PHP：在数据库中查找一组数字，总和为特定数字

Question

首先，我是 php 新手...所以我仍然在程序上编码和理解 php。也就是说，

我有一组数字（数量）存储在数据库中。

问题：使用 PHP 和 mySQL，

1. 从数据库中提取此信息以便金额与其交易 ID 相关联的最佳方法是什么

2. 最重要的是，我需要在数据库中找到一组匹配的数字，等于 29 的总和。

下面是我的数据库mydb的事务表Transaction_tlb

    Transaction_ID |     Name         |       Date      | Amount
    ---------------|------------------|-----------------|------------ 
    11012          | Jonathan May     |   6/12/2016     |     84
    21012          | John Pedesta     |   6/12/2016     |     38
    31012          | Mary Johnson     |   1/01/2017     |     12
    41012          | John Johnson     |   8/01/2017     |     13
    51012          | Keith Jayron     |   8/01/2017     |     17
    61012          | Brenda Goldson   |   8/01/2017     |     2
    71012          | Joshua Traveen   |   8/01/2017     |     78
    81012          | Remy ma Goldstein|   8/01/2017     |     1
    91012          | Barbie Traveen   |   8/01/2017     |     1

现在，我有一个想法..但它没有效率。我将尝试所有可能的情况。这意味着如果我有 n 个值要检查，时间复杂度将约为 2^n。这是非常低效的（另外，我什至不知道我的代码是否有意义。（见下文）

我在这个 YouTube 视频中看到了一个类似的例子：https://www.youtube.com/watch?v=XKu_SEDAykw&t

但是，我不确定如何在 php 中编写代码。

代码：

<?php
  if (!mysql_connect("localhost", "mysql_user", "mysql_password") || !mysql_select_db("mydb")) {
      die("Could not connect: " . mysql_error()); } //End DB Connect

  $capacity = 29; //Knapsack Capacity or Sum

  //Select Transact ID and Value from the Database where Amount is <= Capacity
  $fetchQuery = "SELECT 'Transaction_ID', 'Amount' FROM 'Transaction_tlb' WHERE 'Amount' <= $capacity"; 

  $components = array(); //new array to hold components

  if ($queryResults = mysql_query($fetchQuery)) {

     //check if data was pulled
     if (mysql_num_row($queryResults) != NULL) {
        while ($row = mysqli_fetch_assoc($queryResults) {
           $components[$row['Transaction_ID']] = $row['Amount'];
        }
     }
  }

  /* Correct me if i am wrong, but, Components associative array Should be something like
  $components = array('11012'=> 84, '21012'=> 38, '31012'=> 12, '41012'=> 13, '51012'=> 17, 
                      '61012'=> 2, '71012'=> 78, '81012'=> 1, '91012'=> 1);
  */

  $components = asort($components) // sort array in ascending order
  $componentCount = count($component)


  function match ($componentCount, $capacity) {
              $temp = match (($componentCount - 1), $capacity);
              $temp1 = $component[$componentCount] + match (($componentCount - 1), ($capacity - $component[$componentCount]));
              $result = max($temp, $temp1);
         return $result;
         }
}?>

谁能指出我正确的方向？这段代码不起作用......即使它起作用......该方法根本没有效率。当我有 300 万条记录可供使用时会发生什么？我需要帮助。

Answer 1

您可以根据0/1 Knapsack problem 来表述您的问题。 PHP 中现成的实现是available。

使用链接页面中定义的函数knapSolveFast2，可以按照以下示例进行操作。这里的想法是您将进入背包算法的“权重”设置为等于值本身。

$components = array(84, 38, 12, 13, 17, 2, 78, 1, 1);

$m = array();
list($m4, $pickedItems) = knapSolveFast2($components, $components, sizeof($components)-1, 29, $m);

echo "sum: $m4\n";
echo "selected components:\n";
foreach($pickedItems as $idx){
    echo "\t$idx --> $components[$idx]\n";
}

产生：

sum: 29
selected components:
    2 --> 12
    4 --> 17 

注意事项：

• 您可以修改 SQL 查询以跳过 amount 大于所需总和 (29) 的行
• 上面的函数会选择一个解决方案（假设它存在），它不会提供所有解决方案
• 应该检查返回值$m4 是否确实等于指定的总和 (29) - 当算法起作用时，指定的数量只是不能保证达到的上限（例如 37 代替29，返回值只有 34，因为没有输入数字的组合，其总和将产生 37)

