C# Random Number Creation - c#

Ok so I know to generate a random number one would create a method such as this:
private int RandomNumber(int min, int max)
{
Random random = new Random();
return random.Next(min, max);
}
Now lets say instead of having it take two numbers example:
RandomNumber(4, 25);
I want it to make the max a value that will be determined, example:
RandomNumber(0, Alist.Count);
Is there any way to do this? I tried making AList.Count = another int number and putting that in but to no avail.

Yes, RandomNumber(0, Alist.Count) will work perfectly. However, instead of making a function for that, you should be using one instance of Random:
Random r = new Random();
DoSomething(r.Next(0, Alist.Count));
DoSomething(r.Next(0, Alist.Count));
DoSomething(r.Next(0, Alist.Count));
// etc., whatever you want.
Edit: So you want a random element from a list?
Random r = new Random();
Alist[r.Next(0, Alist.Count)] // There's your random element.

In light of the comments in minitech's answer I thought I'd give my own spin on returning random indexes from a list - feel free to take it or leave it ;).
Personally I'd declare an extension method for O(n) shuffling a list (using a Fisher-Yates shuffle):
public static class ListExtensions
{
public static List<T> Shuffle<T>(this List<T> list, Random random)
{
if (list == null) throw new ArgumentNullException("list");
if (random == null) throw new ArgumentNullException("random");
for (int i = list.Count - 1; i >= 1; i--)
{
int j = random.Next(0, i + 1);
T temp = list[i];
list[i] = list[j];
list[j] = temp;
}
return list; // best made a void method, but for examples I'll return list.
}
}
And then if reordering the original list is acceptable, simply call:
Alist.Shuffle(new Random());
If reordering is not acceptable, and I want a random list of unique indexes I'd use:
List<int> indexes = Enumerable.Range(0, Alist.Count).ToList().Shuffle(new Random());
Or I could create a new list, with the original elements shuffled:
var shuffledList = Alist.ToList().Shuffle(new Random());
It's a pretty versatile extension method, perhaps worth adding to one's arsenal.

No, you would not create a method like that. Consider:
int num1 = RandomNumber(1, 6);
int num2 = RandomNumber(1, 6);
Almost all of the times, the two variables will end up having the same value. If you create Random instances too close in time, they will get the seed from the same clock tick, and end up giving you the same sequence of numbers.
Anyway, you can very well use Alist.Count in a call to a method, but you can't assign a number to it. You would have to add items to the list to change the Count property:
List<int> Alist = new List<int> { 1, 2, 3 };
int index = RandomNumber(0, Alist.Count);

here you go, this should do the trick. this will take your list and sort it in random order:
list = list.OrderBy( itm => Guid.NewGuid() ).ToList();

Related

Why I cannot generate more than 6551 random numbers

I have the following code in C#, inside the Main method of a Simple Console Application.
I have debuged and, after List.Count = 6551 it seems the random values repeat themselves.
List<int> list = new List<int>(9999);
bool waiting = true;
Random random = new Random(DateTime.Today.Milliseconds);
do
{
int units = random.Next(0, 9);
int tens = random.Next(0, 9);
int hundreds = random.Next(0, 9);
int thousands = random.Next(0, 9);
int result = int.Parse(String.Format("{0}{1}{2}{3}", units, tens, hundreds, thousands));
if(list.Contains(result))
{
continue;
}
else
{
list.Add(result);
}
if(list.Count == 9999)
{
waiting = false;
}
}while(waiting);
Console.WriteLine("Finished"):
Console.ReadKey();
Your digts range from 0 to (and excluding!) 9, which makes eight choices per digit (0-8) and thus 6561 combinations (9*9*9*9).
Also, beware that your algorithm is extremely inefficient. Eventually, your list will be very crowded and then, the algorithm will spend most of the time checking whether a given random number is already contained in your result list.
If your goal is to shuffle indices, you can reach that more eficiently by keeping a list of indices you haven't inserted already.
The answer has already be given by Georg, but concerning your efficiency problem: if what you want is a shuffled list of integer, you can do it with an extension method on the list.
var shuffled = Enumerable.Range(0, 10000).Shuffle();
Look here for more info An extension method on IEnumerable needed for shuffling

List or better Algorithm for random C# - Unity [duplicate]

What is the best way to randomize an array of strings with .NET? My array contains about 500 strings and I'd like to create a new Array with the same strings but in a random order.
Please include a C# example in your answer.
The following implementation uses the Fisher-Yates algorithm AKA the Knuth Shuffle. It runs in O(n) time and shuffles in place, so is better performing than the 'sort by random' technique, although it is more lines of code. See here for some comparative performance measurements. I have used System.Random, which is fine for non-cryptographic purposes.*
static class RandomExtensions
{
public static void Shuffle<T> (this Random rng, T[] array)
{
int n = array.Length;
while (n > 1)
{
int k = rng.Next(n--);
T temp = array[n];
array[n] = array[k];
array[k] = temp;
}
}
}
Usage:
var array = new int[] {1, 2, 3, 4};
var rng = new Random();
rng.Shuffle(array);
rng.Shuffle(array); // different order from first call to Shuffle
* For longer arrays, in order to make the (extremely large) number of permutations equally probable it would be necessary to run a pseudo-random number generator (PRNG) through many iterations for each swap to produce enough entropy. For a 500-element array only a very small fraction of the possible 500! permutations will be possible to obtain using a PRNG. Nevertheless, the Fisher-Yates algorithm is unbiased and therefore the shuffle will be as good as the RNG you use.
If you're on .NET 3.5, you can use the following IEnumerable coolness:
Random rnd=new Random();
string[] MyRandomArray = MyArray.OrderBy(x => rnd.Next()).ToArray();
Edit: and here's the corresponding VB.NET code:
Dim rnd As New System.Random
Dim MyRandomArray = MyArray.OrderBy(Function() rnd.Next()).ToArray()
Second edit, in response to remarks that System.Random "isn't threadsafe" and "only suitable for toy apps" due to returning a time-based sequence: as used in my example, Random() is perfectly thread-safe, unless you're allowing the routine in which you randomize the array to be re-entered, in which case you'll need something like lock (MyRandomArray) anyway in order not to corrupt your data, which will protect rnd as well.
Also, it should be well-understood that System.Random as a source of entropy isn't very strong. As noted in the MSDN documentation, you should use something derived from System.Security.Cryptography.RandomNumberGenerator if you're doing anything security-related. For example:
using System.Security.Cryptography;
...
RNGCryptoServiceProvider rnd = new RNGCryptoServiceProvider();
string[] MyRandomArray = MyArray.OrderBy(x => GetNextInt32(rnd)).ToArray();
...
static int GetNextInt32(RNGCryptoServiceProvider rnd)
{
byte[] randomInt = new byte[4];
rnd.GetBytes(randomInt);
return Convert.ToInt32(randomInt[0]);
}
You're looking for a shuffling algorithm, right?
Okay, there are two ways to do this: the clever-but-people-always-seem-to-misunderstand-it-and-get-it-wrong-so-maybe-its-not-that-clever-after-all way, and the dumb-as-rocks-but-who-cares-because-it-works way.
Dumb way
Create a duplicate of your first array, but tag each string should with a random number.
Sort the duplicate array with respect to the random number.
This algorithm works well, but make sure that your random number generator is unlikely to tag two strings with the same number. Because of the so-called Birthday Paradox, this happens more often than you might expect. Its time complexity is O(n log n).
Clever way
I'll describe this as a recursive algorithm:
To shuffle an array of size n (indices in the range [0..n-1]):
if n = 0
do nothing
if n > 0
(recursive step) shuffle the first n-1 elements of the array
choose a random index, x, in the range [0..n-1]
swap the element at index n-1 with the element at index x
The iterative equivalent is to walk an iterator through the array, swapping with random elements as you go along, but notice that you cannot swap with an element after the one that the iterator points to. This is a very common mistake, and leads to a biased shuffle.
