Tag Archives: .net

Taking on the Gilded Rose Kata

I was recently invited to attend the Chicago Software Craftsmanship presentation on “Untangling the Gilded Rose”.  While I was unfortunately unable to attend, I found the topic inspirational enough to write about.  The topic was a code kata, described as taking an existing code base and refactoring it to be more maintainable.  The  kata for this blog post is called the Gilded Rose Kata and is described here with a project template found here.

1.  Write the Specs

We are given the specs, but I’ve copied and consolidated them below.  They seem simple at first, but going through them multiple times reveals some subtleties.

  1. ‘Aged Brie’ increases quality increase by 1 each day.
  2.  ‘+5 Dexterity Vest’ quality increase by 1 each day.
  3. ‘Elixir of the Mongoose’ quality increase by 1 each day.
  4. ‘Sulfuras, Hand of Ragnaros’ quality and sellin stay the same.
  5. ‘Backstage passes to a TAFKAL80ETC concert’ quality goes up by 1 each day when the Sellin is greater than 10.  Quality increases by 2 when there are 10 days or less and by 3 when there are 5 days or less but Quality drops to 0 after the concert.
  6. “Conjured” items degrade in Quality twice as fast as normal items.
  7. All items quality is never negative.
  8. All items quality may never go over 50.

Do we understand the specs?  In this case, a few things weren’t clear from the specs.

  1. Would the item being updated ever be in an invalid state?  If so, how would we handle it?  I’m going to assume no for the purposes of the refactoring since neither the original code nor the specs address this issue.
  2. Should the Quality being calculated before or after the SellIn is reduced?  We can derive this from the current code base, which indicate we should reduce the sellin before calculating the quality.
  3. Should we consider if multiple rules could apply to an item?  For example, what if there was a “Conjured Aged Brie” that required both the rules for “Conjured” and “Aged Brie” be applied?  We’ll assume no, as there no indication from the specs this should be considered.
  4. There are many other possible requirements that we’ll ignore and assume should be up to our discretion since they are not explicitly stated, e.g. performance, development methodology, testing methodology, etc.

2.  Write the tests

One of the purposes of the Kata is to practice TDD, or test driven development. So we’ll first write our unit tests.  However, this leads to a common chicken or egg problem with TDD.  How do I know what classes and methods to test when there is no class design yet? Typically, I will think through the overall class design and write the stubs before writing the units test, so at least there is a starting point.  There will be probably be changes and not all of our assumptions may turn out to be true, but its better to plot a course now and adjust later than to completely miss the mark.

3. What are the design goals?

The kata is pretty open ended as far as design goals. Given this, we’ll come up with our own set of goals to accomplish.

  1. Localize item processing to the item’s type –  The low cohesion of the items type to the calculation for that type’s quality is probably the most serious problem we see with the current design.  If you were trying to follow how the quality was calculated for “Backstage passes to a TAFKAL80ETC concert” it would be difficult as it is in 2 different places, separated by unrelated code, and deeply nested within if/else structures.  We want to have a single place for each items quality calculation.
  2. Reduce nested if/else structures.   Nested if/else constructs make code difficult to understand. Why?  Because you have to understand several different lines of code in conjunction to understand if a particular nested statement will be executed.  Favor flattening conditional statements whenever possible.  This will localize the conditional logic of whether a particular statement is executed.
  3. Make testable – We want all of our code to be easily testable with unit tests so any further refactoring will be easier and have code coverage.

4. The design

So the design I went with is to have a class for each unique update algorithm.  This is going to be represented by a strategy pattern.  This will allow us to iterate through each item and apply the update function to each item independently of knowing the type of each item.  This leaves us with a mapping problem of finding the correct strategy for the type.  We will solve this with a factory pattern to create and return the strategy instance for the unique type.

public interface IUpdateStrategyFactory
    IUpdateStrategy Create(string name);

public interface IUpdateStrategy
    void UpdateQuality(Item item);

5.  Write the code

I won’t go through all the trivial details, but here is the factory, 2 of the strategies, and the greatly simplifed UpdateQuality() loop.

public class UpdateStrategyFactory : IUpdateStrategyFactory
    public IUpdateStrategy Create(string name)
        if (name == "Sulfuras, Hand of Ragnaros")
            return new DoNothingUpdate();
        else if (name == "Aged Brie")
            return new AgedBrieUpdateStrategy();
        else if (name == "Backstage passes to a TAFKAL80ETC concert")
            return new BackstagePassesUpdateStrategy();
        else if (name.Contains("Conjured"))
            return new ConjuredUpdateStrategy();
            return new StandardUpdateStrategy();
public class StandardUpdateStrategy : IUpdateStrategy
    public void UpdateQuality(Item item)
        if (item.Quality <= 0)
        else if (item.Quality == 1)
            item.Quality = 0;
            item.Quality -= item.SellIn < 0 ? 2 : 1;

The final loop is now much easier to read and understand.

public void UpdateQuality()
    foreach (Item item in Items)
        var updateStrategy = _strategyFactory.Create(item.Name);

That’s it!  If you have any feedback, please leave it below in the comments.  The source code is available for download.

Update: I’ve added my solution to github. https://github.com/jacerhea/GildedRose


Project Euler Problem 10 in F#

This problem is very similar to several of the other prime generation problems, but being the dedicated blogger that I am here is problem 10:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

  1. Generate primes using Sieve of Eratosthenes.
  2. Stop before reaching 2 million.
  3. Find the sum of this set of primes.

type public PrimesGenerator() =

member this.getPrimesMax max =

let primes = new BitArray(max+1, true)

let result = new ResizeArray(max/10)

for n = 2 to max do

if primes.[n] then

let start = (int64 n * int64 n)

if start < int64 max then

let i = ref (int start)

while !i <= max do primes.[!i] <- false; i := !i + n

result.Add n


member this.Problem9() =

let primesGen = new PrimesGenerator()

primesGen.getPrimesMax 2000000 |> Seq.map (fun prime -> (int64)prime) |> Seq.sum

Download the source code with unit tests.

Project Euler Problem 9 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 9 with my solution:

A Pythagorean triplet is a set of three natural numbers, a b c, for which,

a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

  1. Enumerate from 1 to 1000.  Create tuples of all sets (m, n) where m > n.
  2. Create Pythagorean triplets using Euclid’s formula from the tuples.
  3. Find the set where a + b + c = 1000.
  4. Find the product of this set.
let problem9 =
    let pythagorean_triplets(m:int, n:int) =
        let a = m*m-n*n
        let b = 2*m*n
        let c = m*m+n*n
    let tops = 1000
    [for m in [1..tops] do
        for n in [1..m-1] do yield (m, n)] |> Seq.map (fun t -> pythagorean_triplets((fst t, snd t)))
            |> Seq.filter (fun x -> x |> Seq.sum = tops) |> Seq.head |> Seq.fold (fun acc elem -> acc * elem) 1

Download the source code with unit tests.

Project Euler Problem 7 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 7:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10,001st prime number?

  1. Generate primes using Sieve of Eratosthenes.
  2. Skip the first ten thousand primes and take the next one.

type public PrimesGenerator() =

member this.PrimesInfinite () =

let rec nextPrime n p primes =

if primes |> Map.containsKey n then

nextPrime (n + p) p primes


primes.Add(n, p)

let rec prime n primes =

seq {

if primes |> Map.containsKey n then

let p = primes.Item n

yield! prime (n + 1) (nextPrime (n + p) p (primes.Remove n))


yield n

yield! prime (n + 1) (primes.Add(n * n, n))


prime 2 Map.empty

member this.Problem7() =

let primes = new PrimesGenerator()

primes.PrimesInfinite() |> Seq.skip 10000 |> Seq.take 1 |> Seq.head

Download the source code with unit tests.

Project Euler Problem 8 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 8 and my solution:

Find the greatest product of five consecutive digits in the 1000-digit number.


  1. Parse the input string to a list of integers
  2. Create lists of each 5 consecutive digits.
  3. Find the product of each list.
  4. Find the max product.
open System

let Problem8 =
    let multiply lst = lst |> Seq.fold (fun acc elem -> acc * elem) 1
    let s = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
    let values = s.ToCharArray() |> Seq.map (fun x -> Int32.Parse(x.ToString())) |> Seq.toList
    [0..(values.Length - 5)]
        |> Seq.map (fun h -> [0..4] |> Seq.map (fun a -> values.[(h + a)]) |> multiply )
        |> Seq.max

Download the source code with unit tests.

Project Euler Problem 6 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 6 and my solution:

The sum of the squares of the first ten natural numbers is,

12 + 22 + … + 102 = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 – 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

  1. Enumerate 1-100 and  square each value.  Add the results together
  2. Enumerate 1-100 and add each value together.  Square the resulting sum.
  3. Find the difference between the two.
let problem6 = 
    let values = [1..100]
    let sumOfSquares = values |> Seq.map (fun x -> pown x 2) |> Seq.sum
    let squareOfSums = values |> Seq.sum |> (fun x -> pown x 2)
    squareOfSums - sumOfSquares

Download the source code with unit tests.

Project Euler Problem 4 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 4 and my solution:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 ×99.

Find the largest palindrome made from the product of two 3-digit numbers.

  1. Create the set of all products of two 3-digit numbers.
  2. Filter the values that are palindromes.
  3. Find the max of the palindrome values.
open System

let problem4 =
    let reverseNumber value = Convert.ToInt32(new string(Array.rev (value.ToString().ToCharArray())))
    [for i in [100..999] do
        for j in [100..999] do yield i * j]
            |> Seq.filter (fun x -> x = reverseNumber(x)) |> Seq.max

Download the source code with unit tests.

Project Euler in F#, Problems 1 and 2

I recently stumbled across the Project Euler website. From wikipedia: “Project Euler (named after Leonhard Euler) is a website dedicated to a series of computational problems intended to be solved with computer programs”.

I thought this would be a good opportunity to continue working with F# and have some material for writing more blog articles.

Problem 1 is very straight forward.

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

I went with the most logical algorithm I could think of.

  1. Enumerate though each of the numbers 1-999.
  2. Filter the values that are multiples of either 3 or 5.
  3. Add up the resulting values.

The resulting F# code is quite elegant:

let problem1 = 
    [1..999] |> Seq.filter (fun i -> i % 3 = 0 || i % 5 = 0) |> Seq.sum

After looking at other algorithms people had used, there are more efficient methods using summations. But not bad for my first solution. On to problem 2

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Again, there is a pretty obvious solution here.

  1. Generate the Fibonacci sequence up to 4 million.
  2. Filter the values that are even.
  3. Add up the resulting values.

The resulting F# code:

let Problem2 =
    Seq.unfold (fun state ->
        if (snd state > 4000000) then None
        else Some(fst state + snd state, (snd state, fst state + snd state))) (1,1)
            |> Seq.filter (fun x -> x % 2 = 0) |> Seq.sum

Again, a more efficient solution can be found but this is at least semantically easy to follow.

I’m planning on continuing this series of blog articles with a problem or 2 for each entry, and seeing how many of the problems I can get through.  I’ll also provide the source code.   If you’re interested in more in this series stay tuned!

Einstein meet F#, part 3

If you read the previous two blog posts we started to solve a riddle Einstein created, using the F# language. In this last post, we’ll finish the solution by defining the set of rules for the remaining set of houses, and hopefully find the answer to the riddle.

let livesNextTo(house1:House, house2:House) =
house1.Number = house2.Number - 1 || house1.Number = house2.Number + 1

let multiRule1(houses:List<House>) =
let greenHouse = houses |> List.find (fun h -> h.Color = ColorHouse.Green)
let whiteHouse = houses |> List.find (fun h -> h.Color = ColorHouse.White)
greenHouse.Number = whiteHouse.Number - 1

let multiRule2(houses:List<House>) =
let blendSmoker = houses |> List.find (fun h -> h.Smoke = Smoke.Blends)
let catPerson = houses |> List.find (fun h -> h.Pet = Pet.Cats)
livesNextTo(blendSmoker, catPerson)

And for the multi rules….

let multiRules = [multiRule1;multiRule2;multiRule3;multiRule4;multiRule5;]

let multiRulesPredicate(houses:List<House>) = multiRules |> List.forall(fun rule -> rule(houses))

Next we are going to group the houses by the position they have been assigned.

