Swapping the GameObject's children in a random way.

Hello, guys. I need your help.

I have a game object, that spawns 9 it's children. Like:

1 2 3
4 5 6
7 8 9

Is there a way to swap them all in random order without getting duplicates. It have to be like this:

5 4 3
1 8 9
2 6 7

Thanks.

 

When you say "it has to be like this", I am assuming that you mean the visual placement of the child objects. Put the children objects in an array/list before spawning them, shuffle the array/list and then spawn them the usual way?

 

Code (CSharp):

1.  
2.     public static void Randomize<T>(T[] items)
3.     {
4.         Random rand = new Random();
5.  
6.         // For each spot in the array, pick
7.         // a random item to swap into that spot.
8.         for (int i = 0; i < items.Length - 1; i++)
9.         {
10.             int j = rand.Next(i, items.Length);
11.             T temp = items[i];
12.             items[i] = items[j];
13.             items[j] = temp;
14.         }
15.     }

Source: http://csharphelper.com/blog/2014/07/randomize-arrays-in-c/

Use https://docs.unity3d.com/ScriptReference/Transform.SetSiblingIndex.html to change the Transform's child index

 

I'm not sure about the array/list thing. That is what I have.

The game has to be a puzzle. I put a square picture into a Picture Slicer that cuts this picture into several pieces and then insert this pieces into game object's children. The thing is that the array can be not only 3x3, but 5x5, 10x10, whatever. It will always be a square. Here is the code, maybe it will give you some information.

Code (CSharp):

1. void CreatePuzzle()
2.     {
3.         blocks = new Block[blocksPerLine, blocksPerLine];
4.         Texture2D[,] imageSlices = ImageSlicer.GetSlices(image, blocksPerLine);
5.         for (int y = 0; y < blocksPerLine; y++)
6.         {
7.             for (int x = 0; x < blocksPerLine; x++)
8.             {
9.                 GameObject blockObject = GameObject.CreatePrimitive(PrimitiveType.Quad);
10.                 blockObject.transform.position = -Vector2.one * (blocksPerLine - 1) * .5f + new Vector2(x, y);
11.                 blockObject.transform.parent = transform;
12.  
13.                 Block block = blockObject.AddComponent<Block>();
14.                 block.OnBlockPressed += PlayerMoveBlockInput;
15.                 block.OnFinishedMoving += OnBlockFinishedMoving;
16.  
17.                 block.Init(new Vector2Int(x, y), imageSlices[x, y]);
18.  
19.                 blocks[x, y] = block;
20.          }
21.       }
22.    }
23.  

 

In the line:

Code (CSharp):

1. block.Init(new Vector2Int(x, y), imageSlices[x, y]);

You can simply call a function which returns the next random value from the imageSlices array instead of manually setting the value there.

For example, I created a simple script to simulate picking random values from a 2D array. You can use a setup similar to it:

Code (CSharp):

1. public class Main : MonoBehaviour
2.     {
3.         #region Public Fields
4.         // 2D array of values
5.         public int[,] ValueArray = new int[4, 4];
6.         #endregion
7.         //
8.         #region Protected Fields
9.         #endregion
10.         //
11.         #region Private Fields
12.         // Random number generator
13.         private System.Random r;
14.         // List of unique values
15.         private List<int> Values = new List<int>();
16.         #endregion
17.  
18.         #region Unity Methods
19.         // Use this for initialization
20.         void Start ()
21.         {
22.             r = new System.Random();
23.  
24.             int x = ValueArray.GetLength(0);
25.             int y = ValueArray.GetLength(1);
26.  
27.             for (int i = 0; i < x; i++)
28.             {
29.                 for (int j = 0; j < y; j++)
30.                 {
31.                     ValueArray[i, j] = UnityEngine.Random.Range(1, 100);
32.                 }
33.             }
34.         }
35.      
36.         // Update is called once per frame
37.         void Update ()
38.         {
39.             if (Input.GetKeyUp(KeyCode.Space))
40.             {
41.                 print("Next Value: " + GetNextValue());
42.             }
43.         }
44.         #endregion
45.  
46.         #region Public Methods
47.         #endregion
48.         //
49.         #region Protected Methods
50.         #endregion
51.         //
52.         #region Private Methods
53.         private int GetNextValue()
54.         {
55.             // limits of the 2D array
56.             int x = ValueArray.GetLength(0);
57.             int y = ValueArray.GetLength(1);
58.             // range of values to randomize against
59.             int values = x * y;
60.             // new random index
61.             int index = r.Next(values);
62.             // new value
63.             int value = ValueArray[index / x, index % y];
64.             // check for uniqueness
65.             if (Values.Contains(value))
66.             {
67.                 return GetNextValue();
68.             }
69.             // add the new value to the list
70.             Values.Add(value);
71.             // return the new value
72.             return value;
73.         }
74.         #endregion
75.     }

In your case, the only things you will need is the Random object, the list which stores the unique values (it will be a list of Texture2D in your example) and the GetNextValue function.

Hope that helps.

 

Simple example of shuffling children in hierarchy would be calling this method on parent:

Code (CSharp):

1. [ContextMenu("Shuffle")]
2.     public void Shuffle()
3.     {
4.         List<int> indexes = new List<int>();
5.         List<Transform> items = new List<Transform>();
6.         for (int i = 0; i < transform.childCount;++i)
7.         {
8.             indexes.Add(i);
9.             items.Add(transform.GetChild(i));
10.         }
11.          
12.         foreach (var item in items)
13.         {
14.             item.SetSiblingIndex(indexes[Random.Range(0, indexes.Count)]);
15.         }
16.     }

 

man you've saved me quite a lot of work. thank you