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I need some help to complete this code of non-duplicate random numbers

I need some help to do my homework. I should write non-duplicate random numbers. I'm able to show random numbers but I don't know about non-duplicate.
Here's my code:
Random r = new Random();
for (int i = 0; i < 40; i++)
{
int temp = r.Next(0, 100);
Console.WriteLine(temp);
}
What do I need to do to generate non-duplicate number?

Note that this answer only deals with (relatively) small, pre-determined sets.
The reason the other (simple) solution is inefficient is this: you want to generate 100 random numbers between 0 and 99. You get to the point where you have generated 90 random numbers, and just need 10 more.
The problem is that you're still generating numbers between 0 and 99 every time, except now your chance of finding a number that hasn't already been generated is 1 in 10. So 9 of every 10 numbers you generate has already been added to the list.
Once you get down to just needing 1 number, your chance of generating the remaining 1 that hasn't already been generated is 1 in 100. So for every 100 numbers you generate, only 1 of them will be the last possible number.
I'm sure this is simplifying things given that the Random class is pseudo-random (i.e. it's an algorithm that appears random), but this does explain your situation and why the other answer will be slower.
An improved solution would be this:
// Add all of the numbers 0 to 100 to a list
var availableNumbers = new List<int>();
for (int i = 0; i < 100; ++i)
{
availableNumbers.Add(i);
}
Random random = new Random();
for (int i = 0; i < 40; ++i)
{
// Choose a random position in the available numbers list
var idx = random.Next(0, availableNumbers.Count);
// Print the number from this position in the list
Console.WriteLine(availableNumbers[idx]);
// Remove the item at this position
availableNumbers.RemoveAt(idx);
}
Because we start with a list of all available numbers, we are able to choose numbers from it at random. Removing items from the available numbers list means that they are not available to be chosen a second time. We no longer have to try many times to find an unused number, as removing them when we select them ensures that all of the numbers in the available numbers list are always only unused numbers.

You may use a HashSet to store the numbers and make sure there are no duplicates. Here's an example:
HashSet<int> numbers = new HashSet<int>();
Random r = new Random();
for (int i = 0; i < 40; i++)
{
int temp;
do
{
temp = r.Next(0, 100);
} while (numbers.Add(temp) == false); // If the `.Add()` method returns false,
// that means the number already exists.
// So, we try to generate another number.
Console.WriteLine(temp);
}

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;
}

Randomly select a specific quantity of indices from an array?

I have an array of boolean values and need to randomly select a specific quantity of indices for values which are true.
What is the most efficient way to generate the array of indices?
For instance,
BitArray mask = GenerateSomeMask(length: 100000);
int[] randomIndices = RandomIndicesForTrue(mask, quantity: 10);
In this case the length of randomIndices would be 10.

There's a faster way to do this that requires only a single scan of the list.
Consider picking a line at random from a text file when you don't know how many lines are in the file, and the file is too large to fit in memory. The obvious solution is to read the file once to count the lines, pick a random number in the range of 0 to Count-1, and then read the file again up to the chosen line number. That works, but requires you to read the file twice.
A faster solution is to read the first line and save it as the selected line. You replace the selected line with the next line with probability 1/2. When you read the third line, you replace with probability 1/3, etc. When you've read the entire file, you have selected a line at random, and every line had equal probability of being selected. The code looks something like this:
string selectedLine = null;
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
double prob = 1.0/numLines;
if (rnd.Next() >= prob)
selectedLine = line;
}
Now, what if you want to select 2 lines? You select the first two. Then, as each line is read the probability that it will replace one of the two lines is 2/n, where n is the number of lines already read. If you determine that you need to replace a line, you randomly select the line to be replaced. You can follow that same basic idea to select any number of lines at random. For example:
string[] selectedLines = new int[M];
int numLines = 0;
Random rnd = new Random();
foreach (var line in File.ReadLines(filename))
{
++numLines;
if (numLines <= M)
{
selectedLines[numLines-1] = line;
}
else
{
double prob = (double)M/numLines;
if (rnd.Next() >= prob)
{
int ix = rnd.Next(M);
selectedLines[ix] = line;
}
}
}
You can apply that to your BitArray quite easily:
int[] selected = new int[quantity];
int num = 0; // number of True items seen
Random rnd = new Random();
for (int i = 0; i < items.Length; ++i)
{
if (items[i])
{
++num;
if (num <= quantity)
{
selected[num-1] = i;
}
else
{
double prob = (double)quantity/num;
if (rnd.Next() > prob)
{
int ix = rnd.Next(quantity);
selected[ix] = i;
}
}
}
}
You'll need some special case code at the end to handle the case where there aren't quantity set bits in the array, but you'll need that with any solution.
This makes a single pass over the BitArray, and the only extra memory it uses is for the list of selected indexes. I'd be surprised if it wasn't significantly faster than the LINQ version.
Note that I used the probability calculation to illustrate the math. You can change the inner loop code in the first example to:
if (rnd.Next(numLines+1) == numLines)
{
selectedLine = line;
}
++numLines;
You can make a similar change to the other examples. That does the same thing as the probability calculation, and should execute a little faster because it eliminates a floating point divide for each item.

There are two families of approaches you can use: deterministic and non-deterministic. The first one involves finding all the eligible elements in the collection and then picking N at random; the second involves randomly reaching into the collection until you have found N eligible items.
Since the size of your collection is not negligible at 100K and you only want to pick a few out of those, at first sight non-deterministic sounds like it should be considered because it can give very good results in practice. However, since there is no guarantee that N true values even exist in the collection, going non-deterministic could put your program into an infinite loop (less catastrophically, it could just take a very long time to produce results).
Therefore I am going to suggest going for a deterministic approach, even though you are going to pay for the guarantees you need through the nose with resource usage. In particular, the operation will involve in-place sorting of an auxiliary collection; this will practically undo the nice space savings you got by using BitArray.
Theory aside, let's get to work. The standard way to handle this is:
Filter all eligible indices into an auxiliary collection.
Randomly shuffle the collection with Fisher-Yates (there's a convenient implementation on StackOverflow).
Pick the N first items of the shuffled collection. If there are less than N then your input cannot satisfy your requirements.
Translated into LINQ:
var results = mask
.Select((i, f) => Tuple.Create) // project into index/bool pairs
.Where(t => t.Item2) // keep only those where bool == true
.Select(t => t.Item1) // extract indices
.ToList() // prerequisite for next step
.Shuffle() // Fisher-Yates
.Take(quantity) // pick N
.ToArray(); // into an int[]
if (results.Length < quantity)
{
// not enough true values in input
}

If you have 10 indices to choose from, you could generate a random number from 0 to 2^10 - 1, and use that as you mask.

C# Random Number Creation

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();