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This question already has answers here:
Produce a random number in a range using C#
(7 answers)
Closed 5 years ago.
So I need to generate random numbers based on user set parameters, and make a basic math expression. They have 3 options; single, double, or triple digit numbers. So based on that I need to generate a random number from 0 to 9, 0 to 99, and 0 to 100.
So far I've read about the Random class. This is my poor attempt:
Random rnd = new Random();
int value = 0;
if (digits == 1)
{
rnd.Next(value);
q1Lbl.Text = value.ToString();
}
You need this overload of Random.Next():
public virtual int Next(
int minValue,
int maxValue
)
Where:
minValue = The inclusive lower bound of the random number returned.
maxValue = The exclusive upper bound of the random number returned.
maxValue must be greater than or equal to minValue.
Note the words inclusive and exclusive in the parameter descriptions. This means the minimum value can be returned in the possible values, while the maximum value will not be possible.
This is further clarified in the description of the return value:
Return Value - A 32-bit signed integer greater than or equal to
minValue and less than maxValue; that is, the range of return values
includes minValue but not maxValue. If minValue equals maxValue,
minValue is returned.
For example, to get single digit values between 0 and 9 (inclusive), you'd use:
int value = rnd.Next(0, 10); // return a value between 0 and 9 inclusive
To get double digits, you'd use:
int value = rnd.Next(10, 100); // return a value between 10 and 99 inclusive
Finally, to get triple digit numbers, you'd use:
int value = rnd.Next(100, 1000); // return a value between 100 and 999 inclusive
Random rnd = new Random();
int single = rnd.Next(1, 10); // 1 ~9
int double = rnd.Next(1, 100); // 1~99
int triple = rnd.Next(1,101); // 1~100
loop it if you want to achieve the values multiple times
There's an overload of the Next method, you only have to pass the range:
Random rnd = new Random();
rnd.Next(1,20); //Gives you a number between 1 and 20
Here you find the entire documentation https://www.google.com.co/search?q=next+random+c%23&oq=next+random&aqs=chrome.1.69i57j0l5.3144j0j7&sourceid=chrome&ie=UTF-8
I need to generate a lot of random numbers which must be anywhere between 1 and int.MaxValue. The arguments passed to the Next method of the Random class are not always the same. In one senario the arguments might be as follows:
Next(1, int.MaxValue);
In another they might as well be:
Next(1, 2);
The issue here is that whenever you pass values like 1 and 2 or 99 and 100, the lower number is always the "random" number returned by the method. The reason for this is because the method subtracts 1 from the maximum value (unless min and max are the same) and then gives you the random number. How would I go about generating a range of random numbers within a range of numbers, as stated above, without getting this predictable outcome?
You would need to pass the inclusive upper bound + 1, ie:
var result = rand.Next(1, 2+1); // Returns 1 or 2
var result2 = rand.Next(99, 101); // Returns 99 or 100
Note that this won't work for int.MaxValue, of course. There is no way to have Random directly return int.MaxValue. To get a [0,int.MaxValue] result, you would need to do:
var result = rand.Next(-1, int.MaxValue) + 1;
The upper bound is exclusive, not inclusive. Given that, the range [1,2) only contains one number, 1, not two.
From the documentation, the first parameter to Next(int, int) is "The inclusive lower bound", while the second is "The exclusive upper bound".
If you want to generate a random number that might be 1 or 2, you should use the following call:
rand.Next(1, 3)
Try this
using System;
namespace ConsoleApplication3
{
class Program
{
static readonly Random r = new Random();
static void Main(string[] args)
{
for (int i = 2; i <= 100; i++)
{
Console.WriteLine(GetRandom(i));
}
}
private static int GetRandom(int i)
{
return 1 + (r.Next(int.MaxValue)%i);
}
}
}
Cheers.
You can get int.MaxValue by adding 1 after getting random int:
var result = rand.Next(1, int.MaxValue) + 1;
How about a method that takes the lower and upper limits and gives you what you're looking for?
Example:
public int GetRandom(int lower, int upper)
{
return upper == int.MaxValue ? rand.Next(lower - 1, upper) + 1 : rand.Next(lower, upper + 1);
}
If the upper limit is int.MaxValue, it shifts the range down by 1 and then adds it back after it gives the random number. Otherwise, it adds 1 to the upper limit and then gives you the random number.
Then when you use it, you'll just do something like:
var randomOne = GetRandom(1, 2);
var randomTwo = GetRandom(99,int.MaxValue);
If I wanted to generate a random number for all possible numbers an Int32 could contain would the following code be a reasonable way of doing so? Is there any reason why it may not be a good idea? (ie. a uniform distribution at least as good as Random.Next() itself anyway)
public static int NextInt(Random Rnd) //-2,147,483,648 to 2,147,483,647
{
int AnInt;
AnInt = Rnd.Next(System.Int32.MinValue, System.Int32.MaxValue);
AnInt += Rnd.Next(2);
return AnInt;
}
You could use Random.NextBytes to obtain 4 bytes, then use BitConverter.ToInt32 to convert those to an int.
Something like:
byte[] buf = new byte[4];
Rnd.NextBytes(buf);
int i = BitConverter.ToInt32(buf,0);
Your proposed solution will slightly skew the distribution. The minValue and maxValue will occur less frequently than the interior values. As an example, assume that int has a MinValue of -2 and a MaxValue of 1. Here are the possible initial values, with each followed by the resulting values after the Random(2):
-2: -2 -1
-1: -1 0
0: 0 1
half of the negative -2 values will get modified up to -1, and only half of 0 will get modified up to 1. So the values -2 and 1 will occur less frequently than -1 and 0.
Damien's solution is good. Another choice would be:
if (Random(2) == 0) {
return Random(int.MinValue, 0);
} else {
return 1 + Random(-1, int.MaxValue);
}
another solution, similar to Damiens approach, and faster than the previous one would be
int i = r.Next(ushort.MinValue, ushort.MaxValue + 1) << 16;
i |= r.Next(ushort.MinValue, ushort.MaxValue + 1);
A uniform distribution does not mean you get each number exactly once. For that you need a permutation
Now, if you need a random permutation of all 4-billion numbers you're a bit stuck. .NET does not allow objects to be larger than 2GBs. You can work around that, but I assume that's not really what you need.
If you less numbers (say, 100, or 5 million, less than a few billions) without repetitions, you should do this:
Maintain a set of integers, starting empty. Choose a random number. If it's already in the set, choose another random number. If it's not in the set, add it and return it.
That way you guarantee each number will be returned only once.
I have a class where I get random bytes into a 8KB buffer and distribute numbers from by converting them from the random bytes. This gives you the full int distribution. The 8KB buffer is used to you do not need to call NextBytes for every new random byte[].
// Get 4 bytes from the random buffer and cast to int (all numbers equally this way
public int GetRandomInt()
{
CheckBuf(sizeof(int));
return BitConverter.ToInt32(_buf, _idx);
}
// Get bytes for your buffer. Both random class and cryptoAPI support this
protected override void GetNewBuf(byte[] buf)
{
_rnd.NextBytes(buf);
}
// cyrptoAPI does better random numbers but is slower
public StrongRandomNumberGenerator()
{
_rnd = new RNGCryptoServiceProvider();
}
This question already has answers here:
Random number generator only generating one random number
(15 answers)
Closed 9 years ago.
I use the method to generate unique number but I always get the same number -2147483648. Even if I stop the program, recompile it and run again I still see the same number.
public static int GetRandomInt(int length)
{
var min = Math.Pow(10, length - 1);
var max = Math.Pow(10, length) - 1;
var random = new Random();
return random.Next((int)min, (int)max);
}
Try externalizing the random instance:
private readonly Random _random = new Random();
public static int GetRandomInt(int length)
{
var min = Math.Pow(10, length - 1);
var max = Math.Pow(10, length) - 1;
return _random.Next((int)min, (int)max);
}
This is not an issue of not reusing Random instance, the results he gets should be random on multiple starts, not always being -(2^32)
This is the issue with length being too big, and casting powers of length to int. If you break the code into following lines:
var min = Math.Pow(10, length - 1);
var max = Math.Pow(10, length) - 1;
var random = new Random();
var a = (int)min;
var b = (int)max;
return random.Next(a, b);
You'll see that a and b are -2147483648, making that the only possible result of Next(min, max) (the doc specifies if min==max, return min).
