Determine Event Recurrence Pattern for a set of dates

Question

I am looking for a pattern, algorithm, or library that will take a set of dates and return a description of the recurrence if one exits, i.e. the set [11-01-2010, 11-08-2010, 11

Answer 1

Grammatical Evolution (GE) is suitable for this kind of problem, because you are searching for an answer that adheres to a certain language. Grammatical Evolution is also used for program generation, composing music, designing, etcetera.

I'd approach the task like this:

Structure the problem space with a grammar.

Construct a Context-free Grammar that can represent all desired recurrence patterns. Consider production rules like these:

datepattern -> datepattern 'and' datepattern
datepattern -> frequency bounds
frequency -> 'every' ordinal weekday 'of the month'
frequency -> 'every' weekday
ordinal -> ordinal 'and' ordinal
ordinal -> 'first' | 'second' | 'third'
bounds -> 'in the year' year

An example of a pattern generated by these rules is: 'every second and third wednesday of the month in the year 2010 and every tuesday in the year 2011'

One way to implement such a grammar would be through a class hierarchy that you will later operate on through reflection, as I've done in the example below.

Map this language to a set of dates

You should create a function that takes a clause from your language and recursively returns the set of all dates covered by it. This allows you to compare your answers to the input.

Guided by the grammar, search for potential solutions

You could use a Genetic algorithm or Simulated Annealing to match the dates to the grammar, try your luck with Dynamic Programming or start simple with a brute force enumeration of all possible clauses.

Should you go with a Genetic Algorithm, your mutation concept should consist of substituting an expression for another one based on the application of one of your production rules.

Have a look at the following GE-related sites for code and information: http://www.bangor.ac.uk/~eep201/jge/
http://nohejl.name/age/
http://www.geneticprogramming.us/Home_Page.html

Evaluate each solution

The fitness function could take into account the textual length of the solution, the number of dates generated more than once, the number of dates missed, as well as the number of wrong dates generated.

Example code

By request, and because it's such an interesting challenge, I've written a rudimentary implementation of the algorithm to get you started. Although it works it is by no means finished, the design should definitively get some more thought, and once you have gleaned the fundamental take-aways from this example I recommend you consider using one the libraries I've mentioned above.

  /// <summary>
  ///  This is a very basic example implementation of a grammatical evolution algorithm for formulating a recurrence pattern in a set of dates.
  ///  It needs significant extensions and optimizations to be useful in a production setting.
  /// </summary>
  static class Program
  {

    #region "Class hierarchy that codifies the grammar"

    class DatePattern
    {

      public Frequency frequency;
      public Bounds bounds;

      public override string ToString() { return "" + frequency + " " + bounds; }

      public IEnumerable<DateTime> Dates()
      {
        return frequency == null ? new DateTime[] { } : frequency.FilterDates(bounds.GetDates());
      }

    }

    abstract class Bounds
    {
      public abstract IEnumerable<DateTime> GetDates();
    }

    class YearBounds : Bounds
    {

      /* in the year .. */
      public int year;

      public override string ToString() { return "in the year " + year; }

      public override IEnumerable<DateTime> GetDates()
      {
        var firstDayOfYear = new DateTime(year, 1, 1);
        return Enumerable.Range(0, new DateTime(year, 12, 31).DayOfYear)
          .Select(dayOfYear => firstDayOfYear.AddDays(dayOfYear));
      }
    }

    abstract class Frequency
    {
      public abstract IEnumerable<DateTime> FilterDates(IEnumerable<DateTime> Dates);
    }

    class WeeklyFrequency : Frequency
    {

      /* every .. */
      public DayOfWeek dayOfWeek;

      public override string ToString() { return "every " + dayOfWeek; }

      public override IEnumerable<DateTime> FilterDates(IEnumerable<DateTime> Dates)
      {
        return Dates.Where(date => (date.DayOfWeek == dayOfWeek));
      }

