Here's a bit of a puzzler: Random.Next() has an overload that accepts a minimum value and a maximum value. This overload returns a number that is greater than or equal to the minimum value (inclusive) and less than the maximum value (exclusive).
I would like to include the entire range including the maximum value. In some cases, I could accomplish this by just adding one to the maximum value. But in this case, the maximum value can be int.MaxValue, and adding one to this would not accomplish what I want.
So does anyone know a good trick to get a random number from int.MinValue to int.MaxValue, inclusively?
UPDATE:
Note that the lower range can be int.MinValue but can also be something else. If I know it would always be int.MinValue then the problem would be simpler.
The internal implementation of Random.Next(int minValue, int maxValue) generates two samples for large ranges, like the range between Int32.MinValue and Int32.MaxValue. For the NextInclusive method I had to use another large range Next, totaling four samples. So the performance should be comparable with the version that fills a buffer with 4 bytes (one sample per byte).
public static class RandomExtensions
{
public static int NextInclusive(this Random random, int minValue, int maxValue)
{
if (maxValue == Int32.MaxValue)
{
if (minValue == Int32.MinValue)
{
var value1 = random.Next(Int32.MinValue, Int32.MaxValue);
var value2 = random.Next(Int32.MinValue, Int32.MaxValue);
return value1 < value2 ? value1 : value1 + 1;
}
return random.Next(minValue - 1, Int32.MaxValue) + 1;
}
return random.Next(minValue, maxValue + 1);
}
}
Some results:
new Random(0).NextInclusive(int.MaxValue - 1, int.MaxValue); // returns int.MaxValue
new Random(1).NextInclusive(int.MaxValue - 1, int.MaxValue); // returns int.MaxValue - 1
new Random(0).NextInclusive(int.MinValue, int.MinValue + 1); // returns int.MinValue + 1
new Random(1).NextInclusive(int.MinValue, int.MinValue + 1); // returns int.MinValue
new Random(24917099).NextInclusive(int.MinValue, int.MaxValue); // returns int.MinValue
var random = new Random(784288084);
random.NextInclusive(int.MinValue, int.MaxValue);
random.NextInclusive(int.MinValue, int.MaxValue); // returns int.MaxValue
Update: My implementation has mediocre performance for the largest possible range (Int32.MinValue - Int32.MaxValue), so I came up with a new one that is 4 times faster. It produces around 22,000,000 random numbers per second in my machine. I don't think that it can get any faster than that.
public static int NextInclusive(this Random random, int minValue, int maxValue)
{
if (maxValue == Int32.MaxValue)
{
if (minValue == Int32.MinValue)
{
var value1 = random.Next() % 0x10000;
var value2 = random.Next() % 0x10000;
return (value1 << 16) | value2;
}
return random.Next(minValue - 1, Int32.MaxValue) + 1;
}
return random.Next(minValue, maxValue + 1);
}
Some results:
new Random(0).NextInclusive(int.MaxValue - 1, int.MaxValue); // = int.MaxValue
new Random(1).NextInclusive(int.MaxValue - 1, int.MaxValue); // = int.MaxValue - 1
new Random(0).NextInclusive(int.MinValue, int.MinValue + 1); // = int.MinValue + 1
new Random(1).NextInclusive(int.MinValue, int.MinValue + 1); // = int.MinValue
new Random(1655705829).NextInclusive(int.MinValue, int.MaxValue); // = int.MaxValue
var random = new Random(1704364573);
random.NextInclusive(int.MinValue, int.MaxValue);
random.NextInclusive(int.MinValue, int.MaxValue);
random.NextInclusive(int.MinValue, int.MaxValue); // = int.MinValue
No casting, no long, all boundary cases taken into account, best performance.
static class RandomExtension
{
private static readonly byte[] bytes = new byte[sizeof(int)];
public static int InclusiveNext(this Random random, int min, int max)
{
if (max < int.MaxValue)
// can safely increase 'max'
return random.Next(min, max + 1);
// now 'max' is definitely 'int.MaxValue'
if (min > int.MinValue)
// can safely decrease 'min'
// so get ['min' - 1, 'max' - 1]
// and move it to ['min', 'max']
return random.Next(min - 1, max) + 1;
// now 'max' is definitely 'int.MaxValue'
// and 'min' is definitely 'int.MinValue'
// so the only option is
random.NextBytes(bytes);
return BitConverter.ToInt32(bytes, 0);
}
}
Well, I have a trick. I'm not sure I'd describe it as a "good trick", but I feel like it might work.
public static class RandomExtensions
{
public static int NextInclusive(this Random rng, int minValue, int maxValue)
{
if (maxValue == int.MaxValue)
{
var bytes = new byte[4];
rng.NextBytes(bytes);
return BitConverter.ToInt32(bytes, 0);
}
return rng.Next(minValue, maxValue + 1);
}
}
So, basically an extension method that will simply generate four bytes if the upper-bound is int.MaxValue and convert to an int, otherwise just use the standard Next(int, int) overload.
Note that if maxValue is int.MaxValue it will ignore minValue. Guess I didn't account for that...
Split the ranges in two, and compensate for the MaxValue:
r.Next(2) == 0 ? r.Next(int.MinValue, 0) : (1 + r.Next(-1, int.MaxValue))
If we make the ranges of equal size, we can get the same result with different math. Here we rely on the fact that int.MinValue = -1 - int.MaxValue:
r.Next(int.MinValue, 0) - (r.Next(2) == 0 ? 0 : int.MinValue)
I'd suggest using System.Numerics.BigInteger like this:
class InclusiveRandom
{
private readonly Random rnd = new Random();
public byte Next(byte min, byte max) => (byte)NextHelper(min, max);
public sbyte Next(sbyte min, sbyte max) => (sbyte)NextHelper(min, max);
public short Next(short min, short max) => (short)NextHelper(min, max);
public ushort Next(ushort min, ushort max) => (ushort)NextHelper(min, max);
public int Next(int min, int max) => (int)NextHelper(min, max);
public uint Next(uint min, uint max) => (uint)NextHelper(min, max);
public long Next(long min, long max) => (long)NextHelper(min, max);
public ulong Next(ulong min, ulong max) => (ulong)NextHelper(min, max);
private BigInteger NextHelper(BigInteger min, BigInteger max)
{
if (max <= min)
throw new ArgumentException($"max {max} should be greater than min {min}");
return min + RandomHelper(max - min);
}
private BigInteger RandomHelper(BigInteger bigInteger)
{
byte[] bytes = bigInteger.ToByteArray();
BigInteger random;
do
{
rnd.NextBytes(bytes);
bytes[bytes.Length - 1] &= 0x7F;
random = new BigInteger(bytes);
} while (random > bigInteger);
return random;
}
}
I tested it with sbyte.
var rnd = new InclusiveRandom();
var frequency = Enumerable.Range(sbyte.MinValue, sbyte.MaxValue - sbyte.MinValue + 1).ToDictionary(i => (sbyte)i, i => 0ul);
var count = 100000000;
for (var i = 0; i < count; i++)
frequency[rnd.Next(sbyte.MinValue, sbyte.MaxValue)]++;
foreach (var i in frequency)
chart1.Series[0].Points.AddXY(i.Key, (double)i.Value / count);
chart1.ChartAreas[0].AxisY.StripLines
.Add(new StripLine { Interval = 0, IntervalOffset = 1d / 256, StripWidth = 0.0003, BackColor = Color.Red });
Distribution is OK.
This is guaranteed to work with negative and non-negative values:
public static int NextIntegerInclusive(this Random r, int min_value, int max_value)
{
if (max_value < min_value)
{
throw new InvalidOperationException("max_value must be greater than min_value.");
}
long offsetFromZero =(long)min_value; // e.g. -2,147,483,648
long bound = (long)max_value; // e.g. 2,147,483,647
bound -= offsetFromZero; // e.g. 4,294,967,295 (uint.MaxValue)
bound += Math.Sign(bound); // e.g. 4,294,967,296 (uint.MaxValue + 1)
return (int) (Math.Round(r.NextDouble() * bound) + offsetFromZero); // e.g. -2,147,483,648 => 2,147,483,647
}
It's actually interesting that this isn't the implementation for Random.Next(int, int), because you can derive the behavior of exclusive from the behavior of inclusive, but not the other way around.
