Algorithm for a poker-style scoring system

Question

What I need is to create five random integer (say rand(1,5)). Then, I generate a score based on these numbers. For instance, if I get a result of 1,2,3,4,5 then that would equal a zero score, but if I got 1,1,3,4,5 that would be 1 as we have a pair. Similar to a poker kind of scoring, so five of the same number would be a "full house" thus resulting in the highest score.

How would I go about the scoring system, even if it is just the mathematical equation?

More detail:

1-5 will hold separate images and then will be fought against "The House" which will have identical code to the user to determine the winner. Here's some example draws and the score they would receive:

1,2,3,4,5 = score 0
1,1,2,3,4 = score 1 (1 pair)
1,1,2,2,4 = score 2 (2 pair)
1,1,1,3,4 = score 3 (3 of a kind)
1,1,1,1,5 = score 4 (4 of a kind)
1,1,1,3,3 = score 5 (full house)
1,1,1,1,1 = score 6 (5 of a kind)

The combination of numbers is irreverent if they score 6 and the house scores 6, it's a tie.

    if (isset($_POST['play'])) {
    $rand1 = rand(1, 5);
    $rand2 = rand(1, 5);
    $rand3 = rand(1, 5);
    $rand4 = rand(1, 5);
    $rand5 = rand(1, 5);
    if ($_POST['bet'] <= $user_data['coins']) {
        if ($_POST['bet'] < 999999999) {
            if ($_POST['bet'] > 0.99) {
                if ($user_data['coins'] >= 1) {
                    $array = array($rand1,$rand2,$rand3,$rand4,$rand5);
                    print_r(array_count_values($array));
                    echo $rand1.', '.$rand2.', '.$rand3.', '.$rand4.', '.$rand5;

                    Array( // Here I don't understand
                    1 => 3,//
                    2 => 1,//
                    3 => 1 //
                    );
                }
            }
        }
    }   
}

This outputs ; Array ( [5] => 2 [4] => 2 [1] => 1 ) 5, 5, 4, 4, 1

Answer 1

Here's the approach I would consider, building on @Lele's answer. Warning: this is a bit confusing, so sit down with a cup of tea for this one.

• Build a set of five buckets, [1] to [5], and scan a player's numbers, so that the count for each number is stored in the corresponding bucket
• Then count the numbers you are left with into a new bucket system, with each position representing the number of counts you have for something.

So, if your score is this:

1 1 2 2 4

Then your first buckets are:

2 2 0 1 0

That's because you have two ones, two twos, and one four. And your second buckets are:

1 2 0 0 0

That's because you have two two-counts, and one one-count. Here, you disregard the first position (since a one-count for something does not score anything) and score for the others. So, test for two twos, and score that two.

If you score is this:

5 5 5 5 1

Then your first buckets are:

1 0 0 0 4

That's one one and four fives. So your second buckets are:

1 0 0 1 0

Your lookup table for this could be:

x 1 0 0 0 -> one pair
x 2 0 0 0 -> two pairs
x 0 1 0 0 -> three of a kind
x 1 1 0 0 -> full house
x 0 0 1 0 -> four of a kind
x 0 0 0 1 -> five of a kind

The 'x' means that you don't match on this. So, your lookup table matches four numbers to a score.

I was rather interested in this problem, so I have written some code to do the above. You'll still need to do the lookup table, but that is relatively trivial, and will be good practice for you. Here is a demo, with comments (run code here):

<?php
    function counting(array $array) {
        // Input figures
        print_r($array);

        // Run the figures twice through the bucket-counter
        $firstBuckets = bucketCounter($array);
        $secondBuckets = bucketCounter($firstBuckets);

        // Ignore counts of 1
        array_shift($secondBuckets);

        // Output, just need to do the lookup now
        echo ' converts to ';
        print_r($secondBuckets);
        echo "<br />";
    }

    /**
     * Bucket counter
     */
    function bucketCounter(array $array) {
        $result = array(0, 0, 0, 0, 0, );
        foreach($array as $value) {
            if ($value > 0) {
                $result[$value - 1]++;
            }
        }

        return $result;
    }

    // Try some demos here!
    counting(array(1, 2, 3, 4, 5));
    counting(array(1, 1, 2, 4, 2));
    counting(array(1, 1, 1, 1, 1));

?>

The demos I've included seem to work, but do hunt for bugs!

Answer 2

Use array_count_value function for this.

$array = array(1,1,1,2,5);

print_r(array_count_values($array));

Array(
1 => 3,
2 => 1,
3 => 1
);

Answer 3

If the range is quite small, you can use counting sort approach. For each number, provide a "bucket" to count how many times a number appear. Scan once to fill in the buckets. Then another scan, but this time against the bucket to get the highest value. That's your score.