    }

    class MonthlyFrequency : Frequency
    {

      /* every .. */
      public Ordinal ordinal;
      public DayOfWeek dayOfWeek;
      /* .. of the month */

      public override string ToString() { return "every " + ordinal + " " + dayOfWeek + " of the month"; }

      public override IEnumerable<DateTime> FilterDates(IEnumerable<DateTime> Dates)
      {
        return Dates.Where(date => (date.DayOfWeek == dayOfWeek) && (int)ordinal == (date.Day - 1) / 7);
      }

    }

    enum Ordinal { First, Second, Third, Fourth, Fifth }

    #endregion

    static Random random = new Random();
    const double MUTATION_RATE = 0.3;
    static Dictionary<Type, Type[]> subtypes = new Dictionary<Type, Type[]>();

    static void Main()
    {

      // The input signifies the recurrence 'every first thursday of the month in 2010':
      var input = new DateTime[] {new DateTime(2010,12,2), new DateTime(2010,11,4),new DateTime(2010,10,7),new DateTime(2010,9,2),
                    new DateTime(2010,8,5),new DateTime(2010,7,1),new DateTime(2010,6,3),new DateTime(2010,5,6),
                    new DateTime(2010,4,1),new DateTime(2010,3,4),new DateTime(2010,2,4),new DateTime(2010,1,7) };


      for (int cTests = 0; cTests < 20; cTests++)
      {
        // Initialize with a random population
        int treesize = 0;
        var population = new DatePattern[] { (DatePattern)Generate(typeof(DatePattern), ref treesize), (DatePattern)Generate(typeof(DatePattern), ref treesize), (DatePattern)Generate(typeof(DatePattern), ref treesize) };
        Run(input, new List<DatePattern>(population));
      }
    }

    private static void Run(DateTime[] input, List<DatePattern> population)
    {
      var strongest = population[0];
      int strongestFitness = int.MinValue;
      int bestTry = int.MaxValue;
      for (int cGenerations = 0; cGenerations < 300 && strongestFitness < -100; cGenerations++)
      {
        // Select the best individuals to survive:
        var survivers = population
            .Select(individual => new { Fitness = Fitness(input, individual), individual })
            .OrderByDescending(pair => pair.Fitness)
            .Take(5)
            .Select(pair => pair.individual)
            .ToArray();
        population.Clear();

        // The survivers are the foundation for the next generation:
        foreach (var parent in survivers)
        {
          for (int cChildren = 0; cChildren < 3; cChildren++)
          {
            int treeSize = 1;
            DatePattern child = (DatePattern)Mutate(parent, ref treeSize); // NB: procreation may also be done through crossover.
            population.Add((DatePattern)child);

            var childFitness = Fitness(input, child);
            if (childFitness > strongestFitness)
            {
              bestTry = cGenerations;
              strongestFitness = childFitness;
              strongest = child;
            }

          }
        }
      }
      Trace.WriteLine("Found best match with fitness " + Fitness(input, strongest) + " after " + bestTry + " generations: " + strongest);

    }

    private static object Mutate(object original, ref int treeSize)
    {
      treeSize = 0;


      object replacement = Construct(original.GetType());
      foreach (var field in original.GetType().GetFields())
      {
        object newFieldValue = field.GetValue(original);
        int subtreeSize;
        if (field.FieldType.IsEnum)
        {
          subtreeSize = 1;
          if (random.NextDouble() <= MUTATION_RATE)
            newFieldValue = ConstructRandomEnumValue(field.FieldType);
        }
        else if (field.FieldType == typeof(int))
        {
          subtreeSize = 1;
          if (random.NextDouble() <= MUTATION_RATE)
            newFieldValue = (random.Next(2) == 0
            ? Math.Min(int.MaxValue - 1, (int)newFieldValue) + 1
            : Math.Max(int.MinValue + 1, (int)newFieldValue) - 1);
        }
        else
        {
          subtreeSize = 0;
          newFieldValue = Mutate(field.GetValue(original), ref subtreeSize); // mutate pre-maturely to find out subtreeSize

          if (random.NextDouble() <= MUTATION_RATE / subtreeSize) // makes high-level nodes mutate less.
          {
            subtreeSize = 0; // init so we can track the size of the subtree soon to be made.
            newFieldValue = Generate(field.FieldType, ref subtreeSize);
          }
        }
        field.SetValue(replacement, newFieldValue);
        treeSize += subtreeSize;
      }
      return replacement;

    }

    private static object ConstructRandomEnumValue(Type type)
    {
      var vals = type.GetEnumValues();
      return vals.GetValue(random.Next(vals.Length));
    }

    private static object Construct(Type type)
    {
      return type.GetConstructor(new Type[] { }).Invoke(new object[] { });
    }

    private static object Generate(Type type, ref int treesize)
    {
      if (type.IsEnum)
      {
        return ConstructRandomEnumValue(type);
      }
      else if (typeof(int) == type)
      {
        return random.Next(10) + 2005;
      }
      else
      {
        if (type.IsAbstract)
        {
          // pick one of the concrete subtypes:
          var subtypes = GetConcreteSubtypes(type);
          type = subtypes[random.Next(subtypes.Length)];
        }
        object newobj = Construct(type);

        foreach (var field in type.GetFields())
        {
          treesize++;
          field.SetValue(newobj, Generate(field.FieldType, ref treesize));
        }
        return newobj;
      }
    }


    private static int Fitness(DateTime[] input, DatePattern individual)
    {
      var output = individual.Dates().ToArray();
      var avgDateDiff = Math.Abs((output.Average(d => d.Ticks / (24.0 * 60 * 60 * 10000000)) - input.Average(d => d.Ticks / (24.0 * 60 * 60 * 10000000))));
      return
        -individual.ToString().Length // succinct patterns are preferred.
        - input.Except(output).Count() * 300 // Forgetting some of the dates is bad.
        - output.Except(input).Count() * 3000 // Spurious dates cause even more confusion to the user.
      - (int)(avgDateDiff) * 30000; // The difference in average date is the most important guide.
    }

    private static Type[] GetConcreteSubtypes(Type supertype)
    {
      if (subtypes.ContainsKey(supertype))
      {
        return subtypes[supertype];
      }
      else
      {

        var types = AppDomain.CurrentDomain.GetAssemblies().ToList()
            .SelectMany(s => s.GetTypes())
            .Where(p => supertype.IsAssignableFrom(p) && !p.IsAbstract).ToArray();
        subtypes.Add(supertype, types);
        return types;
      }
    }
  }

Hope this gets you on track. Be sure to share your actual solution somewhere; I think it will be quite useful in lots of scenarios.

Answer 2

You could access the system date or system dateandtime and construct crude calendar points in memory based on the date and the day of the week as returned by the call or function result. Then use the number of days in relevant months to sum them and add on the number of days of the day variable in the input and/or access the calendar point for the relevant week starting sunday or monday and calculate or increment index forward to the correct day. Construct text string using fixed characters and insert the relevant variable such as the full name of the day of the week as required. There may be multiple traversals needed to obtain all the events of which the occurrences are to be displayed or counted.

Answer 3

I like @arjen answer but I don't think there is any need for complex algorithm. This is so so simple. If there is a pattern, there is a pattern... therefore a simple algorithm would work. First we need to think of the types of patterns we are looking for: daily, weekly, monthly and yearly.

How to recognize?

Daily: there is a record every day Weekly: there is a record every week Monthly: there is a record every month Yearly: there is a record every year

Difficult? No. Just count how many repetitions you have and then classify.

Here is my implementation

RecurrencePatternAnalyser.java

public class RecurrencePatternAnalyser {

    // Local copy of calendars by add() method 
    private ArrayList<Calendar> mCalendars = new ArrayList<Calendar>();

    // Used to count the uniqueness of each year/month/day 
    private HashMap<Integer, Integer> year_count = new HashMap<Integer,Integer>();
    private HashMap<Integer, Integer> month_count = new HashMap<Integer,Integer>();
    private HashMap<Integer, Integer> day_count = new HashMap<Integer,Integer>();
    private HashMap<Integer, Integer> busday_count = new HashMap<Integer,Integer>();

    // Used for counting payments before due date on weekends
    private int day_goodpayer_ocurrences = 0;
    private int day_goodPayer = 0;

