CODEWITHSUNDEEP
phpmath
JREAM
JREAM

Reputation: 5931

Stuck on Simple Math Algorithm

I can't remember what this is called in math, it's not powers though. I'm validating a consecutive order of numbers. They always start at 1 and can go to 3. I really have 3 conditions only.

So my solution was:

$sum = array_sum($group);
count: 1 = (Result) 1
count: 2 = (Result) 3 (1 + 2)
count: 3 = (Result) 6 (1 + 2 + 3)

However, I would like to trim the fat of these 3 if statements into something simpler, can you help me with that math algorithm?

    $winnerCount = [
        [1, 3]
    ];

    foreach ($winnerCount as $_key => $_group)
    {
        $winnerTotal = count($_group);
        $sum = array_sum($winnerCount[$_key]);

        if ($winnerTotal == 1 && $sum != 1) {
            $error = true;
        }
        if ($winnerTotal == 2 && $sum != 3) {
            $error = true;
        }
        if ($winnerTotal == 3 && $sum != 6) {
            $error = true;
        }

        echo $sum;
    }

I do reckon I need some coffee this fine afternoon.

I suppose I could do something kinda like this:

$result = [
    1 => 1,
    2 => 3,
    3 => 6
];

if ($winnerTotal != $result[$winnerTotal]) {
    $error = true
}

Upvotes: 0

Views: 113

Answers (3)

Zzyrk
Zzyrk

Reputation: 905

formula : n = n(n+1)/2

therefore you could try the followong in your loop :

if($winnerTotal != $winnerTotal ($winnerTotal + 1 ) / 2){
    $error = true;
}

Upvotes: 2

Fred the Magic Wonder Dog
Fred the Magic Wonder Dog

Reputation: 9109

The nth triangular number is given by the formula 

g(n) = n(n+1)/2

Upvotes: 2

Alireza Fallah
Alireza Fallah

Reputation: 4607

You can use a loop like this :

function sumNum($num){
    $sum = 0 ;
    for($i=1;$i<=$num;$i++){
        $sum += $i;
    }
    return $sum;
}
$sum = sumNum($count);

OR most simple way :

$sum = ($count*($count+1))/2;

Upvotes: 1

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