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Thread: Calculating Factors

09232005, 06:06 PM #1
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Calculating Factors
Hi, I need a way of working out what two numbers can be multiplied together to create another number, e.g.
If my number was 18
I could do:
1*9
2*9
3*6
Does anyone know of a Javascript function which is able to work this out given a number?

09232005, 06:27 PM #2

09262005, 03:43 PM #3
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Thanks for the link, however it does not seem to work fully, if I type in 100 and ask it to give me the factors it gives:
Code:Number: 100 is Even Factors Are: (50,2)(25,4)(20,5)

09262005, 09:52 PM #4
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Easy enough to fix. Make the changes in red to that script mentioned above.
Code:function calc() { var dnum = ((eval(document.res.inpa.value)) / 2); var i; var pol; var inum = (Math.round(dnum)); if (inum == dnum) { document.res.rses.value="Number: " + (eval(document.res.inpa.value)) + " is Even\n\n"; } else { document.res.rses.value="Number: " + (eval(document.res.inpa.value)) + " is Odd\n\n"; } document.res.rses.value += "Factors Are:\n\n"; var num = Math.round(eval(document.res.inpa.value)); for (i = 1; i < (num / 2); i++) { var chkd = Math.round(num / i); var inn = Math.round(num / i); var outt = (num / i); if (inn == outt && chkd >= i) { document.res.rses.value = document.res.rses.value + "(" + (num/i) + "," + i + ")"; } } }
I was going to point to a prime factorization script that I wrote for a friend a couple years ago but it doesn't do exactly what you want  it only finds prime factors. 100 => 5, 5, 2, 2. But it does it FAST. It works very well for large numbers.

09272005, 05:33 PM #5
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Prime factorization...
This function returns the prime factorization of the number passed into it.
Code:function prime(p){ if(!prime[p]){ var f=2; if(p%2) for(f=3;(p%f)&&(f*f<=p);f+=2); if(f*f>p){ prime[p]=p } else{ prime[p]=(prime[f]=f)+'*'+prime(p/f) } } return prime[p] }

09272005, 08:33 PM #6
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O/t
ca_redwards:
Your function consistently beats mine by a handful of loop iterations and recursive calls.
I tested with this number: 994070203729722
Yours ran 18875 total loop iterations with 5 recursive calls.
Mine ran 18939 loops with 10 recursive calls.
Both functions found these primes: 2, 3, 3, 7, 11149, 707637103
Mine:Code:function prime(number){ var sq = Math.sqrt(number); for(var i = 2; i <= sq; i++){ if(number%i == 0 && number != i){ return prime(i) + " * " + prime(number/i); } if(i>2) i++; } return number; }

09272005, 09:12 PM #7
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Prime factorization...
A few notes about these functions...
Originally Posted by kanselOriginally Posted by ca_redwards
When iterating through all the possible odd factors beyond 2, your code repeatedly compares the trial factor against 2, whereas mine does not.
Your function returns a nonprime as a product of two recursive results, but in the same case, mine employs only the second recursive term. The first term, known to be prime, is stored into an enumerated property of the function itself. And the final result is stored likewise.
Thus, before factoring anything, my function first checks to see if the answer is known. If so, it is summarily returned. My function learns from experience.
There might be an even better solution where possible factors are rejected if they are known to be factorable...
What do you think?Last edited by ca_redwards; 09272005 at 09:41 PM.

09282005, 01:46 PM #8
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I have managed to get the code down to:
Code:function calc() { var i; var num = Math.round(100); for (i = 1; i < (num / 2); i++) { var inn = Math.round(num / i); var outt = (num / i); if (inn == outt && inn >= i) { document.getElementById('fact').value = document.getElementById('fact').value + "(" + (num/i) + "," + i + ")"; } } }

09282005, 02:32 PM #9
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Code:var roundUp = Math.ceil(x); var roundDown = Math.floor(x); var round = Math.round(x);

10192005, 08:29 AM #10
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I realize that you've already resolved your problem, but perhaps this recursive method will be helpful to you or anyone else that was purusing this thread for such. Also, before anyone points it out, this is Java, not Javascript like the rest of the source provided in this thread.
Code:private int d=2; public String primeFactorRec(int n) { if(n==1) {return "";} else if(n%d==0) {n=n/d; return ((String.valueOf(d)+" "+primeFactorRec(n)).trim()).replace(' ','*');} else {d++; return primeFactorRec(n);} }
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