CODEWITHSUNDEEP
phpnumbersfactoring
Belgin Fish
Belgin Fish

Reputation: 19847

Find what 2 numbers add to something and multiply to something

Hey so I'm making a factoring program and I'm wondering if anyone can give me any ideas on an efficient way to find what two numbers multiple to a specified number, and also add to a specified number.

for example I may have

(a)(b) = 6

a + b = 5

So essentially i just need a way to find the a and b values. In this case they would be 2 and 3.

Can anyone give me any ideas on where to start? Negative numbers must also be considered for use.

Upvotes: 2

Views: 225276

Answers (4)

danii
danii

Reputation: 5703

There is no need to loop, just use simple math to solve this equation system:

a*b = i;

a+b = j;

a = j/b;

a = i-b;

j/b = i-b; so:

b + j/b + i = 0

b^2 + i*b + j = 0

From here, its a quadratic equation, and it's trivial to find b (just implement the quadratic equation formula) and from there get the value for a.

There you go:

function finder($add,$product)
{

 $inside_root = $add*$add - 4*$product;

 if($inside_root >=0)
 {

     $b = ($add + sqrt($inside_root))/2;
     $a = $add - $b;

     echo "$a+$b = $add and $a*$b=$product\n";

 }else
 {
   echo "No real solution\n";
 }
}

Real live action:

http://codepad.org/JBxMgHBd

Upvotes: 8

Michael Borgwardt
Michael Borgwardt

Reputation: 346397

That's basically a set of 2 simultaneous equations:

x*y = a
X+y = b

(using the mathematical convention of x and y for the variables to solve and a and b for arbitrary constants).

But the solution involves a quadratic equation (because of the x*y), so depending on the actual values of a and b, there may not be a solution, or there may be multiple solutions.

Upvotes: 1

DaveRandom
DaveRandom

Reputation: 88697

Here is how I would do that:

$sum = 5;
$product = 6;

$found = FALSE;
for ($a = 1; $a < $sum; $a++) {
  $b = $sum - $a;
  if ($a * $b == $product) {
    $found = TRUE;
    break;
  }
}

if ($found) {
  echo "The answer is a = $a, b = $b.";
} else {
  echo "There is no answer where a and b are both integers.";
}

Basically, start at $a = 1 and $b = $sum - $a, step through it one at a time since we know then that $a + $b == $sum is always true, and multiply $a and $b to see if they equal $product. If they do, that's the answer.

See it working

Whether that is the most efficient method is very much debatable.

Upvotes: 3

Brendon Dugan
Brendon Dugan

Reputation: 2168

With the multiplication, I recommend using the modulo operator (%) to determine which numbers divide evenly into the target number like:

$factors = array();
for($i = 0; $i < $target; $i++){
    if($target % $i == 0){
        $temp = array()
        $a = $i;
        $b = $target / $i;
        $temp["a"] = $a;
        $temp["b"] = $b;
        $temp["index"] = $i;
        array_push($factors, $temp);
    }
}

This would leave you with an array of factors of the target number.

Upvotes: 2

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