Erlang Class - 2nd Recursion - Assignment

Tail Recursion Excercise

%tail Recursion
-module(recursion_1_3_4).
-export([sum/1,fib/1,perfect/1,perfect1/1]).

% Try 1
%
%    Sum of numbers
%    F(0)+F(1)+....F(N-1)+F(N)

 sum(N) -> sum(N-1,N).
 
 sum(0,K)->K;
 sum(N,K)->sum(N-1,K+N).
 
% Try 2
%
%    max of numbers
%    F(0)+F(1)+....F(N-1)+F(N)

%fibonnaci
%0,1,2,3,4,5...
%0,1,1,2,3,5,...
fib(N)-> fib(N,0,1).
fib(N,N1,_) when N==0 -> N1;
fib(N,N1,N2) when N>0->fib(N-1,N2,N1+N2).

%fib(4)
%fib(4,0,1)
%fib(3,1,1)
%fib(2,1,2)
%fib(1,2,3)
%fib(0,3,5)
%3

%Perfect numbers

%A positive integer is perfect when it is the sum of its divisors, e.g. 6=1+2+3, 28=1+2+4+7+14.

%Define a function perfect/1 that takes a positive number N and returns a boolean which indicates whether or not the number is perfect. You may well want to %use an accumulating parameter to hold the sum of the divisors “so far”.

perfect(N)->divisor(N,1,0).
divisor(N,Num,Sum) when Num == N -> Sum == N;
divisor(N,Num,Sum) ->
if (N>0) and (N rem Num == 0)  ->  divisor(N,Num+1,Sum+Num);
true -> divisor(N,Num+1,Sum)
end.

perfect1(N)->divisor1(N,1,0).
divisor1(N,Num,Sum) when Num == N -> Sum == N;
divisor1(N,Num,Sum) when(N>0) and (N rem Num == 0) ->  divisor1(N,Num+1,Sum+Num);
divisor1(N,Num,Sum) when(N>0)  -> divisor1(N,Num+1,Sum).