I made a program to generate 7 random numbers for a lottery using a array. I have generated a random number between 1, 50 but every number shows in order and not on the same line. I would also like to store the auto generated numbers in a array to use. I am not sure how to fix this any help would be appreciated
static void AutoGenrateNumbers()
{
int temp;
int number = 0;
int[] lotto = new int[7];
Random rand = new Random();
for (int i = 0; i <= 50; i++)
{
number = 0;
temp = rand.Next(1, 50);
while (number <= i)
{
if (temp == number)
{
number = 0;
temp = rand.Next(1, 50);
}
else
{
number++;
}
}
temp = number;
Console.WriteLine($"the new lotto winning numbers are:{number}Bonus:{number}");
}
}
Is this what you need?
static void AutoGenrateNumbers()
{
int temp;
int[] lotto = new int[7];
Random rand = new Random();
for (int i = 0; i < 7; i++)
{
temp = rand.Next(1, 50);
lotto[i]= temp;
}
Console.Write($"the new lotto winning numbers are: ");
for (int i = 0; i < 6; i++)
{
Console.Write(lotto[i]+" ");
}
Console.Write($"Bonus:{lotto[6]}");
}
edit: if you want the numbers to be unique:
static void AutoGenrateNumbers()
{
int temp;
int[] lotto = new int[7];
Random rand = new Random();
for (int i = 0; i < 7; i++)
{
do
{
temp = rand.Next(1, 50);
}
while (lotto.Contains(temp));
lotto[i]= temp;
}
Console.Write($"the new lotto winning numbers are: ");
for (int i = 0; i < 6; i++)
{
Console.Write(lotto[i]+" ");
}
Console.Write($"Bonus:{lotto[6]}");
}
A better way to do this is just to generate all the numbers 1-50, shuffle them and then just take 7. Using Jon Skeet's Shuffle extension method found here:
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
{
T[] elements = source.ToArray();
for (int i = elements.Length - 1; i >= 0; i--)
{
int swapIndex = rng.Next(i + 1);
yield return elements[swapIndex];
elements[swapIndex] = elements[i];
}
}
Now your code is very simple:
static void AutoGenrateNumbers()
{
var lotto = Enumerable.Range(0, 50).Shuffle(new Random()).Take(7);
Console.WriteLine("the new lotto winning numbers are: {0}", string.Join(",", lotto));
}
Fiddle here
Just to add to the existing answers tried to do that in one LINQ statement:
static void Main(string[] args)
{
var rand = new Random();
Enumerable
.Range(1, 7)
.Aggregate(new List<int>(), (x, y) =>
{
var num = rand.Next(1, 51);
while (x.Contains(num))
{
num = rand.Next(1, 51);
}
x.Add(num);
return x;
})
.ForEach(x => Console.Write($"{x} "));
}
The result is something like:
34 24 46 27 11 17 2
Related
Creating an array of 10 elements and assigning them by counting randomly, assigning a new number if the same numbers are repeated
I tried to use the contains method but it didn't appear in the list after the array, I used the exists method but it didn't work either, what kind of way should I follow? thanks
static void Main(string[] args)
{
Random Rnd = new Random();
int[] Numbers = new int[10];
for (int i = 0; i < Numbers.Length; i++)
{
int rast = Rnd.Next(10);
bool b = Array.Exists(Numbers, element => element == rast);
if (!b)
{
i--;
}
else { Numbers[i] = rast; }
}
foreach (int item in Numbers)
{
Console.WriteLine(item);
}
}
It appears that you want the numbers from 0 to 9 in an array in random order. If that is so:
var rng = new Random();
var numbers = Enumerable.Range(0, 10).OrderBy(n => rng.NextDouble()).ToArray();
You're close, but there are two issues in your algorithm:
if (!b) should be if (b)
Rnd.Next(1,10) gets numbers between 1 and 9 so you will never get a 10 and the algorithm will never finish.
After fixing these two your program will work.
Anyway, here's a version that uses .Contains
Random Rnd = new Random();
int[] Numbers = new int[10];
for (int i = 0; i < Numbers.Length; i++)
{
// + 1 because the range of return values includes minValue but not maxValue.
int r = Rnd.Next(minValue: 1, maxValue: Numbers.Length + 1);
while(Numbers.Contains(r))
{
r = Rnd.Next(minValue: 1, maxValue: Numbers.Length + 1);
}
Numbers[i] = r;
}
Console.WriteLine(string.Join(",", Numbers));
Example output:
9,8,7,1,4,6,3,10,5,2
Here's an example. I hope this is the output you want.
public static void Main()
{
Random Rnd = new Random();
int[] Numbers = new int[10];
for (int i = 0; i < Numbers.Length; i++)
{
int rast = Rnd.Next(10);
bool b = Array.Exists(Numbers, element => element == rast);
if (!b)
{
Numbers[i] = rast;
Console.WriteLine(rast);
}
else {
i--;
}
}
}
As a result, I have found the solution, thanks to everyone who answered and showed me the way
int[] Numbers = new int[10];
int x, y;
Random ran = new Random();
for (x = 0; x < 10; x++)
{
Numbers[x] = ran.Next(1, 10);
for (y = x; y >= 0; y--)
{
if (x == y)
{
continue;
}
if (Numbers[x] == Numbers[y])
{
Numbers[x] = ran.Next(1, 10);
y = x;
}
}
Console.WriteLine(Numbers[x]);
}
when i created new project as Console .net in visual studio 2019 ,
it doesnt bring this libraries automatically ,
Because the following libraries are not attached, the codings I used before did not work, I developed another simple solution, and in this way, I realized that I had a problem due to these libraries not being attached.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
int[] Numbers= new int[10];
Random rnd = new Random();
for (int i = 0; i < Numbers.Length; i++)
{
int Num= rnd.Next(10);
if (Numbers.Contains(Num))
{
i--;
}
else
{
Console.WriteLine(Numbers[i] = Num);
}
}
I have created a simple program which randomly generates 6 winning numbers. While the program works, I would also like for it to ensure that the same number isn't outputted twice as well as sorting them into numerical order when outputted. How would I go about doing such a thing while sticking to similar techniques already used? My code is down below. Any help is very much appreciated.
int temp;
int[] lotto = new int[6];
Random rand = new Random();
for (int i = 0; i < 6; i++)
{
temp = rand.Next(1, 59);
lotto[i] = temp;
}
Console.Write($"The lotterry winning numbers are: ");
for (int i = 0; i < 6; i++)
{
Console.Write(lotto[i] + " ");
}
Console.ReadKey();
Based on a Fisher-Yates shuffle, but saves some work because we know we don't need all the values (if we only need 6 values out of 10 million potentials, we only need to take the first six iterations of the fisher-yates algorithm).
public IEnumerable<int> DrawNumbers(int count, int MaxNumbers)
{
var r = new Random(); //ideally, make this a static member somewhere
var possibles = Enumerable.Range(1, MaxNumbers).ToList();
for (int i = 0; i < count; i++)
{
var index = r.Next(i, MaxNumbers);
yield return possibles[index];
possibles[index] = possibles[i];
}
}
var lottoNumbers = DrawNumbers(6, 59);
Console.Write("The lotterry winning numbers are: ");
Console.WriteLine(string.Join(" ", lottoNumbers.OrderBy(n => n)));
See it work here:
https://dotnetfiddle.net/NXYkpU
You can use Linq to create sequence [1..59] and order it by random to shuffle it.
Random rand = new Random();
var winners = Enumerable.Range(1, 59)
.OrderBy(x => rand.Next())
.Take(6)
.OrderBy(x => x)
.ToList();
Console.WriteLine(String.Join(" ", winners));
int temp;
int[] lotto = new int[6];
Random rand = new Random();
int i = 0;
while(i < 6)
{
temp = rand.Next(1, 59);
//check if lotto contains just generated number, if so skip that number
bool alreadyExist = false;
foreach (int item in lotto)
{
if (item == temp)
{
alreadyExist = true;
break;
}
}
if (alreadyExist)
continue;
lotto[i] = temp;
i++;
}
Console.Write($"The lotterry winning numbers are: ");
// Sort array in ascending order.
Array.Sort(lotto);
for (int j = 0; j < 6; j++)
{
Console.Write(lotto[j] + " ");
}
Console.ReadKey();
I would probably do it Dmitri's way because it is quick and obvious and performance isn't that important with an array this size.
But just for fun, this is slightly more efficient.
IEnumerable<int> GetNumbers(int min, int max, int count)
{
var random = new Random();
var size = max - min + 1;
var numbers = Enumerable.Range(min, size).ToArray();
while (count > 0)
{
size--;
var index = random.Next(0, size);
yield return numbers[index];
numbers[index] = numbers[size];
count--;
}
}
This solution creates an array containing all possible values and selects them randomly. Each time a selection is made, the array is "shrunk" by moving the last element to replace the element that was chosen, preventing duplicates.
To use:
var numbers = GetNumbers(1, 59, 6).ToList();
foreach (var number in numbers.OrderBy(x => x))
{
Console.WriteLine(number);
}
i'm code:
Random rand = new Random();
int[] arr = new int[4];
for (int i = 0; i < 4; i++)
{
for (int k = 0; k < 4; k++)
{
int rad = rand.Next(1, 5);
if (arr[k] != rad)
arr[i] = rad;
}
}
for (int i = 0; i < 4; i++)
MessageBox.Show(arr[i].ToString());
I wanna production numbers from 1 to 4 am and unequal with each other.
tnx.
Think it the other way round:
instead of:
generating a number and then check it if it is not duplicate
you make it such that:
you already have a set of non-duplicate numbers, then you take it
one-by-one - removing the possibilities of duplicates.
Something like this will do:
List<int> list = Enumerable.Range(1, 4).ToList();
List<int> rndList = new List<int>();
Random rnd = new Random();
int no = 0;
for (int i = 0; i < 4; ++i) {
no = rnd.Next(0, list.Count);
rndList.Add(list[no]);
list.Remove(list[no]);
}
The result is in your rndList.
This way, no duplicate will occur.
Create an array with unique elements and then shuffle it, like in the code below, it will shuffle an array in uniformly random order it uses Fisher-Yates shuffle algorithm:
int N = 20;
var theArray = new int[N];
for (int i = 0; i < N; i++)
theArray[i] = i;
Shuffle(theArray);
public static void Shuffle(int[] a) {
if (a == null) throw new ArgumentNullException("Array is null");
int n = a.Length;
var theRandom = new Random();
for (int i = 0; i < n; i++) {
int r = i + theRandom.Next(n-i); // between i and n-1
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
}
Explanation and template version of algorithm could be found in this post with nice answer from Jon Skeet.
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
{
T[] elements = source.ToArray();
// Note i > 0 to avoid final pointless iteration
for (int i = elements.Length-1; i > 0; i--)
{
// Swap element "i" with a random earlier element it (or itself)
int swapIndex = rng.Next(i + 1);
T tmp = elements[i];
elements[i] = elements[swapIndex];
elements[swapIndex] = tmp;
}
// Lazily yield (avoiding aliasing issues etc)
foreach (T element in elements)
{
yield return element;
}
}
Create a list containing the numbers you want, then shuffle them:
var rnd = new Random();
List<int> rndList = Enumerable.Range(1, 4).OrderBy(r => rnd.Next()).ToList();
If you want an array instead of a list:
var rnd = new Random();
int[] rndArray = Enumerable.Range(1, 4).OrderBy(r => rnd.Next()).ToArray();
int[] arr = new int[5];
int i = 0;
while (i < 5)
{
Random rand = new Random();
int a = rand.Next(1,6);
bool alreadyexist = false;
for (int j = 0; j < 5; j++)
{
if (a == arr[j])
{
alreadyexist = true;
}
}
if (alreadyexist == false)
{
arr[i] = a;
i++;
}
}
for (int k = 0; k < 5; k++)
{
MessageBox.Show(arr[k].ToString());
}
This question already has answers here:
Random number generator with no duplicates
(12 answers)
Closed 2 years ago.
Hi everyone I am trying to generate 6 different numbers on the same line in c# but the problem that i face is some of the numbers are repeating on the same line.Here is my code to
var rand = new Random();
List<int> listNumbers = new List<int>();
int numbers = rand.Next(1,49);
for (int i= 0 ; i < 6 ;i++)
{
listNumbers.Add(numbers);
numbers = rand.Next(1,49);
}
somewhere my output is
17 23 23 31 33 48
Check each number that you generate against the previous numbers:
List<int> listNumbers = new List<int>();
int number;
for (int i = 0; i < 6; i++)
{
do {
number = rand.Next(1, 49);
} while (listNumbers.Contains(number));
listNumbers.Add(number);
}
Another approach is to create a list of possible numbers, and remove numbers that you pick from the list:
List<int> possible = Enumerable.Range(1, 48).ToList();
List<int> listNumbers = new List<int>();
for (int i = 0; i < 6; i++)
{
int index = rand.Next(0, possible.Count);
listNumbers.Add(possible[index]);
possible.RemoveAt(index);
}
listNumbers.AddRange(Enumerable.Range(1, 48)
.OrderBy(i => rand.Next())
.Take(6))
Create a HashSet and generate a unique random numbers
public List<int> GetRandomNumber(int from,int to,int numberOfElement)
{
var random = new Random();
HashSet<int> numbers = new HashSet<int>();
while (numbers.Count < numberOfElement)
{
numbers.Add(random.Next(from, to));
}
return numbers.ToList();
}
Make it a while loop and add the integers to a hashset. Stop the loop when you have six integers.
