PHP代碼未能給出corrent結果最有可能的邏輯錯誤

Question

過去幾天我一直在這個問題上絞盡腦汁，對於我的生活似乎可以找到問題代碼。本質上，這段代碼生成一個隨機數量的對象，其中包含一個x和y座標和一個半徑，然後代碼檢查新對象是否與任何其他對象相沖突，如果沒有，那麼它會將它添加到主數組中然後返回調用函數。我的問題是，當我加載頁面時，所有的對象都在那裏，但一些仍然相互碰撞，我不明白爲什麼。任何人都可以看到問題嗎？

public function Generate($chunkX, $chunkY) { 
    if (!(isset($this->ChunkX) && isset($this->ChunkY))) { 
     $this->ChunkX = $chunkX; 
     $this->ChunkY = $chunkY; 
    } 
    $counter = 0; 
    $this->ObjectLocations = array(); 
    $totalAstroids = $this->GetAstroidNo(); 


    while ($counter < $totalAstroids) { 
     $tempObjectLocations = array(); 
     //X and Y Chunk Coordinates 
     $tempObjectLocations['chunkX'] = $chunkX; 
     $tempObjectLocations['chunkY'] = $chunkY; 
     //X and Y coordinates for the object. 
     $tempObjectLocations['coordX'] = rand(4, 60); 
     $tempObjectLocations['coordY'] = rand(4, 60); 
     $tempObjectLocations['radius'] = rand(4, 12); 
     //Checks if objects already exist in array 
     if (count($this->ObjectLocations) > 0) { 

      //if the object does not collide with any other object 
      //the location will be added into the database 
      if ($this->isColliding($tempObjectLocations) == false) { 
       array_push($this->ObjectLocations, $tempObjectLocations); 
       $counter += 1; 
      } 
      // if object is the first created insert into table. 
     } else { 
      array_push($this->ObjectLocations, $tempObjectLocations); 
      $counter += 1; 
     } 
    } 

    return $this->ObjectLocations; 
} 
public function isColliding($obj1) { 
    //Checks if object conflicts with nearby objects 
    $a = count($this->ObjectLocations); 
    for ($i = 0; $i < $a; $i++) { 
     $obj2 = $this->ObjectLocations[$i]; 

     //Calculates the distance between two points 
     $distance = sqrt(($obj1['coordX'] - $obj2['coordX'])^2 + ($obj1['coordY'] - $obj2['coordY'])^2); 

     //Checks if the distance between the two objects is 
     //more than the radius of both objects added together 
     if ($distance < ($obj1['radius'] + $obj2['radius'])) { 
      return true; 
     } 
    } 
    return false; 
} 

JSON結果

parseResponse([ 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 54, 
    "coordY": 17, 
    "radius": 8 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 41, 
    "coordY": 57, 
    "radius": 12 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 42, 
    "coordY": 36, 
    "radius": 8 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 40, 
    "coordY": 58, 
    "radius": 8 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 25, 
    "coordY": 58, 
    "radius": 12 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 57, 
    "coordY": 8, 
    "radius": 10 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 46, 
    "coordY": 17, 
    "radius": 11 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 42, 
    "coordY": 29, 
    "radius": 8 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 18, 
    "coordY": 58, 
    "radius": 11 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 59, 
    "coordY": 5, 
    "radius": 11 
}, 
{ 
    "chunkX": "1", 
    "chunkY": "1", 
    "coordX": 15, 
    "coordY": 56, 
    "radius": 12 
} 

]）;

Answer 1

我有一些建議。在你的isColliding：

public function isColliding($obj1) { 
    //Checks if object conflicts with nearby objects 
    $a = count($this->ObjectLocations); 
    for ($i = 0; $i < $a; $i++) { 
     $obj2 = $this->ObjectLocations[$i]; 

     //Calculates the distance between two points 
     $distance = sqrt(($obj1['coordX'] - $obj2['coordX'])^2 + ($obj1['coordY'] - $obj2['coordY'])^2); 

     //Checks if the distance between the two objects is 
     //more than the radius of both objects added together 
     if ($distance < ($obj1['radius'] + $obj2['radius'])) { // -> Bad idea ! 
      return true; 
     } 
    } 
    return false; 
} 

我已經標記不好的地方。爲什麼？因爲你把小行星看成一個羣衆點，但實際上並沒有。如果他們的半徑總和等於他們之間的距離，他們仍然會相互碰撞。所以這種情況看起來應該是這樣的：

if ($distance <= ($obj1['radius'] + $obj2['radius'])) { // -> Should work :) 
       return true; 
      } 

每個人看起來都看不到。有一些基本的錯誤（我沒有看到這個:)。）。在PHP ^運算符是XOR運算符不電運營商:) 您的腳本，以便正確的符號是：

public function isColliding($obj1) { 
    //Checks if object conflicts with nearby objects 
    $a = count($this->ObjectLocations); 
    for ($i = 0; $i < $a; $i++) { 
     $obj2 = $this->ObjectLocations[$i]; 

     //Calculates the distance between two points 
//correct ^2 to pow function 
     $distance = sqrt(pow($obj1['coordX'] - $obj2['coordX'], 2) + pow($obj1['coordY'] - $obj2['coordY'], 2)); 

     //Checks if the distance between the two objects is 
     //more than the radius of both objects added together 
     if ($distance < ($obj1['radius'] + $obj2['radius'])) { // -> Bad idea ! 
      return true; 
     } 
    } 
    return false; 
} 

Answer 2

也許沒有幫助的答案，但是......我認爲你的IF/ELSE語句應該導致兩種不同的狀態？

 if ($this->isColliding($tempObjectLocations) == false) { 
      array_push($this->ObjectLocations, $tempObjectLocations); 
      $counter += 1; 
     } 
     // if object is the first created insert into table. 
    } else { 
     array_push($this->ObjectLocations, $tempObjectLocations); 
     $counter += 1; 
    } 

正如我看到你把它推到一個數組中是否碰撞？