Need help

A way to speed up the shown algorithm would be to utilize a list of all primes and only test divisibility with those. However, then you have to deal with how to deal with problems that may occur if the input contains a prime divisor that is larger then the largest in your list. However, if you know the input will have a particular bound, you can easily utilize that to your advantage.

jboeinga at 2007-7-7 16:15:18 >

> Can u give it to me in the java language plzz

> Thanks for helping

Can you give it to me in the java language?

If your answer is "no," then you need to talk to your teacher, see if you can't a) get your failing butt up to a B by learning the language and doing some work or b) get transferred out of the class.

Note that c) cheat was stricken from the answer card as it was an innately immoral choice. And we wouldn't want to tempt you with one of those, now would we?

Adeodatusa at 2007-7-7 16:15:18 >

I did do the code and i cant print all the factor number , it just print 2

that is the code , can anyone tell me where is the problem :

public class FactorGenerator

{

public FactorGenerator(int aNumber)

{

number = aNumber;

}

public int factor()

{

if (number==1)

return 1;

for(int i=2;i<number;i++)

{

if(number%i==0)

{

number = number/i;

return i ;

}

}

return number;

}

private int number;

}>

sam_bijania at 2007-7-7 16:15:18 >

> You mean n / 2?

Nope, i mean Math.sqrt(n)

if (n % x == 0) then (n % (n / x) == 0)

Now if x >= sqrt(n) that means n / x <= sqrt(n) and you've already taken care of it earlier in your loop.

public class TestPrimeFactor {

public static void factor(long n) {

if (n == 1) return;

boolean wasFactored = false;

for (long x = 2; x <= (long)Math.sqrt(n); x++) {

if (n % x == 0) {

System.out.print(x + " ");

factor(n / x);

wasFactored = true;

break;

}

}

if (!wasFactored) System.out.print(n + " ");

}

public static void main(String[] args) {

factor(1201254554551233235);

}

}

weebiba at 2007-7-7 16:15:18 >

Two remarks:

* that's not exactly adeodatus' algorithm: if you would leave out the isFactored, you'd never reach 19 for 38, for instance [of course, your algorithm works, but I was replying to in the algorithm posted by Adeodatus, you only need to test integers from 2 to the square root, which is not correct]

* the x++ is unnecessarily inefficient: treat 2 as a separate case and start the loop with 3, increment by two

levi_ha at 2007-7-7 16:15:18 >

Hey guys,

I am not sure I am following this conversation accurately. However,

I dont think the code will work accurately using a for loop. The reason

is that after a factorial is found, the for loop will continue. It seems to me that it

should be stopped after each factorial is calculated.

Adeodatus has the right idea but I suggest using while instead. Try this:

public class TestFactors {

void factor(int n) {

if (n == 1 || n<2) {return;}

int x = 2;

while(n%x != 0){

x++;

} System.out.println(x);

if(n/x!=1){factor (n / x);}

return;

}

public static void main(String[] input) {

TestFactors next = new TestFactors();

next.factor(75);

}

}

azarrabia at 2007-7-7 16:15:18 >

> Hey guys,

> I am not sure I am following this conversation

> accurately. However,

> I dont think the code will work accurately using a

> for loop. The reason

> is that after a factorial is found, the for loop will

> continue.

No, in the version I posted, there is a break statement.

weebiba at 2007-7-20 1:47:41 >