我正在编写一个脚本，将任意数字$epi均分为任意数量的二进制位$dpi。 epi代表Ends per Inch。 dpi表示每英寸凹痕。有3个要求：
这是我到目前为止所拥有的。当运行space()方法时，它主要执行我需要的但，如果其foreach循环必须执行多次，则$ epi的分布不均匀。
<?php
class ReedSubstitution{
public $epi;
public $dpi;
function substitute($e,$d){
$this->epi=is_numeric($e)?$e:0;
$this->dpi=is_numeric($d)?$d:0;
if(empty($this->epi)||empty($this->dpi)) throw new Exception('Both desired ends per unit and available dents per unit must be specified.');
if($this->epi%$this->dpi ==0) return array($this->epi/$this->dpi);//short circuit for easy case
$this->unfraction();//make equivalent integers if dpi or epi are fractional
$gcd= ($this->epi < $this->dpi) ? $this->euclid($this->epi,$this->dpi) : $this->euclid($this->dpi,$this->epi);
$e=$this->epi/$gcd;
$d=$this->dpi/$gcd;
$q=floor($e/$d);//count that every dent gets
$r=$e%$d;//extra to be spread out over array
$reed=array_fill(0,$d,$q);
$this->space($reed,$r);
return $reed;
}
protected function unfraction(){
$emult=1;
$dmult=1;
if($this->epi-intval($this->epi)){ //epi has a fractional component
list($tmp,$er)=explode('.',$this->epi);
$emult=pow(10,strlen($er));
}
if($this->dpi-intval($this->dpi)){//dpi has a fractional component
list($tmp,$dr)=explode('.',$this->dpi);
$dmult=pow(10,strlen($dr));
}
$mult=($emult>$dmult)?$emult:$dmult;
$this->epi*=$mult;
$this->dpi*=$mult;
}
/**
* @desc evenly distribute remaining ends over entirety of dents
* @param Array $reed, dents in one pattern repeat
* @param Integer $r, remaining ends to be distributed
*/
protected function space(&$reed,$r){
$c=count($reed);
$i=0;
$jump=($r < $c-$r) ? $r : $c-$r;//use the smallest jump possible to reduce the looping
do{
for($i; $i<$c; $i=$i+$jump, $r--){
if($r==0)break;
$reed[$i]++;
}
$i=array_search(min($reed),$reed);//begin next loop (if necessary) at position with fewest ends
}while($r>0);
}
/**
* @desc return greatest common divisor as determined by Euclid's algorithm
* @param integer $large
* @param integer $small
* @return integer
*/
protected function euclid($large, $small){
$modulus=$large%$small;
if($modulus==0) {
return $small;
} else if($modulus==1){
return 1;
} else {
return $this->euclid($small,$modulus);//recursion
}
}
}
?>
输出错误：
$r=new ReedSubstitution();
$arr=$r->substitute(9,28);
/* BAD DISTRIBUTION
Array
(
[0] => 1
[1] => 1
[2] => 1
[3] => 0
[4] => 0
[5] => 0
[6] => 0
[7] => 0
[8] => 0
[9] => 1
[10] => 1
[11] => 1
[12] => 0
[13] => 0
[14] => 0
[15] => 0
[16] => 0
[17] => 0
[18] => 1
[19] => 1
[20] => 0
[21] => 0
[22] => 0
[23] => 0
[24] => 0
[25] => 0
[26] => 0
[27] => 1
)
*/
以上分布应如下所示：
/* GOOD DISTRIBUTION
Array
(
[0] => 1
[1] => 0
[2] => 0
[3] => 1
[4] => 0
[5] => 0
[6] => 1
[7] => 0
[8] => 0
[9] => 1
[10] => 0
[11] => 0
[12] => 1
[13] => 0
[14] => 0
[15] => 1
[16] => 0
[17] => 0
[18] => 1
[19] => 0
[20] => 0
[21] => 1
[22] => 0
[23] => 0
[24] => 1
[25] => 0
[26] => 0
[27] => 0
)
*/
如何修复space()方法，以便需要多个循环的分箱才能产生可接受的分布？
答案 0 :(得分：2)
我希望这有助于或至少指出正确的方向：
protected function space(&$reed, $r)
{
$totalReeds = count($reed);
$f = floor($r/$totalReeds);
$d = ($r % $totalReeds) / $totalReeds;
$c = 0;
for ($i = 0; $i < $totalReeds; $i++)
{
$add = round($f + $d + $c);
$reed[$i] += $add;
$c = $c + $f + $d - $add;
}
}
然而，它可能不会产生完全你可能期望的东西：
Array
(
[0] => 2
[1] => 1
[2] => 2
[3] => 2
[4] => 1
[5] => 2
[6] => 1
[7] => 2
)
虽然结果是均匀分布。
P.S。我不太了解你所处理的真实网站相关问题，但我希望我正确掌握了数学概念。
答案 1 :(得分：0)
您可以尝试以下空间功能。
protected function space(&$reed, $r)
{
$totalReeds = count ($reed);
$postion = floor($totalReeds/$r);
for ($i=0; $i<=$totalReeds; $i=$i+$postion, $r--) {
if ($r <= 0) break;
$reed[$i]++;
}
}
我只是尝试了一些输入，它对我有用。而且它给了我前面给出的例子的正确输出。
您可以尝试使用此功能，让我知道它是否适用于您的所有输入。
- 编辑 -
试试这段代码。
protected function space(&$reed, $r)
{
$totalReeds = count ($reed);
$postion = floor($totalReeds/$r);
$postion = $postion == 1? 2 : $postion;
for ($i=0; $i<=$totalReeds; $i=$i+$postion, $r--) {
if ($r <= 0) break;
if (isset($reed[$i])) $reed[$i]++;
}
}
我希望这有效。我不确定您在评论中说的其他输入的预期结果。所以我假设以下是正确的。如果我错了，请发布其他输入的预期结果。
此致
答案 2 :(得分：-1)
老实说，我没有得到关于它的数学问题，因为我发现的关于箱子和凹痕的一切都在编织（我不是母语，所以我可能错过了一些明显的东西），但我想我理解你的要求（如果这不是一个严格的数学问题）。
无论如何，我把它清理了一下，所以我发布完整的类代码，但如果你不喜欢它，你可以使用space（）separetely。我的算法可以进行很多优化（第二个循环可以删除），但我懒得去做。
class ReedSubstitution {
private $epi;
private $dpi;
public function substitute($e, $d) {
$this->epi = floatval($e);
$this->dpi = floatval($d);
if (empty($this->epi) || empty($this->dpi)) {
throw new Exception('Both desired ends per unit and available dents per unit must be specified.');
}
//make equivalent integers if dpi or epi are fractional
$this->unfraction();
if ($this->epi % $this->dpi == 0) {
return array($this->epi / $this->dpi);
}
$gcd = $this->euclid($this->epi, $this->dpi);
$e = $this->epi / $gcd;
$d = $this->dpi / $gcd;
$r = $e % $d; //extra to be spread out over array
$q = ($e - $r) / $d; //count that every dent gets
$reed = array_fill(0, $d, $q);
$this->space($reed, $r);
return $reed;
}
protected function unfraction() {
//Find fraction start position
$epi_fract_pos = strpos($this->epi, '.');
$dpi_fract_pos = strpos($this->dpi, '.');
//Find fraction length
$epi_fract_len = $epi_fract_pos ? (strlen($this->epi) - $epi_fract_pos - 1) : 0;
$dpi_fract_len = $dpi_fract_pos ? (strlen($this->dpi) - $dpi_fract_pos - 1) : 0;
//Calculate max fraction length
$fraction_len = max($epi_fract_len, $dpi_fract_len);
//Unfraction variables
$mult = pow(10, $fraction_len);
$this->epi*=$mult;
$this->dpi*=$mult;
}
/**
* @desc evenly distribute remaining ends over entirety of dents
* @param Array $reed, dents in one pattern repeat
* @param Integer $r, remaining ends to be distributed
*/
protected function space(&$reed, $r) {
$c = count($reed);
$base = $reed[0];
for ($i = 0; $i < $r; $i++) {
$reed[$i]++;
}
while ($reed[$c - 1] === $base) {
$reed[$c] = $base;
//Find the longest $base+1 array with least $base behind it
$max_base_plus_size = -$c;
$cur_base_plus_size = 0;
$cur_base_plus_pos = array(NULL, NULL);
$max_base_plus_pos = NULL;
for ($i = 0; $i <= $c; $i++) {
if ($reed[$i] != $base) {
if($cur_base_plus_pos[0]===NULL) {
$cur_base_plus_pos[0] = $i;
}
if ($reed[$i + 1] == $base) {
$cur_base_plus_pos[1]=$i;
if ($max_base_plus_size < $cur_base_plus_size) {
$max_base_plus_size = $cur_base_plus_size;
$max_base_plus_pos = $cur_base_plus_pos;
}
$cur_base_plus_size = 0;
$cur_base_plus_pos = array(NULL, NULL);
}
else {
$cur_base_plus_size++;
}
} else {
$cur_base_plus_size--;
$cur_base_plus_pos[1] = $i - 1;
}
}
$insert_pos=ceil(($max_base_plus_pos[1]+$max_base_plus_pos[0])/2);
array_splice($reed,$insert_pos,0,$base);
}
array_splice($reed,$c);
$reed=array_reverse($reed);//Optional, just to get the same output as you submitted
}
/**
* @desc return greatest common divisor as determined by Euclid's algorithm
* @param integer $x
* @param integer $y
* @return integer
*/
protected function euclid($x, $y) {
while ($x) {
$temp = $x;
$x = $y % $x;
$y = $temp;
}
return $temp;
}
}