LTD Puzzle 1: Friday the Thirteenths

[AlexCuse]

With another friday the thirteenth fast approaching, now seems like a good time for a themed programming puzzle. The goal is to identify all friday the thirteenths for a given timeframe. We'll use a relatively small number of years, like 10. This should make it a little easier in procedural languages.

The rules:
When posting code, please use the [hide]some code here[/hide] tags, so that you don't ruin the fun for others.
Be creative.
Be sure to mention what language it is!
Don't do drugs.

Points will be awarded for "cool code", as identified by whoever chooses to do so via LessThanDot's rating system (http://wiki.lessthandot.com/index.php/F ... ing_system). The winner will be everyone who learns something by either reading or writing the code presented.

Have fun!

Alex

[damber]

Here is the short version (i.e. doesn't bother with options, summaries, etc and is less efficient (cycles through every date, instead of just the 13th of every month)

csharpLine number Off | Show | Select all

1. using System;
2.  
3. namespace F13s
4. {
5.     class MainClass
6.     {
7.         public static void Main(string[] args)
8.         {
9.             DateTime startDate = DateTime.Parse(args[0]);
10.             DateTime endDate = DateTime.Parse(args[1]);    
11.  
12.             while ( ! startDate.Equals(endDate) )
13.             {
14.                 if (startDate.DayOfWeek == DayOfWeek.Friday && startDate.Day == 13) Console.WriteLine(startDate.ToShortDateString() + " is Friday 13th");      
15.                 startDate = startDate.AddDays(1);   
16.             }
17.         }
18.     }
19. }

Code is hidden, SHOW

[HarleyQuinn]

Here's a first go in VB6 (it's not particularly optimised at this stage)

vbLine number Off | Show | Select all

1. Public Function GetFriday13ths(vEndDate As Date) As Long
2. Dim StartDate As Date
3. Dim intDateAdd As Integer
4.  
5. intDateAdd = IIf(Day(Date) > 13, 1, 0)
6.  
7. StartDate = DateSerial(Year(Date), Month(Date) + intDateAdd, 13)
8.  
9. Do Until StartDate >= vEndDate
10.     If Weekday(StartDate) = 6 Then
11.         Debug.Print StartDate
12.         GetFriday13ths = GetFriday13ths + 1
13.     End If
14.     StartDate = DateSerial(Year(StartDate), Month(StartDate) + 1, 13)
15. Loop
16.  
17. End Function

Code is hidden, SHOW

[chrissie1]

Hers is one that uses VB.Net and linq and the console.

vbnetLine number Off | Show | Select all

1. Imports System
2.  
3. Module Module1
4.  
5.     Sub Main()
6.         Console.WriteLine("Give the startdate")
7.         Dim input As String = Console.ReadLine
8.         Dim startdate As DateTime = DateTime.Parse(input)
9.         Console.WriteLine("Give the enddate")
10.         input = Console.ReadLine
11.         Dim enddate As DateTime = DateTime.Parse(input)
12.         Dim _dates As IList(Of DateTime) = New List(Of DateTime)
13.         While (Not startdate.Equals(enddate))
14.             _dates.Add(startdate)
15.             startdate = startdate.AddDays(1)
16.         End While
17.         Dim dates = _
18.         From _Date In _dates _
19.         Where _Date.DayOfWeek = DayOfWeek.Friday And _Date.Day = 13 _
20.         Order By _Date.Year, _Date.Month, _Date.Day Descending _
21.         Select _Date
22.         For Each _date As Date In dates
23.             Console.WriteLine(_date.ToString)
24.         Next
25.         Console.WriteLine("Press any key to continue")
26.         Console.ReadLine()
27.     End Sub
28.  
29. End Module

Code is hidden, SHOW

this one takes 708 milliseconds to spit out all the 13ths from 12/02/200 (yes 200) until 12/02/5000 yes 5000.

[AlexCuse]

here's a neat solution using C# 3.0

csharpLine number Off | Show | Select all

1. using System;
2. using System.Collections.Generic;
3. using System.Linq;
4. using System.Text;
5.  
6. namespace FridayThireenths
7. {
8.     class Program
9.     {
10.         struct Thirteenth
11.         {
12.             private DateTime _dt;
13.  
14.             public Thirteenth(int month, int year)
15.             {
16.                 _dt = new DateTime(year, month, 13);
17.             }
18.  
19.             public bool IsFriday()
20.             {
21.                 return (_dt.DayOfWeek == DayOfWeek.Friday);
22.             }
23.  
24.             public override string ToString()
25.             {
26.                 return _dt.ToString("MM/dd/yyyy");
27.             }
28.  
29.             public static Thirteenth operator +(Thirteenth t, int monthAdd)
30.             {
31.                 DateTime temp = t._dt.AddMonths(monthAdd);
32.                 return new Thirteenth(temp.Month, temp.Year);
33.             }
34.         }
35.         static void Main(string[] args)
36.         {
37.             int yearsToLook;
38.  
39.             Console.WriteLine("How Many Years ahead to look?");
40.  
41.             while (!int.TryParse(Console.ReadLine(), out yearsToLook))
42.                 Console.WriteLine("Please enter a numeric value");
43.  
44.             var tlist = new List<Thirteenth>();
45.  
46.             int monthsToCount = (yearsToLook * 12) + (12 - DateTime.Now.Month);
47.             Thirteenth baseThirteenth = new Thirteenth(DateTime.Now.Month, DateTime.Now.Year);
48.             tlist.Add(baseThirteenth);
49.  
50.             for (int i = 1; i <= monthsToCount; i++)
51.             {
52.                 tlist.Add(baseThirteenth + i);
53.             }
54.              
55.             var fridayThirteenths = from t in tlist where t.IsFriday() select t;
56.  
57.             foreach (Thirteenth t in fridayThirteenths)
58.                 Console.WriteLine(t.ToString());
59.  
60.             Console.WriteLine("there are {0} friday the thirteenths between {2} and the end of {1}",
61.                 new object[]{fridayThirteenths.Count<Thirteenth>().ToString() , DateTime.Now.Year + yearsToLook, DateTime.Now.ToString("dd/MM/yyyy")});
62.  
63.         }
64.     }
65. }

Code is hidden, SHOW

[SQLDenis]

SQL server, tested on 2000 and 2005



Code

tsqlLine number Off | Show | Select all

1. SELECT DATEADD(m, number,'1998-01-13')
2.  FROM  master..spt_values WHERE type = 'P'
3. and DATENAME(dw,DATEADD(m, number,'1998-01-13')) = 'friday'

Code is hidden, SHOW

[sean]

I think you want to check the and in your where.... try this instead:

tsqlLine number Off | Show | Select all

1. SELECT DATEADD(m, number,'1998-01-13')
2.  FROM  master..spt_values WHERE type = 'P'
3. and DATENAME(dw,DATEADD(m, number,'1998-01-13')) = 'friday'

Code is hidden, SHOW

[SQLDenis]

Sean I looked at that code for 5 minutes until I found what you were talking about

yes that is a typo I forgot to change it in the WHERE clause (that was from a 1st version I posted, I have changed it now)

The code I ran was actually identical to what you posted

[tarwn]

And of course I have to post an ASP/VBScript solution. I tried to do it differently then everyone else (at most it loops n+2 where n is number of 13's):

aspLine number Off | Show | Select all

1. Function GetThirteens(stDate, enDate)
2.     Dim stI,  enI,  cI
3.     Dim t : t = Array(3,2,1,0,2,1,0,13,12,11,10,9,8,7,6,5,4,3,2,1,0,2,1,0,2,1,0,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,10,9,8,7,6,5,4,3,2,1,0,0,8,7,6,5,4,3,2,1,0,5,4,3,2,1,0,0,8,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,13,12,11,10,9,8,7,6,5,4,3,2,1,0,5,4,3,2,1,0,2,1,0,13,12,11,10,9,8,7,6,5,4,3,2,1,0,2,1,0,5,4,3,2,1,0,7,6,5,4,3,2,1,0,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,5,4,3,2,1,0,2,1,0,10,9,8,7,6,5,4,3,2,1,0,7,6,5,4,3,2,1,0,0,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,7,6,5,4,3,2,1,0,2,1,0,2,1,0,13,12,11,10,9,8,7,6,5,4,3,2,1,0,2,1,0,5,4,3,2,1,0,7,6,5,4,3,2,1,0,0,7,6,5,4,3,2,1,0,5,4,3,2,1,0,7,6,5,4,3,2,1,0,8,7,6,5,4,3,2,1,0,5,4)
4.     stI = ((Year(stDate) - 2) mod 28) * 12 + Month(stDate) - 1 + Abs(cInt(Day(stDate) > 13))
5.     enI = (Year(enDate) - Year(stDate)) * 12 + Month(enDate) - 1 - Abs(cInt(Day(enDate) < 13)) + stI
6.  
7.     For cI = stI To enI
8.         cI = cI + t(cI Mod 336)
9.         If cI <= enI Then GetThirteens = GetThirteens & MonthName(Month(DateAdd("m", cI - stI, stDate))) & "-" & Year(DateAdd("m", cI - stI, stDate)) & ", "
10.     Next
11. End Function

Code is hidden, SHOW

Apparently we need to work on the line wrap for the Hide function Copy it into a text editor if you want to see the couple lines that are getting cut off.

[xtcsonik]

Faster, better, stronger...
http://weblogs.asp.net/blogs/mikebosch/friday.png

[cLFlaVA]

JavaScript :)

javascriptLine number Off | Show | Select all

1. <script type="text/javascript">
2. <!--
3. function findUnluckies( f ) {
4.     document.open();
5.     var t = new Date();
6.     var c = new Date();
7.     var cy = t.getFullYear();
8.     for ( var i = 0; i < f; i++ ) {
9.         m = ( i != 0 ) ? 0 : ( t.getDate() <= 13 ) ? t.getMonth() : t.getMonth() + 1;
10.         for ( var x = m; x < 12; x++ ) {
11.             c.setFullYear( cy+i, x, 13 );
12.             if ( c.getDay() == 5 ) document.write (c + " is a Friday the 13th<br />");
13.         }
14.     }
15.     document.close();
16. }
17.  
18. findUnluckies ( 10 );
19. //-->
20. </script>

Code is hidden, SHOW

[eksimba]

C# with a lambda expression:

csharpLine number Off | Show | Select all

1. using System;
2. using System.Collections.Generic;
3. using System.Linq;
4.  
5. namespace FridaysTheThirteenth
6. {
7.     class Program
8.     {
9.         static void Main(string[] args)
10.         {
11.             ShowFridayTheThirteenthsBetweenDates(DateTime.Parse("01/01/2000"), DateTime.Parse("12/31/2010"));
12.         }
13.  
14.         private static void ShowFridayTheThirteenthsBetweenDates(DateTime start, DateTime end)
15.         {
16.             foreach (var date in fillDates(start, end).Where(date => date.DayOfWeek.Equals(DayOfWeek.Friday)
17.                                                                                       && date.Day.Equals(13)))
18.             {
19.                 Console.WriteLine(date.ToLongDateString());
20.             }
21.             Console.ReadLine();
22.         }
23.  
24.         private static IList<DateTime> fillDates(DateTime start, DateTime end)
25.         {
26.             IList<DateTime> dates = new List<DateTime>();
27.             while (start < end)
28.             {
29.                 dates.Add(start);
30.                 start = start.AddDays(1);
31.             }
32.             return dates;
33.         }
34.     }
35. }
36.  

Code is hidden, SHOW

[spoulson]

Old school Perl 5

Usage: Perl f13.pl <start year> [end year]
Omit end year and it'll default to 10 years up

perlLine number Off | Show | Select all

1. ##use Time::HiRes 'gettimeofday';
2. use strict;
3.  
4. ##my $starttime = gettimeofday;       # start benchmark timer
5. my ($startyear, $endyear) = @ARGV;  # get command line
6.  
7. # if $endyear is null (or zero), default to 10 years ahead
8. $endyear ||= $startyear + 10;
9.  
10. # Check all years and months in range
11. for my $y ($startyear..$endyear) {
12.    for (0..11) {
13.       if (GetDOW($y, $_, 13) == 5) {
14.          # print matching date in RFC 3339 format
15.          printf("%4d-%02d-13\\n", $y, $_+1);
16.       }
17.    }
18. }
19.  
20. # Display elapsed time
21. ##printf("%dms elapsed\\n", int((gettimeofday - $starttime)*1000));
22.  
23. # Usage: $dow = GetDOW($year, $month, $day);
24. # $year = year, $month = 0-11, $day = 1-31
25. # $dow = 0-6 (0=Sunday)
26. sub GetDOW {
27.    my $y = $_[0] % 100;
28.    (
29.       (3 - (int($_[0] / 100) % 4)) * 2
30.       + $y + ($y >> 2)
31.       + (0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5)[$_[1]]
32.       + $_[2]
33.       # Fix for leap years
34.       + (($_[1] < 2 && ($_[0] % 4 == 0 && $_[0] % 100 != 0
35.          || $_[0] % 400 == 0)) and -1)
36.    ) % 7;
37. }

Code is hidden, SHOW

Runs in 477ms from year 200 to 5000 with the assumption we only care about the Gregorian calendar. If I take out the printf, it runs in 147ms. My PC is a 1.6GHz dual core Dell Optiplex 745 running XP.

I double commented the time measurement stuff because the module Time::HiRes is not installed by default and will fail to run on plain installs of Perl.

It's funny that in doing this exercise, I found an error in the DOW formula on the WikiPedia page. I commented my code change as "Fix for leap years".

[webbes]

C# 2.0
There's no need to waist memory on a creating a list or to loop through ALL dates in range..

csharpLine number Off | Show | Select all

1. using System;
2. using System.Collections.Generic;
3.  
4. namespace FridayThe13th {
5.     class Program {
6.         static void Main() {
7.             DateTime startDate = DateTime.Now;
8.             DateTime endDate = startDate.AddYears(10);
9.  
10.             DayOfWeek dayOfWeek = DayOfWeek.Friday;
11.             int day = 13;
12.  
13.             foreach (DateTime date in DayOfWeekAtMonthDay(startDate, endDate, dayOfWeek, day)) {
14.                 Console.WriteLine(date.ToLongDateString());
15.             }
16.  
17.             Console.ReadLine();
18.         }
19.  
20.         private static IEnumerable<DateTime> DayOfWeekAtMonthDay(DateTime startDate, DateTime endDate, DayOfWeek dayOfWeek, int day) {
21.             DateTime checkDate = startDate;
22.  
23.             int dayDiff = day - checkDate.Day;
24.             if (dayDiff < 0) {
25.                 checkDate = checkDate.AddMonths(1);
26.             }
27.  
28.             checkDate = checkDate.AddDays(dayDiff);
29.  
30.             while (checkDate <= endDate) {
31.                 if (checkDate.DayOfWeek == dayOfWeek) {
32.                     yield return checkDate;
33.                 }
34.                 checkDate = checkDate.AddMonths(1);
35.             }
36.         }
37.     }
38. }
39.  

Code is hidden, SHOW

[SQLDenis]

I would love to see a Python, Ruby, F# or Haskell solution

[ilitirit]

csharpLine number Off | Show | Select all

1. using System;
2.  
3. public class MyClass
4. {
5.     public static void Main()
6.     {
7.         // Go 10 years back
8.         DateTime workingDate = DateTime.Today.AddYears(-10);
9.  
10.         // Figure out if where we are relative to the 13th of this month
11.         int difference = 13 - workingDate.Day;
12.  
13.         // Compensate for the difference
14.         workingDate = workingDate.AddDays(difference);
15.  
16.         // If we are past the 13th, skip to the next month
17.         if (difference < 0)
18.             workingDate = workingDate.AddMonths(1);
19.  
20.         // Loop through all the months until today
21.         while (workingDate <= DateTime.Today)
22.         {
23.             // Print the date if we land on Friday the 13th.
24.             if (workingDate.DayOfWeek == DayOfWeek.Friday)
25.                 Console.WriteLine(workingDate.ToString("yyyy/MM/dd"));
26.  
27.             // Increment by a month
28.             workingDate = workingDate.AddMonths(1);
29.         }
30.     }
31. }

Code is hidden, SHOW

[traingamer]

The first time I wrote a function like this, it was in Pick basic. Haven't used that for a number of years, so here's one in VBA. Rather than cycling through all the months:

