Archive for the ‘.Net’ Category

Coin Flips with Go

Thursday, September 15th, 2016

Following on from my little experiment with flipping coins millions of times, I thought that it would be interesting to write the same program in Go for comparison.

func main() {
    rand.Seed(time.Now().UTC().UnixNano())
 
    var competitionSize, runs int
    flag.IntVar(&competitionSize, "competitionSize", 10, "The size of each coin tossing competition.")
    flag.IntVar(&runs, "runs", 10, "The number of runs of the competition")
 
    flag.Parse()
 
    fmt.Println("heads, tails, flips")
    for i := 0; i < runs; i++ {
        countHeads := 0
        for j := 0; j < competitionSize; j++ {
            countHeads += rand.Intn(2)
        }
 
        fmt.Printf("%d, %d, %d\n", countHeads, competitionSize-countHeads, competitionSize)
    }
}

The basic idea of how to write the program is unchanged in this language. The output showed a similar distribution to the C# program.

However, Go has straightforward command line arguments parsing built into the standard library (like Python or Perl). In C#, there are a number of third party libraries that do this that can be installed via NuGet. I went with Command Line Parser.

To use this, I needed to add a class with properties with attributes to describe my expected options:

class Options
{
    // Taken from http://www.bbc.co.uk/news/politics/eu_referendum/results
    [Option('s', "competitionSize", DefaultValue = 33551983, HelpText = "The size of each coin tossing competition.")]
    public int CompetitionSize { get; set; }
 
    [Option('r', "runs", DefaultValue = 1000, HelpText = "The number of runs of the competition")]
    public int Runs { get; set; }
}

The command line arguments are then parsed in the main method:

var commandLineArgs = new Options();
if (Parser.Default.ParseArguments(args, commandLineArgs))
{
    var runs = commandLineArgs.Runs;
    var flips = commandLineArgs.CompetitionSize;
    ...
}
else
{
    WriteLine("Unable to parse the command line arguments!");
}

While this is not difficult to understand, it is more complicated than it needs to be and requires quite a bit more typing than in the Go program.

10 runs of the referendum simulation in both programs on Windows 10 with an Intel i7 4500U running at 1.8GHz took 7 seconds for the C# program and 13 seconds for the Go program. I wouldn’t take such simple programs very seriously for drawing conclusions about the relative speed of these languages.

The updated C# program and the complete Go program are on GitHub:
https://github.com/robert-impey/CodingExperiments/blob/master/C-Sharp/CoinFlipsAreRandom/CoinFlipsAreRandom/Program.cs
https://github.com/robert-impey/CodingExperiments/blob/master/Go/coinToss.go

Was the EU Referendum Random?

Monday, September 5th, 2016

One of the claims that I saw on social media in the aftermath of the recent EU referendum here in the UK was that the result (52% to 48%) was so close that it was little different from tossing a coin.

Without getting bogged down in the politics of that referendum, or the various campaigns that led up to it; I want to consider whether this claim holds any water. How similar to millions people each tossing a coin and voting accordingly was the result?

According to the BBC, there were 17,410,742 votes to leave and 16,141,241 votes to remain, giving a total of 33,551,983 votes. If we were to make each of these people toss a coin and count up the results, the ratio of heads/tails or remains/leaves could be anywhere between all heads and all tails. However, we would expect the counts to be about equal if the coins were all fair. Of course, any ratio is possible, but if we were to run the coin tossing game repeatedly, we would expect to mean ratio to converge on 1:1. How likely would a 52:48 ratio be?

The leave side got a share of the vote equal to 0.51891842. Therefore, their absolute deviation from the expectation (the mean, or 0.5) is 0.01891842. The difference between the two counts is 1,269,501. Would we expect to see a deviation of this magnitude in a coin tossing competition?

Being a computer programmer rather than a mathematician, I’m going to look at this using a simple program.

WriteLine("heads, tails, flips, heads share");
 
var runs = 1000;
 
// Taken from http://www.bbc.co.uk/news/politics/eu_referendum/results
var flips = 33551983;
 
var randomNumberGenerator = new Random();
 
for (var run = 0; run < runs; run++)
{
    var heads = 0;
 
    for (var coinToss = 0; coinToss < flips; coinToss++)
    {
        var toss = randomNumberGenerator.Next(2);
 
        if (toss > 0)
        {
            heads++;
        }
    }
 
    WriteLine($"{heads}, {flips - heads}, {flips}, {(0.0 + heads) / flips}");
}

This program simulates 1,000 coin tossing competitions with 33,551,983 players and writes the counts as comma separated values.

