What's wrong with my code?

[ILIKEMILK]

What is it you are trying to achieve?
I need to write a program in which I find out if a number is a perfect number or not. A perfect number means that if you add up all the factors of the number, their sum will be the number itself. For example 6 is a perfect number because the sum of its factors (1+2+3) = 6.
What is the problem you are having?
I'm not sure how I can find the factors of a number. What would the code for that be?
Describe what you have tried to solve this problem
This is my code at the moment.
Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)
var userNum : int
var numoriginal : int := 0
var factoredNum : int := 1
var factoredSum : int := 0
put "Please enter a positive integer. " ..
get userNum
for i : 1 .. userNum
userNum / i
put i
end for

Turing:

<Add your code here>

Please specify what version of Turing you are using
<Answer Here>

[apython1992]

Have you checked out Turing's mod operator?

[Raknarg]

I would use a function actually. I'll explain it in a short sense.
So a function is like a set of steps used to get a value. If you've worked with procedures before, they're very similar. You call the name, give it a parameter to do stuff with, and you tell it what kind of result you're looking for. You use it by calling the name and having the value in brackets next to it.
For instance:

Turing:

function square (x : int) : int     result x * x end square put square (5)

In this case you would get an output of 25, or x times x.
In your case, it would be a bit different. You see, your code seems to be tying to find an answer with an integer; what you need is a boolean, a.k.a. true or false. Therefore, you would use a fuction with the result of a boolean.

Turing:

function perfect (x : int) : boolean     var total : int := 0     for i : 1 .. x - 1         if x / i = x div i then             total := total + i         end if     end for     if total = x then         result true     else         result false     end if end perfect

if x / i = x div checks to see if they both divide to produce am integer (means its a factor). If it is, it adds it to a total.
Try using that.

[apython1992]

The mod operator does that in a cleaner, more efficient, and more readable way. It produces the remainder of the division. For example, 3 mod 2 is 1, because when you divide 3 by 2, you get 1 remainder 1. If the remainder is 0, you have yourself a factor.

[Raknarg]

Damn, I forgot about that -.-'
Would've read the original comment, but it took me 30 mins to make a reply. So distracted

[Raknarg]

In that case...

Turing:

function perfect (x : int) : boolean     var total : int := 0     for i : 1 .. x - 1         if x mod i = 0 then             total := total + i         end if     end for     if total = x then         result true     else         result false     end if end perfect

[apython1992]

Much better We should try to let the OP figure out as much as he can without giving too much code, but +1 karma for helping.

[Raknarg]

Lol. I don't like to give out code, but it's habit from grade 10 with too many noobs
Plus, I find that I learn much more when I have an expert to show me the best way to do something.

[Zren]

A factor is never more than half itself (other than itself of course).
Eg: The factors of 12 are 1, 2, 3, 4, 6, and 12.

1 x n = n
The next possible factor is 2.
2 x n/2 = ?

Thus you can limit the range of numbers you have to check.

So you could do:
Add 1 and n as factors. Since 1 times itself will always be itself.
Check the range of 2 -> n/2 for other factors.

Efficiency might not seem so bad when n is a low number like 12, but when your checking number bigger than that, the loop tends to chug.

[Raknarg]

Yeah, I know, I just didn't feel like putting it in -.-' If you have a small code like that, it wont really make a speed difference anyways.

Turing:

function perfect (x : int) : boolean     var total : int := 0     for i : 1 .. x div 2         if x mod i = 0 then             total := total + i         end if     end for     if total = x then         result true     else         result false     end if end perfect

[ILIKEMILK]

Here is my code now, but it still doesnt work.


var userNum, factoredNum : int
var factoredSum : int := 0

put "Please enter a positive integer. " ..
get userNum
for decreasing i : userNum .. 1
factoredNum := userNum mod i
if
userNum mod i = 0 then
factoredSum := factoredSum + i
end if
end for
if factoredSum = userNum then
put userNum, " is a perfect number."
else
put
userNum, " is not a perfect number."
end if

[apython1992]

In figuring out if a number is perfect or not, you only care about its proper divisors. The proper divisors of say, 6, are 1, 2, and 3. Your program should then test all numbers from 1 to the inputted number, but not including that number. Think you can figure out how to change that?

[apython1992]

Oh and as Zren has mentioned, you really only need to test up to half of the inputted number.

[whoareyou]

I did a program like this a while back. You can take a look at the help the other users had given me when I had troubles with it.

http://compsci.ca/v3/viewtopic.php?t=27673

You really only need two variables, one to get the number from the user and one to store the sum of the factors.

Then you'd use a FOR loop to calculate any factors from the 1 to 1 less than the number (since the number itself cannot be a factor); IF you've found a factor, it should be added to the "sum of factors variable". (you could also implement the range limit as mentioned my the previous posters when searching for factors.)

Lastly, you need to check IF the sum of factors variables equals the number that was entered. IF it is, then it is a perfect number. ELSE, the number that the user entered is not a perfect number.

You could also LOOP the program so that the program can check more than one number.

[ILIKEMILK]

^ Thanks, that helped out a lot. I got it now.