php – 消除不可能的选择

发布时间:2022-04-30 发布网站:脚本宝典
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我试图以编程方式解决这个问题时遇到了一些麻烦.这不是我正在做的事情,但为了简化事情,我们说有一定数量的球和一定数量的人.每个人必须选择一个球,人们可能只限于他们可以选择的球类型.目标是确定在消除所有不可能的组合后人们必须选择的选项.

例1:

一个简单的例子中,说我们有两个人,一个红球和一个绿球.人1可以选择任一球,但是人2只能选择绿球.这可以说明如下:

PErson 1: RG
Person 2: G

因为我们知道人2必须选择绿球,这意味着人1不能选择那个球,因此必须选择红球.所以这可以简化为:

Person 1: R
Person 2: G

所以在这种情况下,我们确切地知道每个人会选择什么.

例2:

现在让我们添加一些复杂性.现在我们有4个人需要从2个红球,1个绿球和4个蓝球中选择.人1可以选择任何球,人2和3可以选择红色或绿色球,人4必须选择红色球.所以我们有以下选择:

Person 1: RRGBBBB
Person 2: RRG
Person 3: RRG
Person 4: RR

由于人4只有一种选择,我们知道那个人必须选择一个红球.因此,我们可以从所有其他人中消除1个红球:

Person 1: RGBBBB
Person 2: RG
Person 3: RG
Person 4: RR

然而,这是它变得非常棘手的地方.我们可以看到,第2和第3人必须选择一个红球和一个绿球,我们只是不知道哪一个会选择哪一个.但是,由于我们每个球只剩下1个,因此红色和绿色也可以作为选项从第1个人中删除

Person 1: BBBB
Person 2: RG
Person 3: RG
Person 4: RR

现在,我们可以使用以下选项从每个条目中删除重复项:

Person 1: B
Person 2: RG
Person 3: RG
Person 4: R

我们知道人1和人4的选择,而人2和3可以选择红色和绿色.

为了解决这个问题,我所做的就是循环人们,首先如果人们只有一个球类型,从阵列中删除那个人并将其放入结果数组中(因为我知道这是那个人必须选择的)然后如果存在,则从阵列中的每个其他人移除该球类型中的一个.

在此之后,在我看来,规则是:

所以我所做的就是再次遍历所有人,并检查具有相同选项的其他人(或这些选项的子集),如果这等于该人的选项总数,请从选项中删除所有其他人.

这是我到目前为止解决这两种情况的方法

// The quantITy of each ball
$balls = array(
    'r' => 1,'g' => 1,'b' => 1,'k' => 1,);
// The options available for each person
$options = array(
    array('r','g','b','k'),array('r','g'),'b'),array('b',);

// Put both of these together into one array
$people = [];
foreach ($options as $option) {
    $person = [];
    foreach ($option as $ball_key) {
        $person[$ball_key] = $balls[$ball_key];
    }
    $people[] = $person;
}

PRint_r($people);
// This produces an array like:
// Array
// (
//     [0] => Array
//         (
//             [r] => 2
//             [g] => 1
//             [b] => 4
//         )
//
//     [1] => Array
//         (
//             [r] => 2
//             [g] => 1
//         )
//
//     [2] => Array
//         (
//             [r] => 2
//             [g] => 1
//         )
//
//     [3] => Array
//         (
//             [r] => 2
//         )
//
// )

// This will be used to hold the final result
$results = [];

do {
    // If anything changes,this needs to be set to true. Any time anything
    // changes we loop through everything again in case it caused a casCADing
    // effect
    $has_change = false;

    // Step 1:
    // Find out if there are any people who have only one option and remove it
    // From the array and add it to the result and suBTract one from all other
    // people with this option
    foreach ($people as $p_index => $p_options) {
        if (count($p_options) === 1) {
            $color = key($p_options);

            foreach ($people as $p_index_tmp => $p_options_tmp) {
                // It's the current person,so skip it
                if ($p_index_tmp === $p_index) {
                    continue;
                }

                if (isset($p_options_tmp[$color])) {
                    // Subtract 1 from this color from this person and if zero,// remove it.
                    if (--$people[$p_index_tmp][$color] === 0) {
                        unset($people[$p_index_tmp][$color]);
                    }

                    $has_change = true;
                }
            }

            // Add to results array and remove from people array
            $results[$p_index] = array($color => 1);
            unset($people[$p_index]);
        }
    }