Time complexity is O(n).
This algorithm is simple but not efficient, O(N2). All the "order by" algorithms are typically O(N log N). It probably doesn't make a difference below hundreds of thousands of elements but it would for large lists.
var stringlist = ... // add your values to stringlist
var r = new Random();
var res = new List<string>(stringlist.Count);
while (stringlist.Count >0)
{
var i = r.Next(stringlist.Count);
res.Add(stringlist[i]);
stringlist.RemoveAt(i);
}
The reason why it's O(N2) is subtle: List.RemoveAt() is a O(N) operation unless you remove in order from the end.
You can also make an extention method out of Matt Howells. Example.
namespace System
{
public static class MSSystemExtenstions
{
private static Random rng = new Random();
public static void Shuffle<T>(this T[] array)
{
rng = new Random();
int n = array.Length;
while (n > 1)
{
int k = rng.Next(n);
n--;
T temp = array[n];
array[n] = array[k];
array[k] = temp;
}
}
}
}
Then you can just use it like:
string[] names = new string[] {
"Aaron Moline1",
"Aaron Moline2",
"Aaron Moline3",
"Aaron Moline4",
"Aaron Moline5",
"Aaron Moline6",
"Aaron Moline7",
"Aaron Moline8",
"Aaron Moline9",
};
names.Shuffle<string>();
Just thinking off the top of my head, you could do this:
public string[] Randomize(string[] input)
{
List<string> inputList = input.ToList();
string[] output = new string[input.Length];
Random randomizer = new Random();
int i = 0;
while (inputList.Count > 0)
{
int index = r.Next(inputList.Count);
output[i++] = inputList[index];
inputList.RemoveAt(index);
}
return (output);
}
Randomizing the array is intensive as you have to shift around a bunch of strings. Why not just randomly read from the array? In the worst case you could even create a wrapper class with a getNextString(). If you really do need to create a random array then you could do something like
for i = 0 -> i= array.length * 5
swap two strings in random places
The *5 is arbitrary.
public static void Shuffle(object[] arr)
{
Random rand = new Random();
for (int i = arr.Length - 1; i >= 1; i--)
{
int j = rand.Next(i + 1);
object tmp = arr[j];
arr[j] = arr[i];
arr[i] = tmp;
}
}
Generate an array of random floats or ints of the same length. Sort that array, and do corresponding swaps on your target array.
This yields a truly independent sort.
Ok, this is clearly a bump from my side (apologizes...), but I often use a quite general and cryptographically strong method.
public static class EnumerableExtensions
{
static readonly RNGCryptoServiceProvider RngCryptoServiceProvider = new RNGCryptoServiceProvider();
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> enumerable)
{
var randomIntegerBuffer = new byte[4];
Func<int> rand = () =>
{
RngCryptoServiceProvider.GetBytes(randomIntegerBuffer);
return BitConverter.ToInt32(randomIntegerBuffer, 0);
};
return from item in enumerable
let rec = new {item, rnd = rand()}
orderby rec.rnd
select rec.item;
}
}
Shuffle() is an extension on any IEnumerable so getting, say, numbers from 0 to 1000 in random order in a list can be done with
Enumerable.Range(0,1000).Shuffle().ToList()
This method also wont give any surprises when it comes to sorting, since the sort value is generated and remembered exactly once per element in the sequence.
Random r = new Random();
List<string> list = new List(originalArray);
List<string> randomStrings = new List();
while(list.Count > 0)
{
int i = r.Random(list.Count);
randomStrings.Add(list[i]);
list.RemoveAt(i);
}
Jacco, your solution ising a custom IComparer isn't safe. The Sort routines require the comparer to conform to several requirements in order to function properly. First among them is consistency. If the comparer is called on the same pair of objects, it must always return the same result. (the comparison must also be transitive).
Failure to meet these requirements can cause any number of problems in the sorting routine including the possibility of an infinite loop.
