drupol\phpartition\Linear::fillPartitions()

Complexity

Conditions	9
Paths	128

Size

Total Lines	83
Code Lines	42

Duplication

Lines	0
Ratio	0 %

Importance

Changes

0

<?php
    protected function fillPartitions(Partitions $partitions, Partition $dataset, int $chunks): void
    {
        $dataset = $dataset->getArrayCopy();

        // An array S of non-negative numbers {s1, ... ,sn}
        $s = \array_merge([null], $dataset); // adapt indices here: [0..n-1] => [1..n]
        // Integer K - number of ranges to split items into
        $k = $chunks;
        $n = \count($dataset);
        // Let M[n,k] be the minimum possible cost over all partitionings of N elements to K ranges
        $m = [];
        // Let D[n,k] be the position of K-th divider
        // which produces the minimum possible cost partitioning of N elements to K ranges
        $d = [];
        // Note: For code simplicity we don't use zero indices for `m` and `d`
        //       to make code match math formulas
        // Let pi be the sum of first i elements (cost calculation optimization)
        $p = [];
        // 1) Init prefix sums array
        //    pi = sum of {s1, ..., si}
        $p[0] = $this->getPartitionItemFactory()::create(0);
        for ($i = 1; $i <= $n; ++$i) {
            $p[$i] = $this->getPartitionItemFactory()::create($p[$i - 1]->getWeight() + $s[$i]->getWeight());
        }
        // 2) Init boundaries
        for ($i = 1; $i <= $n; ++$i) {
            // The only possible partitioning of i elements to 1 range is a single all-elements range
            // The cost of that partitioning is the sum of those i elements
            $m[$i][1] = $p[$i]; // sum of {s1, ..., si} -- optimized using pi
        }
        for ($j = 1; $j <= $k; ++$j) {
            // The only possible partitioning of 1 element into j ranges is a single one-element range
            // The cost of that partitioning is the value of first element
            $m[1][$j] = $s[1];
        }
        // 3) Main recurrence (fill the rest of values in table M)
        for ($i = 2; $i <= $n; ++$i) {
            for ($j = 2; $j <= $k; ++$j) {
                $solutions = [];
                for ($x = 1; ($i - 1) >= $x; ++$x) {
                    $solutions[] = [
                        0 => $this->getPartitionItemFactory()::create(
                            \max(
                                $m[$x][$j - 1]->getWeight(),
                                $p[$i]->getWeight() - $p[$x]->getWeight()
                            )
                        ),
                        1 => $x,
                    ];
                }

                \usort(
                    $solutions,
                    static function (array $x, array $y) {
                        return $x[0] <=> $y[0];
                    }
                );

                $best_solution = $solutions[0];
                $m[$i][$j] = $best_solution[0];
                $d[$i][$j] = $best_solution[1];
            }
        }

        // 4) Reconstruct partitioning
        $i = $n;
        $j = $k;
        $partition = [];
        while (0 < $j) {
            // delimiter position
            $dp = $d[$i][$j] ?? 0;
            // Add elements after delimiter {sdp, ..., si} to resulting $partition.
            $partition[] = \array_slice($s, $dp + 1, $i - $dp);
            // Step forward: look for delimiter position for partitioning M[$dp, $j-1]
            $i = $dp;
            --$j;
        }

        // Fix order as we reconstructed the partitioning from end to start
        $partition = \array_reverse($partition);

        foreach ($partition as $i => $p) {
            $partitions->partition($i)->exchangeArray($p);
        }