    // Step 2:
    // If there are $n number of people who all have the same $n number of
    // options (or a subset of them),these options can be removed
    // from all other people,where $n is less than the total number of people
    foreach ($people as $p_index => $p_options) {
        $num_options = array_sum($p_options);
        if ($num_options < count($people)) {
            // Look for other people with no different options from the ones
            // that this person has
            $people_with_same_options = [];
            foreach ($people as $p_index_tmp => $p_options_tmp) {
                foreach (array_keys($p_options_tmp) as $color) {
                    if (array_seArch($color,array_keys($p_options)) === false) {
                        // This color was not found in the options,so we can
                        // skip this person.
                        // (Make sure we break out of both foreach loops)
                        continue 2;
                    }
                }
                // This person has all the same options,so append to the array
                $people_with_same_options[] = $p_index_tmp;
            }

            // Remove these options from the other people if the number of
            // people with only these options is equal to the number of options
            if (count($people_with_same_options) === $num_options) {
                foreach ($people as $p_index_tmp => $p_options_tmp) {
                    if (array_search($p_index_tmp,$people_with_same_options) === false) {
                        foreach (array_keys($p_options) as $option) {
                            unset($people[$p_index_tmp][$option]);

                            $has_change = true;
                        }
                    }
                }
            }
        }
    }
}
while ($has_change === true);

// Combine any remaining people into the result and sort it
$results = $results + $people;
ksort($results);

print_r($results);

这会产生以下结果:

Array
(
    [0] => Array
        (
            [b] => 1
        )

    [1] => Array
        (
            [r] => 1
            [g] => 1
        )

    [2] => Array
        (
            [r] => 1
            [g] => 1
        )

    [3] => Array
        (
            [r] => 1
        )

)

例3:

此示例不适用于上述代码.假设有1个红球,1个绿球,1个蓝球,1个黄球和4个人.人1可以选择任何球,人2可以选择红色或绿色,人3可以选择绿色或蓝色,人4可以选择红色或蓝色.

在视觉上这看起来像:

Person 1: RGBY
Person 2: RG
Person 3: GB
Person 4: RB

由于红色,绿色和蓝色3种颜色是2,3和4人的唯一选择,因此它们完全包含在这3个人中,因此它们都可以从人1中消除,这意味着人1必须选择黄色.如果第一个人选择黄色的东西,那么其他人就不可能选择他们的球.

把它放到我的PHP程序中,我有这些输入值:

// The quantity of each ball
$balls = array(
    'r' => 1,'y' => 1,'y'),);

但是我想不出如何循环查找这样的案例而不迭代每个可能的人组合.有什么想法可以做到一点

我更喜欢采用类似OOP的方法.所以我基本上从头开始.我希望你没关系.

所以,以下看起来(不可否认)非常丑陋,除了你的三个例子之外,我还没有测试过它,但是在这里

class Ball {
    private $color;
    public function __construct($color) {
        $this->color = $color;
    }
    public function getColor() {
        return $this->color;
    }
}
class Ball_resource extends Ball {
    private $num_available;

    public function __construct($color,$number) {
        parent::__construct($color);
        $this->num_available = $number;
    }
    public function take() {
        $this->num_available--;
    }

    public function isExhausted() {
        return $this->num_available <= 0;
    }
}
class Person {
    /**
     *
     * @VAR Ball
     */
    private $Allowed_balls = array();

    public function addConstraint($color) {
        $this->allowed_balls[$color] = new Ball($color);
        return $this;
    }
    public function getConstraints() {
        return $this->allowed_balls;
    }

    public function getNumberOfconstraints() {
        return count($this->allowed_balls);
    }

    /**
     * return true if removal was successful; false otherwise
     */
    public function removeConstraint(Ball $ball) { // todo remove
        if (isset ($this->allowed_balls [$ball->getColor()])) {
            unset ($this->allowed_balls [$ball->getColor()]);
            return true;
        }
        else {
            // this means our puzzle isn't solvable
            return false;
        }
    }
}
class Simplifier {
    /**
     *
     * @var Person
     */
    private $persons = array ();
    /**
     *
     * @var Ball_resource
     */
    private $availableBalls = array ();

    public function addPerson(Person $person) {
        $this->persons[] = $person;
        return $this;
    }
    public function addBallRessource(Ball_resource $ball_resource) {
        $this->availableBalls[] = $ball_resource;
        return $this;
    }


    public function getChoices() {      
        $queue = $this->persons; // shallow copy

        while (count($queue) > 0) {
            // find most constrained person(s)
            $numberOfConstraints = 1; // each person must have at least one constraint
            while (true) {
                $resolve_queue = array();
                foreach($queue as $person) {
                    if ($person->getNumberOfConstraints() === $numberOfConstraints) {
                        $resolve_queue[] = $person;
                    }
                }
                // find mutually dependent constraint groups connected with a person
                $First_run = true;
                foreach ($resolve_queue as $startPerson) {
                    // check if we havent already been removed via dependencies
                    if ($first_run || !self::contains($queue,$startPerson)) {
                        $dependent_persons = $this->findMutuallyDependentPersons($startPerson,$resolve_queue);
                        // make a set out of their combined constraints
                        $combinedConstraints = $this->getConstraintsSet($dependent_persons);
                        $this->adjustResources($dependent_persons);
                        $this->removeFromQueue($dependent_persons,$queue);
                        // substract each ball of this set from all less constrained persons
                        $this->substractConstraintsFroMLessConstrainedPersons($queue,$combinedConstraints,$numberOfConstraints);
                        $first_run = false;
                        continue 3;
                    }
                }
                $numberOfConstraints++;
            }