Regarding the solutions that associate a random numeric value with each entry and then sort by that value, these are lead to an inherent bias in the output because any time two entries are assigned the same numeric value, the randomness of the output will be compromised. (In a "stable" sort routine, whichever is first in the input will be first in the output. Array.Sort doesn't happen to be stable, but there is still a bias based on the partitioning done by the Quicksort algorithm).
You need to do some thinking about what level of randomness you require. If you are running a poker site where you need cryptographic levels of randomness to protect against a determined attacker you have very different requirements from someone who just wants to randomize a song playlist.
For song-list shuffling, there's no problem using a seeded PRNG (like System.Random). For a poker site, it's not even an option and you need to think about the problem a lot harder than anyone is going to do for you on stackoverflow. (using a cryptographic RNG is only the beginning, you need to ensure that your algorithm doesn't introduce a bias, that you have sufficient sources of entropy, and that you don't expose any internal state that would compromise subsequent randomness).
This post has already been pretty well answered - use a Durstenfeld implementation of the Fisher-Yates shuffle for a fast and unbiased result. There have even been some implementations posted, though I note some are actually incorrect.
I wrote a couple of posts a while back about implementing full and partial shuffles using this technique, and (this second link is where I'm hoping to add value) also a follow-up post about how to check whether your implementation is unbiased, which can be used to check any shuffle algorithm. You can see at the end of the second post the effect of a simple mistake in the random number selection can make.
You don't need complicated algorithms.
Just one simple line:
Random random = new Random();
array.ToList().Sort((x, y) => random.Next(-1, 1)).ToArray();
Note that we need to convert the Array to a List first, if you don't use List in the first place.
Also, mind that this is not efficient for very large arrays! Otherwise it's clean & simple.
This is a complete working Console solution based on the example provided in here:
class Program
{
static string[] words1 = new string[] { "brown", "jumped", "the", "fox", "quick" };
static void Main()
{
var result = Shuffle(words1);
foreach (var i in result)
{
Console.Write(i + " ");
}
Console.ReadKey();
}
static string[] Shuffle(string[] wordArray) {
Random random = new Random();
for (int i = wordArray.Length - 1; i > 0; i--)
{
int swapIndex = random.Next(i + 1);
string temp = wordArray[i];
wordArray[i] = wordArray[swapIndex];
wordArray[swapIndex] = temp;
}
return wordArray;
}
}
int[] numbers = {0,1,2,3,4,5,6,7,8,9};
List<int> numList = new List<int>();
numList.AddRange(numbers);
Console.WriteLine("Original Order");
for (int i = 0; i < numList.Count; i++)
{
Console.Write(String.Format("{0} ",numList[i]));
}
Random random = new Random();
Console.WriteLine("\n\nRandom Order");
for (int i = 0; i < numList.Capacity; i++)
{
int randomIndex = random.Next(numList.Count);
Console.Write(String.Format("{0} ", numList[randomIndex]));
numList.RemoveAt(randomIndex);
}
Console.ReadLine();
Could be:
Random random = new();
string RandomWord()
{
const string CHARS = "abcdefghijklmnoprstuvwxyz";
int n = random.Next(CHARS.Length);
return string.Join("", CHARS.OrderBy(x => random.Next()).ToArray())[0..n];
}
Here's a simple way using OLINQ:
// Input array
List<String> lst = new List<string>();
for (int i = 0; i < 500; i += 1) lst.Add(i.ToString());
// Output array
List<String> lstRandom = new List<string>();
// Randomize
Random rnd = new Random();
lstRandom.AddRange(from s in lst orderby rnd.Next(100) select s);
private ArrayList ShuffleArrayList(ArrayList source)
{
ArrayList sortedList = new ArrayList();
Random generator = new Random();
while (source.Count > 0)
{
int position = generator.Next(source.Count);
sortedList.Add(source[position]);
source.RemoveAt(position);
}
return sortedList;
}

