Re: help with starting a program (newbie)

["Björn Persson" <spam-away@nowhere.nil>]

You seem to be on the right track. What's missing is mostly the code for 
printing the chains when they're found. Your approach seems to be that 
for each number in the range you calculate a sequence of sums of 
divisors and see whether it wraps around before it exceeds the maximum 
length. This might not be the fastest method and it will find the same 
chain multiple times, but I see that the sample run you posted does list 
  the same chains multiple times, so apparenty this is acceptable.

>    Minnumber : Natural range 0 .. Integer'Last;

This is exactly the same as "Minnumber : Natural;".

>    Maxnumber : Natural range Minnumber .. Integer'Last;

You're trying to codify that Maxnumber must not be less than Minnumber. 
This is not a bad idea, but then you have to know the value of Minnumber 
before you can declare Maxnumber. You have to do it like this:

procedure Sociablechain is

    -- other code here

    Minnumber : Natural;
    -- We can't reference Minnumber yet.
begin
    Put(Item => " Please enter your starting integer. ");
    Get(Item => Minnumber);
    New_Line;
    declare
       -- Now Minnumber has a value.
       Maxnumber : Natural range Minnumber .. Integer'Last;
    begin
       Put(Item => " Please enter your ending integer. ");
       Get(Item => Maxnumber);
       New_Line;

       -- other code here

    end;
end Sociablechain;

>     Length : Natural range 1 .. 30;

Is it specified that the program shouldn't accept lengths greater than 
30? If not, you should probably allow much greater lengths.

>    function Sum_Of_Divisors (
[...]
>       for I in 1 .. N - 1 loop

Numbers greater than half of N can't very well be divisors of N. You 
could cut the execution time in half by looping over 1 .. N / 2 instead.

>    procedure Sociable_Chain (
>          Minnumber,
>          Maxnumber  : Natural ) is

You need to think more about the parameters to this procedure. What do 
you need to know to start calculating a chain?

>       A : array (1 .. 30) of Positive;
>       I : Positive;

You could let the upper bound of A be the maximum length that the user 
entered instead of a fixed number. Then only the computer's memory would 
limit how long chains you could handle.

A and I should have more descriptive names.

> 
>    begin
>       A(1) :=Minnumber;

Here Minnumber seems to be a starting point rather than a minimum.

>       A(2) := Sum_Of_Divisors (A(1));
>       while I < Length and A (1) /= A(I) loop

What value will I have the first time you reach this point?

>          I:= I + 1;
>          A(I) := Sum_Of_Divisors (A(I - 1));
> 
> 
>       end loop;

Now you need to find out whether the loop ended because a sociable chain 
was found or because the maximum length was exceeded � and if a chain 
was found you know what to do with it, don't you?

> 
> end Sociable_Chain;

-- 
Bj�rn Persson                              PGP key A88682FD
                    omb jor ers @sv ge.
                    r o.b n.p son eri nu