[Type B] Chapter– 6 Class 12 CS – Sumita Arora Assignment | Q/A

Solution:

(c) compute(3) = compute(2) + 2*3-1

Solution:

(b)
def compute(N):
    if N == 1:
        return 1
    else:
       return compute(N-1)+2*N-1

(D) 6

Solution:

(a)
15
12
9
6
3
Finally

(b)
Infinite loop

(c)
10
8
6
4
2
Finally

(d)
Infinite loop

def express(x, n):
    if n == 0:
        return 1
    elif n%2 == 0:
        return express(xx, n/2) 
    else: 
        return express(x, n-1)

express(2,5)

Solution:

Nothing will be displayed on the output screen as absence of print() function.
Hence, no output.

Note: But beside this fact 32 will be finally result of the above code as the express() is called.

def check(n):
  if n<=1:
    return True
  elif n%2 == 0:
    return check(n/2)
  else:
    return check(n/1)

What is the value returned by check(8)?

Solution:

True

Yes, the last condition will make an infinite recursion to occur. If any number odd and greater than 1 is passed than the code will enter in infinite recursion.

Hence, in the case of check(3) an infinite loop will occur.

def Fn(n):
  print(n, end=" ")
  if n<3:
    return n
  else:
    return Fn(n//2) - Fn(n//3)

What will be the result produced by following function calls?
(a) Fn(12)      (b) Fn(10)   (c) Fn(7)

Solution:

Output:
(a) 12 6 3 1 1 2 4 2 1 -3
(b) 10 5 2 1 3 1 1 1
(c) 7 3 1 1 2 -2

def recur(p):
  if p==0:
    print("##")
  else:
    recur(p)
    p=p-1
recur(5)

Solution:

This code will act as infinite recursion as the p=p-1 will never going to run as of positioning.

#the corrected code will be
def recur(p):
  if p==0:
    print("##")
  else:
    p=p-1
    recur(p)

recur(5)