When a superclass variable refers to a subclass object and a method is called on that object, the proper implementation is deter
2 answers:
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Answer:
def fizzbuzz (num):
for item in range(num):
if item % 2 == 0 and item % 3 == 0:
print("fizzbuzz")
elif item % 3 == 0:
print("buzz")
elif item % 2 == 0:
print("fizz")
else:
print (item)
fizzbuzz(20)
Explanation:
Using Python programming Language
Use a for loop to iterate from 0 up to the number using the range function
Within the for loop use the modulo (%) operator to determine divisibility by 2 and 3 and print the required output
see attached program output screen.
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that returns indeed the sum of the first n-1 odd numbers, and we have to proof that
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So,
is the sum of the first n odd numbers, as required: