How to count occurrences of Friday 13th
9
$begingroup$
I would like to find a function that will count the number of times Friday 13th happens in a particular calendar year.
Does anybody have any hints ?
Thank you
date-and-time
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
$endgroup$
add a comment |
9
$begingroup$
I would like to find a function that will count the number of times Friday 13th happens in a particular calendar year.
Does anybody have any hints ?
Thank you
date-and-time
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
$endgroup$
1
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11
add a comment |
9
9
9
$begingroup$
I would like to find a function that will count the number of times Friday 13th happens in a particular calendar year.
Does anybody have any hints ?
Thank you
date-and-time
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
$endgroup$
I would like to find a function that will count the number of times Friday 13th happens in a particular calendar year.
Does anybody have any hints ?
Thank you
date-and-time
date-and-time
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
edited Mar 5 at 2:20
J. M. is slightly pensive♦
97.8k10304464
97.8k10304464
asked Mar 4 at 19:23
user63250
1
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11
add a comment |
1
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11
1
1
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11
add a comment |
2 Answers
2
active
oldest
votes
10
$begingroup$
Select[
Table[DateObject@{2019, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
{Day: Fri 13 Sep 2019,Day: Fri 13 Dec 2019}
countFri13[year_Integer]:=Length @ Select[
Table[DateObject@{year, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
$endgroup$
add a comment |
13
$begingroup$
I worked on this problem in 2015. Here is part on my notebook from that time.
A not so good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @
DateRange[DateObject[{year, 1, 13}], DateObject[{year, 12, 13}], {1, "Month"}]
A good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Table[DateObject[{year, i, 13}], {i, 12}]
A better algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Array[DateObject[{year, #, 13}] &, 12]
Using the better algorithm, I got (at the time I created the notebook)
friday13th @ 2014
friday13th @ 2015
And for this year, I get
friday13th @ 2019
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
10
$begingroup$
Select[
Table[DateObject@{2019, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
{Day: Fri 13 Sep 2019,Day: Fri 13 Dec 2019}
countFri13[year_Integer]:=Length @ Select[
Table[DateObject@{year, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
$endgroup$
add a comment |
10
$begingroup$
Select[
Table[DateObject@{2019, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
{Day: Fri 13 Sep 2019,Day: Fri 13 Dec 2019}
countFri13[year_Integer]:=Length @ Select[
Table[DateObject@{year, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
$endgroup$
add a comment |
10
10
10
$begingroup$
Select[
Table[DateObject@{2019, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
{Day: Fri 13 Sep 2019,Day: Fri 13 Dec 2019}
countFri13[year_Integer]:=Length @ Select[
Table[DateObject@{year, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
$endgroup$
Select[
Table[DateObject@{2019, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
{Day: Fri 13 Sep 2019,Day: Fri 13 Dec 2019}
countFri13[year_Integer]:=Length @ Select[
Table[DateObject@{year, m, 13}, {m, 12}],
DateString[#, "DayName"] === "Friday" &
]
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
answered Mar 4 at 19:35
Kuba♦Kuba
106k12207529
106k12207529
add a comment |
add a comment |
13
$begingroup$
I worked on this problem in 2015. Here is part on my notebook from that time.
A not so good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @
DateRange[DateObject[{year, 1, 13}], DateObject[{year, 12, 13}], {1, "Month"}]
A good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Table[DateObject[{year, i, 13}], {i, 12}]
A better algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Array[DateObject[{year, #, 13}] &, 12]
Using the better algorithm, I got (at the time I created the notebook)
friday13th @ 2014
friday13th @ 2015
And for this year, I get
friday13th @ 2019
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
$endgroup$
add a comment |
13
$begingroup$
I worked on this problem in 2015. Here is part on my notebook from that time.
A not so good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @
DateRange[DateObject[{year, 1, 13}], DateObject[{year, 12, 13}], {1, "Month"}]
A good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Table[DateObject[{year, i, 13}], {i, 12}]
A better algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Array[DateObject[{year, #, 13}] &, 12]
Using the better algorithm, I got (at the time I created the notebook)
friday13th @ 2014
friday13th @ 2015
And for this year, I get
friday13th @ 2019
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
$endgroup$
add a comment |
13
13
13
$begingroup$
I worked on this problem in 2015. Here is part on my notebook from that time.
A not so good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @
DateRange[DateObject[{year, 1, 13}], DateObject[{year, 12, 13}], {1, "Month"}]
A good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Table[DateObject[{year, i, 13}], {i, 12}]
A better algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Array[DateObject[{year, #, 13}] &, 12]
Using the better algorithm, I got (at the time I created the notebook)
friday13th @ 2014
friday13th @ 2015
And for this year, I get
friday13th @ 2019
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
$endgroup$
I worked on this problem in 2015. Here is part on my notebook from that time.
A not so good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @
DateRange[DateObject[{year, 1, 13}], DateObject[{year, 12, 13}], {1, "Month"}]
A good algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Table[DateObject[{year, i, 13}], {i, 12}]
A better algorithm.
friday13th[year_Integer] :=
Select[DayName[#] === Friday &] @ Array[DateObject[{year, #, 13}] &, 12]
Using the better algorithm, I got (at the time I created the notebook)
friday13th @ 2014
friday13th @ 2015
And for this year, I get
friday13th @ 2019
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
answered Mar 4 at 19:52
m_goldbergm_goldberg
87.4k872198
87.4k872198
add a comment |
add a comment |
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1
$begingroup$
I fell into a delightful rabbit hole of day-counting algorithms on Wikipedia. I wanted to leave a link to the Doomsday algorithm for mental calculation of the day of the week, for fun: Doomsday rule on Wiki.
$endgroup$
– MarcoB
Mar 4 at 19:54
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is slightly pensive♦
Mar 5 at 2:21
$begingroup$
Sounds like one for the code golf boys and girls
$endgroup$
– Strawberry
Mar 5 at 13:05
$begingroup$
Oh, of course they did it already: codegolf.stackexchange.com/questions/69510/…
$endgroup$
– Strawberry
Mar 5 at 13:11