Consider the recursive functions represented by the following code segment:int bar(int n) { if (n == 1) return 0; else return 1 + bar(n/2); } int foo(int n) { if (n == 1) return 1; else return 1 + foo(bar(n)); }The smallest positive integer $n$ for which foo(n) returns 5 is ________. (answer in integer) Note: Ignore syntax errors (if any) in the function.

Let us analyze the behavior of the two recursive functions:

1. Function bar(n):

   • Base case: bar(1) = 0.

   • Recursive case: for integer division n / 2, bar(n) = 1 + bar(n / 2).

   • This computes $\lfloor \log_2 n \rfloor$ for any integer $n \ge 1$.

   • To get bar(n) = k, the smallest positive integer is $n = 2^k$.

2. Function foo(n):

   • Base case: foo(1) = 1.

   • Recursive case: foo(n) = 1 + foo(bar(n)) for $n > 1$.

3. Finding the smallest $n$ for which foo(n) = 5:

   • foo(n) = 1 + foo(bar(n)) = 5$\implies$foo(bar(n)) = 4.

   • Let $n_1 = \text{bar}(n)$, so foo(n_1) = 4$\implies$foo(bar(n_1)) = 3.

   • Let $n_2 = \text{bar}(n_1)$, so foo(n_2) = 3$\implies$foo(bar(n_2)) = 2.

   • Let $n_3 = \text{bar}(n_2)$, so foo(n_3) = 2$\implies$foo(bar(n_3)) = 1.

   • For foo(bar(n_3)) = 1, the base case requires $\text{bar}(n_3) = 1$.

4. Working backwards to minimize $n$:

   • Smallest $n_3$ such that $\text{bar}(n_3) = 1$ is $n_3 = 2^1 = 2$.

   • Smallest $n_2$ such that $\text{bar}(n_2) = n_3 = 2$ is $n_2 = 2^2 = 4$.

   • Smallest $n_1$ such that $\text{bar}(n_1) = n_2 = 4$ is $n_1 = 2^4 = 16$.

   • Smallest $n$ such that $\text{bar}(n) = n_1 = 16$ is $n = 2^{16} = 65536$.

Thus, the smallest positive integer $n$ is 65536.