Alternative way to write the function

[AGNuke]

Hi all!

I wrote a program which factorises a number to its prime factors. Therefore what I obviously needed is a function which can do so...

Here is what I want:

1.) Divide the number entered with 2(d). If its divide, print 2 and if n/d is not 1, then print X (multiplication sign) and repeat the function with n/=d.

2.) If the number is not divisible with 2(d), increase d by one and repeat the function (ie with 3)

Here's what I wrote:

void factor(int n, int d)// Heart of the program
{
	if(n % d == 0)// if n is divisible by d
	{
		n = n/d;
		cout << d;
		if(n == 1)
			cout << endl;
		else
		{
			cout << " X ";
			factor(n,d);
		}
	}
	else
	{
		d++;
		factor(n,d);
	}
}

NOTE: I defined int d = 2 up in the main function itself. That's one of the thing I want to get rid of.

What I am asking for is I tried to write this as "for" loop before but I was unsuccessful.

Can anyone suggest that how can I rewrite this function as "for" loop, as I can learn from that.

Thanks!

[hanst99]

Well... to rewrite this to use a for loop:

drop the d from the argument list, you don't really need it.
Instead, make d a local variable (and maybe rename n and d to something more readible).

make a for loop, initial action is d=2, continue condition is n>1, no continue action.

Inside that for loop- well, do pretty much the same things you are doing now, just without the recursive calls.

PS: instead of printing the factors out, you might wanna store them in a map or something like that and return the container.

[ultifinitus]

for (int mytempvariable = 0; mytempvariable < 100; mytempvariable++)

one step at a time:

the first section of the for loop int mytempvariable = 0; makes a new integer variable and stores 0 in it.

the second section of the for loop mytempvariable < 100; is the condition of the for loop. In other words, as long as this statement returns true, run my for loop.

the third section of the for loop mytempvariable++ adds one to the variable every time a loop ends.

This example of a for loop will start at 0, run through the loop with the number being 0, then add 1 to mytempvariable. It will continue on with that pattern until it gets to 99, once it runs through 99, it will add one to the variable, making it 100, which will make the condition (mytempvariable < 100) false. So it will end the loop.

*phew

[AGNuke]

Here's how I interpreted :-

void factor(int number)
{
	for(int divisor = 2; divisor <= number; divisor++)
	{
		if(number%divisor == 0)
		{
			number /= divisor;
			cout << divisor;
			if (number == 1)
				cout << endl;
			else
			{
				cout << "x";
				factor(number);
			}
		}
	}
}

The problem I was, and still encountering is the output.

For eg. If I enter the number 8, it should return the output :

2 x 2 x 2

But it is returning :

2 x 2 x 2
4

How's that 4 is coming? And that is a problem with quite all the number with few exceptions such as primes and 4 and 9...

[Mathhead200]

divisor <= number
Is not going to work. You are changing the value of number in the loop: number /= divisor;

Try:

for(int divisor = 2; number > 1; divisor++) {
   ...
}

[AGNuke]

Now I crashed into another problem... :( (IDK if my luck is kidding me)

@Mathhead200

I tried this as your recommendation:-

void factor(int n)
{
	for(int d = 2; n > 1; d++)
	{
		if(n%d == 0)
		{
			n /= d;
			cout << d;
			if (n == 1)
				cout << endl;
			else
			{
				cout << " x ";
				factor(n);
			}
		}
	}
}

Now my problem :-
If I input 8, the output is

2 x 2 x 2
-2 x 4

Am I missing something?

BTW, this is just for educational purpose, so don't take headache... :P

I already made two functions performing this, but I need the one I was first attempting for and to find out where I was mistaken...

[Mathhead200]

void factor(int n)
{
	for(int d = 2; n > 1; /*d++*/)
	{
		if(n%d == 0)
		{
			n /= d;
			cout << d;
			// if (n == 1)
			// 	cout << endl;
			// else
			// {
			if( n > 1 ) //to avoid the trailing " x "
				cout << " x ";
			// factor(n); //recursion! Why do you need this, you are already in a loop?
			// }
		} else //only increment d if not a factor
			d++;
	}
	cout << endl; //the loop ends when n == 1
}

Perhaps better written as a while loop:

void printFactors(int n) {
   int d = 2;
   while( n > 1 ) {
      if( d % n == 0 ) {
        //then d is a factor of n
         cout << d;
         if( n > 1 )
            cout << " x ";
         n /= d;
      } else
         d++; //look for the next factor
   }
   cout << endl; //optional
}

[AGNuke]

@Mathhead
Thanks buddy for telling me where I was fooling around with my first attempt. I am very much grateful to you...

Now I learnt 4 function for the same output - 2 Original, 1 corrected and 1 suggested.

And morever they all use different sets of loops... :D

Thanks to everybody once more!