Tuesday 18 June 2013

Use Teamcity as the Continuous Integration and Build Management Server

I've been using Teamcity as the continuous integration and build management server for the last few years. Not until recently I tried out Hudson (or Jenkins) as the alternative solution. It is quite frustrated to learn that Hudson, being an open-sourced project, doesn't even support remote build (where I can submit my local changes to Teamcity and run a selected builds, and commit once all the builds are successful).

In the scenario where the project has quite a number of builds for various tests, e.g. unit tests, integration tests, external tests, smoke tests, etc., running these tests with my local changes to check whether I break any existing tests or not is rather crucial in agile development approach, where I always want to have a feedback in the shortest amount of time. Of course, without being able to do a remote run with multiple test suites running in parallel, you may argue that I can still run the test suites locally one by one, and only check in when all the tests pass locally, in which case, Hudson is sufficient enough and it is FREE!

I don't intend to devote my blog to compare Teamcity and Hudson, but given the experiences of using both, I can conclude that Teamcity is no doubt the first choice in terms of CI and build management, as its feature is a perfect match for a developer with TDD (Test-Driven Development) and agile mind set. Having said that, I am gonna demonstrate how easy it is to setup the Teamcity for my project hosted on Git. JetBrains does offer a Teamcity professional version which is FREE and comes with a limit of 20 build configurations and 3 build agents.

Installation

  • Download the latest version for Linux (or whichever suits your OS)

  • Unzip the downloaded file, e.g. tar -zxvf TeamCity-7.1.5.tar.gz
    youyang@monkey-dev-01:/app> ll
    total 12
    drwxr-xr-x  6 youyang users 4096 Mar 22 02:31 monkey
    drwxr-xr-x 12 youyang users 4096 Jun 14 09:31 TeamCity
    drwxr-xr-x 13 youyang users 4096 Jun 15 09:11 TeamCity-BuildAgent
    youyang@monkey-dev-01:/app> 
    

  • start up teamcity, e.g. ./app/TeamCity/bin/startup.sh , on port 8111 by default


  • Configuration

  • browse to http://localhost:8111 to verify that Teamcity has started up successfully, which will take a minute or two before it asks you to setup an admin account

  • once the admin account is created and logged in, go to the Administration menu to create a new Project. Specify name and description on the General tab, shown as below

  • create a new VCS root on the VCS Roots tab, and choose Git (or whatever VCS you use), then fill in the information as shown below

  • go back to the project config, and create a build configuration

  • complete a minimum 3-step configuration for the build. Fill in the build name and description for Step 1; Choose the VCS root specified before in Step 2; Specify the build step with Ant (or other build runners the project is using), as shown below

  • this is the minimum configuration for a project with only one build configuration, which is good enough for the demo

  • Teamcity Agent

  • With a project setup on Teamcity server, we still need to install agents so that we can run the builds on them. Go Administration->Install Build Agents, notice that the link for the zip file distribution is http://monkey-dev-01:8111/update/buildAgent.zip

  • On the machine that you want to install your build agent, wget the zip file from, e.g. http://monkey-dev-01:8111/update/buildAgent.zip, unzip to, e.g. /app/TeamCity-BuildAgent folder

  • create the buildAgent.properties from the template file buildAgent.dist.properties under, e.g. /app/TeamCity-BuildAgent/conf folder

  • setup the following properties in the buildAgent.properties file

  • # name of the agent appearing on Teamcity server
    name=Monkey-Dev-01
    # setup the JDK
    env.JAVA_HOME=/app/monkey/java
    
  • start up the agent in the background by typing "/app/TeamCity-BuildAgent/bin/agent.sh start"

  • Wait a few minutes before your agent shown under the Agents->Connected tab on Teamcity server, as shown below

  • Once the agent is connected, you are ready to run your first build on Teamcity:

  • If you are using Intellij IDE, install the Teamcity plugin so that you can send a remote run to the Teamcity server by selecting which builds to run.


  • Friday 10 May 2013

    Bitwise Operations in Java

    The bitwise operations in Java are rarely discussed, and you can sense that by reading the official tutorials on this topic. However, there are interesting usages that you should probably be aware of.

    The bitwise operators operates on integral types (int and long). Before explaining how bitwise operators work, you should have an idea of how integers are represented in the binary form. As mentioned by the official primitive data type tutorials, the int data type is a 32-bit signed 2's complement integer, whose value ranges from -2^32 to 2^32-1, inclusively. On the other hand, the long data type is a 64-bit signed 2's complement integer, whose value ranges from -2^64 to 2^64-1, inclusively. In both cases, the highest order bit of int and long shows the sign of the value and is not part of the numeric value, where 0 being positive and 1 being negative.

    For example, an int value of 1 in binary form looks like this
    00000000000000000000000000000001
    
    while an int value of -1 in binary form looks like this
    11111111111111111111111111111111
    
    The negative value is transformed in 2 steps:
    Step 1: bitwise complement of +1, such that
    00000000000000000000000000000001 
    becomes
    11111111111111111111111111111110
    
    Step 2: add 1, such that
    11111111111111111111111111111110 
    becomes
    11111111111111111111111111111111
    
    Given a negative value, where the highest (leftmost) bit is 1, you can work out what value the binary form represents in 2 steps:
    Step 1: bitwise complement of the binary value, such that
    11111111111111111111111111111011
    becomes
    00000000000000000000000000000100
    
    Step 2: add 1, such that
    00000000000000000000000000000100
    becomes
    00000000000000000000000000000101
    
    Therefore, the original binary form represents -5.

    The bitwise operators

    OperatorNameExampleResultDescription
    a & band6 & 901 if both bits are 1, or 0 if either bit is 0
    a | bor 6 | 9151 if either bit is 1, or 0 if both bits are 0
    a ^ bxor 6 ^ 9151 if both bits are different, 0 if both bits are the same
    ~anot ~9 -10Inverts the bits. (equivalent to +1 and then change sign)
    n << pleft shift60 << 2240Shifts the bits of n left p positions. Zero bits are shifted into the low-order positions.
    n >> pright shift60 >> 215Shifts the bits of n right p positions. If n is a 2's complement signed number, the sign bit is shifted into the high-order positions.
    n >>> punsigned right shift-60 >>> 204095Shifts the bits of n right p positions. Zeros are shifted into the high-order positions.

    Packing and Unpacking

    Multiple int values of limited range can be "packed" into one int, by using the bitwise operators. For example, we have the following integer variables age (range 0~255, or 8-bit), gender (0~1, or 1-bit), weight (range 0~255, or 8-bit), and height (range 0-255, or 8-bit). These integers can be packed and unpacked into/from a 32-bit integer like this:

    Represent boolean values in a byte

    Suppose we have 8 boolean values that are true or false, and we want to store them in a single byte, and sent it across the network, and then unpack it and use it to toggle 8 different things:


    Multiply and Divide

    When using x << 3, it is equivalent to multiply x by 8 (or 2^3); similarly, when using x >> 2, it is equivalent to divide x by 4 (or 2^2). In fact, a smart enough compiler will replace your multiplication or division expressions using bit shift operators such that it is more compact and faster. So there is no need to explicitly use bit shift operator if you want multiply or divide by a power of 2!

    Thursday 18 April 2013

    I/O Redirection in Bash

    Since Bash is a superset of Bourne shell, the IO redirection is the same across the Bourne shell family. To find out exactly which shell you are using in Unix or Linux, typing in command
    echo $SHELL
    
    shall display an executable command with its path, for example
    PathShell
    /bin/shBourne shell
    /bin/bashBourne Again SHell
    /bin/cshC SHell
    /bin/kshKorn SHell

    Bash takes input from standard input (stdin), writes output to standard output (stdout), and writes error output to standard error (stderr). Standard input is connected to the terminal keyboard, while standard output and error are connected to the terminal screen, by default. Redirection of I/O is accomplished by using a redirection metacharacter followed by the desired destination (stdout & stderr) or source (stdin). Bash uses file descriptor numbers to refer to the standard I/O, where 0 is standard input, 1 is standard output, and 2 is standard error. Below are the common redirection for the Bourne shell family:

    Meta CharacterAction
    >Redirect standard output
    >>Append to standard output
    2>Redirect standard error
    2>&1Redirect standard error to standard ouput
    2>&1|Redirect standard error to standard ouput, and pipe them to another command
    <Redirect standard input
    |Pipe standard output to another command

    Please note that because < and > are referring to standard input and output respectively, the numbers 0 and 1 are not required in this case.

    Thursday 4 April 2013

    Sorting Algorithm Performance Comparisons with Caliper's Micro Benchmarking

    After inspired by Martin Thompson's presentation on Performance Testing Java Application, I decided to try out the micro benchmarking framework Caliper on the sorting algorithms that I recently wrote in order to understand better how different sorting algorithms behaves given an array of randomly distributed values.

    It was actually a bit involving to find out where I could download the caliper JAR file, as it was not available on their google code base. You can download it from the maven repository, or grab it from my git described at the bottom of this article.

    I created a SortingAlgorithmsBenchmark class that extends SimpleBenchmark, and some public methods, whose names have prefix 'time'. Then use the caliper Runner to run the benchmarks on the SortingAlgorithmsBenchmark class, as shown below
    import com.google.caliper.Runner;
    import com.google.caliper.SimpleBenchmark;
    import sorting.*;
    
    import java.util.Arrays;
    import java.util.Random;
    
    public class SortingAlgorithmsBenchmark extends SimpleBenchmark {
    
        private static final int SIZE = 10000;
        private static final int MAX_VALUE = 10000;
        private int[] values;
    
        @Override
        protected void setUp() throws Exception {
            values = new int[SIZE];
            Random generator = new Random();
            for (int i = 0; i < values.length; i++) {
                values[i] = generator.nextInt(MAX_VALUE);
            }
        }
    
        public void timeBubbleSort(int reps) {
            for (int i = 0; i < reps; i++) {
                BubbleSort.sort(values);
            }
        }
    
        public void timeInsertionSort(int reps) {
            for (int i = 0; i < reps; i++) {
                InsertionSort.sort(values);
            }
        }
    
        public void timeSelectionSort(int reps) {
            for (int i = 0; i < reps; i++) {
                SelectionSort.sort(values);
            }
        }
    
        public void timeMergeSort(int reps) {
            for (int i = 0; i < reps; i++) {
                MergeSort.sort(values);
            }
        }
    
        public void timeShellSort(int reps) {
            for (int i = 0; i < reps; i++) {
                ShellSort.sort(values);
            }
        }
    
        public void timeArraysSort(int reps) {
            for (int i = 0; i < reps; i++) {
                Arrays.sort(values);
            }
        }
    
        public static void main(String[] args) {
            Runner.main(SortingAlgorithmsBenchmark.class, args);
        }
    
    }
    

    Here are the results when running the benchmarks with a random array of size 10,000. Apparently, bubble sort and selection sort are the slowest of all, with selection sort behaving rather unstable (you will notice the selection sort benchmark is quite different across different runs. It is pleased to see that the sorting algorithm (Dual-Pivot Quick Sort) used by Arrays performs fairly well with a medium size array.

    0% Scenario{vm=java, trial=0, benchmark=BubbleSort} 75703105.85 ns; σ=3428843.80 ns @ 10 trials
    17% Scenario{vm=java, trial=0, benchmark=InsertionSort} 73305.66 ns; σ=712.37 ns @ 6 trials
    33% Scenario{vm=java, trial=0, benchmark=SelectionSort} 49624130.05 ns; σ=345774.06 ns @ 3 trials
    50% Scenario{vm=java, trial=0, benchmark=MergeSort} 478932.15 ns; σ=683.99 ns @ 3 trials
    67% Scenario{vm=java, trial=0, benchmark=ShellSort} 262061.27 ns; σ=11605.29 ns @ 10 trials
    83% Scenario{vm=java, trial=0, benchmark=ArraysSort} 8063.53 ns; σ=20167.07 ns @ 10 trials
    
        benchmark       us linear runtime
       BubbleSort 75703.11 ==============================
    InsertionSort    73.31 =
    SelectionSort 49624.13 ===================
        MergeSort   478.93 =
        ShellSort   262.06 =
       ArraysSort     8.06 =
    
    vm: java
    trial: 0
    

    Increasing the array size to 100,000, and excluding the bubble sort and selection sort in the benckmarks, we have another set of results, where insertion sort starts to behave terribly in this case, and the dual-pivot quick sort is still the quickest.

    0% Scenario{vm=java, trial=0, benchmark=InsertionSort} 2031873339.00 ns; σ=3247182.51 ns @ 3 trials
    25% Scenario{vm=java, trial=0, benchmark=MergeSort} 5701257.43 ns; σ=37183.00 ns @ 3 trials
    50% Scenario{vm=java, trial=0, benchmark=ShellSort} 2277851.07 ns; σ=21077.62 ns @ 3 trials
    75% Scenario{vm=java, trial=0, benchmark=ArraysSort} 210179.59 ns; σ=164843.22 ns @ 10 trials
    
        benchmark      us linear runtime
    InsertionSort 2031873 ==============================
        MergeSort    5701 =
        ShellSort    2278 =
       ArraysSort     210 =
    
    vm: java
    trial: 0
    

    You can find the source code for the sorting algorithms here, and the benchmarks runner here, all of which are part of my github project DataStructureAndAlgorithms

    Wednesday 3 April 2013

    A Different Builder Pattern Example in Java

    Builder design pattern is often used to separate the construction of a complex object from its various representations. I will show you a nicer way of writing a Builder pattern, rather than the example shown in the wiki.

    First of all, I created a POJO called Vehicle, which has some simple properties about its model, its make, etc. Then, nested within this POJO, I created a Builder that holds some default properties which can be used to build a vehicle. The builder can be created by a static method newBuilder(), and we can override the vehicle properties by chaining the with* methods, and finally return the constructed vehicle using the build() method.

    public class Vehicle {
    
        private String model;
        private String make;
        private int numberOfWheels;
        private double height;
        private double width;
        private double length;
        private int numberOfDoors;
    
        private Vehicle(String model, String make, int numberOfWheels, double height, double width, double length, int numberOfDoors) {
            this.model = model;
            this.make = make;
            this.numberOfWheels = numberOfWheels;
            this.height = height;
            this.width = width;
            this.length = length;
            this.numberOfDoors = numberOfDoors;
        }
    
        public static class Builder {
    
            private String model = "M5";
            private String make = "BMW";
            private int numberOfWheels = 4;
            private double height = 1.45;
            private double width = 1.9;
            private double length = 4.62;
            private int numberOfDoors = 5;
    
            public static Builder newBuilder() {
                return new Builder();
            }
    
            public Builder withModel(String model) {
                this.model = model;
                return this;
            }
    
            public Builder withMake(String make) {
                this.make = make;
                return this;
            }
    
            public Vehicle build() {
                return new Vehicle(model, make, numberOfWheels, height, width,length, numberOfDoors);
            }
        }
    }
    

