def f(a, b): if (a == 0): return b if (a % 2 == 1): return 2 * f((a - 1) / 2, b) return b + f(a - 1, b) print(f(15, 10))The value printed by the code snippet is 160 (Answer in integer).
Step 1: Understanding the function behavior.
The function f(a, b)
follows a recursive pattern.
If \( a = 0 \), it returns \( b \).
If \( a \) is odd, it halves \( a - 1 \) and multiplies the result by 2.
Otherwise, it reduces \( a \) by 1 and adds \( b \).
Step 2: Evaluating f(15, 10)
.
1. \( f(15,10) = 2 \times f(7,10) \)
2. \( f(7,10) = 2 \times f(3,10) \)
3. \( f(3,10) = 2 \times f(1,10) \)
4. \( f(1,10) = 2 \times f(0,10) \)
5. \( f(0,10) = 10 \)
6. \( f(1,10) = 2 \times 10 = 20 \)
7. \( f(3,10) = 2 \times 20 = 40 \)
8. \( f(7,10) = 2 \times 40 = 80 \)
9. \( f(15,10) = 2 \times 80 = 160 \)
Thus, the output of the program is: 160.
A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius \( r \) cm as shown in the figure. The side of the dodecagon is \( d \) cm. All the triangles (numbered 1 to 12 in the figure) are used to form squares of side \( r \) cm, and each numbered triangle is used only once to form a square. The number of squares that can be formed and the number of triangles required to form each square, respectively, are:
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
Create empty stack S Set x = 0, flag = 0, sum = 0 Push x onto S while (S is not empty){ if (flag equals 0){ Set x = x + 1 Push x onto S } if (x equals 8): Set flag = 1 if (flag equals 1){ x = Pop(S) if (x is odd): Pop(S) Set sum = sum + x } } Output sumThe value of \( sum \) output by a program executing the above pseudocode is:
Consider a directed graph \( G = (V,E) \), where \( V = \{0,1,2,\dots,100\} \) and
\[ E = \{(i,j) : 0 < j - i \leq 2, \text{ for all } i,j \in V \}. \] Suppose the adjacency list of each vertex is in decreasing order of vertex number, and depth-first search (DFS) is performed at vertex 0. The number of vertices that will be discovered after vertex 50 is: