Question

What is the worst-case asymptotic time complexity of the following divide-andconquer algorithm (give a Θ-bound). The input is an array A of size n. You may assume that n is a power of 2.

(NOTE: It doesn’t matter what the algorithm does, just analyze its complexity).

Assume that the non-recursive function call, bar(A1,A2,A3,n) has cost 3n.

Show your work! Next to each statement show its cost when the algorithm is executed on an imput of size n abd give the expression for the run time and derive its complexity in term of Θ.

foo(n,A){

if (n == 1)

return A[0]

Let A1,A2,A3 to be of size n/2

for ( i=0; i<=n/2-1; i++) {

A1[i] = A[i];

A2[i] = A[n/2+i]

   A3[i] = A[2i];

bar(A1,A2,A3,n);

}

x = foo(n/2,A1);

y = foo(n/2,A2);

z = foo(n/2,A3);

return (x+y)*z;

}

Answer 1

`Hey,

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foo(n,A){--------------------------T(n)

if (n == 1)----------------------------cost 1

return A[0]----------------------------cost 1

Let A1,A2,A3 to be of size n/2

for ( i=0; i<=n/2-1; i++) {----------------------------cost n+1

A1[i] = A[i];----------------------------cost 1*n/2

A2[i] = A[n/2+i]----------------------------cost 1*n/2

   A3[i] = A[2i];----------------------------cost 1*n/2

bar(A1,A2,A3,n);----------------------------cost 3n*n/2

}

x = foo(n/2,A1);----------------------------cost T(n/2)+1

y = foo(n/2,A2);----------------------------cost T(n/2)+1

z = foo(n/2,A3);----------------------------cost T(n/2)+1

return (x+y)*z;----------------------------cost 1

}

So, total cost recurrence is defined as

T(n)=3*T(n/2)+1+(n+1)+3*n/2+3*n*n/2+3

So,

T(n)=3*T(n/2)+(3*n^2)/2 + (5*n)/2 + 5

for all n>1

T(0)=2

So, Applying masster theorem a=3, b=2 p=0.

So, it is theta(n^2)

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Thanks.