Create array without duplicates c#

So this is my code right now, and I need help so it won't makes duplicates. I need this for school so if you could explain a little bit too it would be helpful. Btw don't care about the comments it's on Swedish
int temp;
int[] myArray = new int[20]; // Array med 20 tal
Random random = new Random(); // Skapar metoden "Random"
for (int i = 0; i < myArray.Length; i++) // Forloop med 20 positioner
{
myArray[i] = random.Next(1, 100); // Ger random värden till dessa 20 positionerna
for (int j = 0; j < myArray.Length; j++)
{
if (myArray[i] > myArray[j]) // Array [i] större än Array [j]
{
//temp = myArray[j];
//myArray[j] = myArray[i];
//myArray[i] = temp;
}
}
Console.Write(myArray[i] + " ");
}
you can try linq to .Distinct() and to convert it to array use .ToArray()
var s = { 5, 7, 7, 4, 3};
var q = s.Distinct().ToArray();
At it's simplest, it looks like you probably want to
private static Random rng = new Random(); //class level definition
var myArray = Enumerable.Range(1, 20).OrderBy(X => rng.Next()).ToArray();
Alternatively, if this is for school and you need to justify your code... fill an array with the numbers 1 to 20, then use a Fisher-Yates shuffle to randomise the order of the array.
See: https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
Since the array has exatly the size of the random value range (which is 20), you will get every number exactly once. It's easiest to create every number once using Enumerable.Range and only change the order in which they appear.
Changing the order can be done by OrderBy(), while random is used here.
This is all based on IEnumerable<T>. So it needs to be put into an array, which is simply done by calling ToArray().
public int[] RandomlyOrderedValues()
{
Random random = new Random();
return Enumerable.Range(1, 20).OrderBy(x => random.Next()).ToArray();
}
I'm not your teacher but hope you still play around by yourself, read the docs and finally express it in your own words.
Changed question, now the random range is larger than the array size.
It's always best to work with IEnumerable<T>, there you get the most powerful tools.
// let's create inifite number of random values:
// note that the Random instance needs to be created only once,
// so it's put into a field.
private Random random = new Random();
public IEnumerable<int> InfiniteRandomValues()
{
while (true)
{
yield return random.Next(1, 100);
}
}
public int[] GetUniqueRandomValues(int numberOfValues)
{
return InfiniteRandomValues()
// only uninque values
.Distinct()
// generate the requested number of distinct values
.Take(numberOfValues)
// put it into an array.
.ToArray();
}
How does it work? When you create random values, you don't know how many it will be, because you cannot know how many duplicates it will create. A generator for an infinite number of values has certainly enough values. Think of it as a factory. Only when the IEnumerable is iterated, the values are created.
This is called "deferred execution". Only when you iterate the IEnumerable, the values are requested by the source.
Distinct works like this. It returns only as many distinct values as are requested by its caller.
Which is Take. This one reduces the number of items that are taken, but still doesn't iterate itselves.
ToArray finally iterates its source and pulls as many values as there are. Read it backwards now: It takes all values from Take, which returns 20. Itselves it takes 20 values from Distinct, which iterates its source until it got 20 distinct values. Distinct takes its values from the InifiteRandomNumbers factory and can take as many as it needs.
When you finally understand how these things work, you can use it quite intuitively.
Another, more classic implemenation
private int[] GetRandomValues()
{
Random random = new Random();
int[] values = new int[20];
for(int i = 0; i < 20; i++)
{
// create random values until you found a distinct oune.
int nextValue;
do
{
nextValue = random.Next(1, 100);
} while (ContainsValue(values, i, nextValue))
values[i] = nextValue;
}
return values;
}
// When adding the values to a List instead of an array, it would be
// much easier, but need copying the vlaues to the array at the end.
// When using the array directly, you have to know which values you
// already generated, because it's initialized with zero.
// This method checks wether the value is in the array within
// the items until endIndex.
private bool ContainsValue(int[] values, int endIndex, int valueToFind)
{
// simple linq way:
// return values.Take(endIndex).Contains(valueToFind);
// classic way:
for(int i = 0; i < endIndex; i++)
{
if (values[i] = valueToFind) return true;
}
return false;
}

