C-Programming

Question 25

Consider the following C program

int f1 (int n)
{
  if(n == 0||n == 1)
    return n;
  else
    return (2*f1(n-1)+3*f1(n-2));
}
 
int f2 (int n)
{
  int i;
  int X[N], Y[N], Z[N];

  X[0]=Y[0]=Z[0]=0;
  X[1]=1; Y[1]=2; Z[1]=3;
  for(i=2; i<=n; i++) {
    X[i] = Y[i-1] + Z[i-2];
    Y[i] = 2*X[i];
    Z[i] = 3*X[i];
  }
  return X[n];
}

f1(8) and f2(8) return the values

A	1661 and 1640
B	59 and 59
C	1640 and 1640
D	1640 and 1661

ProgrammingC-ProgrammingGATE 2008

Question 25 Explanation:

Both functions perform same operation, so output is same, means either (B) or (C) is correct.
f1(2) = 2*f1(1) + 3*f1(0) = 2
f1(3) = 2*f1(2) + 3*f1(1) = 2*2 + 3*1 = 7
f1(4) = 2*f1(3) + 3*f1(2) = 2*7 + 3*2 = 20
f1(5) = 2*f1(4) + 3*f1(3) = 2*20 + 3*7 = 40 + 21 = 61
We can skip after this as the only remaining choice is (C).
f1(6) = 2*f1(5) + 3*f1(4) = 2*61 + 3*20 = 122 + 60 = 182
f1(7) = 2*f1(6) + 3*f1(5) = 2*182 + 3*61 = 364 + 183 = 547
f1(8) = 2*f1(7) + 3*f1(6) = 2*547 + 3*182 = 1094 + 546 = 1640

Correct Answer: C

Question 25 Explanation:

Both functions perform same operation, so output is same, means either (B) or (C) is correct.
f1(2) = 2*f1(1) + 3*f1(0) = 2
f1(3) = 2*f1(2) + 3*f1(1) = 2*2 + 3*1 = 7
f1(4) = 2*f1(3) + 3*f1(2) = 2*7 + 3*2 = 20
f1(5) = 2*f1(4) + 3*f1(3) = 2*20 + 3*7 = 40 + 21 = 61
We can skip after this as the only remaining choice is (C).
f1(6) = 2*f1(5) + 3*f1(4) = 2*61 + 3*20 = 122 + 60 = 182
f1(7) = 2*f1(6) + 3*f1(5) = 2*182 + 3*61 = 364 + 183 = 547
f1(8) = 2*f1(7) + 3*f1(6) = 2*547 + 3*182 = 1094 + 546 = 1640

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C-Programming