public static class RandomExtensions
{
private const long IntegerRange = (long)int.MaxValue - int.MinValue;
public static int NextInclusive(this Random random, int minValue, int maxValue)
{
if (minValue > maxValue)
{
throw new ArgumentOutOfRangeException(nameof(minValue));
}
var buffer = new byte[4];
random.NextBytes(buffer);
var a = BitConverter.ToInt32(buffer, 0);
var b = a - (long)int.MinValue;
var c = b * (1.0 / IntegerRange);
var d = c * ((long)maxValue - minValue + 1);
var e = (long)d + minValue;
return (int)e;
}
}
new Random(0).NextInclusive(int.MaxValue - 1, int.MaxValue); // returns int.MaxValue
new Random(1).NextInclusive(int.MaxValue - 1, int.MaxValue); // returns int.MaxValue - 1
new Random(0).NextInclusive(int.MinValue, int.MinValue + 1); // returns int.MinValue + 1
new Random(1).NextInclusive(int.MinValue, int.MinValue + 1); // returns int.MinValue
new Random(-451732719).NextInclusive(int.MinValue, int.MaxValue); // returns int.MinValue
new Random(-394328071).NextInclusive(int.MinValue, int.MaxValue); // returns int.MaxValue
As I understand it you want Random to put out a value between -2.147.483.648 and +2.147.483.647. But the problem is that Random given those values will only give values from -2.147.483.648 to +2.147.483.646, as the maximum is exclusive.
Option 0: Take the thing away and learn to do without it
Douglas Adams was not a Programmer AFAIK, but he has some good advice for us: "The technology involved in making anything invisible is so infinitely complex that nine hundred and ninety-nine billion, nine hundred and ninety-nine million, nine hundred and ninety-nine thousand, nine hundred and ninety-nine times out of a trillion it is much simpler and more effective just to take the thing away and do without it."
This might be such a case.
Option 1: We need a bigger Random!
Random.Next() uses Int32 as Argument. One option I can think off would be to use a different Random Function Which can take the next higher level of Integers (Int64) as input. An Int32 is implicitly cast into an Int64. Int64 Number = Int64(Int32.MaxValue)+1;
But afaik, you would have to go outside of .NET libraries to do this. At that point, you might as well look for a Random that is inclusive of the Max.
But I think there is a mathematical reason it had to exclude one value.
Option 2: Roll more
Another way is to use two calls of Random - each for one half of the range - and then add them.
Number1 = rng.Next(-2.147.483.648, 0);
Number2 = rng.Next(0, 2.147.483.647);
resut = Number1 + Number2;
However, I am 90% certain that will ruin the random distribution. My P&P RPG experience gave me some experience with dice chances and I know for a fact rolling 2 dice (or the same 2 times) will get you a very different result distribution than one specific die. If you do not need this random distribution, that is an option. But if you do not care too much about the distribution it is worth a check.
Option 3: Do you need the full range? Or do you just care for min and max to be in it?
I assume you are doing some form of testing and you need both Int.MaxValue and Int.MinValue to be in the range. But do you need every value in between as well, or could you do without one of them?
If you have to lose value, would you prefer loosing 4 rather then Int.MaxValue?
Number = rng.Next(Int.MinValue, Int.MaxValue);
if(Number > 3)
Number = Number +1;
code like this would get you every number between MinValue and MaxValue, except 4. But in most cases code that can deal with 3 and 5 can also deal with 4. There is no need to explicitly test 4.
Of course, that assumes 4 is not some important test number that has to be run (I avoided 1 and 0 for those reasons). You could also decide the number to "skip" Randomly:
skipAbleNumber = rng.Next(Int.MinValue +1, Int.MaxValue);
And then use > skipAbleNumber rather than > 4.
You can not use Random.Next() to achieve what you want, because you can not correspond sequence of N numbers to N+1 and not miss one :). Period.
But you can use Random.NextDouble(), which returns double result:
0 <= x < 1 aka [0, 1)
between 0, where [ is inclusive sign and ) exclusive
How do we correspond N numbers to [0, 1)?