    // Add a new calendar to the analysis
    public void add(Calendar date)
    {
        mCalendars.add(date);
        addYear( date.get(Calendar.YEAR) );
        addMonth( date.get(Calendar.MONTH) );
        addDay( date.get(Calendar.DAY_OF_MONTH) );
        addWeekendDays( date );
    }

    public void printCounts()
    {
        System.out.println("Year: " +  getYearCount() + 
                " month: " +  getMonthCount() + " day: " +  getDayCount());
    }

    public RecurrencePattern getPattern()
    {
        int records = mCalendars.size();
        if (records==1)
            return null;

        RecurrencePattern rp = null;

        if (getYearCount()==records)
        {
            rp = new RecurrencePatternYearly();
            if (records>=3)
                rp.setConfidence(1);
            else if (records==2)
                rp.setConfidence(0.9f);
        }
        else if (getMonthCount()==records)
        {
            rp = new RecurrencePatternMonthly();
            if (records>=12)
                rp.setConfidence(1);
            else
                rp.setConfidence(1-(-0.0168f * records + 0.2f));
        } 
        else 
        {
            calcDaysRepetitionWithWeekends();
            if (day_goodpayer_ocurrences==records)
            {
                rp = new RecurrencePatternMonthly();
                rp.setPattern(RecurrencePattern.PatternType.MONTHLY_GOOD_PAYER);
                if (records>=12)
                    rp.setConfidence(0.95f);
                else
                    rp.setConfidence(1-(-0.0168f * records + 0.25f));
            }
        }

        return rp;
    }

    // Increment one more year/month/day on each count variable
    private void addYear(int key_year)  { incrementHash(year_count, key_year); }
    private void addMonth(int key_month)    { incrementHash(month_count, key_month); }
    private void addDay(int key_day)    { incrementHash(day_count, key_day); }

    // Retrieve number of unique entries for the records
    private int getYearCount() { return year_count.size(); }
    private int getMonthCount() { return month_count.size(); }
    private int getDayCount() { return day_count.size(); }

    // Generic function to increment the hash by 1  
    private void incrementHash(HashMap<Integer, Integer> var, Integer key)
    {
        Integer oldCount = var.get(key);
        Integer newCount = 0;
        if ( oldCount != null ) {
            newCount = oldCount;
        }
        newCount++;
        var.put(key, newCount);
    }

    // As Bank are closed during weekends, some dates might be anticipated
    // to Fridays. These will be false positives for the recurrence pattern.
    // This function adds Saturdays and Sundays to the count when a date is 
    // Friday.
    private void addWeekendDays(Calendar c)
    {
        int key_day = c.get(Calendar.DAY_OF_MONTH);
        incrementHash(busday_count, key_day);
        if (c.get(Calendar.DAY_OF_WEEK) == Calendar.FRIDAY)
        {
            // Adds Saturday
            c.add(Calendar.DATE, 1);
            key_day = c.get(Calendar.DAY_OF_MONTH);
            incrementHash(busday_count, key_day);
            // Adds Sunday
            c.add(Calendar.DATE, 1);
            key_day = c.get(Calendar.DAY_OF_MONTH);
            incrementHash(busday_count, key_day);
        }
    }

    private void calcDaysRepetitionWithWeekends()
    {               
        Iterator<Entry<Integer, Integer>> it =
                busday_count.entrySet().iterator();
        while (it.hasNext()) {
            @SuppressWarnings("rawtypes")
            Map.Entry pair = (Map.Entry)it.next();
            if ((int)pair.getValue() > day_goodpayer_ocurrences)
            {
                day_goodpayer_ocurrences = (int) pair.getValue();
                day_goodPayer = (int) pair.getKey();
            }
            //it.remove(); // avoids a ConcurrentModificationException
        }
    }
}

RecurrencePattern.java

public abstract class RecurrencePattern {

    public enum PatternType {
        YEARLY, MONTHLY, WEEKLY, DAILY, MONTHLY_GOOD_PAYER 
    }   
    public enum OrdinalType {
        FIRST, SECOND, THIRD, FOURTH, FIFTH 
    }

    protected PatternType pattern;
    private float confidence;
    private int frequency;

    public PatternType getPattern() {
        return pattern;
    }

    public void setPattern(PatternType pattern) {
        this.pattern = pattern;
    }

    public float getConfidence() {
        return confidence;
    }
    public void setConfidence(float confidence) {
        this.confidence = confidence;
    }
    public int getFrequency() {
        return frequency;
    }
    public void setFrequency(int frequency) {
        this.frequency = frequency;
    }   
}

