Reputation: 1512
This is more of a puzzle than anything. I've actually found a solution but it is so slow I thought I lost my internet connection (see below).
Let's say I have an array of numbers, like so:
$numbers_array = array(1, 2, 3, 4, 5, 6, 7, 8, 9);
Let's also say that I have a number some numbers, stored in variables like so:
$sum = 15;
$sum2 = 24;
$sum3 = 400;
I am trying to create a function that will return true if any of the numbers in $numbers_array
can be added together (each only used once) to form the sums:
function is_summable($array_of_nums, $sum_to_check) {
//What to put here?
}
var_dump(is_summable($numbers_array, $sum));
var_dump(is_summable($numbers_array, $sum2));
var_dump(is_summable($numbers_array, $sum3));
The above should output:
bool(true)
bool(true)
bool(false)
Because 7 + 8 = 15, 7 + 8 + 9 = 24, but no combination of 1-9 can create 200.
Here's my EXTREMELY slow solution:
function is_summable($numbers, $sum) {
//Sort provided numbers and assign numerical keys.
asort($numbers);
$numbers = array_values($numbers);
//Var for additions and var for number of provided numbers.
$total = 0;
$numbers_length = count($numbers);
//Empty var to fill below.
$code = '';
//Loop and add for() loops.
for ($i = 0; $i < $numbers_length; $i++) {
$code .= 'for ($n' . $i . ' = 0; $n' . $i . ' < ' . $numbers_length . '; $n' . $i . '++) {';
if ($i != 0) {
$code .= 'if ($n' . $i . ' != $n' . ($i - 1) . ') {';
}
$code .= '$total += intval($numbers[$n' . $i . ']);';
$code .= 'if ($total == $sum) {';
$code .= 'return true;';
$code .= '}';
}
//Add ending bracket for for() loops above.
for ($l = 0; $l < $numbers_length; $l++) {
$code .= '$total -= intval($numbers[$n' . $i . ']);';
if ($l != 0) {
$code .= '}';
}
$code .= '}';
}
//Finally, eval the code.
eval($code);
//If "true" not returned above, return false.
return false;
}
$num_arr = array(1,2,3,4,5,6,7,8,9);
var_dump(is_summable($num_arr, 24));
As always, help is appreciated!!
Upvotes: 8
Views: 451
Reputation: 4475
Your problem is in fact a standard algorithmic problem (as Jon mentioned knapsack problem), more specifically Subset sum problem. It can be solved in polynomial time (have look on wiki page).
Pseudocode:
initialize a list S to contain one element 0.
for each i from 1 to N do
let T be a list consisting of xi + y, for all y in S
let U be the union of T and S
sort U
make S empty
let y be the smallest element of U
add y to S
for each element z of U in increasing order do
//trim the list by eliminating numbers close to one another
//and throw out elements greater than s
if y + cs/N < z ≤ s, set y = z and add z to S
if S contains a number between (1 − c)s and s, output yes, otherwise no
Upvotes: 3