Assume that \( C(n) \) is the total number of function calls made when \( \texttt{mystery}(n) \) is called (including the initial call).
For \( n > 0 \), the function counts the current call → 1, then calls \( \texttt{mystery}(n-1) \), and calls \( \texttt{mystery}(n-2) \).
So, the recurrence becomes: \( C(n) = 1 + C(n-1) + C(n-2) \)
Now, notice the base case: when \( n \le 0 \), it simply returns 1. Hence, \( C(n) = 1 \quad \text{for } n \le 0 \).
\[ C(0) = 1, \quad C(-1) = 1 \]
Let us fill in the entries in the following table:
\[ \begin{array}{|c|c|c|c|c|c|} \hline n & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline C(n) & & & & & & \\ \hline \end{array} \]
Filling in the base cases first:
\[ \begin{array}{|c|c|c|c|c|c|} \hline n & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline C(n) & 1 & 1 & & & & \\ \hline \end{array} \]
Now, filling in the remaining entries. Each entry is the sum of the previous two entries plus 1:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \color{#0d47a1}{n} & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline \color{#0d47a1}{C(n)} & 1 & 1 & 3 & 5 & 9 & \color{#d32f2f}{\boxed{15}} \\ \hline \end{array} \]
Correct Option: A
If they ask for \( \texttt{mystery}(10) \), you can simply extend the table:
\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & -1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline C(n) & 1 & 1 & 3 & 5 & 9 & 15 & 25 & 41 & 67 & 109 & 177 & 287 \\ \hline \end{array} \]
Also, I noticed that some students made a silly mistake and read the base case in the function as \( \texttt{n} \le 1 \) instead of \( \texttt{n} \le 0 \). If you make this mistake, the answer will be 9, because the base cases would then be \( C(0) = 1 \) and \( C(1) = 1 \).
\[ \begin{array}{|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & 4 \\ \hline C(n) & 1 & 1 & 3 & 5 & 9 \\ \hline \end{array} \]