Putting the output into Excel, the largest deviation was 0.000275081 (or a difference between counts of 18,459) and the smallest was 1.49022E-08 (which was 16,775,992 heads and 16,775,991 tails. This happened twice in 1,000 runs!) The largest difference between heads and tails was 69 times smaller than the result from the referendum.

Plotting the shares in decreasing order, we see how quickly larger deviations fall off:

deviation

Putting the deviations into bins and counting the competitions by deviation from the expectation, we see that smaller deviations are more common:

bins

Whatever else we might say about the result, we cannot seriously claim that the result was random.

The full program can be found here:

https://github.com/robert-impey/CodingExperiments/blob/master/C-Sharp/CoinFlipsAreRandom/CoinFlipsAreRandom/Program.cs

The Excel file can be found here:
http://www.reversing-entropy.com/wp-content/uploads/2016/09/EuRef.xlsx

Finding Reversed Words

Sunday, November 23rd, 2014

Following on from coding experiments with using the decorator pattern to reverse strings in Java and with palindromes and string reversal in C#, I decided to search for the words in the English language that are the reverse of other words. The Devil lived, but Black Sabbath gave us Live Evil. To beat Sega takes ages. You’re upping the ante living near Etna.

I had a copy of a dictionary file from Ubuntu on my hard drive. To make life a little simpler, I used some PowerShell magic to remove the many words that end in “‘s” (there are no words that start “s'”) and to make the list all lower case:

PS M:data> Get-Content .\british-english.csv | ?{ $_ -NotMatch "'s$" } | %{ $_.ToLower() } | Sort-Object | Get-Unique | Set-Content british-english-without-apostrophe-s.txt

The simplest algorithm for finding words that are the reverse of other words is to simply go through the whole list and then check whether any of the other words are the reverse of that word. To do this, we need a method to test if two strings are a reversed pair. Pushing the characters of the first string onto a stack and then popping them off as you check them against each character of the second string can achieve this:

public bool AreReversedPair(string firstString, string secondString)
{
    if (firstString.Length != secondString.Length)
    {
        return false;
    }
 
    var firstStringStack = new Stack<char>();
 
    foreach (var c in firstString)
    {
        firstStringStack.Push(c);
    }
 
    foreach (var currentCharFromSecond in secondString)
    {
        var currentCharFromFirst = firstStringStack.Pop();
 
        if (currentCharFromFirst != currentCharFromSecond)
        {
            return false;
        }
    }
 
    return true;
}

Next, the code for finding the pairs:

public IEnumerable<ReversedWordPair> FindReversedWords(IQueryable<string> allWords)
{
    var reversedWords = new LinkedList<ReversedWordPair>();
    var reversedStringChecker = new StackReversedStringChecker();
 
    foreach (var firstWord in allWords)
    {
        foreach (var secondWord in allWords)
        {
            if (reversedStringChecker.AreReversedPair(firstWord, secondWord))
            {
                reversedWords.AddLast(new ReversedWordPair(firstWord, secondWord));
            }
        }
    }
 
    return reversedWords;
}

This code is very straightforward, but not fast at all. The time complexity of the algorithm is O(n^2) because for each word, we need to check every other word. For a list with almost 73,000 words, this takes a long time. In spite of this, it does finish. I left the program running, went to dinner and found a list waiting for me when I came back.

reversed-word-pairs.txt

As the list of words is not going to change very often and I only need to run the algorithm once, I could just leave the program as it is. However, it would be nice to reduce the time complexity of the algorithm. The simplest way to do this would be to reverse each word and then check whether the list of all the words contains the reversed word (readers who are familiar with .Net may have noticed that I passed the list of words as an IQueryable rather than an IEnumerable for the list of words). Depending on the data structure, checking whether the reversed string is a member of the set of words should be less than O(n). For example, retrieval from a Trie takes O(m) where m is the length of the string. This would leave us with O(nm) time. As with all optimisations, you’ve got to time the code to see whether you get a boost or not.

My next task is to come up with a meaningful sentence with all words the reverse of the corresponding word at the other end of the sentence, with a palindrome at the centre. There must be something that can be made of Dennis, who sinned in the straw, only to get warts to stun his nuts.

See if you can make a mirror sentence of your own.

The complete code, along with some tests, can be found at:

https://github.com/robert-impey/CodingExperiments/tree/master/C-Sharp/FindReversedWords

Fizz Buzz with Functional Programming

Saturday, October 11th, 2014

Fizz Buzz is a counting game where players count upwards saying either the number, “Fizz!” if the number is a multiple of 3, “Buzz!” if a multiple of 5, or “Fizz! Buzz!” if the a multiple of 3 and 5.

It’s also a simple programming problem that can be solved in a number of ways.

In C#, the solution could be as simple as:

var max = 30;
 
Console.WriteLine("With a for loop:");
 
for (var i = 1; i <= max; i++)
{
    Console.Write(i + ": ");
 
    if (i % 3 == 0)
    {
        Console.Write("Fizz! ");
    }
 
    if (i % 5 == 0)
    {
        Console.Write("Bang!");
    }
 
    Console.WriteLine();
}

href=”https://github.com/robert-impey/CodingExperiments/blob/master/C-Sharp/FizzBang/FizzBang/Program.cs

I’m going to show a way of solving this problem using F#. This is not the simplest way to solve this problem. However, the F# solution below does demonstrate the use of higher order functions, an important concept in functional programming.

F# is a strongly typed language. This means that the compiler checks that the values that we bind and pass to and return from functions conform to its expectations. This catches a huge number of programmer errors almost as soon as they are made. It also allow the programmer to state his or her intentions by defining types. In this program, we are dealing with checking numbers that may or may not have the special property of causing us to say a message. We are going to need a function that can check numbers, so we define a type that describes such functions:

type NumberChecker = int -> string option

An example of such a function is

let isEven x = 
    if x % 2 = 0 then Some "That number's even!"
    else None

This takes x and checks whether the remainder after dividing by 2 is 0. If so, it returns something (the message). Otherwise, it returns nothing.

We can test this function using a match statement. We call isEven with a number. If the function returns a message (the case where there is Some message) we can print it. If the function returned None, then we’ll just print the number.

let number = 4
match isEven number with
| Some message -> printfn "%s" message
| None -> printfn "%d" number

This code is fine if we only want to check one number with one number checker. We want to be able to check numbers with different checker functions.

For example, we might also define:

let isOdd x =
    if x % 2 = 1 then Some "That number's odd!"
    else None

To test this, we could duplicate our code:

match isOdd number with
| Some message -> printfn "%s" message
| None -> printfn "%d" number

Nobody likes duplicated effort. The only difference between these lines of code and the isEven printing code is the checker function. This is where a higher order functions come in. A higher order function takes a function as an input or returns a function as the output.

let numberCheckerPrinter (numberChecker : NumberChecker) number =
    match numberChecker number with
    | Some message -> printfn "%s" message
    | None -> printfn "%d" number

The first argument to this function is annotated as a NumberChecker, which we defined at the top of our program. We can pass into this function any number checker function we like and it will either print the message (if one is returned) or the number on its own.

numberCheckerPrinter isEven 2
numberCheckerPrinter isEven 3
numberCheckerPrinter isOdd 2
numberCheckerPrinter isOdd 3

This has this output:

That number's even!
3
2
That number's odd!

Note that the first argument (which happens to be a function) is passed to the function in the same way as the second argument (which happens to be an integer).

We could go on defining number functions in the way, but we’re already beginning to repeat ourselves. The definition of the higher order function was a function that took a function as an input or returned a function. It would nice if the programming language could some of the work for us and make number checkers. All that stays the same from number checker to number checker is that they all must accept an integer and return either a message or nothing.

We might have a number of number checkers that vary only in that they check a different divisor and have a different message.

let divisorNumberChecker (divisor, message) number = 
    if number % divisor = 0 then
        Some message
    else    
        None

Note that this function does not conform to the NumberChecker type that is defined at the top as it takes two arguments: a tuple of the divisor and the message and the number. We can use this function to make new number checkers by fixing the first input with partial function application.

let fizz = divisorNumberChecker (3, "Fizz!")

Even though divisorNumberChecker takes two arguments, we only supplied the first. This causes F# to return a new function which accepts the remaining argument and has fixed the first argument.

Running this in F# interactive gives the type of fizz, which matches NumberChecker.

> 

val fizz : (int -> string option)

Making the “Buzz!” checker is simply:

let buzz = divisorNumberChecker (5, "Buzz!")

We have functions for the “Fizz!” and the “Buzz!” part of the game, we now need to deal with the case when the number is divisible by 3 and 5. We could define a function to do this, but this a duplicated effort that we want to avoid. After all we have fizz and buzz, we want to reuse them.

let andNumberChecker (f : NumberChecker, g : NumberChecker) x  = 
    match f x, g x with
    | Some messageF, Some messageG -> messageF + " " + messageG |> Some
    | _ -> None

The andNumberChecker lets us combine two number checker functions into one. If both functions return a message for the given number, then it joins the messages together and returns them. Otherwise, it returns None.