Answer 2

这确实是一个背包问题，但我会尝试给出一个不是最优的完整解决方案，但会说明解决问题的完整策略。

首先，您只需对数字数组进行一次迭代即可完成此操作，无需递归和预排序。您只需要动态编程，跟踪所有以前可能的部分和“路径”。这个想法有点类似于您描述的递归方法，但我们可以迭代地完成它而无需预排序。

假设输入数组为 [84, 38, 12, 13, 17, 2, 78, 1, 1]，目标为 29，我们循环遍历这些数字，如下所示：

* 84 - too big, move on
* 38 - too big, move on
* 12 - gives us a subtarget of 29-12 = 17
            subtargets:
              17 (paths: 12)
* 13 - gives us a subtarget of 29-13=16
            subtargets:
              16 (paths: 13)
              17 (paths: 12)
* 17 - is a subtarget, fulfilling the '12' path;
   and gives us a subtarget of 29-17=12
            subtargets:
              12 (paths: 17)
              16 (paths: 13)
              17 (paths: 12)
            solutions:
              12+17
etc.

这里的诀窍是，在循环数字时，我们会保留一个subTargets 的查找表，这些数字可以使用先前看到的数字的一个或多个组合（“路径”）为我们提供解决方案。如果新数字是子目标，我们将添加到解决方案列表中；如果不是，那么我们追加到现有路径 num<subTarget 并继续前进。

一个快速而肮脏的 PHP 函数来做到这一点：

// Note: only positive non-zero integer values are supported
// Also, we may return duplicate addend sets where the only difference is the order
function findAddends($components, $target)
{
    // A structure to hold our partial result paths
    // The integer key is the sub-target and the value is an array of string representations
    // of the 'paths' to get to that sub-target. E.g. for target=29
    // subTargets = {
    //   26: { '=3':true },
    //   15: { '=12+2':true, '=13+1':true }
    // }
    // We are (mis)using associative arrays as HashSets
    $subTargets = array();

    // And our found solutions, stored as string keys to avoid duplicates (again using associative array as a HashSet)
    $solutions = array();

    // One loop to Rule Them All
    echo 'Looping over the array of values...' . PHP_EOL;
    foreach ($components as $num) {
        echo 'Processing number ' . $num . '...' . PHP_EOL;

        if ($num > $target) {
            echo $num . ' is too large, so we skip it' . PHP_EOL;
            continue;
        }

        if ($num == $target) {
            echo $num . ' is an exact match. Adding to solutions..' . PHP_EOL;
            $solutions['='.$num] = true;
            continue;
        }

        // For every subtarget that is larger than $num we get a new 'sub-subtarget' as well
        foreach ($subTargets as $subTarget => $paths) {
            if ($num > $subTarget) { continue; }

            if ($num == $subTarget) {
                echo 'Solution(s) found for ' . $num . ' with previous sub-target. Adding to solutions..' . PHP_EOL;
                foreach ($paths as $path => $bool) {
                    $solutions[$path . '+' . $num] = true;
                }
                continue;
            }

            // Our new 'sub-sub-target' is:
            $subRemainder = $subTarget-$num;
            // Add the new sub-sub-target including the 'path' of addends to get there
            if ( ! isset($subTargets[$subRemainder])) { $subTargets[$subRemainder] = array(); }

            // For each path to the original sub-target, we add the $num which creates a new path to the subRemainder
            foreach ($paths as $path => $bool) {
                $subTargets[$subRemainder][$path.'+'.$num] = true;
            }
        }

        // Subtracting the number from our original target gives us a new sub-target
        $remainder = $target - $num;

        // Add the new sub-target including the 'path' of addends to get there
        if ( ! isset($subTargets[$remainder])) { $subTargets[$remainder] = array(); }
        $subTargets[$remainder]['='.$num] = true;
    }
    return $solutions;
}

像这样运行代码：

$componentArr = array(84, 38, 12, 13, 17, 2, 78, 1, 1);
$addends = findAddends($componentArr, 29);

echo 'Result:'.PHP_EOL;
foreach ($addends as $addendSet => $bool) {
    echo $addendSet . PHP_EOL;
}

哪个输出：

Looping over the array of values...
Processing number 84...
84 is too large, so we skip it
Processing number 38...
38 is too large, so we skip it
Processing number 12...
Processing number 13...
Processing number 17...
Solution(s) found for 17 with previous sub-target. Adding to solutions..
Processing number 2...
Processing number 78...
78 is too large, so we skip it
Processing number 1...
Processing number 1...
Solution(s) found for 1 with previous sub-target. Adding to solutions..

Result:
=12+17
=12+13+2+1+1