RecurrencePatternMonthly.java

public class RecurrencePatternMonthly extends RecurrencePattern {
    private boolean isDayFixed;
    private boolean isDayOrdinal;
    private OrdinalType ordinaltype;

    public RecurrencePatternMonthly()
    {
        this.pattern = PatternType.MONTHLY;
    }
}

RecurrencePatternYearly.java

public class RecurrencePatternYearly extends RecurrencePattern {
    private boolean isDayFixed;
    private boolean isMonthFixed;
    private boolean isDayOrdinal;
    private OrdinalType ordinaltype;

    public RecurrencePatternYearly()
    {
        this.pattern = PatternType.YEARLY;
    }
}   

Main.java

public class Algofin {

    static Connection c = null;

    public static void main(String[] args) {
        //openConnection();
        //readSqlFile();

        RecurrencePatternAnalyser r = new RecurrencePatternAnalyser();

        //System.out.println(new GregorianCalendar(2015,1,30).get(Calendar.MONTH));
        r.add(new GregorianCalendar(2015,0,1));
        r.add(new GregorianCalendar(2015,0,30));
        r.add(new GregorianCalendar(2015,1,27));
        r.add(new GregorianCalendar(2015,3,1));
        r.add(new GregorianCalendar(2015,4,1));

        r.printCounts();

        RecurrencePattern rp;
        rp=r.getPattern();
        System.out.println("Pattern: " + rp.getPattern() + " confidence: " + rp.getConfidence());
    }
}

Answer 4

If your purpose is to generate human-readable descriptions of the pattern, as in your "Every Monday in November", then you probably want to start by enumerating the possible descriptions. Descriptions can be broken down into frequency and bounds, for example,

Frequency:

• Every day ...
• Every other/third/fourth day ...
• Weekdays/weekends ...
• Every Monday ...
• Alternate Mondays ...
• The first/second/last Monday ...
• ...

Bounds:

• ... in January
• ... between 25 March and 25 October
• ...

There won't be all that many of each, and you can check for them one by one.

Answer 5

First, find a sequence, if it exists:

step = {day,month,year}
period=0
for d = 1 to dates.count-1
    interval(d,step)=datedifference(s,date(d),date(d+1))
next
' Find frequency with largest interval
for s = year downto day
    found=true
    for d = 1 to dates.count-2
        if interval(d,s)=interval(d+1,s) then
            found=false
            exit for
        end if
    next
    if found then
        period=s
        frequency=interval(1,s)
        exit for
    end if
next

if period>0
    Select case period
      case day
        if frequency mod 7 = 0 then
          say "every" dayname(date(1))
        else
          say "every" frequency "days"
        end if
      case month
        say "every" frequency "months on day" daynumber(date(1))
      case years
        say "every" frequency "years on" daynumber(date(1)) monthname(date(1))
    end select
end if

Finally, deal with "in November", "from 2007 to 2010" etc., should be obvious.

HTH

Answer 6

I think you'll have to build it, and I think it will be a devil in the details kind of project. Start by getting much more thorough requirements. Which date patterns do you want to recognize? Come up with a list of examples that you want your algorithm to successfully identify. Write your algorithm to meet your examples. Put your examples in a test suite so when you get different requirements later you can make sure you didn't break the old ones.

I predict you will write 200 if-then-else statements.

OK, I do have one idea. Get familiar with the concepts of sets, unions, coverage, intersection and so on. Have a list of short patterns that you search for, say, "Every day in October", "Every day in November", and "Every day in December." If these short patterns are contained within the set of dates, then define a union function that can combine shorter patterns in intelligent ways. For example, let's say you matched the three patterns I mention above. If you Union them together you get, "Every day in October through December." You could aim to return the most succinct set of unions that cover your set of dates or something like that.