Instead of using a List, you should use an HashSet. The HashSet<> prohibites multiple identical values. And the Add method returns a bool that indicates if the element was added to the list, Please find the example code below.
public static IEnumerable<int> GetRandomNumbers(int count)
{
HashSet<int> randomNumbers = new HashSet<int>();
for (int i = 0; i < count; i++)
while (!randomNumbers.Add(random.Next()));
return randomNumbers;
}
I've switched your for loop with a do...while loop and set the stopping condition on the list count being smaller then 6.
This might not be the best solution but it's the closest to your original code.
List<int> listNumbers = new List<int>();
do
{
int numbers = rand.Next(1,49);
if(!listNumbers.Contains(number)) {
listNumbers.Add(numbers);
}
} while (listNumbers.Count < 6)
The best approach (CPU time-wise) for such tasks is creating an array of all possible numbers and taking 6 items from it while removing the item you just took from the array.
Example:
const int min = 1, max = 49;
List<int> listNumbers = new List<int>();
int[] numbers = new int[max - min + 1];
int i, len = max - min + 1, number;
for (i = min; i < max; i++) numbers[i - min] = i;
for (i = 0; i < 6; i++) {
number = rand.Next(0, len - 1);
listNumbers.Add(numbers[number]);
if (number != (len - 1)) numbers[number] = numbers[len - 1];
len--;
}
If you are not worried about the min, max, and range then you can use this.
var nexnumber = Guid.NewGuid().GetHashCode();
if (nexnumber < 0)
{
nexnumber *= -1;
}
What you do is to generate a random number each time in the loop. There is a chance of course that the next random number may be the same as the previous one. Just add one check that the current random number is not present in the sequence. You can use a while loop like: while (currentRandom not in listNumbers): generateNewRandomNumber
Paste the below in the class as a new method
public int randomNumber()
{
var random = new Random();
int randomNumber = random.Next(10000, 99999);
return randomNumber;
}
And use the below anywhere in the tests wherever required
var RandNum = randomNumber();
driver.FindElement(By.CssSelector("[class='test']")).SendKeys(**RandNum**);
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
int[] que = new int[6];
int x, y, z;
Random ran = new Random();
for ( x = 0; x < 6; x++)
{
que[x] = ran.Next(1,49);
for (y = x; y >= 0; y--)
{
if (x == y)
{
continue;
}
if (que[x] == que[y])
{
que[x] = ran.Next(1,49);
y = x;
}
}
}
listBox1.Items.Clear();
for (z = 0; z < 6; z++)
{
listBox1.Items.Add(que[z].ToString());
}
}
}
This question already has answers here:
Randomize a List<T>
(28 answers)
Closed 9 years ago.
I need to print numbers from 1 to 50 in random order without repeating
it .
static void Main(string[] args)
{
ArrayList r = new ArrayList();
Random ran = new Random();
for (int i = 0; i < 50; i++)
{
r.Add(ran.Next(1,51));
}
for (int i = 0; i < 50; i++)
Console.WriteLine(r[i]);
Console.ReadKey();
}
What you want here is the Fisher Yates Shuffle
Here is the algorithm as implemented by Jeff Atwood
cards = Enumerable.Range(1, 50).ToList();
for (int i = cards.Count - 1; i > 0; i--)
{
int n = ran.Next(i + 1);
int temp = cards[i];
cards[i] = cards[n];
cards[n] = temp;
}
If you don't want to repeat the numbers between 1 and 50, your best bet is to populate a list with the numbers 1 to 50 and then shuffle the contents. There's a good post on shuffling here:
Randomize a List<T>
All you need to do is this check if the number already exists in the list and if so get another one:
static void Main(string[] args)
{
ArrayList r = new ArrayList();
Random ran = new Random();
int num = 0;
for (int i = 0; i < 50; i++)
{
do { num = ran.Next(1, 51); } while (r.Contains(num));
r.Add(num);
}
for (int i = 0; i < 50; i++)
Console.WriteLine(r[i]);
Console.ReadKey();
}
Edit: This will greatly increase the effeciency, preventing long pauses waiting for a non-collision number:
static void Main(string[] args)
{
List<int> numbers = new List<int>();
Random ran = new Random();
int number = 0;
int min = 1;
int max = 51;
for (int i = 0; i < 50; i++)
{
do
{
number = ran.Next(min, max);
}
while (numbers.Contains(number));
numbers.Add(number);
if (number == min) min++;
if (number == max - 1) max--;
}
for (int i = 0; i < 50; i++)
Console.WriteLine(numbers[i]);
Console.ReadKey();
}