From: AA2e72E on 15 Jun 2010 11:14 I could say 10?20 but that would more than likely mean nothing to you. Perhaps http://stackoverflow.com/questions/254844/random-array-using-linq-and-c might provide another insight. From: Peter Duniho on 15 Jun 2010 11:23 Luigi wrote:> I'm trying in this way: > > int[] result = new int[10]; > Random random = new Random(); > int adding; > > for (var i = 0; i < 10; i++) > { > adding = random.Next(1, 20); > if (!result.Contains(adding)) > result[i] = adding; > } > > but it seems that give me also the number zero. Your original post states you want to also include the number zero. Do you or don't you want zero in the output? The correct (most efficient) way to accomplish what you want to do is to initialize an array of numbers that is exactly the range of numbers you want. For example, if you really do want the numbers "from 0 to 20 inclusive", then you need 21 elements, initialized starting at 0 and ending with 20. Whatever the range, then the next step is to "shuffle" the array. There have been a number of discussions in this newsgroup already, which you can review using Google Groups, or you can just look up "shuffle" algorithms on Wikipedia. The basic idea is to select a random element from the array to swap into the current position of the array, starting with the first element; with each iteration, increment the current position, and select a random element only from the range of that position through to the end of the array. Pete From: Peter Duniho on 15 Jun 2010 11:25 Peter Duniho wrote:> [...] > Whatever the range, then the next step is to "shuffle" the array. There > have been a number of discussions in this newsgroup already, which you > can review using Google Groups, or you can just look up "shuffle" > algorithms on Wikipedia. The basic idea is to select a random element > from the array to swap into the current position of the array, starting > with the first element; with each iteration, increment the current > position, and select a random element only from the range of that > position through to the end of the array. Oh, and it should go without saying, but just in case: If you only want 10 numbers from that range, then you simply choose the first 10 from the shuffled array. Pete From: Fred Mellender on 15 Jun 2010 11:41 Put the integers 0-20 in a list. Shuffle it. Take the first 10 from the list. To shuffle a list (not tested): public static List shuffledList(List listToShuffle, Random rand) { /* * Make a new list of elements picked from listToShuffle * in a random order. */ List randList = new List(listToShuffle); for (int k = randList.Count-1; k >= 0; k--) { int randIndx = rand.Next(k); T temp = randList[k]; randList[k] = randList[randIndx]; randList[randIndx] = temp; } return randList; } "Luigi" wrote in message news:FCD10C7D-3A0B-4909-8E05-84D257BE8206(a)microsoft.com...> Hello, > how can I write a method that returns me 10 random numbers from 0 to 20 > (included), without repetitions? > > Thanks a lot. > > Luigi > From: rossum on 15 Jun 2010 12:55 On Tue, 15 Jun 2010 03:10:16 -0700, Luigi wrote: >Hello, >how can I write a method that returns me 10 random numbers from 0 to 20 >(included), without repetitions? > >Thanks a lot. > >Luigi As others have said, put the numbers you want into some container or array and shuffle. The standard shuffling algorithm is in Knuth, Algorithm 3.4.2 P. It is covered on Wikipedia: http://en.wikipedia.org/wiki/Fisherâ€“Yates_shuffle#The_modern_algorithm The article contains Java code which should be easily convertible to C#. rossum