You need to split [0, 1) to N equal segments:
[0, 1/N), [1/N, 2/N), ... [N-1/N, 1)
And here is where it becomes important that one border is inclusive and another is exclusive - all N segments are absolutely equal!
Here is my code: I made it as a simple console program.
class Program
{
private static Int64 _segmentsQty;
private static double _step;
private static Random _random = new Random();
static void Main()
{
InclusiveRandomPrep();
for (int i = 1; i < 20; i++)
{
Console.WriteLine(InclusiveRandom());
}
Console.ReadLine();
}
public static void InclusiveRandomPrep()
{
_segmentsQty = (Int64)int.MaxValue - int.MinValue;
_step = 1.0 / _segmentsQty;
}
public static int InclusiveRandom()
{
var randomDouble = _random.NextDouble();
var times = randomDouble / _step;
var result = (Int64)Math.Floor(times);
return (int)result + int.MinValue;
}
}
This method can give you a random integer within any integer limits. If the maximum limit is less than int.MaxValue, then it uses the ordinary Random.Next(Int32, Int32) but with adding 1 to upper limit to include its value. If not but with lower limit greater than int.MinValue, it lowers the lower limit with 1 to shift the range 1 less that add 1 to the result. Finally, if both limits are int.MinValue and int.MaxValue, it generates a random integer 'a' that is either 0 or 1 with 50% probability of each, then it generates two other integers, the first is between int.MinValue and -1 inclusive, 2147483648 values, and the second is between 0 and int.MaxValue inclusive , 2147483648 values also, and using them with the value of 'a' it select an integer with totally equal probability.
private int RandomInclusive(int min, int max)
{
if (max < int.MaxValue)
return Random.Next(min, max + 1);
if (min > int.MinValue)
return Random.Next(min - 1, max) + 1;
int a = Random.Next(2);
return Random.Next(int.MinValue, 0) * a + (Random.Next(-1, int.MaxValue) + 1) * (1 - a);
}
What about this?
using System;
public class Example
{
public static void Main()
{
Random rnd = new Random();
int min_value = max_value;
int max_value = min_value;
Console.WriteLine("\n20 random integers from 10 to 20:");
for (int ctr = 1; ctr <= 20; ctr++)
{
Console.Write("{0,6}", rnd.Next(min_value, max_value));
if (ctr % 5 == 0) Console.WriteLine();
}
}
}
You can try this. A bit hacky but can get you both min and max inclusive.
static void Main(string[] args)
{
int x = 0;
var r = new Random();
for (var i = 0; i < 32; i++)
x = x | (r.Next(0, 2) << i);
Console.WriteLine(x);
Console.ReadKey();
}
You can add 1 to generated number randomly so it still random and cover full range integer.
public static class RandomExtension
{
public static int NextInclusive(this Random random, int minValue, int maxValue)
{
var randInt = random.Next(minValue, maxValue);
var plus = random.Next(0, 2);
return randInt + plus;
}
}
Will this work for you?
int random(Random rnd, int min, int max)
{
return Convert.ToInt32(rnd.NextDouble() * (max - min) + min);
}
Related
I have a function (f) the takes a number of items (n) and a number of columns (c) and returns the optimal layout as an array of items per column. I define optimal as being as square as possible. So f(4,4) would return [4,4,4,4], f(17,4) would return [5,4,4,4], and f(1,4) would return [1,0,0,0]. My function works correctly in all my tests, but I am looking to alter it. My desire to do this is not because I am looking increase performance. I just want to do this, because I am experimenting and want to learn different techniques.
Here is the code:
public static int[] f(int n, int c){
int[] a = new int[c];
if(c>0 && n>=0){
int opt = (n-(n%c))/c;
n = n - (opt*c);
for(int i = 0;i<a.Length;i++){
a[i] = opt;
if(n>0){
a[i]++;
n--;
}
}
}
return a;
}
The function works by first determining the optimal number of items per col:
int opt = (n-(n%c))/c;
So f(17,4) would yield 4, f(19,4) would also yield 4, and f(3,4) would yield 0. Then the reminder is calculated:
n = n - (opt*c);
I then loop through the array (of length c) and assign a[i] equal to the optimal value. Finally, if the reminder is greater than 0 I add 1 to a[i]. This equally distributes the reminder across the array. This is the part I would like to alter.
