1. Give the computational complexity of the following piece of code in Big-Oh no

Question

1. Give the computational complexity of the following piece of code in Big-Oh notation and explain how you arrived at your result:

for (i=1; i<=n; i=i+2){

// constant number of operations

}

2. Give the computational complexity of the following piece of code in Big-Oh notation and explain how you arrived at your result:

for (i=1; i<=n; i++){

for (j=i; j<=n; j=j+2){

// constant number of operations

}

}

3. Give the computational complexity of the following piece of code in Big-Oh notation and explain how you arrived at your result:

for (j=n; j>1; j=j/2){

// constant number of operations

}

4. Give the computational complexity of the following piece of code in Big-Oh notation and explain how you arrived at your result:

for (i=1; i<=n; i++){

for (j=n; j>1; j=j/2){

// constant number of operations

}

}

5. Is n 2 + 100 of complexity O(n 2 )? Use the formal definition of Big-Oh to answer this question.

6. Is n 2 + 100 of complexity O(n 3 )? Use the formal definition of Big-Oh to answer this question.

7. Is 2n+1 of complexity O(2n)? Use the formal definition of Big-Oh to answer this question.

8. Is 22n of complexity O(2n)? Use the formal definition of Big-Oh to answer this question.

9. Explain why the statement “The running time of the algorithm is at least O(n 2 )” does not provide any information.

Explanation / Answer

ANS(1) since it is known that Big-Oh gives the asymptotic upper bound of a function

as given for(i=1;i<=n;i=i+2)

{ // constant number of operation}

this for loop runs for f(n)=n/2 times because i is incremented by 2 each time

by defintion Big-Oh 0 f(n) cg(n) here c is a positive constant

f(n)=O(g(n)) so 2(n/2)=n , where c=2  

computational complexity =O(n)  

ANS(2)

loop 1 for (i=1; i<=n; i++){

loop2   for (j=i; j<=n; j=j+2){

// constant number of operations}}

lets see how the loop run

loop1 value of i number of times loop 2 runs( rounded off to floor inetger value)

1 n/2

2 (n-1)/2

3 (n-2)/2

-- -----

n-1 (n-(n-1))/2=1/2= 0 (rounded off value)

n 0

SUM= n/2 + (n-1)/2 + ------------------ + 0

SUM =(1/2)[ n+ (n-1)+ ------------- +1]= (n(n+1))/4

let sum be a function of n f(n)= (n2+n)/4

now using the definition Big -Oh c(n2+n)/4 where c = 4 we get n2+n in which dominating term is n2

computational complexity =O(n2)  

ANS(3) for (j=n; j>1; j=j/2){

// constant number of operation }

lets see how the loop runs

every time loop executes value of j is halved  

so there comes a point where n/2k =1 now evaluating the value of k we get k=log2n

   this k is the number of times loop executed

computational complexity =O(log2n)

ANS(4)

loop1 for (i=1; i<=n; i++){

loop 2   for (j=n; j>1; j=j/2){

// constant number of operation }}

lets see how the loop run

inner loop i,e, loop2 executes

every time loop2 executes value of j is halved  

so there comes a point where n/2k =1 now evaluating the value of k we get k=log2n

   this k is the number of times loop executed

loop1 value of i number of times loop 2 runs

1 log2n

2   log2n

3 log2n

-- -----

n   log2n

SUM= nlog2n

thus SUM gives the value number of times both loop executed

computational complexity =O(nlog2n)  

ANS(5) f(n)=n2+100 of complexity O(n2)

By definition of big oh notation f(n)<=c.g(n)

then in Big-Oh notation it can be represented by O(g(n))

now if we choose an appropriate c =2 and g(n) =n2 we can satisfy the condition so ,

f(n)=n2+100 of complexity O(n2)

ANS(6) f(n)=n2+100 of complexity O(n3)

By definition of big oh notation f(n)<=c.g(n) for c>0

since the function g(n) is itself greater than f(n)

then in Big-Oh notation it can be represented by O(g(n))

now if we choose an appropriate c =1 and  g(n) =n3 we can satisfy the condition so ,

f(n)=n2+100 of complexity O(n3)

ANS(7) f(n)=2n +1 of complexity O(2n)

By definition of big oh notation f(n)<=c.g(n) for c>0,

then in Big-Oh notation it can be represented by O(g(n))

now if we choose an appropriate c (say2) and g(n) =2n we can satisfy the condition so ,

f(n)=2n+1 of complexity O(2n)   

ANS(8) f(n)=22n of complexity O(2n)

By definition of big oh notation f(n)<=c.g(n) for c>0,

then in Big-Oh notation it can be represented by O(g(n))

now if we choose an appropriate c (say 12) and g(n) =2n  we can satisfy the condition so ,

f(n)=22n of complexity O(2n)   

ANS(9)  “The running time of the algorithm is at least O(n2 )” does not provide any information.

T(n) : running time of the algorithm A . we are more concerned over the upper bound and lower bound of T (n)

But the statement   T(n)>=O(n2 )

UPPER BOUND : Because T(n)>=O(n2 ) then there's no information about upper bound of T(n)

LOWER BOUND: assume f(n)=n2 then the statement T(n)>=f(n) , but f(n) could be anything smaller than n2   for example- constant , n ........ thus there no conclusion on the lower bound T(n) either

Thus it can be concluded that "The running time of the algorithm is at least O(n2)" is meaningless.

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