That should beif n<= 0
then sum of integer(n) := 0
else sum of integer(n) := n+ sum of integer(n)
else sum of integer(n+1) = sum of integer(n) + 1
Give a recursive algorithm for finding the sum of the first n positive integer.
Procedure sum of integer ( n : positive integer )
if n<= 0
then sum of integer(n) := 0
else sum of integer(n) := n+ sum of integer(n)
is it correct or not ?? some one please help me ??
You have to have express sum_of_integer(n) in terms of sum_of_integer(n-1) so:
Also as n is declared as a positive integer you cannot test if it is <=0
CBCode:Procedure sum_of_integer ( n : positive integer ) if n== 1 then sum_of_integer := 1 else sum_of_integer := 1+ sum_of_integer(n-1)
Thank you very much for your reply but i do not understand the logic
Procedure sum_of_integer ( n : positive integer )
if n== 1
then
sum_of_integer(n) := 1
else
sum_of_integer(n) := 1+ sum_of_integer(n-1)
what is the value of sum_of_integer(n-1) ?? if we take n = 2 if you can describe little bit ?? appreciate your reply.
Thanks
Give a recursive algorithm for finding the maximum of the finite set of integer, making use of the fact that the maximum of n integer is the larger of the last integer in the list and the maximum of the first n-1 integer in the list.
procedure largest(a1,a2,.....,an: integer)
if n=1
then largest (a1,a2,.....,an: integer) : = a1
else largest(a1,a2,.....,an: integer) : = max(largest(a1,a2,.....,an-1),an)
is the answer correct or not ?? if not please suggest the correct answer. appreciate your reply.
Thanks
I think your pseudo-code use is not acceptable, something more like:
would be better.Code:procedure largest(n:unsigned, a[1:n]: integer) if n=1 then return a[1] else return max(largest(n-1, a[1:n-1]),a[n])
Alternativly can you give a link to a site that specified the syntax for your pseudo-code.
CB