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I am using Random to generate a sequence of random number. I am constructing the random object just once and then inside the loop generating the random values (300 of them). The problem is that once I get all the values and do a sort on them I realize that some of them are equal and/or sequential: I am generating numbers from 0 to 50000.

This is my script:

Random rnd = new Random(); for (int n=0; n < 300; n++) { int RndNumber = rnd.Next(0, 50000); System.Threading.Thread.Sleep(3); }

Can someone have a clue on why this is happening, and how can I improve this to make it more random?

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  • You need to be specific about how you define "more random" Commented Feb 1, 2011 at 4:12
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    why are you sorting random numbers? doesn't that defeat the purpose of generating random numbers? Commented Feb 1, 2011 at 4:14
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    I'm no mathematician, but it seems to me that it would be highly unlikely that, after sorting a list of 300 random numbers between 0 and 50000, at least a couple of them wouldn't be equal or sequential. Commented Feb 1, 2011 at 4:15
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    Why are you sorting the random values? Is it just to test their distribution? This might have something to do with the Birthday paradox Commented Feb 1, 2011 at 4:15
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    Do you expect the numbers to be equally spaced across the range - 10, 110, 210, 310, etc? That seems far less random. Commented Feb 1, 2011 at 4:18

3 Answers 3

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So this is the birthday paradox*. When you draw 300 numbers from 50000 the approximate probability that at least two of them are equal is

p(300) = 1 - exp(-300 * 300 / (2 * 50000)) = 0.59

(I could work out the exact probability but I'm lazy!.)

So, chances are more likely than not that you'll have a collision. Sequential is even more likely (now you don't need a collision, you just need n - 1 and n or n and n + 1 to be hit for some n).

Random is fickle.

*: In case you're not familiar with it, it says that if you have twenty-three people in a room, it is more likely than not that at least two people in the room share the same birthday.

!: Okay, I worked it out. It's 0.5953830515549951746819986449....

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Research:

If you use the constructor without parameters new Random() the seed is depending on the current servertime.

Random(): "Initializes a new instance of the Random class, using a time-dependent" http://msdn.microsoft.com/en-us/library/system.random.aspx

So, if I try it like this:

for(int i = 0; i < 1000; i++) { Random ran = new Random(); Console.WriteLine(ran.Next(50001)); }

I only get 3 different numbers about 300 times within a thousand calls! Not that random...

Setting the seed in the constructor new Random(0) returns a fix serie of numbers.

e.g. new Random(0).Next(50) always! returns 36. Try it yourself, if you don't trust me;

What we need for "real" random numbers is a changing seed, that's independent of time.

I'm using Hashcode of changing values:

e.g. Guid.NewGuid().GetHashCode() or DateTime.Now.GetHashCode()

Result:

Random ran = new Random(Guid.NewGuid().GetHashCode()); for(int i = 0; i < 1000; i++) { Console.WriteLine(ran.Next(50001)); }

or (for better performance):

for(int i = 0; i < 1000; i++) { int val = Guid.NewGuid().GetHashCode() % 50001; val = val > 0 ? val : -val; Console.WriteLine(val); }

PS: The maximum of the Next(max)-method is always max - 1;

-> ran.Next(11) can return 0,1,2,...,8,9,10. Not 11!

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5 Comments

DateTime.Now.GetHashCode() is no better than the default seed.
+1 This was "random" enough for my liking :) No clumps of numbers when I tried to generate 2000 of them.
@CodesInChaos Your right. I don't know exactly if the default constructor of Random uses the DateTimes hashcode or milliseconds or whatever, but it ain't random at all
I love it, a very simple and modular solution
@Bahamut Do you realize that it wasn't your new seed value that fixed it, it was the fact that you moved the initialization of the Random instance out of the loop? Experiments are most effective when only one variable changes between trials.
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As an explanation of why you're seeing the occasional duplicate, Jason's answer is right on.

If what you want is 300 distinct random numbers, what about something like this?

static IEnumerable<int> GetRandoms(int min, int max) { var rand = new Random(); while (true) { yield return rand.Next(min, max); } } var distinctRandoms = GetRandoms(0, 50000).Distinct().Take(300);
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5 Comments

Yup it gives distinct number but it still have conjunction numbers, this is due to the Birthday paradox I believe. This seems to be fulfilling what i need to do. Thanks Dan Tao
Keep in mind that by forcing distinction, you are in fact harming the "Randomness" - the distribution is no longer (pseudo) uniform.
@bavaza: Not sure if that comment is aimed at me or the OP, but I think the OP does not actually want true randomness; otherwise he wouldn't be asking this question. Of course you're correct that eliminating duplicates compromises the randomness of the data, just as sorting does.
@Dan: It was aimed at the OP and other SO junkies out there. I decided to add it because I often found that people tend to abuse statistics. Anyways, thanks for the answer :-).
perfect though for psuedo random lists thanks DanTao

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