Can Random.Next() ever return int.MaxValue? - c#

The documentation for the Random.Next() method states that it returns:
A 32-bit signed integer that is greater than or equal to 0 and less than MaxValue.
But, I took a peek at the implementation, and while I don't understand the algorithm (a quick Google search suggests that it is a subtractive generator), I can't see any way in which a value of exactly int.MaxValue is ruled out.
If, for pedantic reasons, someone wants a random number across the entire range of 32-bit integers, does Random.Next() alone suffice, or does it become necessary to do something like assemble two separate 16-bit samples?

It will always be less than int.MaxValue.
In your linked source code it explicitly handles int.MaxValue:
if (retVal == MBIG) retVal--;
MBIG is defined earlier:
private const int MBIG = int.MaxValue;
https://github.com/dotnet/runtime/blob/master/src/libraries/System.Private.CoreLib/src/System/Random.cs#L105

Related

Difference between two large numbers C#

There are already solutions to this problem for small numbers:
Here: Difference between 2 numbers
Here: C# function to find the delta of two numbers
Here: How can I find the difference between 2 values in C#?
I'll summarise the answer to them all:
Math.Abs(a - b)
The problem is when the numbers are large this gives the wrong answer (by means of an overflow). Worse still, if (a - b) = Int32.MinValue then Math.Abs crashes with an exception (because Int32.MaxValue = Int32.MinValue - 1):
System.OverflowException occurred
HResult=0x80131516
Message=Negating the minimum value of a twos complement number is
invalid.
Source=mscorlib
StackTrace: at
System.Math.AbsHelper(Int32 value) at System.Math.Abs(Int32 value)
Its specific nature leads to difficult-to-reproduce bugs.
Maybe I'm missing some well known library function, but is there any way of determining the difference safely?
As suggested by others, use BigInteger as defined in System.Numerics (you'll have to include the namespace in Visual Studio)
Then you can just do:
BigInteger a = new BigInteger();
BigInteger b = new BigInteger();
// Assign values to a and b somewhere in here...
// Then just use included BigInteger.Abs method
BigInteger result = BigInteger.Abs(a - b);
Jeremy Thompson's answer is still valid, but note that the BigInteger namespace includes an absolute value method, so there shouldn't be any need for special logic. Also, Math.Abs expects a decimal, so it will give you grief if you try to pass in a BigInteger.
Keep in mind there are caveats to using BigIntegers. If you have a ludicrously large number, C# will try to allocate memory for it, and you may run into out of memory exceptions. On the flip side, BigIntegers are great because the amount of memory allotted to them is dynamically changed as the number gets larger.
Check out the microsoft reference here for more info: https://msdn.microsoft.com/en-us/library/system.numerics.biginteger(v=vs.110).aspx
The question is, how do you want to hold the difference between two large numbers? If you're calculating the difference between two signed long (64-bit) integers, for example, and the difference will not fit into a signed long integer, how do you intend to store it?
long a = +(1 << 62) + 1000;
long b = -(1 << 62);
long dif = a - b; // Overflow, bit truncation
The difference between a and b is wider than 64 bits, so when it's stored into a long integer, its high-order bits are truncated, and you get a strange value for dif.
In other words, you cannot store all possible differences between signed integer values of a given width into a signed integer of the same width. (You can only store half of all of the possible values; the other half require an extra bit.)
Your options are to either use a wider type to hold the difference (which won't help you if you're already using the widest long integer type), or to use a different arithmetic type. If you need at least 64 signed bits of precision, you'll probably need to use BigInteger.
The BigInteger was introduced in .Net 4.0.
There are some open source implementations available in lower versions of the .Net Framework, however you'd be wise to go with the standard.
If the Math.Abs still gives you grief you can implement the function yourself; if the number is negative (a - b < 0) simply trim the negative symbol so its unsigned.
Also, have you tried using Doubles? They hold much larger values.
Here's an alternative that might be interesting to you, but is very much within the confines of a particular int size. This example uses Int32, and uses bitwise operators to accomplish the difference and then the absolute value. This implementation is tolerant of your scenario where a - b equals the min int value, it naturally returns the min int value (not much else you can do, without casting things to the a larger data type). I don't think this is as good an answer as using BigInteger, but it is fun to play with if nothing else:
static int diff(int a, int b)
{
int xorResult = (a ^ b);
int diff = (a & xorResult) - (b & xorResult);
return (diff + (diff >> 31)) ^ (diff >> 31);
}
Here are some cases I ran it through to play with the behavior:
Console.WriteLine(diff(13, 14)); // 1
Console.WriteLine(diff(11, 9)); // 2
Console.WriteLine(diff(5002000, 2346728)); // 2655272
Console.WriteLine(diff(int.MinValue, 0)); // Should be 2147483648, but int data type can't go that large. Actual result will be -2147483648.

Increment forever and you get -2147483648?

For a clever and complicated reason that I don't really want to explain (because it involves making a timer in an extremely ugly and hacky way), I wrote some C# code sort of like this:
int i = 0;
while (i >= 0) i++; //Should increment forever
Console.Write(i);
I expected the program to hang forever or crash or something, but, to my surprise, after waiting for about 20 seconds or so, I get this ouput:
-2147483648
Well, programming has taught me many things, but I still cannot grasp why continually incrementing a number causes it to eventually be negative...what's going on here?
In C#, the built-in integers are represented by a sequence of bit values of a predefined length. For the basic int datatype that length is 32 bits. Since 32 bits can only represent 4,294,967,296 different possible values (since that is 2^32), clearly your code will not loop forever with continually increasing values.
Since int can hold both positive and negative numbers, the sign of the number must be encoded somehow. This is done with first bit. If the first bit is 1, then the number is negative.
Here are the int values laid out on a number-line in hexadecimal and decimal:
Hexadecimal Decimal
----------- -----------
0x80000000 -2147483648
0x80000001 -2147483647
0x80000002 -2147483646
... ...
0xFFFFFFFE -2
0xFFFFFFFF -1
0x00000000 0
0x00000001 1
0x00000002 2
... ...
0x7FFFFFFE 2147483646
0x7FFFFFFF 2147483647
As you can see from this chart, the bits that represent the smallest possible value are what you would get by adding one to the largest possible value, while ignoring the interpretation of the sign bit. When a signed number is added in this way, it is called "integer overflow". Whether or not an integer overflow is allowed or treated as an error is configurable with the checked and unchecked statements in C#. The default is unchecked, which is why no error occured, but you got that crazy small number in your program.
This representation is called 2's Complement.
The value is overflowing the positive range of 32 bit integer storage going to 0xFFFFFFFF which is -2147483648 in decimal. This means you overflow at 31 bit integers.
It's been pointed out else where that if you use an unsigned int you'll get different behaviour as the 32nd bit isn't being used to store the sign of of the number.
What you are experiencing is Integer Overflow.
In computer programming, an integer overflow occurs when an arithmetic operation attempts to create a numeric value that is larger than can be represented within the available storage space. For instance, adding 1 to the largest value that can be represented constitutes an integer overflow. The most common result in these cases is for the least significant representable bits of the result to be stored (the result is said to wrap).
int is a signed integer. Once past the max value, it starts from the min value (large negative) and marches towards 0.
Try again with uint and see what is different.
Try it like this:
int i = 0;
while (i >= 0)
checked{ i++; } //Should increment forever
Console.Write(i);
And explain the results
What the others have been saying. If you want something that can go on forever (and I wont remark on why you would need something of this sort), use the BigInteger class in the System.Numerics namespace (.NET 4+). You can do the comparison to an arbitrarily large number.
It has a lot to do with how positive numbers and negative numbers are really stored in memory (at bit level).
If you're interested, check this video: Programming Paradigms at 12:25 and onwards. Pretty interesting and you will understand why your code behaves the way it does.
This happens because when the variable "i" reaches the maximum int limit, the next value will be a negative one.
I hope this does not sound like smart-ass advice, because its well meant, and not meant to be snarky.
What you are asking is for us to describe that which is pretty fundamental behaviour for integer datatypes.
There is a reason why datatypes are covered in the 1st year of any computer science course, its really very fundamental to understanding how and where things can go wrong (you can probably already see how the behaviour above if unexpected causes unexpected behaviour i.e. a bug in your application).
My advice is get hold of the reading material for 1st year computer science + Knuth's seminal work "The art of computer pragramming" and for ~ $500 you will have everything you need to become a great programmer, much cheaper than a whole Uni course ;-)

C# Random.Next - never returns the upper bound?

random.Next(0,5)
It never returns the 5 (but sometimes returns the 0.) Why? I thought these are just boundary values that can be returned. Thanks
The maxValue for the upper-bound in the Next() method is exclusive—the range includes minValue, maxValue-1, and all numbers in between.
The documentation says the upper bound is exclusive. Exclusive means that it is not included in the possible return set. In a more mathematical notation 0 <= x < 5 in this case.
Straight from the documentation:
Summary:
Returns a random number within a specified range.
Parameters:
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.
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.
If you look at the parameters, you will see that minValue is inclusive (which is why your 0 occurs) and maxValue is exclusive (your 5 never occurs).
Good way to remember it is to consider max as amount of numbers from which it takes random number.
So random.Next(0,2) means that it takes random out of 2 numbers starting from 0: 0 and 1.
This has been written a long time ago but I will comment anyway. I think the main reason for that design decision is that most if not all random number generator at their core generate numbers from 0 to 2^32-1. So if you specify Int32.MaxValue you will never get that number. Having an exception for one number must have been not acceptable to the designers so they decided to have the bracket exclusive. Problem solved!
When you just look in Google for "c# random" and follow the first links to the method of desire you get here: http://msdn.microsoft.com/en-us/library/aa329893(v=vs.71).aspx
And there is no hint about the exclusiveness of the upper bound. They must have found the mistake in the code and corrected it with documentation.
So it is important to always check the version of the framework when looking at the documentation. Even when working with old versions of the framework, it is worth to have a look at the newer documentation.