let groupedHouses = housesPassedRules |> Seq.groupBy (fun (house:House) -> house.Number) |> Seq.toList

Now we just need to iterate through the values of each group, which will give us a set that correctly describes the index of each house. Now we just need to apply our filtering rules that apply to the sets of houses, and only a single set should be remaining.

let finalSets : List<List<House>> =

[for g1 in snd(groupedHouses.[0]) do

for g2 in snd(groupedHouses.[1]) do

for g3 in snd(groupedHouses.[2]) do

for g4 in snd(groupedHouses.[3]) do

for g5 in snd(groupedHouses.[4]) do

let houseGroup = [g1;g2;g3;g4;g5;]

if validHouseSet(houseGroup) && multiRulesPredicate(houseGroup) then yield houseGroup]

Finally, we’ll make sure we’ve only have a single set and print out the answer to the riddle with the house number of the fish.

let answerCheck = if finalSets.Length <> 1 then raise (new Exception("Answer not found."))

let finalSet = finalSets |> List.head

let fishHouse = finalSet |> List.find (fun house -> house.Pet = Pet.Fishes)

printf "%s" ("Fish is owned by: House:" ^ fishHouse.Number.ToString())

This problem was a great introduction to learning some of the concepts of functional programming as well as the features of the F#.  In another blog series I’ll be taking a look at some of the distinguishing language  features F# has compared to C# and other .NET languages.

Download the source code for this post.

Einstein meets F# Part 2

In part 1, I described a riddle Einstein had come up with, and a proposed algorithm for programmatically finding the solution.  In this post, we will take the next steps of  writing our F# program, including defining the domain objects, creating a list of all the possibilities of homes, and begin filtering down this list to find the answer.

First I wanted to define a set of the distinct types for each of the houses properties as described in the riddle, e.g. colors, nationalities, beverages, smokes, and pets.  An enum seemed like the most natural fit:

type ColorHouse = Red = 1 | Green = 2 | White = 3 | Yellow = 4 | Blue = 5

This was my first conceptual change from the C# world. I had originally (unintentionally) created a discriminated union such as this…

type ColorHouse = Red | Green | White | Yellow | Blue //discriminated union

So what’s the difference between an enum and discriminated union?  You can read about discriminated unions here as well as some of the differences with enums here.  For our purposes we need something easy to enumerate over, which .NET provides many utilities for enumerating over enums but not discriminated unions.

Next, I defined a House type with members for each of our enum types.  This is a little verbose, and there may be more concise way to define a simple type in F# but it suits our purposes for now:

type House(number:int, color:ColorHouse, nationality: Nationality, beverages:Beverages, smoke:Smoke, pet:Pet) =

member this.Number = number

member this.Color = color

member this.Nationality = nationality

member this.Beverages = beverages

member this.Smoke = smoke

member this.Pet = pet

Next, we iterate through each of the available values of each property type, and create a home with a distinct set of property values.  Now we have created a set homes with all combinations of property values represented.

let houses = [for i in [1..5] do

for color in colors do

for nationality in nationalities do

for beverage in beverages do

for smoke in smokes do

for pet in pets do

yield new House(i, color, nationality, beverage, smoke, pet)]

Now that we have all possibilities of house values (15625!) we are going to filter out the houses that do not pass the rules applicable to a single home given to us in the riddle.  Lets create the set of functions that determine a valid home…

let rule1(house:House) =  exnor((house.Nationality = Nationality.Brit), (house.Color = ColorHouse.Red))

let rule10(house:House) = exnor((house.Nationality = Nationality.German), (house.Smoke = Smoke.Prince))

Great, now lets create a list of those rules and apply them to all the house combinations.

let singleRuleSet = [rule1;rule2;rule3;rule4;rule5;rule6;rule7;rule8;rule9;rule10;]

let rulesPredicate(house:House) = singleRuleSet |> List.forall(fun rule -> rule(house))

let housesPassedRules = houses |> List.filter rulesPredicate

We now have the set of homes that pass the ‘single home’ rules. For the final part of this blog series, we will create sets of  sets of homes and a list of ‘multi home’ rules, and apply this last rule filtering.   This should leave a single set of 5 homes, and reveal the answer to the riddle!

To be continued…part 3.

Download the source code for this post.