Instead of checking if(n>0) and adding 1 to the array is there a formula I could use that might look like:
a[i] = opt + n*?????;
So n*??? would always equal 1 if n is greater than 0 and 0 if n is 0 or less?
The simple answer to your question is to use an expression with the conditional operator:
a[i] = opt + (n > 0 ? 1 : 0);
(n > 0 ? 1 : 0) will be 1 if n is greater than 0, and 0 otherwise.
On that note, there is a clearer and more concise way to implement your algorithm.
Determine the total number of items that can be distributed evenly between the slots (call this average). This has the value n / c (using integer division).
Determine the remainder that would be left after those are evenly distributed (call this remainder). This has the value n % c.
Put the value average + 1 in the first remainder slots, and put average in the rest.
The implementation for this would be:
public static int[] Distribute(int total, int buckets)
{
if (total < 0) { throw new ArgumentException("cannot be less than 0", "total"); }
if (buckets < 1) { throw new ArgumentException("cannot be less than 1", "buckets"); }
var average = total / buckets;
var remainder = total % buckets;
var array = new int[buckets];
for (var i = 0; i < buckets; i++)
{
array[i] = average + (i < remainder ? 1 : 0);
}
return array;
}
And the obligatory Linq version:
public static int[] DistributeLinq(int total, int buckets)
{
if (total < 0) { throw new ArgumentException("cannot be less than 0", "total"); }
if (buckets < 1) { throw new ArgumentException("cannot be less than 1", "buckets"); }
var average = total / buckets;
var remainder = total % buckets;
return Enumerable.Range(1, buckets)
.Select(v => average + (v <= remainder ? 1 : 0))
.ToArray();
}
If you want to use a formula:
Math.Max(n - Math.Abs(n - 1), 0)
should do the trick.
Your code should look like:
a[i] = opt + Math.Max(n - Math.Abs(n - 1), 0)
Another option for a formula would be
Math.Max(Math.Sign(n), 0)
If you are looking for a mathematical formula, I'm not sure you're going to find it as the function is discontinuous at n = 0.
How about a simple function which outputs int on a bool expression?
int IsPositive(int number)
{
//if number is > 0 return integer one (1), else return integer zero (0)
return number > 0 ? 1 : 0;
}
You can then use this in your code as such:
a[i] = opt + IsPositive(n);
//opt + 1 if n > 0, opt + 0 if n <= 0
Update: per your comment, you can just move the evaluation inline:
a[i] = opt + (n > 0 ? 1 : 0);
As an aside: you should make #BradleyDotNET's comment one of your programming mottos.
I'm trying to generate a number based on a seed in C#. The only problem is that the seed is too big to be an int32. Is there a way I can use a long as the seed?
And yes, the seed MUST be a long.
Here's a C# version of Java.Util.Random that I ported from the Java Specification.
The best thing to do is to write a Java program to generate a load of numbers and check that this C# version generates the same numbers.
public sealed class JavaRng
{
public JavaRng(long seed)
{
_seed = (seed ^ LARGE_PRIME) & ((1L << 48) - 1);
}
public int NextInt(int n)
{
if (n <= 0)
throw new ArgumentOutOfRangeException("n", n, "n must be positive");
if ((n & -n) == n) // i.e., n is a power of 2
return (int)((n * (long)next(31)) >> 31);
int bits, val;
do
{
bits = next(31);
val = bits % n;
} while (bits - val + (n-1) < 0);
return val;
}
private int next(int bits)
{
_seed = (_seed*LARGE_PRIME + SMALL_PRIME) & ((1L << 48) - 1);
return (int) (((uint)_seed) >> (48 - bits));
}
private long _seed;
private const long LARGE_PRIME = 0x5DEECE66DL;
private const long SMALL_PRIME = 0xBL;
}
For anyone seeing this question today, .NET 6 and upwards provides Random.NextInt64, which has the following overloads:
NextInt64()
Returns a non-negative random integer.
NextInt64(Int64)
Returns a non-negative random integer that is less than the specified maximum.
NextInt64(Int64, Int64)
Returns a random integer that is within a specified range.