Calculate factorials in C#

How can you calculate large factorials using C#? Windows calculator in Win 7 overflows at Factorial (3500). As a programming and mathematical question I am interested in knowing how you can calculate factorial of a larger number (20000, may be) in C#. Any pointers?
[Edit] I just checked with a calc on Win 2k3, since I could recall doing a bigger factorial on Win 2k3. I was surprised by the way things worked out.
Calc on Win2k3 worked with even big numbers. I tried !50000 and I got an answer, 3.3473205095971448369154760940715e+213236
It was very fast while I did all this.
The main question here is not only to find out the appropriate data type, but also a bit mathematical. If I try to write a simple factorial code in C# [recursive or loop], the performance is really bad. It takes multiple seconds to get an answer. How is the calc in Windows 2k3 (or XP) able to perform such a huge factorial in less than 10 seconds? Is there any other way of calculating factorial programmatically in C#?
Have a look at the BigInteger structure:
http://msdn.microsoft.com/en-us/library/system.numerics.biginteger.aspx
Maybe this can help you implement this functionality.
CodeProject has an implementation for older versions of the framework at http://www.codeproject.com/KB/cs/biginteger.aspx.
If I try to write a simple factorial code in C# [recursive or loop], the performance is really bad. It takes multiple seconds to get an answer.
Let's do a quick order-of-magnitude calculation here for a naive implementation of factorial that performs n multiplications. Suppose we are on the last step. 19999! is about 218 bits. 20000 is about 25 bits; we'll assume that it is a 32 bit integer. The final multiplication therefore involves the addition of up to 25 partial results each roughly 218 bits long. The number of bit operations will therefore be on the order of 223.
That's for the last stage; there will be 20000 = 216 such operations at each stage, so that is a total of about 239 operations. Some of them will of course be cheaper, but we're going for an order of magnitude here.
A modern processor does about 232 operations per second. Therefore it will take about 27 seconds to get the result.
Of course, the big integer library writers were not naive; they take advantage of the ability of the chip to do many bit operations in parallel. They're probably doing the math in 32 bit chunks, giving speedups of a factor of 25. So our total order-of-magnitude calculation is that it should take about 22 seconds to get a result.
22 is 4. So your observation that it takes a few seconds to get a result is expected.
How is the calc in Windows 2k3 (or XP) able to perform such a huge factorial in less than 10 seconds?
I don't know. Extreme cleverness in exploiting the math operations on the chip probably. Or, using a non-naive algorithm for calculating factorial. Or, possibly they are using Stirling's Approximation and getting an inexact result.
Is there any other way of calculating factorial programmatically in C#?
Sure. If all you care about is the order of magnitude then you can use Stirling's Approximation. If you care about the exact value then you're going to have to compute it.
There exist sophisticated computational algorithms for efficiently computing the factorials of large, arbitrary precision numbers. The Schönhage–Strassen algorithm, for instance, allows you to perform asymptotically fast multiplication for arbitrarily large integers.
Case in point, Mathematica computes 22000! on my machine in less than 1 second. The Implementation Notes page at reference.wolfram.com states:
(Mathematica's) n! uses an O(log(n) M(n)) algorithm of Schönhage based on dynamic decomposition to prime powers.
Unfortunately, the implementation of such algorithms is both complicated and error prone. Rather than trying to roll your own implementation, it may be wiser for you to license a copy of Mathematica (or a similar product that meets your functional and performance needs) and either use it, or a .NET programming interface to it, to perform your computation.
Have you looked at System.Numerics.BigInteger?
Using System.Numerics BigInteger
var bi = new BigInteger(1);
var factorial = 171;
for (var i = 1; i <= factorial; i++)
{
bi *= i;
}
will be calculated to
1241018070217667823424840524103103992616605577501693185388951803611996075221691752992751978120487585576464959501670387052809889858690710767331242032218484364310473577889968548278290754541561964852153468318044293239598173696899657235903947616152278558180061176365108428800000000000000000000000000000000000000000
For 50000! it takes a couple seconds to calculate but it seems to work and the result is a 213237 digit number and that's also what Wolfram says.
You will probably have to implement your own arbitrary precision numeric type.
There are various approaches. probably not the most efficient, but perhaps the simplest is to have variable length arrays of byte (unsigned char). Each element represents a digit. ideally this would be included in a class, and you can then add a method which let's you multiply the number with another arbitrary precision number. A multiply with a standard C# integer would probably also be a good idea, but a little trickier to implement.
Since they don't give you the result down to the last digit, they may be "cheating" using some approximation.
Check out http://mathworld.wolfram.com/StirlingsApproximation.html
Using Stirling's formula you can calculate (an approximation of) the factorial of n in logn time. Of course, they might as well have a dictionary with pre-calculated values of factorial(n) for every n up to one million, making the calculator show the result extremely fast.
This answer covers limits for basic .Net types to compute and represent n!
Basic code to calculate factorial for "SomeType" that supports multiplication:
SomeType factorial = 1;
int n = 35;
for (int i = 1; i <= n; i++)
{
factorial *= i;
}
Limits for built in number types:
short - correct results up to 7!, incorrect results afterwards, code returns 0 starting 18 (similar to int)
int - correct results up to 12!, incorrect results afterwards, code returns 0 starting at 34 (Why computing factorial of realtively small numbers (34+) returns 0)
float - precise results up to 14!, correct but not precise afterwards, returns infinity starting at 35
long - correct results up to 20!, incorrect results afterwards, code returns 0 starting at 66 (similar to int)
double - precise results up to 22!, correct but not precise afterwards, returns infinity starting at 171
BigInteger - precise and upper limit is set by memory usage only.
Note: integer types overflow pretty quickly and start producing incorrect results. Realistically if you need factorials for any practical usage long is the type to go (up to 20!), if you can't expect limited numbers - BigInteger is the only type provided in .Net Framework to provide precise results (albeit slow for large numbers as there is no built-in optimized n! method)
You need a special big-number library for this. This link introduces the System.Numeric.BigInteger class, and incidentally has an example program that calculates factorials. But don't use the example! If you recurse like that, your stack will grow horribly. Just write a for-loop to do the multiplication.
I don't know how you could do this in a language without arbitrary precision arithmetic. I guess a start could be to count factors of 5 and 2, removing them from the product, and add on these zeroes at the end.
As you can see there are many.
>>> factorial(20000)
<<non-zeroes removed>>0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000L

