution

Explanation of Solution

The reverse operation is performed by using two arrays. Consider another array and copy the elements in the reverse order to the new array from the original array. The operation is performed as swapping the values with the same index in the both array.

After swapping the element to current index it required to set the entry in the second array corresponding to the bit-reverses.

Since the swapping of the element time is depends upon the number of element so suppose there are n -elements in the function of k bit reversed algorithm then the running time is equal to the product of n and k .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

Answer 13

(b)

Program Plan Intro

To describe an implementation of the BIT-REVERSED-INCREMENT procedure that allows the bit-reversal permutation on an n -element to perform in O(n) time.

Answer 14

(b)

Expert Solution

Explanation of Solution

The pseudo-code of the BIT-REVERSED-INCREMENT implementation is given below:

REVERSED-INCREMENT( a )

Set m=1 followed by k zeroes.

while m bitwise-AND ais not zero do

a=abitwise−XOR m .

shift m right by 1.

end while.

return m bitwise-OR a.

end

The algorithm swaps the value of binary counter and reversed counter so it updates the values by incrementing the counter variable.

The operation is performed is carried out bit by bit so it consider one bit at the time to compared and reversed the values.

The BIT-REVERSED-INCREMENT operation will take constant amortized time and to perform the bit-reversed permutation requires a normal binary counter and a reversed counter.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.

Answer 19

(c)

Program Plan Intro

To explain how shifting any word shift left or right by one bit in unit time without changing the run time of the implementation.

Answer 20

(c)

Expert Solution

Explanation of Solution

The procedure uses single shift so the complexity of the procedure is not changed as any words shift left with unit time.

The procedural is based on the number of bit swapped not on the operation therefore any change in the operation such as type of shift does not affect the time complexity of the procedure.

Since, the operation time is depends upon the number of element so the running time is equal to the product of n and k , where n is the size and k is shift bit.

Hence, there is no change in the running time of implementation.