Yesterday is Blog Action day and tons of bloggers posted about single topic – poverty. It is a topic very close to my heart for various reasons. It is a very sad thing not to have food or shelter or healthy living conditions. But man has thrived in all those situations just because he learned how to become better. In my opinion lack of knowledge is much more serious than lack of food. That is one reason why even the poorest parents in the world would love to see their kids going to school, so that they can learn and become better. I want to sensitize my readers about this issue: the new poor – people without information.. Internet penetration can be considered as one parameter to measure information hunger in the world. So I spent sometime yesterday gathering the data about this and created a heat map.
Mapping Internet Penetration Rates in World Countries
I took the data about internet penetration from Internet World Stats and used this in the World heat map excel that my friend Robert (of Excel Dashboards fame) once shared with me to create a map like this:
Download the excel file with the map of world internet penetration rates
Darker countries have less internet penetration. Sadly this map correlates highly with a world poverty map. I guess attacking the information access issues can eventually solve the poverty.
Also checkout Nathan’s visualization on poverty rates in US
14 Responses to “How many ‘Friday the 13th’s are in this year? [Formula fun + challenge]”
in C3=2016
in C4=3
in C5=1 (the first next year with three Friday the 13ths)
=SMALL(IF(MMULT(--(MOD(DATE(C3+ROW(1:1000),COLUMN(A:L),13),7)=6),ROW(1:12)^0)=C4,C3+ROW(1:1000)),C5)
formula check in the next 1000 years
This will generate a table of counts of Friday the 13th's by year. If I didn't screw it up the next year with three is 2026.
I created a simple parameter table with a start date and end date that I wanted to evaluate. That calculates the number of days and generates a list of those days. Then filter and group. The generation of the list in power query (i.e. without populating a date table in excel) is pretty cool, otherwise this isn't really doing anything than creating a big date and filtering/counting.
let
Source = List.Dates(StartDateAsDate, Days2, #duration(1,0,0,0)),
ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
AddDayOfMonthColumn = Table.AddColumn(ConvertDateListToTable, "DayOfMonth", each Date.Day([Column1])),
AddYearColumn = Table.AddColumn(AddDayOfMonthColumn, "Year", each Date.Year([Column1])),
AddDayOfWeekColumn = Table.AddColumn(AddYearColumn, "Day of Week", each Date.DayOfWeek([Column1])),
FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
in
Friday13thsByYear
With the parameters replaced by values should you want to play along at home. This runs for 20 years starting on 1/1/2016.
let
Source = List.Dates(#date(2016,1,1), 7300, #duration(1,0,0,0)),
ConvertDateListToTable = Table.FromList(Source, Splitter.SplitByNothing(), null, null, ExtraValues.Error),
AddDayOfMonthColumn = Table.AddColumn(ConvertDateListToTable, "DayOfMonth", each Date.Day([Column1])),
AddYearColumn = Table.AddColumn(AddDayOfMonthColumn, "Year", each Date.Year([Column1])),
AddDayOfWeekColumn = Table.AddColumn(AddYearColumn, "Day of Week", each Date.DayOfWeek([Column1])),
FilterFriday13 = Table.SelectRows(AddDayOfWeekColumn, each ([DayOfMonth] = 13) and ([Day of Week] = 5)),
Friday13thsByYear = Table.Group(FilterFriday13, {"Year"}, {{"Number of Friday the 13ths!", each Table.RowCount(_), type number}})
in
Friday13thsByYear
=MATCH(3,MMULT(N(WEEKDAY(DATE(C3+ROW(1:100)-1,COLUMN(A:L),13))=6),1^ROW(1:12)),)+C3-1
It should be pointed out that Alex's solution, unlike some others, has the additional advantage of being non-array. My solution was nearly identical but with -- and SIGN instead of N and 1^.
=C3-1+MATCH(3,MMULT(--(WEEKDAY(DATE(C3-1+ROW(1:25),COLUMN(A:L),13))=6),SIGN(ROW(1:12))),0)
Sub Friday13()
Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim r As Long
Range("C7:C12").ClearContents
StartDate = CDate("01/01/" & Range("C3"))
EndDate = CDate("31/12/" & Range("C3"))
r = 7
For x = StartDate To EndDate
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Cells(r, 3) = Month(x)
r = r + 1
End If
Next
End Sub
Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3
Sub ThreeFriday13()
Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim WhatYear As Integer
Dim Counter As Integer
Range("E7").ClearContents
StartDate = CDate("01/01/" & Range("C3") + 1)
EndDate = CDate("31/12/" & Range("C3") + 100)
Counter = 0
For x = StartDate To EndDate
If WhatYear Year(x) Then
WhatYear = Year(x)
'Different year so reset counter
Counter = 0
End If
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Counter = Counter + 1
If Counter = 3 Then
WhatYear = Year(x)
Exit For
End If
End If
Next
Range("E7") = WhatYear
End Sub
*RE-POST as not equal did not show earliuer
Calculate next year with 3 Friday 13th. Good for 100 years different from year entered in cell C3
Sub ThreeFriday13()
Dim StartDate As Date
Dim EndDate As Date
Dim x As Long
Dim WhatYear As Integer
Dim Counter As Integer
Range("E7").ClearContents
StartDate = CDate("01/01/" & Range("C3") + 1)
EndDate = CDate("31/12/" & Range("C3") + 100)
Counter = 0
For x = StartDate To EndDate
If WhatYear NE Year(x) Then
WhatYear = Year(x)
'Different year so reset counter
Counter = 0
End If
If Day(x) = 13 And Weekday(x, vbMonday) = 5 Then
Counter = Counter + 1
If Counter = 3 Then
WhatYear = Year(x)
Exit For
End If
End If
Next
Range("E7") = WhatYear
End Sub
earlier*
I've a doubt with using array formula here.
In sample workbook, I tried to replicate the formula again.
=IFERROR(SMALL(IF(WEEKDAY(DATE($C$3,ROW($A$1:$A$12),13))=6,ROW($A$1:$A$12)),$B7),"")
For this I selected C7 to C12, and typed the same formula and pressed ctrl+alt+Enter. But in all cells it is taking $B7 (and not $B7, $B8, $B9.... etc)
and since it is array formula I can't edit individual cell.
Please guide.
Thanks
Hi Chandoo,
Cool stuff. You need to clarify that the answer of 5 represents the 1st month in the year that has a Friday the 13th, and not the number of Fridays the 13th in the year. Subtle, but important difference.
Thanks,
Pablo
I like the MMULT() function far more, but here's how I would have tackled it. It uses an EDATE() base and MODE() over 100 years. I'm assuming that 100 years is enough time to catch the next year with 3 friday 13th's. Array entered, of course.
{=MODE(IFERROR(YEAR(IF((WEEKDAY(EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))))=6), EDATE(DATE(C3, 1, 13), ROW(INDIRECT("1:1200"))), "")), ""))}
Finding all the Friday the 13ths in a Year:
=SUMPRODUCT((DAY(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))))=13)*(TEXT(ROW(INDIRECT(DATE(C3,1,1)&":"&DATE(C3,12,31))),"ddd")="Fri"))
{=sum(if(day.of.week(DATe($YEAR;{1;2;3;4;5;6;7;8;9;10;11;12};13);1)=6;1;0))}
just list the years