I'd go for the answer provided here by #Dyppl: Random number in long range, is this the way?
Put this function where it's accessible to the code that needs to generate the random number:
long LongRandom(long min, long max, Random rand)
{
byte[] buf = new byte[8];
rand.NextBytes(buf);
long longRand = BitConverter.ToInt64(buf, 0);
return (Math.Abs(longRand % (max - min)) + min);
}
Then call the function like this:
long r = LongRandom(100000000000000000, 100000000000000050, new Random());
I want to make a method that takes a variable of type int or long and returns an array of ints or longs, with each array item being a group of 3 digits. For example:
int[] i = splitNumber(100000);
// Outputs { 100, 000 }
int[] j = splitNumber(12345);
// Outputs { 12, 345 }
int[] k = splitNumber(12345678);
// Outputs { 12, 345, 678 }
// Et cetera
I know how to get the last n digits of a number using the modulo operator, but I have no idea how to get the first n digits, which is the only way to make this method that I can think of. Help please!
Without converting to string:
int[] splitNumber(int value)
{
Stack<int> q = new Stack<int>();
do
{
q.Push(value%1000);
value /= 1000;
} while (value>0);
return q.ToArray();
}
This is simple integer arithmetic; first take the modulo to get the right-most decimals, then divide to throw away the decimals you already added. I used the Stack to avoid reversing a list.
Edit: Using log to get the length was suggested in the comments. It could make for slightly shorter code, but in my opinion it is not better code, because the intent is less clear when reading it. Also, it might be less performant due to the extra Math function calls. Anyways; here it is:
int[] splitNumber(int value)
{
int length = (int) (1 + Math.Log(value, 1000));
var result = from n in Enumerable.Range(1,length)
select ((int)(value / Math.Pow(1000,length-n))) % 1000;
return result.ToArray();
}
By converting into a string and then into int array
int number = 1000000;
string parts = number.ToString("N0", new NumberFormatInfo()
{
NumberGroupSizes = new[] { 3 },
NumberGroupSeparator = "."
});
By using Maths,
public static int[] splitNumberIntoGroupOfDigits(int number)
{
var numberOfDigits = Math.Floor(Math.Log10(number) + 1); // compute number of digits
var intArray = new int[Convert.ToInt32(numberOfDigits / 3)]; // we know the size of array
var lastIndex = intArray.Length -1; // start filling array from the end
while (number != 0)
{
var lastSet = number % 1000;
number = number / 1000;
if (lastSet == 0)
{
intArray[lastIndex] = 0; // set of zeros
--lastIndex;
}
else if (number == 0)
{
intArray[lastIndex] = lastSet; // this could be your last set
--lastIndex;
}
else
{
intArray[lastIndex] = lastSet;
--lastIndex;
}
}
return intArray;
}
Try converting it to string first and do the parsing then convert it back to number again
Convert to string
Get length
If length modulus 3 == 0
String substring it into ints every 3
else if
Find remainder such as one or two left over
Substring remainder off of front of string
Then substring by 3 for the rest
You can first find out how large the number is, then use division to get the first digits, and modulo to keep the rest:
int number = 12345678;
int len = 1;
int div = 1;
while (number >= div * 1000) {
len++;
div *= 1000;
}
int[] result = new int[len];
for (int i = 0; i < result.Length; i++) {
result[i] = number / div;
number %= div;
div /= 1000;
}
You can use this with the System.Linq namespace from .NET 3.5 and above:
int[] splitNumber(long value)
{
LinkedList<int> results = new LinkedList<int>();
do
{
int current = (int) (value % 1000);
results.AddFirst(current);
value /= 1000;
} while (value > 0);
return results.ToArray();// Extension method
}
I use LinkedList<int> to avoid having to Reverse a list before returning. You could also use Stack<int> for the same purpose, which would only require .NET 2.0:
int[] splitNumber(long value)
{
Stack<int> results = new Stack<int>();
do
{
int current = (int) (value % 1000);
results.Push(current);
value /= 1000;
} while (value > 0);
return results.ToArray();
}
I need to get a random number in C#, within (-15, 15) but without generate values between (-10, 10) as float numbers. Like the random should come with in (-15,-10) and (10,15).