Why is Count not an unsigned integer? [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicates:
Why does .NET use int instead of uint in certain classes?
Why is Array.Length an int, and not an uint
I've always wonder why .Count isn't an unsigned integer instead of a signed one?
For example, take ListView.SelectedItems.Count. The number of elements can't be less then 0, so why is it a signed int?
If I try to test if there are elements selected, I would like to test
if (ListView.SelectedItems.Count == 0) {}
but because it's a signed integer, I have to test
if (ListView.SelectedItems.Count <= 0) {}
or is there any case when .Count could be < 0 ?
Unsigned integer is not CLS-compliant (Common Language Specification)
For more info on CLS compliant code, see this link:
http://msdn.microsoft.com/en-us/library/bhc3fa7f.aspx
Mabye because the uint data type is not part of the CLS (common language specification) as not all .Net languages support it.
Here is very similar thread about arrays:
Why is Array.Length an int, and not an uint
It's not CLS compliant, largely to allow wider support from different languages.
A signed int offers ease in porting code from C or C++ that uses pointer arithmetic.
Count can be part of an expression where the overall value can be negative. In particular, count has a direct relationship to indices, where valid indices are always in the range [0, Count - 1], but negative results are used e.g. by some binary search methods (including those provided by the BCL) to reflect the position where a new item should be inserted to maintain order.
Let’s look at this from a practical angle.
For better or worse, signed ints are the normal sort of ints in use in .NET. It was also normal to use signed ints in C and C++. So, most variables are declared to be int rather than unsigned int unless there is a good reason otherwise.
Converting between an unsigned int and a signed int has issues and is not always safe.
On a 32 bit system it is not possible for a collection to have anywhere close to 2^^32 items in it, so a signed int is big enough in all cases.
On a 64 bit system, an unsigned int does not gain you much, in most cases a signed int is still big enough, otherwise you need to use a 64 bit int. (I expect that none of the standard collection will cope well with anywhere near 2^^31 items on a 64 system!)
Therefore given that using an unsigned int has no clear advantage, why would you use an unsigned int?
In vb.net, the normal looping construct (a "For/Next loop") will execute the loop with values up to and including the maximum value specified, unlike C which can easily loop with values below the upper limit. Thus, it is often necessary to specify a loop as e.g. "For I=0 To Array.Length-1"; if Array.Length were unsigned and and zero, that could cause an obvious problem. Even in C, one benefits from being able to say "for (i=Array.Length-1; i GE 0; --i)". Sometimes I think it would be useful to have a 31-bit integer type which would support widening casts to both signed and unsigned int, but I've never heard of a language supporting such.

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