vbaLine number Off | Show | Select all

1. Public Function FriThirteenth()
2. Dim intCtr As Integer
3. Dim strFri(14) As String
4.  
5. ' Set up an array - a year can start on any of seven days
6. ' which determines all Friday 13ths + a second set for leap years
7.  
8. strFri(1) = "January and October"   ' Sunday is first day of year
9. strFri(2) = "April and July"        ' Monday
10. strFri(3) = "September and December"    ' Tuesday
11. strFri(4) = "June"                  ' Wednesday
12. strFri(5) = "February, March and November"  ' Thursday
13. strFri(6) = "August"                ' Friday
14. strFri(7) = "May"                   ' Saturday
15. strFri(8) = "January, April and July"   ' Leap Year Sunday
16. strFri(9) = "September and December"    ' etc.
17. strFri(10) = "June"
18. strFri(11) = "March and November"
19. strFri(12) = "February and August"
20. strFri(13) = "May"
21. strFri(14) = "October"
22.  
23. '  Since the combination of the starting day of the year and whether
24. '  or not it is a leap year determine all of the Friday the 13ths,
25. '  simply cycle through by years. I leave it as an exercise
26. '  for someone else to add start and stop year parameters.
27.  
28. For intCtr = 1900 To 2018
29.     Debug.Print "Friday the 13th falls in " & strFri(DatePart("w", "1/1/" & intCtr) + _
30.        IIf((intCtr Mod 4 = 0) And (Not (intCtr Mod 100 = 0)) Or (intCtr Mod 400 = 0), 7, 0)) _
31.         & " in " & intCtr
32. Next intCtr
33. End Function

Code is hidden, SHOW

[coneybeare]

RUBY
-------

Quick implementation using a for loop over days

rubyLine number Off | Show | Select all

1. #!/usr/local/bin/ruby
2. def usage
3.   puts ''
4.   puts 'findfri13ths will find the number of Friday the 13ths in a given timeframe from now.  It takes one argument: a positive integer of days (from today) to create a date-range.'
5.   puts ''
6.   puts 'USAGE: findfri13ths days'
7.   puts '   EX: findfri13ths 3'
8.   puts '   EX: findfri13ths 1286'
9.   puts ''
10. end
11.  
12. now = Time.now
13. count = 0
14. days = ARGV[0]
15. if days.nil? || !days.instance_of?(Fixnum) || days.to_i < 1
16.   usage && exit
17. end
18.  
19.  
20. secs_per_day = 86400
21. for day in (0..days.to_i)
22.   time = now + day*secs_per_day
23.   if time.day == 13 && time.wday == 5
24.     puts time
25.     count += 1
26.   end
27. end
28.  
29. puts count.to_s + " Friday the 13th's in the next " + days.to_s + " days."

Code is hidden, SHOW

[mbaird]

Java -- Just a quick implementation that prints every Friday 13th for the next 10 years (from whenever you run it).

javaLine number Off | Show | Select all

1. import java.text.DateFormat;
2. import java.text.SimpleDateFormat;
3. import java.util.Calendar;
4. import java.util.GregorianCalendar;
5.  
6. public class Test
7. {
8.    private static final DateFormat format = new SimpleDateFormat("EEE MMM dd yyyy");
9.    
10.    public static void main(String... args) {
11.       GregorianCalendar cal = new GregorianCalendar();
12.       GregorianCalendar stopDate = new GregorianCalendar();
13.       stopDate.add(Calendar.YEAR, 10);
14.        
15.       // Move ahead to the next Friday
16.       while (cal.get(Calendar.DAY_OF_WEEK) != 6) cal.add(Calendar.DATE, 1);
17.        
18.       while (cal.before(stopDate)) {
19.          if (cal.get(Calendar.DAY_OF_MONTH) == 13)
20.             System.out.println(format.format(cal.getTime()));
21.            
22.             cal.add(Calendar.DATE, 7);
23.       }
24.    }
25. }

Code is hidden, SHOW

[fullpunk]

An implementation in python using tables to find friday the 13th according to the first day of the year:

pythonLine number Off | Show | Select all

1. import calendar
2.  
3. # The years to search for friday the 13th
4. years = range(2000, 2100)
5.  
6. # normal[x] contains the months where there are friday the 13th for a year
7. # beginning on weekday x (where 0 means sunday).
8. normal = ((1, 10), (4, 7), (9, 12), (6,), (2, 3), (8,), (5,))
9.  
10. # same as normal, but for leap years.
11. leap = ((1, 4, 7), (9, 12), (6,), (3, 11), (8, 2), (5,), (10,))
12.  
13. for y in years:
14.     # compute the weekday of january 1st (0 for sunday)
15.     offset = (calendar.weekday(y, 1, 1) + 1) % 7
16.  
17.     # choose good table
18.     if calendar.isleap(y):
19.         months = leap[offset]
20.     else:
21.         months = normal[offset]
22.  
23.     # print results
24.     print '%d: %s' % (y, ', '.join([calendar.month_name[m] for m in months]))
25.  

