Generate random numbers in loop without repeating.

Solution 4

The first solution may work, but it will start to become slower the more numbers were already chosen. (after all, after having 4999 numbers selected there is only one possibility, but it may be generating any value, needing to try again, and again, and again).

One thing that may make it faster is using a HashSet instead a list, as searching in a HashSet is faster. But my real solution is different.

First we create a list which all possible values.

List<int> available = new List<int>(5000);
for(int i=1; i<=5000; i++)
  available.Add(i);

Then, we will keep generating random indexes (instead of random values). Such indexes will be used to get the value from the available list, put it in the result and then remove it from the available list.

List<int> result = new List<int>(5000);
while(available.Count > 0)
{
  int index = random.Next(availableCount);
  result.Add(available[index]);
  available.RemoveAt(index);
}
 
return result;

This way, if you create a list with 1 million values, you will have only 1 million calls to random.Next();
Surely the list itself has its own problems, but it is better than keep waiting until the last available number is guess.

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Posted 24 Aug '11 - 12:29

Paulo Zemek52.2K

Comments

dr.m-rostami - 27 Sep '12 - 17:20

this is the best solution thx all

Solution 6

I believe the tip linked below covers what you are after.

How to generate many random various numbers?

I recommend looking at alternative 1

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Posted 1 Sep '11 - 1:31

GParkings3.9K

Solution 2

See my tip: "Random extraction of 5 cards from a deck"[^], it is C++ but you should get the idea.

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Posted 24 Aug '11 - 2:18

CPallini246K

Solution 3

Try this snippet

private static List<int> Randomize(int count, int min, int max)
        {
            Random r = new Random();
            List<int> result = new List<int>();
            if (count < max - min)
            {
                while (result.Count <= count)
                {
                    int number = r.Next(min, max);
                    if (!result.Contains(number))
                    {
                        result.Add(number);
                    }
                }
            }
            else
            {
                Console.WriteLine("Select another boundaries or number count");
            }
            return result;
        }

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Posted 24 Aug '11 - 2:47

pavel.bidenko549

Solution 1

Link[^]

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Posted 24 Aug '11 - 1:54

mottudeepu624

Comments

Simon Bang Terkildsen - 24 Aug '11 - 8:29

This doesn't solve the OP's problem.

Solution 7

i'm colecting all solution and this is the result:

using System;
using System.Collections;
using System.Configuration;
using System.Data;
using System.Linq;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.HtmlControls;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Xml.Linq;
using System.Collections.Generic;
 
namespace WebApplication1
{
    public partial class _Default : System.Web.UI.Page
    {
        public static Random r=new Random();
        public static int number;
        protected void Page_Load(object sender, EventArgs e)
        {
 
        }
 
        protected void ASPxButton1_Click(object sender, EventArgs e)
        {
           //creat a text box in desing mode and call it ASPxTextBox1
            if (ASPxTextBox1.Text == "")
            {
                Response.Write("first enter you count number");
            }
            else
            {
                number = Convert.ToInt32(ASPxTextBox1.Text);
                List<int> available = new List<int>(number);
                for (int i = 1; i <= number; i++)
                    available.Add(i);
                List<int> result = new List<int>(number);
                while (available.Count > 0)
                {
                    int index = r.Next(available.Count);
                    result.Add(available[index]);
                    available.RemoveAt(index);
                }
                for (int i = 0; i < result.Count; i++)
                {
                    Response.Write(result[i] + "-");
                }
            }
        }
 
        protected int Pro_R(int min, int max)
        {
            return (r.Next(min, max));
        }
    }
}

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Posted 27 Sep '12 - 11:24

dr.m-rostami399

Edited 27 Sep '12 - 11:58

Kenneth Haugland63.4K

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

0	Sergey Alexandrovich Kryukov	259
1	OriginalGriff	220
2	Mahesh Bailwal	159
3	CPallini	143
4	Maciej Los	140

0	Sergey Alexandrovich Kryukov	10,214
1	OriginalGriff	7,819
2	CPallini	4,181
3	Rohan Leuva	3,522
4	Maciej Los	3,089

Generate random numbers in loop without repeating.

6 solutions

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Solution 2

Solution 3

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Solution 7

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Generate random numbers in loop without repeating.

6 solutions

Solution 4

Solution 6

Solution 2

Solution 3

Solution 1

Solution 7

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