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vladimir2022 [97]
2 years ago
8

Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\

Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Computers and Technology
1 answer:
AleksandrR [38]2 years ago
4 0

Answer:

Check the explanation

Explanation:

1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

Note : Log (m *n) = Log m + Log n : this is property of logarithm

3) f(n) = O( n^2 ), since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n

for n times, this makes the complexity – n * n = n2

there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.

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Answer:

d = {"R":2.89,"P":3.09, "S":3.39, "N":0}

noofgallons = input("Enter number of Gallons:")

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cwr = input("Is Car Was Requieed")

if(gastype == "R"):

   gal == nooffgallons * d['R']

   if (cwr=="y" and gal<10):

       gal+=3.00;

       elif(cwr=="y" and gal>=10):

           gal+=1.25

           elif(cwr=="n"):

               gal=gal

   elif(gastype==P):

       gal== nooffgallons * d['P']

       if(cwr="y" and gal<10):

           gal+=3.00;

           elif(cwr=="y"and gal>=10):

           gal+=1.25

           elif(cwr=="n"):

               gal=gal

       elif(gastype ==S):

           gal=noofgallons *d['S']

           if(cwr="y" and gal<10):

               gal+=3.00;

               elif(cwr=="y"and gal>=10):

                   gal+=1.25

                   elif(cwr=="n"):

                       gal=gal

                       

                       

print("tptal cost paid"+ gal)

The above program uses a dictionary for storing gasoline types and their cost, and dictionary is always constant, and hence its fulfilled the requirement of keeping it constant. And car wash is treated accordingly. And if elif else ladder has been used to calculate the total number of gallons.

Rest part is self explanatory.

Explanation:

The code is self explanatory.

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