    Let me use a unit test, written in Groovy, to demonstrate exactly how we use the builder. I created a test to show how we can build a default vehicle without providing any overrides to the default vehicle properties, and another test to show how we can override the default vehicle properties to create a different vehicle.

    import org.junit.Test
    
    class VehicleTest {
    
        @Test
        public void canBuildADefaultVehicle() {
            def defaultVehicle = Vehicle.Builder.newBuilder().build()
            assert defaultVehicle.model == "M5"
            assert defaultVehicle.make == "BMW"
            assert defaultVehicle.numberOfWheels == 4
            assert defaultVehicle.height == 1.45
            assert defaultVehicle.width == 1.9
            assert defaultVehicle.length == 4.62
            assert defaultVehicle.numberOfDoors == 5
        }
    
        @Test
        public void canBuildADifferentVehicle() {
            def defaultVehicle = Vehicle.Builder.newBuilder().withModel("M3").build()
            assert defaultVehicle.model == "M3"
            assert defaultVehicle.make == "BMW"
            assert defaultVehicle.numberOfWheels == 4
            assert defaultVehicle.height == 1.45
            assert defaultVehicle.width == 1.9
            assert defaultVehicle.length == 4.62
            assert defaultVehicle.numberOfDoors == 5
        }
    }
    

    Some says the chaining syntax makes the building block easier to read. :)

    Tuesday 2 April 2013

    Solving Farmer-Wolf-Goat-Cabbage riddle with Groovy/Java


  • The Riddle - Farmer Wolf Goat Cabbage
  • A farmer is on the west bank of a river with a wolf, a goat and a cabbage in his care. Without his presence the wolf would eat the goat or the goat would eat the cabbage. The wolf is not interested in the cabbage. The farmer wishes to bring his three charges across the river. However the boat available to him can only carry one of the wolf, goat and cabbage besides himself. The puzzle is: is there a sequence of river crossings so that the farmer can transfer himself and the other three all intact to the east bank?
  • The Solution
  • It is a task that given an initial state, a final state, and a set of rules, one will start from the initial state and try to reach to the final state by passing through some intermediate states. Each move will transform from one state to the other, and there might be multiple valid moves from a given state. Such a collection of interconnected states can be represented by State Space Graph.
    Let's use 'f', 'c', 'g', 'w' to denote the farmer, the cabbage, the goat, and the wolf, and use '|' to separate the river where left of the '|' denotes west bank and right of the '|' denotes east bank. Initially, they are all at the west bank of the river, which is represented as 'fcgw |' as shown below. We can solve the riddle by figuring out what the possible and valid moves are, using either Breadth-First Search or Depth-First Search, on a state space graph shown below
       

  • Depth First Search
  • I use DFS to find the first possible solution to the riddle, where it looks like this

    or with a possibility of backtracking like this

  • Breadth First Search
  • I can build a complete state space graph using BFS, excluding the red states as shown below

  • Sample Source Code
  • Implemented in a mix of Groovy and Java, where Java is mainly used for the POJOs, the sample source code, including the unit test, demonstrates how to build a complete state space graph (which is also a directed graph) and how to find all possible solutions based on the state space graph.
    DirectedGraph.java
    import java.util.*;
    
    public class DirectedGraph<T> implements Iterable<T> {
    
        // key is a Node, value is a set of Nodes connected by outgoing edges from the key
        private final Map<T, Set<T>> graph = new HashMap<T, Set<T>>();
    
        public boolean addNode(T node) {
            if (graph.containsKey(node)) {
                return false;
            }
    
            graph.put(node, new HashSet<T>());
            return true;
        }
    
        public void addNodes(Collection<T> nodes) {
            for (T node : nodes) {
                addNode(node);
            }
        }
    
        public void addEdge(T src, T dest) {
            validateSourceAndDestinationNodes(src, dest);
    
            // Add the edge by adding the dest node into the outgoing edges
            graph.get(src).add(dest);
        }
    
        public void removeEdge(T src, T dest) {
            validateSourceAndDestinationNodes(src, dest);
    
            graph.get(src).remove(dest);
        }
    
        public boolean edgeExists(T src, T dest) {
            validateSourceAndDestinationNodes(src, dest);
    
            return graph.get(src).contains(dest);
        }
    
        public Set<T> edgesFrom(T node) {
            // Check that the node exists.
            Set<T> edges = graph.get(node);
            if (edges == null)
                throw new NoSuchElementException("Source node does not exist.");
    
            return Collections.unmodifiableSet(edges);
        }
    
        public Iterator<T> iterator() {
            return graph.keySet().iterator();
        }
    
        public int size() {
            return graph.size();
        }
    
        public boolean isEmpty() {
            return graph.isEmpty();
        }
    
        private void validateSourceAndDestinationNodes(T src, T dest) {
            // Confirm both endpoints exist
            if (!graph.containsKey(src) || !graph.containsKey(dest))
                throw new NoSuchElementException("Both nodes must be in the graph.");
        }
    
    }
    
    BoatLocation.java
    public enum BoatLocation {
        WestBank, EastBank
    }
    
    RiverRole.java, implemented Comparable interface in order to be used in a SortedSet
    public class RiverRole implements Comparable<RiverRole> {
    
        public final String name;
        public final boolean canSailTheBoat;
    
        public RiverRole(String name, boolean canSailTheBoat) {
            this.name = name;
            this.canSailTheBoat = canSailTheBoat;
        }
    
        @Override
        public int compareTo(RiverRole o) {
            return this.name.compareTo(o.name);
        }
    
        @Override
        public String toString() {
            return name;
        }
    }
    
    RiverState.java
    import java.util.Collection;
    import java.util.SortedSet;
    import java.util.TreeSet;
    
    public class RiverState {
    
        private SortedSet<RiverRole> westBank;
        private SortedSet<RiverRole> eastBank;
        private BoatLocation boatLocation;
    
        public RiverState(Collection<RiverRole> westBank, Collection<RiverRole> eastBank, BoatLocation boatLocation) {
            this.westBank = new TreeSet<RiverRole>(westBank);
            this.eastBank = new TreeSet<RiverRole>(eastBank);
            this.boatLocation = boatLocation;
        }
    
        public SortedSet<RiverRole> getWestBank() {
            return new TreeSet<RiverRole>(westBank);
        }
    
        public SortedSet<RiverRole> getEastBank() {
            return new TreeSet<RiverRole>(eastBank);
        }
    
        public BoatLocation getBoatLocation() {
            return boatLocation;
        }
    
        @Override
        public String toString() {
            return "State{westBank=" + westBank + ", eastBank=" + eastBank + ", boatLocation=" + boatLocation + "}";
        }
    
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof RiverState)) return false;
    
            RiverState riverState = (RiverState) o;
    
            return eastBank.equals(riverState.eastBank) &&
                   westBank.equals(riverState.westBank) &&
                   boatLocation.equals(riverState.boatLocation);
        }
    
        public int hashCode() {
            int result;
            result = westBank.hashCode();
            result = 31 * result + eastBank.hashCode();
            result = 31 * result + boatLocation.hashCode();
            return result;
        }
    }
    
    FarmerWolfGoatRiddle.groovy
    import graph.DirectedGraph
    import org.codehaus.groovy.runtime.MethodClosure
    
    import static fun.rivercrossing.BoatLocation.EastBank
    import static fun.rivercrossing.BoatLocation.WestBank
    
    public class FarmerWolfGoatRiddle {
    
        /*  Farmer Wolf Goat Puzzle
            A farmer is on the east bank of a river with a wolf, goat and cabbage in his care. Without his presence the wolf would eat
            the goat or the goat would eat the cabbage. The wolf is not interested in the cabbage. The farmer wishes to bring his three
            charges across the river. However the boat available to him can only carry one of the wolf, goat and cabbage besides
            himself. The puzzle is: is there a sequence of river crossings so that the farmer can transfer himself and the other three all
            intact to the west bank?
        */
    
        RiverRole farmer = new RiverRole("farmer", true);
        RiverRole wolf = new RiverRole("wolf", false);
        RiverRole goat = new RiverRole("goat", false);
        RiverRole cabbage = new RiverRole("cabbage", false);
    
        RiverState initialState = new RiverState([farmer, wolf, goat, cabbage], [], BoatLocation.WestBank)
        RiverState finalState = new RiverState([], [farmer, wolf, goat, cabbage], EastBank)
    
        Closure<Boolean> areRolesHarmonisedOnRiverBank = { Collection<RiverRole> roles ->
            // describe clearly what the rules are
            if (!roles.contains(farmer)) {
                if (roles.contains(wolf) && roles.contains(goat)) return false
                if (roles.contains(goat) && roles.contains(cabbage)) return false
            }
            return true
        }
    
    
        public Collection<Deque<RiverState>> findAllPossibleSolutions() {
            DirectedGraph<RiverState> graph = buildCompleteGraph()
    
            Collection<Deque<RiverState>> solutions = []
            Deque<RiverState> currentSolution = new ArrayDeque<RiverState>()
    
            explore(graph, initialState, currentSolution, solutions)
    
            return solutions
        }
    
        // using Depth First Search (Stack Implementation) to identify a solution
        private void explore(DirectedGraph<RiverState> graph, RiverState currentNode, Deque<RiverState> currentSolution, Collection<Deque<RiverState>> solutions) {
            currentSolution.push(currentNode)
            graph.edgesFrom(currentNode).each { RiverState node ->
                if (node == finalState) {
                    // add the current solution, including the final node, to the solutions
                    currentSolution.push(node)
                    solutions.add(new ArrayDeque<RiverState>(currentSolution))
                    currentSolution.pop()
                } else {
                    // recursively explore
                    explore(graph, node, currentSolution, solutions)
                }
            }
            currentSolution.pop()
        }
    
        // using Breath First Search to build the complete graph
        private DirectedGraph<RiverState> buildCompleteGraph() {
            DirectedGraph<RiverState> graph = new DirectedGraph<RiverState>()
            graph.addNode(initialState)
    
            Set<RiverState> visitedNodes = new HashSet<RiverState>()
    
            // use a queue to keep the states to be visited in sequence
            Deque<RiverState> currentStates = new ArrayDeque<RiverState>()
            currentStates.add(initialState)
    
            while (!currentStates.isEmpty()) {
                if (currentStates.peek() != finalState) {
                    def currentState = currentStates.removeFirst()
                    if (!visitedNodes.contains(currentState)) {
                        visitedNodes.add(currentState)
                        def possibleStates = findNextPossibleStates(currentState)
                        def validStates = possibleStates.findAll { RiverState possibleState ->
                            // matched state should not be one of the visited states, and it should pass the validations
                            return !visitedNodes.contains(possibleState) && passRuleValidations(possibleState)
                        }
                        // add successors and edges
                        graph.addNodes(validStates)
                        validStates.each { graph.addEdge(currentState, it) }
    
                        // add to current states so that we can visit them next
                        currentStates.addAll(validStates)
                    }
                } else {
                    currentStates.removeFirst() // remove the final state from the currentStates queue
                }
            }
    
            return graph
        }
    
        private boolean passRuleValidations(RiverState state) {
            return areRolesHarmonisedOnRiverBank(state.eastBank) && areRolesHarmonisedOnRiverBank(state.westBank)
        }
    
        private List<RiverState> findNextPossibleStates(RiverState currentState) {
            // from the last visited state, find out who can sail the boat, and where is the sailor
            List<RiverState> possibleStates = []
    
            transitFrom(currentState, possibleStates)
    
            return possibleStates
        }
    
        private void transitFrom(RiverState currentState, List<RiverState> possibleStates) {
            // check where the boat is
            BoatLocation boatOrigin = currentState.boatLocation
            BoatLocation boatDestination
    
            // we use method closure to get a copy of west bank or east bank roles
            MethodClosure origin, destination
    
            // multiple assignments
            (origin, destination, boatDestination) = boatOrigin == WestBank ?
                [currentState.&getWestBank, currentState.&getEastBank, EastBank] :
                [currentState.&getEastBank, currentState.&getWestBank, WestBank]
    
            def possibleSailors = origin().findAll { RiverRole role -> role.canSailTheBoat }
    
            possibleSailors?.each { RiverRole sailor ->
                // pick 0 passengers onto the boat
                Collection<RiverRole> onBoat = [sailor]
                possibleStates.add(createNextPossibleState(origin, destination, onBoat, boatDestination))
    
                Collection<RiverRole> possiblePassengers = origin() as Collection<RiverRole>
                possiblePassengers.remove(sailor)
    
                // pick 1 passenger onto the boat
                possiblePassengers.each { RiverRole passenger ->
                    onBoat = [passenger, sailor]
                    possibleStates.add(createNextPossibleState(origin, destination, onBoat, boatDestination))
                }
            }
        }
    
        private RiverState createNextPossibleState(MethodClosure origin, MethodClosure destination, Collection<RiverRole> onBoat, BoatLocation boatDestination) {
            Collection<RiverRole> newDestination = destination() as Collection<RiverRole>
            Collection<RiverRole> newOrigin = origin() as Collection<RiverRole>
    
            // boat transits from origin to destination
            newOrigin.removeAll(onBoat)
            newDestination.addAll(onBoat)
    
            return boatDestination == WestBank ? new RiverState(newDestination, newOrigin, boatDestination) : new RiverState(newOrigin, newDestination, boatDestination)
        }
    
    
    }
    
    FarmerWolfGoatRiddleTest.groovy
    import org.junit.Test
    
    import static fun.rivercrossing.BoatLocation.EastBank
    import static fun.rivercrossing.BoatLocation.WestBank
    
    
    class FarmerWolfGoatRiddleTest {
    
        FarmerWolfGoatRiddle riddle = new FarmerWolfGoatRiddle()
    