sorting a list in linq

I have to make a list containing objects. And the objects need to be in random order.
Here I give them random numbers:
Random tal = new Random();
list1[i].nummer = tal.Next(list1.Count);
listGold.Add(list1[i]);
And now i just need to order them by number.
Which I thought linq could do for me. But it can't :S
I am trying this:
RepeaterSponsorGold.DataSource = listGold.OrderBy(n => n.nummer);
RepeaterSponsorGold.DataBind();
to order my list by nummer and to put the list into my repater. But the lsit doesn't seem to be sorted... or doesn't seem to get random numbers. i don't know which.
Can anybody see what i am doing wrong??
Try
RepeaterSponsorGold.DataSource = listGold.OrderBy(n => n.nummer).ToList();
If you need to sort in random order you could try:
var listGold = list1.OrderBy(n => Guid.NewGuid());
without the need to use Random.
You should't use Random() to shuffle a list, as it can produce determanistic distributions.
Instead, use a suffle:
Randomize a List<T>
public static void Shuffle<T>(this IList<T> list)
{
Random rng = new Random();
int n = list.Count;
while (n > 1) {
n--;
int k = rng.Next(n + 1);
T value = list[k];
list[k] = list[n];
list[n] = value;
}
}
....
listGold.Shuffle();
How about this:
List<string> list = new List<string>()
{
"Blue",
"Brown",
"Beige",
"Red",
"Black"
};
Random random = new Random();
var orderedList = list.Select(c => new {Item = c, Order = random.Next()})
.OrderBy(i=>i.Item)
.Select(x=>x.Item).ToList();
orderedList.ForEach(x=>Console.WriteLine(x));
You have a couple of issues.
The most important is that you should only ever have a single instance of Random, and use it repeatedly, rather than invoking a new one every time.
Second, by using the list count as your seed, you'll get a roughly sequential list.
list1[i].nummer = tal.Next(list1.Count);
Because you're using your list.Count as your seed you'll get a list that is roughly in the order they were added.
Instead try:
Enumerable.Range(0,100)
.ToDictionary(k=> Guid.NewGuid())
.OrderBy(o => o.Key).Select(s => s.Value).ToList()
Using the Guids will give you a more natural randomization than Random.

C# code only gives expected results on step through?

Ok so I have a dice throw app...
When I step through the code it functions normally and 'results' contains the correct number of throw results and they appear to be random, when I leave the code to run and do exactly the same thing it produces a set of identical numbers.
I'm sure this is a logical error I cannot see but fiddling with it for hours hasnt improved the situation, so any help is much appriciated. :)
class Dice
{
public int[] Roll(int _throws, int _sides, int _count)
{
Random rnd = new Random();
int[] results = new int[_throws];
// for each set of dice to throw pass data to calculate method
for (int i = 0; i < _throws; i++)
{
int thisThrow = Calculate(_sides, _count);
//add each throw to a new index of array... repeat for every throw
results[i] = thisThrow;
}
return results;
}
private int Calculate(int _sides, int _count)
{
Random rnd = new Random();
int[] result = new int[_count];
int total = 0;
//for each dice to throw put data into result
for (int i = 0; i < _count; i++)
{
result[i] = rnd.Next(1, _sides);
}
//count the values in result
for (int x = 0; x < _count; x++)
{
total = total + result[x];
}
//return total of all dice to Roll method
return total;
}
}
First mistake: Never use multiple instances of Random, use a single instance, and pass that along with the other parameters.
When you create "Random rnd = new Random();" it is seeded by the current time. When you debug your code (which takes time) it will be seeded differently each time.
Create 1 instance of Random, and reference that everywhere.
You're creating a random class every time you need to create a number. Doing this will give you the nutty results.
See here: FROM MSDN
This problem can be avoided by creating a single Random object rather than multiple ones.
To improve performance, create one Random object to generate many random numbers over time, instead of repeatedly creating a new Random objects to generate one random number.
E.g. create a private instance of Random...
In addition to what has been mentioned before...
Use Random for things like dice, card games, choosing random images and so forth. If you ever need to create a random number for security sake, use System.Security.Cryptography.RandomNumberGenerator. This simple example shows creating a random integer.
RandomNumberGenerator gen = RandomNumberGenerator.Create();
byte[] myBytes = new byte[4];
gen.GetBytes(myBytes);
int myValue = (BitConverter.ToInt32(myBytes, 0));
DO NOT use this unless you have a security need. The performance is less than that of the Random class. I suppose you could use this to seed Random but that might be overkill.
EDIT: It occurred to me that I had never tested this. A quick performance test showed the following:
1,000,000 random numbers:
RandomNumberGenerator: 2.6 seconds
Random: .015 seconds.
So Random is about 150 times faster.
Give the constructor Random a seed. That's the problem.
http://msdn.microsoft.com/en-us/library/aa329890%28VS.71%29.aspx
Random r = new Random(DateTime.Now.Millisecond);

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