Again, I’ve use partial function application to fix the fizz and buzz functions to make fizzBuzz.

let fizzBuzz = andNumberChecker (fizz, buzz)

So far, I’ve been passing single functions as arguments to other functions. Functions can be treated as other objects in F#. Therefore, we can put them into a list and pass that list to a function. This way we can keep checking the same number with different checkers and use the message from the first one that returns something rather than nothing:

let doManyChecks (numberCheckers : NumberChecker list) number =
    let message = 
        List.fold (fun message numberChecker ->
            match message with 
            | Some _ -> message
            | None -> numberChecker number
        ) None numberCheckers
    match message with
    | Some message' -> printfn "%s" message'
    | None -> printfn "%d" number   
 
let doFizzBuzzChecks = doManyChecks [ fizzBuzz; fizz; buzz ]

We can use this to play the game:

doFizzBuzzChecks 3
doFizzBuzzChecks 4
doFizzBuzzChecks 5
doFizzBuzzChecks 15

In FSI:

Fizz!
4
Buzz!
Fizz! Buzz!

Or to play the game with many numbers:

[ 1 .. 30 ] |> List.iter doFizzBuzzChecks

The complete code for this can be found here:

https://github.com/robert-impey/CodingExperiments/blob/master/F-Sharp/Loose/FizzBuzz.fsx

Functional programming has been described as a non-solution to a non-problem. Compared to the for loop in C#, this looks like a lot of work. If you have a problem as simple as Fizz! Buzz!, then it’s best to stick to the simplest solution. However, many interesting problems are much more complicated.

Higher order functions allow for abstracting away the details and generalising code. We might add a rule of saying “Boom!” for primes. The code such a checker might be complicated and require extensive testing. The code that makes use of the checkers does not need to change in order to use this new checker, we simply add the function to the list of checkers. We can also combine it with other checkers, enabling code reuse. Functional programming is getting increasing attention because it allows programmers to closely focus their attention on just their problem and decouple that code from the part of the program that will make use of it.

Using F# to minimise a function

Thursday, August 22nd, 2013

Finding the minimum of functions is at the heart of optimisation. Mathematicians, engineers and programmers have come up with a large number of approaches to solving this problem, including differentiation, genetic algorithms and even exhaustive search.

Consider a quadratic function such that could be written in F# as

let f x = (x ** 2.0) - (2.0 * x) + 1.0

Finding the minimum means finding the input value for the function that returns to lowest value. If we plot the curve, then the minimum is the lowest point of the curve.

One way to find this is find the derivative of the function:

(2.0 * x) - 2.0

and solve the equation where the derivative is equal to 0, which is when x is 1. This probably the best way to solve this problem in practice.

However, not all functions have derivatives that are easy to find. Therefore, an alternative way to minimise a function is an exhaustive search of the inputs in some range. This is not an elegant or generally efficient solution, but it works.

In F#, we might proceed as follows.

We define the range that we want to search and the step between candidate inputs:

let min = 0.0
let max = 10.0
let step = 0.01

We know where to start searching- with the lowest candidate input:

let firstCandidate = f min, min

This creates a tuple of two floats, the first being the output and the second being the lowest candidate input.

We also want a sequence of tuples of two floats that are the remaining candidate solutions:

let remainingCandidates = seq { for c in min + step .. step .. max do yield f c, c }

Note that at this point the sequence has not been enumerated and the function has not been run with each of the candidate inputs. Because sequences are evaluated lazily, the function will not be run for each of the candidate inputs until it is asked for.

We are trying to minimise the function, therefore we need a function that can compare the first elements of two candidate solutions:

let findMin currentMinSln candidateSln =
    match fst currentMinSln < fst candidateSln with
    | true -> currentMinSln
    | false -> candidateSln

Running this code in fsi shows its type to be:

val findMin : 'a * 'b -> 'a * 'b -> 'a * 'b when 'a : comparison

The F# compiler can infer that the function takes in two tuples, each with two elements. The tuples must be of the same type and the first element of each tuple must be comparable. Way to go, type inference!

Everything has been set up at this point. We just need to run the code:

let minSln = Seq.fold findMin firstCandidate remainingCandidates

The fold function is to reduce a sequence to a single value, starting with a given accumulator. In this case, the first candidate solution that we created earlier is our starting point. The output of the function for each candidate input is compared to that accumulator.

Running the code finds the same answer (1) that we found with calculus.

