Select random number in series (620699)

Unity Engine
1

I want to know logic over how to select six random numbers between 0 to 8 ,where if (0,1) or (1,2) or (0,2) got select then the remaining one number out of 0 to 2 shall not be selected again same goes for 3 to 5 series if (3,5) or (3,4) or (4,5) got selected then remaining one number shall not be selected.

So for example in series 0 to 8 six number can be (0,1,3,5,6,8) or (1,2,4,5,7,8) etc

2

To do this you need to track that over time. So some sort of collection will be needed.

Easiest route would be a List of the range of values. Randomly grab a value from the List, removing it in the process. When the list is empty, the sequence is complete.

//create sequence
var lst = new List<int>(9);
for(int i = 0; i < 9; i++)
{
    lst.Add(i);
}

//extract value
int i = Random.Range(0, lst.Count);
int result = lst[i];
lst.RemoveAt(i);
1 Like
3

Wow that seems to be new approach, This numbers are kind of spawn points, so if all 0-2 points get filled then path will be blocked for player to move ahead. How would I know and avoid with your way that 0 to 2 ( 3 to 5,6 to 8) all numbers has got generated?

Basically I don;t want to generate more than 2 number between 0 to 2 or 3 to 5 or 6 to 8 range.

Here is what I hard coded…

 public static int[] getPointsToSpawn(int howManyToSpawn, int tillWhereToSpawn)
        {
   
            int[] howMany = new int[ howManyToSpawn];
            List<int> randomCOllected = new List<int>();
   
            for (int i = 0;i < howManyToSpawn; i++)
            {
   
   
        int randomPoint = generateUniqueRandomNumber(0, tillWhereToSpawn, randomCOllected);
                randomCOllected.Add(randomPoint);
   
   
                if(randomCOllected.Contains(0) && randomCOllected.Contains(1) ||
                    randomCOllected.Contains(0) && randomCOllected.Contains(2)
                    || randomCOllected.Contains(1) && randomCOllected.Contains(2)
                    )
                {
   
   
                    randomCOllected.Add(0);
                    randomCOllected.Add(1);
                    randomCOllected.Add(2);
   
   
   
                }
                else if (randomCOllected.Contains(3) && randomCOllected.Contains(4) ||
                  randomCOllected.Contains(3) && randomCOllected.Contains(5)
                  || randomCOllected.Contains(4) && randomCOllected.Contains(5)
                  )
                {
   
   
                    randomCOllected.Add(3);
                    randomCOllected.Add(4);
                    randomCOllected.Add(5);
   
   
   
                }
                else if (randomCOllected.Contains(6) && randomCOllected.Contains(7) ||
                  randomCOllected.Contains(7) && randomCOllected.Contains(8)
                  || randomCOllected.Contains(6) && randomCOllected.Contains(8)
                  )
                {
   
   
                    randomCOllected.Add(6);
                    randomCOllected.Add(7);
                    randomCOllected.Add(8);
   
   
   
                }
  
   
   
                howMany[i] = randomPoint;
   
   
   
   
            }
            return howMany;
   
   
        }
4

so you want to guarantee a gap? simple, 3 ranges in a similar fashion as above. Randomly pick one from each range, as above. Don’t do anything with that number, just remove it.
Combine the three ranges and do as above to select the ones you want to use.

5

Remove 2 random values from the list (1…8):

List<int> result = Enumerable.Range(1, 8).ToList();
result.RemoveAt(Random.Range(0, result.Count));
result.RemoveAt(Random.Range(0, result.Count));
6

Ok, you dont want to have something like (0, 1, 2)
so:

List<int> result = Enumerable.Range(0, 9).ToList();
result.RemoveAt(Random.Range(0, 2));
result.RemoveAt(Random.Range(2, 4));

Also, please note that if you select 6 in 8, and will not allow (0, 1, 2) or (3, 4, 5), then 6, 7 and 8 must be always present.

7

@L-Tyrosine not seen Enumerable.Range before, did some googling; do you need System.Linq namespace for that or is there another one?

8

Sorry, I was wrong, there are 9 numbers (not 8).
So:

            // Generate (0, 1, 2, 3, 4, 5, 6, 7, 8) list
            List<int> result = Enumerable.Range(0, 9).ToList();

            // Remove something in 0..2 range
            result.RemoveAt(Random.Range(0, 2));

            // Remove something in 3..5 range
            result.RemoveAt(Random.Range(2, 4));

            // Remove one more random value (now anything goes)
            result.RemoveAt(Random.Range(0, result.Count));

This is problem that I would expect to see in Google Jam =D

1 Like
9

you’d think they’d start with collab on making a more interesting webpage for it :hushed::p:smile:

10

Thank you all for support …WIth above code if I increase range from 0 -8 to 0-14 repetition will get increase, so I end up with following code,

  public static int[] getPointsToSpawn(int howManyToSpawn, int tillWhereToSpawn,int howManyToChangeLane)
        {

            int[] howMany = new int[howManyToSpawn];

              
            List<int> randomCOllected = new List<int>();

            for(int k = 3; k <= tillWhereToSpawn; k += 3)
            {
                randomCOllected.Add(provideRnd(k - 3, k));
            }

            for (int i = 0; i < howManyToSpawn; i++)
            {


                int randomPoint = generateUnique(0, tillWhereToSpawn, randomCOllected);
                randomCOllected.Add(randomPoint);






                howMany[i] = randomPoint;




            }
            return howMany;


        }

I pre added ignore values to collected random list,so new random will never spawn that number…Thank you again for your responses…

@L-Tyrosine Glad you liked this :slight_smile: