Below is an algorithm which computes (n)Enumeration(i = 0) (n(n+1))/2 where n is an integer
and n 0:
Algorithmn 1 int rec( int n )
1: if ( n = 0 ) then
2: return 0;
3: end if
4: if ( n%2 == 1 ) then .......... (note: i.e n is odd)
5: return rec(n 1) + n;
6: else
7: return 4 rec(n/2) n/2;
8: end
if Prove that rec(n) = n(n+1)/2 using strong induction.
(a) (3 points) Base case: n = Complete the base case:
(b) Recursive case:
i. (2 points) What are you assuming is true:
ii. (2 points) What are you proving is true:
iii. (3 points) Complete the proof:
case 1: n is odd rec(n) = rec(n 1) + n = (n1)n 2 + n = n 2n+2n 2 = n 2+n 2 = n(n+1) 2 (The
easier case is done for you ,)
case 2: n is even.
Below is an algorithm which computes (n)Enumeration(i = 0) (n(n+1)).pdf
1. Below is an algorithm which computes (n)Enumeration(i = 0) (n(n+1))/2 where n is an integer
and n 0:
Algorithmn 1 int rec( int n )
1: if ( n = 0 ) then
2: return 0;
3: end if
4: if ( n%2 == 1 ) then .......... (note: i.e n is odd)
5: return rec(n 1) + n;
6: else
7: return 4 rec(n/2) n/2;
8: end
if Prove that rec(n) = n(n+1)/2 using strong induction.
(a) (3 points) Base case: n = Complete the base case:
(b) Recursive case:
i. (2 points) What are you assuming is true:
ii. (2 points) What are you proving is true:
iii. (3 points) Complete the proof:
case 1: n is odd rec(n) = rec(n 1) + n = (n1)n 2 + n = n 2n+2n 2 = n 2+n 2 = n(n+1) 2 (The
easier case is done for you ,)
case 2: n is even