Is it possible to get?
This can be done in one line, but I separated it out for clarity.
public double GetRandomNumber()
{
//Between 0 and 1
Random random = new Random();
double randomNumber = random.NextDouble();
//Between -0.5 and 0.5;
randomNumber -= 0.5;
//Between -5.0 and 5.0;
randomNumber *= 10.0;
//Between [-15.0, -10.0] or [10.0, 15.0]
randomNumber += Math.Sign(randomNumber) * 10.0;
return randomNumber;
}
One way would be to use this:
var random = new Random();
var result = random.NextDouble();
if(result < 0.5)
result = -15 + result * 10;
else
result = 5 + result * 10;
Random.NextDouble generates a number between 0.0 and 1.0.
If it is less than 0.5 we treat this as the indicator to create a negative number.
Another approach, assuming that you do want a decimal place in your result (since it isn't entirely clear).
Random rand = new Random();
var intValue = rand.Next(10,15);
var decimalValue = rand.NextDouble();
var sign = rand.Next(0,1);
if (sign == 0) sign = -1;
return (intValue + decimalValue) * sign;
Generate two random numbers. The first random number will choose whether you're in the -15 to -10 range, or the 10 to 15 range, the second - how far along in that range you want to be.
Try:
Random random = new Random();
float value = (float) (Math.Sign((random.NextDouble() - 1)) * (10 + 5 * random.NextDouble()));
static Random r = new Random(2);
static void Main(string[] args)
{
int d = r.Next(-15, 15);
while ((d >= -15 && d <= -10) || (d >= 10 && d <= 15))
Console.WriteLine(d);
Console.ReadLine();
}
If you want integers and you want it to be performant, I would use:
var rand = new Random();
int value = rand.Next(-5, 7);
return value > 0 ? value + 9 : value - 10;
I am assuming you want numbers from the ranges: [-15, -10] and [10, -15]. Adjust hard-coded values above appropriately. To make it more performant, there may be a way to do some bit arithmetic and avoid the comparison but I'll leave that for the comments.
Here is a static method to accomplish that:
public static int RandomRange(int min, int max, bool includeNegatives = false)
{
if( min >= max) throw new Exception("min can't be greater than max");
Random rdm = new Random();
int num = 0;
while (num < min) num = rdm.Next(max + 1);
return num * (rdm.Next() % 2 == 0 ? -1 : 1);
}
To call it:
var TenToFifteenNegatives = RandomRange(10, 15, true);
var TenToFifteenNoNegatives = RandomRange(10, 15);
I'm creating a function where returns multiple of 10?
public static int NextInt(this Random rnd, int min = 0, int max = 1)
{
if (rnd == null) throw new ArgumentNullException("rnd");
if (min >= max) throw new InvalidOperationException();
var delta = max - min;
return min + (int)(rnd.NextDouble() * delta + min);
}
public static int MultipleOf10(this Random rnd, int minZeros = 1,
int maxZeros = 10)
{
if (rnd == null) throw new ArgumentNullException("rnd");
int pow = NextInt(rnd, minZeros, maxZeros);
return (int)Math.Pow(10, pow);
}
I've got doubt about the these two functions. The first one must be similar like Next() and the another one must return multiples of ten.
Can you point me where I fix it? Because I'm almost sure of that.
I think the last line of NextInt() should be
return min + (int)(rnd.NextDouble() * delta);
Or:
return min + rnd.Next(delta);
Or even better:
return rnd.Next(min, max);
Also, you should keep in mind that the upper bounds of Random methods are exclusive. So, if rnd.NextInt(i, j) should return numbers between i and j inclusive, you probably want to change the computation of delta to:
int delta = max - min + 1;
or if you used the last option above, change it to:
return rnd.Next(min, max + 1);
I think you need to define the MultipleOf10 method like so:
public static int MultipleOf10
(this Random rnd, int minZeros = 1, int maxZeros = 10)
{
if (rnd == null) throw new ArgumentNullException("rnd");
if (minZeros >= maxZeros) throw new InvalidOperationException();
return (int)Math.Pow(10, rnd.Next(minZeros, maxZeros));
}
And you don't need the NextInt method.