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 def palindrome(s,n,r):     if n==0:         return r      else:         r+=s[n-1]         return palindrome(s,n-1,r) s="malayalam" n=len(s) r='' res=palindrome(s,n,r) if s==res:     print(res,"is palindrome string") else:     print(res,"is not palindrome string")

  def palindrome(n,r):     if n==0:         return r     else:         r=n%10+r*10         return palindrome(n//10,r) n=1001 r=0 res=palindrome(n,r) if n==res:     print(res,"is palindrome") else:     print(res,"is not palindrome")

  for n in range(2,101):     t=0     for i in range(2,n):         if n%i==0:             t=1             break     if t==0:         print(n,end=" ")

 def fib(n):     if n==0:         return 0     elif n==1:         return 1      else:         return fib(n-1) + fib(n-2) n=int(input()) for i in range(n):     print(fib(i),end=" ")

a=0 b=1 print(a,b,end=" ") for i in range(10):     c=a+b      a=b      b=c     print(c,end=" ")

  Given a     2D Array ,   : 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in   is a subset of values with indices falling in this pattern in  's graphical representation: a b c d e f g There are   hourglasses in  . An  hourglass sum  is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in  , then print the  maximum  hourglass sum. The array will always be  . Example -9 -9 -9 1 1 1 0 -9 0 4 3 2 -9 -9 -9 1 2 3 0 0 8 6 6 0 0 0 0 -2 0 0 0 0 1 2 4 0 The   hourglass sums are: -63, -34, -9, 12, -10, 0, 28, 23, -27, -11, -2, 10, 9, 17, 25, 18 The highest hourglass sum is   from the hourglass beginning at row  , column  : 0 4 3 1 8 6 6 Note:  If you have already solved the Java domain's  Java 2D Array  challenge, you may wish to skip this challenge. Function Description Comp...

 class Library:     def__init__(self,listofbooks,libraryname):         self.lst = listofbooks         self.name = libraryname     def displaybook(self):         return f"\t\t\t\tWelcome to {self.name} \         \navaliable books are -\n {self.lst}"     def lendbook(self,lendname,bookname):         if bookname in lst and bookname not in d:             print(f"{bookname} lend to {lendname}")             d.__setitem__(bookname,lendname)         elif bookname in d:             lst.remove(bookname)             print("Book is not available lend to-",d[bookname],"\nwe only have-",lst)         else:             print("Book is not in the library we only have-",lst)   ...