        @Test
        public void findAllPossibleSolutions() {
            def solutions = riddle.findAllPossibleSolutions()
            // the ordering of solutions will be random
            assert solutions.size() == 2
            assert solutions[0].size() == 8
            assert solutions[1].size() == 8
    
            def assertAllSolutions = { Deque<RiverState> solution1, Deque<RiverState> solution2 ->
                assertFirstSolution(solution1)
                assertSecondSolution(solution2)
            }
    
            // assert with the correct expectations regardless of which is the first solution
            if (solutions[0].contains(new RiverState([riddle.wolf], [riddle.farmer, riddle.goat, riddle.cabbage], EastBank))) {
                assertAllSolutions(solutions[0], solutions[1])
            } else {
                assertAllSolutions(solutions[1], solutions[0])
            }
        }
    
        private void assertFirstSolution(Deque<RiverState> solution) {
            assert solution.poll() == new RiverState([riddle.farmer, riddle.wolf, riddle.goat, riddle.cabbage], [], WestBank)
            assert solution.poll() == new RiverState([riddle.wolf, riddle.cabbage], [riddle.farmer, riddle.goat], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.wolf, riddle.cabbage], [riddle.goat], WestBank)
            assert solution.poll() == new RiverState([riddle.wolf], [riddle.farmer, riddle.goat, riddle.cabbage], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.wolf, riddle.goat], [riddle.cabbage], WestBank)
            assert solution.poll() == new RiverState([riddle.goat], [riddle.farmer, riddle.wolf, riddle.cabbage], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.goat], [riddle.wolf, riddle.cabbage], WestBank)
            assert solution.poll() == new RiverState([], [riddle.farmer, riddle.wolf, riddle.goat, riddle.cabbage], EastBank)
            assert solution.poll() == null
        }
    
        private void assertSecondSolution(Deque<RiverState> solution) {
            assert solution.poll() == new RiverState([riddle.farmer, riddle.wolf, riddle.goat, riddle.cabbage], [], WestBank)
            assert solution.poll() == new RiverState([riddle.wolf, riddle.cabbage], [riddle.farmer, riddle.goat], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.wolf, riddle.cabbage], [riddle.goat], WestBank)
            assert solution.poll() == new RiverState([riddle.cabbage], [riddle.farmer, riddle.goat, riddle.wolf], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.cabbage, riddle.goat], [riddle.wolf], WestBank)
            assert solution.poll() == new RiverState([riddle.goat], [riddle.farmer, riddle.wolf, riddle.cabbage], EastBank)
            assert solution.poll() == new RiverState([riddle.farmer, riddle.goat], [riddle.wolf, riddle.cabbage], WestBank)
            assert solution.poll() == new RiverState([], [riddle.farmer, riddle.wolf, riddle.goat, riddle.cabbage], EastBank)
            assert solution.poll() == null
        }
    }
    

    Sunday 31 March 2013

    Fibonacci numbers in Python and Java

    Fibonacci numbers,  introduced by Leonardo Pisano Bigollo nearly a thousand years ago, were used to solve a fascinating puzzle about the growth of an idealised rabbit population. Below are the first 11 Fibonacci numbers Fn where n=0, 1, 2,..., 10:

    sequenceF0F1F2F3F4F5F6F7F8F9F10
    number011235813213455

    or we can describe the Fibonacci numbers as
    Fn = Fn-1 + Fn-2 (when n>=2)
    
    To write a program to return a Fibonacci number given a sequence, there are many possible implementations. I shall present 2 solutions implemented in both Python and Java, where the first one is using recursion, and the second one is using a while loop, however, the performances of these 2 solutions are drastically different when the sequence is bigger.
    Below are the results when running a Python test script. The recursion way is already showing very poor performance comparing to the while loop way.
    > python test_fib.py 
    .Working out Fibonacci(40)=102334155 recursively took 67.310534 seconds
    .Working out Fibonacci(40)=102334155 with while loop took 1.8e-05 seconds
    Working out Fibonacci(50)=12586269025 with while loop took 1.6e-05 seconds
    Working out Fibonacci(60)=1548008755920 with while loop took 1.8e-05 seconds
    ..
    ----------------------------------------------------------------------
    Ran 4 tests in 67.312s
    
    OK
    
    If we look at the results when running a similar timed test in Java, the same recursion is not as bad as in Python, though still far poorer than the while loop implementation.
    > /somewhere/jdk1.6.0_33/bin/java -classpath ".:" fun.Fib
    Working out Fibonacci(40)=102334155 recursively took 408 ms
    Working out Fibonacci(40)=102334155 with while loop took 2033 ns
    Working out Fibonacci(50)=12586269025 with while loop took 2115 ns
    Working out Fibonacci(60)=1548008755920 with while loop took 2368 ns
    
    The reason why recursion is slow in this case is because there are a lot of duplicated operations as the sequence grows. For example, to solve F(5), we need to solve F(4) once, F(3) twice, F(2) three times, and F(1) twice, as shown below

    or the number of recursion required given a sequence n is
    [1+(n-2)]*(n-2)/2 + (n-3) or n2/2-n/2-2  (when n>=4)
    