This code runs pretty quickly, but this approach is generally slow. Our range is only 10 wide and the step is 0.01, so there are only 1000 candidate inputs. Increasing the size of the range or decreasing the step would increase the number of inputs. More worryingly, the size of the search space increases exponentially as the number of inputs to the function increases. So a function that took two inputs for the same range for each input would have a million candidate inputs, three inputs would take a billion and so on.

However, it is also quite straightforward to parallelise this approach. The sequence of candidate solutions could be split up into partitions and each partition could be sent off to a different computer. Each batch would find a local minimum for the range it was given. Finding the minimum for the whole of our range of inputs is simply a matter of find the minimum in the returned list.

The complete code for this can be found here:

https://github.com/robert-impey/CodingExperiments/blob/master/F%23/Loose/MinimiseQuadratic.fs

Project Euler 1 in F# using Sequences

Wednesday, August 21st, 2013

I have written previously about solving the first problem on the Project Euler website using F# and Haskell.

The problem is to find the sum of the natural numbers that are divisible by 3 or 5 that are less than 1000. To solve this in F#, my first solution looked like this:

[0..999]
|> List.filter (fun x -> x % 3 = 0 || x % 5 = 0)
|> List.sum

This works. However, it was pointed out to me that using lists to generate the numbers below 1000 in my F# solution was memory hungry as the list will be created and stored in memory then piped to the filter and so on. The list is evaluated eagerly. This isn’t a problem with a small list like this, but it could create issues with more elements.

I took a look at sequences in F#, which are evaluated lazily. To get an idea of what this means, consider the sequence of natural numbers:

let countingNumbers = 
    seq { 
        for i in 1 .. System.Int32.MaxValue do 
            printfn "Yielding: %d" i
            yield i 
    }

If you are unfamiliar with lazy evaluation, it might look like this will create all the natural numbers up to System.Int32.MaxValue (2147483647) and print each number as it goes. However, if you run this with F# Interactive (aka fsi), it creates a seq<int> without printing anything out. At this point, the sequence is yet to be enumerated.

The ingenious thing about sequences is that we can work with them without enumerating them. Consider these lines:

let max = 999
let firstUpToMax = Seq.take max countingNumbers

The Seq.take function allows us to take the first elements in a sequence. However, running these lines in fsi does not cause the for loop in the sequence above to be run. The print and yield statements are not run.

F#’s ability to delay enumeration is not limited to sequences. Queries also allow us to describe a sequence of numbers without enumerating them.

I define the following function that will be used to filter the sequence:

let isDivisibleBy3Or5 x = 
    x % 3 = 0 || x % 5 = 0

and then create this query:

let divisibleBy3Or5 = 
    query {
        for countingNumber in firstUpToMax do
        where (isDivisibleBy3Or5 countingNumber)
        select countingNumber
    }

Programmers who are familiar with SQL or LINQ should have no problem understanding this query. In plain English, we might say “we will go through the sequence and find the values that match the given criterion.” Note the use of the future tense. Running this code in fsi does not result in enumeration of our sequence and the printing of the “Yielding…” lines.

I could find the sum of the numbers matching our criteria using a built in function like this:

let seqSum = Seq.sum divisibleBy3Or5

This will cause the sequence to be enumerated and finds the same sum as the list method above.

However, by defining the following function:

let noisySum total next =
    printfn "Adding %d to %d" next total
    total + next

I will be able to see the order in which the sequence is enumerated and when the values are pumped down the pipeline to be folded into the sum.

let seqFoldSum = 
    divisibleBy3Or5
    |> Seq.fold noisySum 0

At this point we get the following out put at fsi:

Yielding: 1
Yielding: 2
Yielding: 3
Adding 3 to 0
Yielding: 4
Yielding: 5
Adding 5 to 3
Yielding: 6
...
Yielding: 996
Adding 996 to 231173
Yielding: 997
Yielding: 998
Yielding: 999
Adding 999 to 232169
 
val seqFoldSum : int = 233168

Which clearly demonstrates the order of execution and that only the numbers that are divisible by 3 or 5 are added to the sum.

Does this make much difference to the memory usage? For this problem, this is not the bottle neck. The integers storing the sum of the numbers overflow before the input lists become large enough to use a lot of memory. However, passing values down the pipeline as they are yielded is somehow more pleasing to this programmer. Also, lazy evaluation means that enumerating the numbers and performing the mathematics do not take place until the result is actually needed. By creating a sequence and query, the programmer can create the code and pass it around the program without doing any heavy computation or using up much memory.

The complete listing of this code can be found at GitHub:

https://github.com/robert-impey/CodingExperiments/blob/master/F%23/Loose/Euler1.fs