Code is hidden, SHOW

[Emtucifor]

VB6. This function is VERY fast.

I wrote a VB6 app using this that calculates the years 1900 through 4899 (including displaying it in a textbox) 100 times. It takes about 0.72 seconds so that is 0.0072 seconds to calculate 5000 years worth of Friday the 13ths.

vbLine number Off | Show | Select all

1. Private Function FridayTheThirteenths(ByVal StartYear As Long, ByVal EndYear As Long) As String
2.     Dim Map(13) As String
3.     Dim Result() As String
4.    
5.     Map(0) = " x        x  "
6.     Map(1) = "    x  x     "
7.     Map(2) = "         x  x"
8.     Map(3) = "      x      "
9.     Map(4) = "  xx       x "
10.     Map(5) = "        x    "
11.     Map(6) = "     x       "
12.     Map(7) = " x  x  x     "
13.     Map(8) = "         x  x"
14.     Map(9) = "      x      "
15.     Map(10) = "   x       x "
16.     Map(11) = "  x     x    "
17.     Map(12) = "     x       "
18.     Map(13) = "          x  "
19.     ReDim Result(StartYear To EndYear)
20.     For StartYear = StartYear To EndYear
21.         Result(StartYear) = StartYear & Map((StartYear + (StartYear - 1) \ 4 - (StartYear - 1) \ 100 + (StartYear - 1) \ 400) Mod 7 _
22.             - 7 * ((StartYear Mod 4 = 0) - (StartYear Mod 100 = 0) + (StartYear Mod 400 = 0)))
23.     Next
24.     FridayTheThirteenths = Join(Result, vbCrLf)
25. End Function

Code is hidden, SHOW

[Emtucifor]

And now that I've looked at all your code, I see that traingamer had the same idea and so did a couple of other people.

In fact, his solution's output is quite a bit more readable than mine!

I also am doing some extra stuff to calculate the correct index offset, but he just reordered the array list. So his way is superior if it's giving the right output.

[spoulson]

And now that I've looked at all your code, I see that traingamer had the same idea and so did a couple of other people.

I wasn't aware of this until after participating in this puzzle, but there's a theorum of Babwani's Congruence that describes formulae for functionally computing a missing datetime field when the others are known. In his paper he also discusses the algorithm that traingamer, fullpunk, and ErikE used in their submissions. It's worth a read if you're ever tasked with coding fast datetime calculations.

[Emtucifor]

I used Excel for my analysis in a sort of brute-force graphical way:

1. Generate a list of all the Friday the 13ths for every year from 1900 through 9999.
2. Calculate the first day of the year (represented as 1 - 7 or 0 - 6) for each Fri 13th and whether the year is a leap year.
3. Use a pivot table to move months to their own columns at the top, including year and leap year information on the left.
4. Do an advanced data filter to get only unique variations of leap year + first day of year + Fri 13th pattern.
5. Notice there are only 14 variations, one for each day of the week in a leap and non-leap year.
6. Try as I might, I couldn't render the resulting pattern in a way that compressed easily or decompressed cheaply, so I finally just stuck the entire thing in an array, of sorts.
7. The list can be reduced to only 7 variations with the understanding that in leap years the "first day of the year" calculation will be advanced one for month 3 and up. However in my algorithm the extra logic for "month 3 and up" made the whole thing slower rather than just using the full 14-day map. For small ranges of years, the 7-variation list with the month 3 logic (or think of it as month 1 & 2 logic if you like) is faster.

After seeing traingamer's solution, I realized I could have simulated the output of a brute-force approach with:

Map(0) = "January 13, <year>" & vbcrlf & "October 13, <year>"

And the output would involve Replace(Map(index), "<year>", StartYear)

Now my somewhat cryptic and table-bound result (I used a textbox with JFMAMJJASOND at the top lined up with the x's) can be friendly looking but still just as fast.

Also, reading through the document found at spoulson's link, I wonder if calculating the day of the week for the month rather than the entire year could yield some optimization for my code. Obviously I'm doing more calculation than traingamer.

[spoulson]

From the looks of it, if you wanted to generalize the day of week calculation, you should be able to generate a lookup table containing 4 calendar years of day/month/dow, because it just repeats. Then lookup by day, month, and year mod 4. No runtime calculation.