How to count occurrences of Friday 13thDetermining the week of a year from a given dateJewishNewYear date...
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How to count occurrences of Friday 13th
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How to count occurrences of Friday 13th
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$begingroup$
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 33 mins ago
J. M. is computer-less♦
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
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 33 mins ago
J. M. is computer-less♦
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
6 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago
add a comment |
$begingroup$
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 33 mins ago
J. M. is computer-less♦
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 33 mins ago
J. M. is computer-less♦
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 33 mins ago
J. M. is computer-less♦
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 33 mins ago
J. M. is computer-less♦
97.1k10303463
edited 33 mins ago
J. M. is computer-less♦
97.1k10303463
edited 33 mins ago
J. M. is computer-less♦
97.1k10303463
97.1k10303463
asked 7 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
BradPeterson87BradPeterson87
333
asked 7 hours ago
BradPeterson87BradPeterson87
333
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
6 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago
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
6 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 7 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
$begingroup$
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 7 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 7 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
$begingroup$
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 7 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
$begingroup$
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 7 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
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 7 hours ago
Kuba♦Kuba
106k12205527
answered 7 hours ago
Kuba♦Kuba
106k12205527
answered 7 hours ago
Kuba♦Kuba
106k12205527
answered 7 hours ago
Kuba♦Kuba
106k12205527
106k12205527
add a comment |
add a comment |
$begingroup$
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 7 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
add a comment |
$begingroup$
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 7 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
add a comment |
$begingroup$
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 7 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
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 7 hours ago
m_goldbergm_goldberg
87.2k872197
answered 7 hours ago
m_goldbergm_goldberg
87.2k872197
answered 7 hours ago
m_goldbergm_goldberg
87.2k872197
answered 7 hours ago
m_goldbergm_goldberg
87.2k872197
87.2k872197
add a comment |
add a comment |
BradPeterson87 is a new contributor. Be nice, and check out our Code of Conduct.
BradPeterson87 is a new contributor. Be nice, and check out our Code of Conduct.
BradPeterson87 is a new contributor. Be nice, and check out our Code of Conduct.
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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
6 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
33 mins ago