Recursion
Question 1 |
FORTRAN implementation do not permit recursion because
they use static allocation for variables | |
they use dynamic allocation for variables | |
stacks are not available on all machines | |
it is not possible to implement recursion on all machines |
Question 1 Explanation:
FORTRAN implementation do not permit recursion because they use the static allocation for variables.
→ Recursion requires dynamic allocation of data.
→ Recursion requires dynamic allocation of data.
Question 2 |
The following recursive function in C is a solution to the Towers of Hanoi problem.
Void move (int n, char A, char B, char C)
{
if (…………………………………) {
move (…………………………………);
printf(“Move disk %d from pole %c to pole %c\n”, n,A,C);
move (………………………………….);
Fill in the dotted parts of the solution.
Theory Explanation is given below. |
Question 2 Explanation:
move (disk-1, source, aux, dest) //Step-1
move disk from source to dest //Step-2
move (disk-1, aux, dest, source) //Step-3
Recurrence: 2T(n - 1) + 1
T(n) = 2T (n - 1) + 1
= 2[2T(n - 2) + 1] + 1
= 22T(n - 2) + 3
⁞
2k T(n - k) + (2k - 1)
= 2n-1T(1) + (2n-1 - 1)
= 2n-1 + 2n-1 - 1
= 2n - 1
≅ O(2n)
void move (int n, char A, char B, char C) {
if (n>0)
move(n-1, A, C, B);
printf("Move disk%d from pole%c to pole%c\n", n,A,C);
move(n-1, B, A, C);
}
}
move disk from source to dest //Step-2
move (disk-1, aux, dest, source) //Step-3
Recurrence: 2T(n - 1) + 1
T(n) = 2T (n - 1) + 1
= 2[2T(n - 2) + 1] + 1
= 22T(n - 2) + 3
⁞
2k T(n - k) + (2k - 1)
= 2n-1T(1) + (2n-1 - 1)
= 2n-1 + 2n-1 - 1
= 2n - 1
≅ O(2n)
void move (int n, char A, char B, char C) {
if (n>0)
move(n-1, A, C, B);
printf("Move disk%d from pole%c to pole%c\n", n,A,C);
move(n-1, B, A, C);
}
}
Question 3 |
Consider the following recursive definition of fib:
fib (n) : = if n = 0 then 1
else if n = 1 than 1
else fib (n – 1) + fib (n – 2)
The number of times fib is called (including the first call) for an evaluation of fib (7) is ___________
41 |
Question 3 Explanation:
The recurrence relation for the no. of calls is
T(n) = T(n-1) + T(n-2) + 2
T(0) = T(1) = 0 (for fib(0) and fib(1), there are no extra recursive calls)
T(2) = 2
T(3) = 4
T(4) = 8
T(5) = 14
T(6) = 24
T(7) = 40
Counting the initial call, we get
40+1 = 41
T(n) = T(n-1) + T(n-2) + 2
T(0) = T(1) = 0 (for fib(0) and fib(1), there are no extra recursive calls)
T(2) = 2
T(3) = 4
T(4) = 8
T(5) = 14
T(6) = 24
T(7) = 40
Counting the initial call, we get
40+1 = 41
There are 3 questions to complete.