    The complexity of using recursion is then actually O(n2), where, on the other hand, the while loop is O(n).
    Below are the source code that are used in the test.
    fib.py
    import datetime
    
    
    class Fib:
        def recursively(self, n):
            if n == 0:
                return 0
            elif n <= 2:
                return 1
            else:
                return self.recursively(n - 1) + self.recursively(n - 2)
    
        def whileLoop(self, n):
            if n == 0:
                return 0
            previous, fib, currentIndex = 0, 1, 1
            while currentIndex < n:
                previous, fib = fib, previous + fib
                currentIndex += 1
    
            return fib
    
        def timedRecursively(self, n):
            start = datetime.datetime.now()
            result = self.recursively(n)
            end = datetime.datetime.now()
            elapsed = end - start
            print "Working out Fibonacci({})={} recursively took {} seconds".format(str(n), str(result), str(elapsed.total_seconds()))
            return result
    
        def timedWhileLoop(self, n):
            start = datetime.datetime.now()
            result = self.whileLoop(n)
            end = datetime.datetime.now()
            elapsed = end - start
            print "Working out Fibonacci({})={} with while loop took {} seconds".format(str(n), str(result), str(elapsed.total_seconds()))
            return result
    

    test_fib.py
    import unittest
    from fib import Fib
    
    
    class TestFib(unittest.TestCase):
        def setUp(self):
            self.fib = Fib()
    
        def test_recursively(self):
            self.assertEqual(self.fib.recursively(0), 0)
            self.assertEqual(self.fib.recursively(1), 1)
            self.assertEqual(self.fib.recursively(2), 1)
            self.assertEqual(self.fib.recursively(3), 2)
            self.assertEqual(self.fib.recursively(4), 3)
            self.assertEqual(self.fib.recursively(5), 5)
            self.assertEqual(self.fib.recursively(6), 8)
            self.assertEqual(self.fib.recursively(7), 13)
            self.assertEqual(self.fib.recursively(8), 21)
            self.assertEqual(self.fib.recursively(9), 34)
            self.assertEqual(self.fib.recursively(10), 55)
    
        def test_whileLoop(self):
            self.assertEqual(self.fib.whileLoop(0), 0)
            self.assertEqual(self.fib.whileLoop(1), 1)
            self.assertEqual(self.fib.whileLoop(2), 1)
            self.assertEqual(self.fib.whileLoop(3), 2)
            self.assertEqual(self.fib.whileLoop(4), 3)
            self.assertEqual(self.fib.whileLoop(5), 5)
            self.assertEqual(self.fib.whileLoop(6), 8)
            self.assertEqual(self.fib.whileLoop(7), 13)
            self.assertEqual(self.fib.whileLoop(8), 21)
            self.assertEqual(self.fib.whileLoop(9), 34)
            self.assertEqual(self.fib.whileLoop(10), 55)
    
        def test_timedRecursively(self):
            self.assertEqual(self.fib.timedRecursively(40), 102334155)
    
        def test_timedWhileLoop(self):
            self.assertEqual(self.fib.timedWhileLoop(40), 102334155)
            self.assertEqual(self.fib.timedWhileLoop(50), 12586269025)
            self.assertEqual(self.fib.timedWhileLoop(60), 1548008755920)
    
    unittest.main()
    

    Fib.java
    package fun;
    
    public class Fib {
    
    
        public static void main(String[] args) {
            recursively(1);
            whileLoop(1);
    
            long start = System.nanoTime();
    
            long result = recursively(40);
            long elapsed = (System.nanoTime() - start) / 1000000;
            System.out.println(String.format("Working out Fibonacci(%s)=%s recursively took %s ms", 40, result, elapsed));
    
            start = System.nanoTime();
            result = whileLoop(40);
            elapsed = (System.nanoTime() - start);
            System.out.println(String.format("Working out Fibonacci(%s)=%s with while loop took %s ns", 40, result, elapsed));
    
            start = System.nanoTime();
            result = whileLoop(50);
            elapsed = (System.nanoTime() - start);
            System.out.println(String.format("Working out Fibonacci(%s)=%s with while loop took %s ns", 50, result, elapsed));
    
            start = System.nanoTime();
            result = whileLoop(60);
            elapsed = (System.nanoTime() - start);
            System.out.println(String.format("Working out Fibonacci(%s)=%s with while loop took %s ns", 60, result, elapsed));
        }
    
        static long recursively(long i) {
            if (i == 0) return 0;
            if (i <= 2) return 1;
            else return recursively(i - 1) + recursively(i - 2);
        }
    
        static long whileLoop(long i) {
            long previous = 0, fib = 1, currentIndex = 1;
            while (currentIndex < i) {
                long newFib = previous + fib;
                previous = fib;
                fib = newFib;
                currentIndex++;
            }
            return fib;
        }
    }