The largest length you can safely use with this method is 9. For a length of 10 you'll get System.ArgumentOutOfRangeException, for length > 10 you'll get the -2147483648 result.
You have three problems with your code.
You should externalize your random variable.
You have a problem with truncation error.
The range between min and max is way to large.
The first problem is because you may not have enough time to advance the seed when reinitializing your random variable. The second error comes from truncating your (what would b very large) numbers down to ints. Finally, your biggest problem is your range between your min and your max. Consider finding the range between min and max (as defined in your code) with inputs 1->20:
length max-min
1 8
2 89
3 899
4 8999
5 89999
6 899999
7 8999999
8 89999999
9 899999999
10 8,999,999,999
11 89999999999
12 899999999999
13 8999999999999
14 89999999999999
15 899999999999999
16 9E+15
17 9E+16
18 9E+17
19 9E+18
And keep in mind that the maximum integer is 2,147,483,647, which is passed on any number greater than 9.
You should keep an instance of Random and not new() it up all the time, that should give you better results.
Also check for what length actually is. It may be giving you funny results as to the limits.
I think the problem is the calculation of min and max. They will be greater than Int32.MaxValue pretty fast...
In your class, have one instance of Random, e.g.:
public class MyClass
{
private readonly Random random = new Random();
public static int GetRandomInt(int length)
{
var min = Math.Pow(10, length - 1);
var max = Math.Pow(10, length) - 1;
return random.Next((int)min, (int)max);
}
}
The fact that random always returns the same values only exists for testing purposes.
Random classes usually use a seed to initialize themselves, and will usually return the same sequence provided the seed is the same one :
Always reuse the same Random() instance instead of recreating one over and over again
if you want unpredictable results, use a time-dependent seed rather than an hard-coded one
It's very difficult to code a truly random number generator. Most methods use external entropy generators (such as mouse movement, cpu temperature, or even complex physical mechanisms such as helium balloons colliding one another...).
The Random instance should be created only once and then reused. The reason for this is that the RNG is by default seeded with the current system time. If you rapidly create new Random instances (and pull one value from it) then many of them will be seeded with the same timestamp, because the loop probably executes faster than the system clock advances.
Remember, a RNG initialized by seed A will always return sequence B. So if you create three Random instances all seeded with for example 123, these three instances will always return the same number on the same iteration.
I'm working in Microsoft Visual C# 2008 Express.
I found this snippet of code:
public static int RandomNumber(int min, int max)
{
Random random = new Random();
return random.Next(min, max);
}
the problem is that I've run it more than 100 times, and it's ALWAYS giving me the same answer when my min = 0 and max = 1. I get 0 every single time. (I created a test function to run it - really - I'm getting 0 each time). I'm having a hard time believing that's a coincidence... is there something else I can do to examine or test this? (I did re-run the test with min = 0 and max = 10 and the first 50ish times, the result was always "5", the 2nd 50ish times, the result was always "9".
?? I need something a little more consistently random...
-Adeena
The problem with min = 0 and max = 1 is that min is inclusive and max is exclusive. So the only possible value for that combination is 0.
random = new Random();
This initiates random number generator with current time (in sec). When you call your function many times before system clock changed, the random number generator is initiated with the same value so it returns same sequence of values.
Don't create a wrapper method for Next. It wastes cycles creating a new instance of the Random class. Just use the same one!
Random myRand = new Random();
for(int i = 0; i < 10; i++)
{
Console.WriteLine(myRand.Next(0, 10).ToString());
}
That should give you ten random values.
As has been said--Random is pseudo-random (as all implementations are), and if you create 100 instances with the same seed, you'll get 100 instances of the same results. Make sure that you're reusing the class.
Also, as folks have said, beware that MinValue is inclusive and MaxValue is exclusive. For what you want, do myRand.Next(0, 2).
That overload of Next() returns:
A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not MaxValue. If minValue equals maxValue, minValue is returned.
