How to count occurrences of Friday 13th
6
$begingroup$
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 4 hours ago
J. M. is computer-less♦
97.1k10303463
asked 11 hours ago
BradPeterson87BradPeterson87
333
New contributor
BradPeterson87 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
6
$begingroup$
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 4 hours ago
J. M. is computer-less♦
97.1k10303463
asked 11 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
10 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours ago
add a comment |
6
6
6
$begingroup$
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 4 hours ago
J. M. is computer-less♦
97.1k10303463
asked 11 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 4 hours ago
J. M. is computer-less♦
97.1k10303463
asked 11 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 4 hours ago
J. M. is computer-less♦
97.1k10303463
asked 11 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 4 hours ago
J. M. is computer-less♦
97.1k10303463
edited 4 hours ago
J. M. is computer-less♦
97.1k10303463
edited 4 hours ago
J. M. is computer-less♦
97.1k10303463
97.1k10303463
asked 11 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 11 hours ago
BradPeterson87BradPeterson87
333
asked 11 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
10 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours 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
10 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours 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
10 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
10 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
7
$begingroup$
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 10 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
9
$begingroup$
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 10 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
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
7
$begingroup$
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 10 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
7
$begingroup$
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 10 hours ago
Kuba♦Kuba
106k12205527
$endgroup$
add a comment |
7
7
7
$begingroup$
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 10 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 10 hours ago
Kuba♦Kuba
106k12205527
answered 10 hours ago
Kuba♦Kuba
106k12205527
answered 10 hours ago
Kuba♦Kuba
106k12205527
answered 10 hours ago
Kuba♦Kuba
106k12205527
106k12205527
add a comment |
add a comment |
9
$begingroup$
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 10 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
add a comment |
9
$begingroup$
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 10 hours ago
m_goldbergm_goldberg
87.2k872197
$endgroup$
add a comment |
9
9
9
$begingroup$
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 10 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 10 hours ago
m_goldbergm_goldberg
87.2k872197
answered 10 hours ago
m_goldbergm_goldberg
87.2k872197
answered 10 hours ago
m_goldbergm_goldberg
87.2k872197
answered 10 hours ago
m_goldbergm_goldberg
87.2k872197
87.2k872197
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
10 hours ago
$begingroup$
Wolfram Challenges.
$endgroup$
– J. M. is computer-less♦
4 hours ago