A recursive function is given:
def mystery(n):
if n $<=$ 0:
return 1
else:
return mystery(n-1) + mystery(n-2)
Find the value of mystery(4).
def fun(L, i=0):
if i $>=$ len(L) - 1:
return 0
if L[i]> L[i+1]:
L[i+1], L[i] = L[i], L[i+1]
return 1 + fun(L, i+1)
else:
return fun(L, i+1)
data = [5, 3, 4, 1, 2]
count = 0
for _ in range(len(data)):
count += fun(data)
print(count)
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 sum
The 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:
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 sum
The value of \( sum \) output by a program executing the above pseudocode is:Consider two distinct positive numbers \( m, n \) with \( m > n \). Let \[ x = n^{\log_n m}, \quad y = m^{\log_m n}. \] The relation between \( x \) and \( y \) is -