0 is the only possible value for it to return. Perhaps you want random.NextDouble(), which will return a double between 0 and 1.
The min is inclusive, but the max is exclusive. Check out the API
You're always getting 0 because Random.Next returns integers. You need to call Random.NextDouble, which will return a number between 0 and 1. Also, you should reuse your Random instance, like this:
[ThreadStatic]
static Random random;
public static Random Random {
get {
if (random == null) random = new Random();
return random;
}
}
public static int RandomInteger(int min, int max)
{
return Random.Next(min, max);
}
public static double RandomDouble() //Between 0 and 1
{
return Random.NextDouble();
}
If you want cryptographically secure random numbers, use the RNGCryptoServiceProvider class; see this article
EDIT: Thread safety
Besides the 0-1 issue already noted in other answers, your problem is a real one when you're looking for a 0-10 range and get identical results 50 times in a row.
new Random() is supposed to return a random number with a seed initialized from the timer (current second), but apparently you're calling this code 50 times a second. MSDN suggests: "To improve performance, create one Random to generate many random numbers over time, instead of repeatedly creating a new Random to generate one random number.". If you create your random generator once outside the method, that should fix your "non-randomness" problem as well as improving performance.
Also consider this post for a better pseudo-random number generator than the system-supplied one, if you need "higher quality" pseudo-random numbers.
As others have mentioned, the Random being built multiple times per second uses the same second as the seed, so I'd put the Random constructor outside your loop, and pass it as a parameter, like this:
public static int RandomNumber(Random random, int min, int max)
{
return random.Next(min, max);
}
Also as mentioned by others, the max is exclusive, so if you want a 0 or 1, you should use [0,2] as your [min,max], or some larger max and then do a binary AND with 1.
public static int RandomOneOrZero(Random random)
{
return random.Next(0, int.MaxValue) & 1;
}
This is an addendum to any answers, as the answer to this specific question is the bounds should be (0, 2) not (0, 1).
However, if you want to use a static wrapper method, then you must remember that Random is not thread-safe, so you either need to provide your own synchronization mechanism or provide a per-thread instance. Here is a largely non-blocking implementation which uses one generator to seed each per-thread generator:
public static class ThreadSafeRandom
{
private static readonly Random seed = new Random();
[ThreadStatic]
private static Random random;
public static int Next(int min, int max)
{
if (random == null)
{
lock (seed)
{
random = new Random(seed.Next());
}
}
return random.Next(min, max);
}
// etc. for other members
}
You're misunderstanding the line "random.Next(min, max)". "min" is in the place of the lowest number allowed to be generated randomly. While "max" is in the place of the lowest number NOT allowed to be generated it is not in the place of the largest number allowed to be drawn. So when the line is random.Next(0, 1) you are basically only allowing 0 to be drawn.
Several posters have stated that Random() uses a seed based on the current second on the system clock and any other instance of Random created in the same second will have the same seed. This is incorrect. The seed for the parameterless constructor of Random is based on the tick count, or number of milliseconds since boot time. This value is updated on most systems approximately every 15 milliseconds but it can vary depending on hardware and system settings.
I found a very simple, but effective way to generate random numbers by just taking the last two digits of the current datetime milliseconds:
int seed = Convert.ToInt32(DateTime.Now.Millisecond.ToString().Substring(1, 2));
int cnr = new Random(seed).Next(100);
It is crude, but it works! :-)
Of course it would statistically generate the same number every hundred times. Alternatively, you could take all three digits or concatenate with other datetime values like seconds or so.
Your range is not correct. The minValue is inclusive in the range and the maxValue is exclusive in the range(meaning it won't be included in the range). So that's why it returns only 0.
Another useful note: having a Random instance be created in the method is not ideal as it might get the same seed on calling. So instead i would say use:
static Random gen = new Random();
public static int RandomNumber(int minValue, int maxValue){
return gen.Next(minValue, maxValue);
}
in VB i always start with the Randomize() function. Just call Randomize() then run your random function. I also do the following:
Function RandomInt(ByVal lower As Integer, ByVal upper As Integer) As Integer
Return CInt(Int((upper - lower + 1) * Rnd() + lower))
End Function
Hope this helps! :)