        }
        return $this->persons; // has been altered implicitly   
    }

    private static function contains(array $haystack,Person $needle) {
        foreach ($haystack as $person) {
            if ($person === $needle) return true;
        }
        return false;
    }

    private function findMutuallyDependentPersons(Person $startPerson,array $persons) {
        // add recursion
        $output = array();
        //$output[] = $startPerson;
        foreach($persons as $person) {
            foreach ( $person->getConstraints () as $constraint ) {
                foreach ( $startPerson->getConstraints () as $targetConstraint ) {
                    if ($constraint->getColor () === $targetConstraint->getColor ()) {
                        $output [] = $person;
                        continue 3;
                    }
                }
            }   
        }
        return $output;
    }

    private function getConstraintsSet(array $persons) {
        $output = array();
        foreach ($persons as $person) {
            foreach ($person->getConstraints() as $constraint) {
                foreach($output as $savedConstraint) {
                    if ($savedConstraint->getColor() === $constraint->getColor()) continue 2;                   
                }
                $output[] = $constraint;
            }
        }
        return $output;
    }

    private function substractConstraintsFromLessConstrainedPersons(array $persons,array $constraints,$constraintThreshold) {
        foreach ($persons as $person) {
            if ($person->getNumberOfConstraints() > $constraintThreshold) {
                foreach($constraints as $constraint) {
                    foreach($this->availableBalls as $availableBall) {
                        if ($availableBall->isExhausted()) {
                            $person->removeConstraint($constraint);
                        }
                    }                   
                }
            }
        }
    }

    private function adjustResources(array $persons) {
        foreach($persons as $person) {
            foreach($person->getConstraints() as $constraint) {
                foreach($this->availableBalls as &$availableBall) {
                    if ($availableBall->getColor() === $constraint->getColor()) {
                        $availableBall->take();
                    }
                }
            }
        }
    }

    private function removeFromQueue(array $persons,array &$queue) {       
        foreach ($persons as $person) {
            foreach ($queue as $key => &$availablePerson) {
                if ($availablePerson === $person) {
                    unset($queue[$key]);
                }
            }
        }
    }
}

整个事情被称为这样:

// Example 2
{
    $person1 = new Person();
    $person1->addConstraint("R")->addConstraint("R")->addConstraint("G")->addConstraint("B")->addConstraint("B")->addConstraint("B")->addConstraint("B");
    $person2 = new Person();
    $person2->addConstraint("R")->addConstraint("R")->addConstraint("G");
    $person3 = new Person();
    $person3->addConstraint("R")->addConstraint("R")->addConstraint("G");
    $person4 = new Person();
    $person4->addConstraint("R")->addConstraint("R");

    $redBalls = new Ball_resource("R",2);
    $greenBalls = new Ball_resource("G",1);
    $blueBalls = new Ball_resource("B",4);

    $simplifier = new Simplifier();
    $simplifier->addPerson($person1)->addPerson($person2)->addPerson($person3)->addPerson($person4);
    $simplifier->addBallRessource($redBalls)->addBallRessource($greenBalls)->addBallRessource($blueBalls);
    $output = $simplifier->getChoices();

    print_r($output);
}

同样对于例3:

// Example 3
{
    $person1 = new Person();
    $person1->addConstraint("R")->addConstraint("G")->addConstraint("B")->addConstraint("Y");
    $person2 = new Person();
    $person2->addConstraint("R")->addConstraint("G");
    $person3 = new Person();
    $person3->addConstraint("G")->addConstraint("B");
    $person4 = new Person();
    $person4->addConstraint("R")->addConstraint("B");

    $redBalls = new Ball_resource("R",1);
    $greenBalls = new Ball_resource("G",1);
    $yellowBalls = new Ball_resource("Y",1);   

    $simplifier = new Simplifier();
    $simplifier->addPerson($person1)->addPerson($person2)->addPerson($person3)->addPerson($person4);
    $simplifier->addBallRessource($redBalls)->addBallRessource($greenBalls)->addBallRessource($blueBalls)->addBallRessource($yellowBalls);
    $output = $simplifier->getChoices();

    print_r($output);
}

为简洁起见,我省略了原始输出.但是对于第二个例子,它基本上等于你在问题中的最后一个列表,而对于例3,它产生相当于:

Person 1: Y
Person 2: RG
Person 3: GB
Person 4: RB

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