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lisov135 [29]
4 years ago
7

An algorithm processes an array of size n by operating on its first one-third, its second one-third, its third one-third, and th

en operating on its third one-third again, recursively. It theån combines the solutions in 21gn time. Derive a recurrence for the running time of above algorithm. You may assume that n=3k for some positive integer k. Use an appropriate method (just pick one method) to solve the recurrence by finding a tight upper and lower bound solution for the recurrence. You must show the procedure of calculation.

Computers and Technology
1 answer:
oee [108]4 years ago
4 0

Answer and Explanation:

The answer is attached below

You might be interested in
A network administrator is configuring an ACL with the command access-list 10 permit 172.16.32.0 0.0.15.255. Which IPv4 address
lord [1]

Answer:

For this wild card mask 0.0.15.255 the ACE IP address will be 172.16.47.254

Explanation:

ACL is the access control list that is used to enlist the ip addresses that allowed or restricted to access the network. ACE is an IP address from the list ACL that has all rules and regulations related to access of network. The ACE could be in the range of IP address in ACL. ACL can be calculated with the help of initial IP address adding with wild card mask.

So

Initial IP address is = 172.16.32.0

Wild card mask =0.0.15.255

by adding above values we can find the last IP address of ACL.

after addition

Final IP address is = 172.16.47.255

The options that are available with question, Only option between the range is  172.16.47.254. So we can say that This is the only ACE IP address in options.

7 0
3 years ago
Write a program with 2 separate functions which compute the GCD (Greatest Common Denominator) and the LCM (Lowest Common Multipl
Maru [420]

Answer:

The program written in Python is as follows

def GCD(num1, num2):

    small = num1

    if num1 > num2:

         small = num2

    for i in range(1, small+1):

         if((num1 % i == 0) and (num2 % i == 0)):

              gcd = i

    print("The GCD is "+ str(gcd))

def LCM(num1,num2):

    big = num2  

    if num1 > num2:

         big = num1

    while(True):

         if((big % num1 == 0) and (big % num2 == 0)):

              lcm = big

              break

         big = big+1

     print("The LCM is "+ str(lcm))

 print("Enter two numbers: ")

num1 = int(input(": "))

num2 = int(input(": "))

GCD(num1, num2)

LCM(num1, num2)

Explanation:

This line defines the GCD function

def GCD(num1, num2):

This line initializes variable small to num1

    small = num1

This line checks if num2 is less than num1, if yes: num2 is assigned to variable small

<em>     if num1 > num2: </em>

<em>          small = num2 </em>

The following iteration determines the GCD of num1 and num2

<em>     for i in range(1, small+1): </em>

<em>          if((num1 % i == 0) and (num2 % i == 0)): </em>

<em>               gcd = i </em>

This line prints the GCD

    print("The GCD is "+ str(gcd))

   

This line defines the LCM function

def LCM(num1,num2):

This line initializes variable big to num2

    big = num2  

This line checks if num1 is greater than num2, if yes: num1 is assigned to variable big

<em>     if num1 > num2: </em>

<em>          big = num1 </em>

The following iteration continues while the LCM has not been gotten.

    while(True):

This if statement determines the LCM using modulo operator

<em>          if((big % num1 == 0) and (big % num2 == 0)): </em>

<em>               lcm = big </em>

<em>               break </em>

<em>          big = big+1 </em>

This line prints the LCM of the two numbers

     print("The LCM is "+ str(lcm))

The main starts here

This line prompts user for two numbers

print("Enter two numbers: ")

The next two lines get user inputs

num1 = int(input(": "))

num2 = int(input(": "))

This calls the GCD function

GCD(num1, num2)

This calls the LCM function

LCM(num1, num2)

<em></em>

<em>See attachment for more structured program</em>

Download txt
5 0
3 years ago
What is the output of the following code segment?
Aliun [14]

Answer:

o

Explanation:

8 0
3 years ago
On early computers, every byte of data read or written was handled by the CPU (i.e., there was no DMA). What implications does t
grandymaker [24]

Answer:

Multiprogramming will be extremely difficult to be achieved.

Explanation:

If every byte of data read or written is handled by the CPU the implications this will have for multiprogramming are not going to be satisfactory.

This is because, unlike before, after the successful completion of the input and output process, the CPU of a computer is not entirely free to work on other instructions or processes.

5 0
3 years ago
Create a simple program of your own using a Loop that counts and displays numbers 1-10.
avanturin [10]

Answer:

I think

Explanation:

you....

8 0
3 years ago
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