The symmetry of months

Question

Introduction

Some months are completely symmetric, meaning they have central symmetry as well as reflection symmetry, like February of 2010:

 February 2010 ┌──┬──┬──┬──┬──┬──┬──┐ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ └──┴──┴──┴──┴──┴──┴──┘

Some months have only central symmetry, like February of 1996 or current month, the April of 2018:

 February 1996 ┌──┬──┬──┬──┐ │ │ │ │ │ ┌──┬──┬──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┴──┴──┘ │ │ │ │ │ └──┴──┴──┴──┘ April 2018 ┌──┐ │ │ ┌──┬──┬──┬──┬──┬──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┴──┴──┴──┴──┴──┘ │ │ └──┘

And some are asymmetric, like the previous month, the March of 2018:

 March 2018 ┌──┬──┬──┬──┐ │ │ │ │ │ ┌──┬──┬──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┤ │ │ │ │ │ │ │ │ ├──┼──┼──┼──┼──┼──┼──┘ │ │ │ │ │ │ │ └──┴──┴──┴──┴──┴──┘

Task

Take an input in form of a date, e.g.:

2018.04
2018.03
2010.02
1996.02

Output the corresponding symmetry, e.g.

2018.04 -> centrally symmetric
2018.03 -> asymmetric
2010.02 -> symmetric
1996.02 -> centrally symmetric

Rules

This is code golf, so the smallest number of bytes wins.
Standard loopholes are obviously not allowed.
Assume that the week starts with Monday (thanks to Angs and Arnauld for suggestion).
Consider only years between 1900 and 2100 (inclusive).
The input and output formatting rules are permissive, meaning you can use any equivalent format that is native to the language of your choice.
Base your solution on the Gregorian calendar.

Consider that dates are weird, you may want to specify exactly the rules, or limit the possible input to a small range (say, 1901-2099) — user202729
– user202729, Commented Apr 1, 2018 at 8:05
Things to avoid when writing challenges/Adding unnecessarily stuff includes "Making answers calculate f(x) for every x in a list." What about "take an input in form of a date"? — user202729
– user202729, Commented Apr 1, 2018 at 8:08
Welcome to PPCG, and nice first challenge! Although this challenge is good, in the future if you want some feedback on the challenge before posting, you can post it in the sandbox. — user202729
– user202729, Commented Apr 1, 2018 at 8:11
Should the output be strictly the mentioned strings, or any 3 distinct values? — Uriel
– Uriel, Commented Apr 1, 2018 at 9:03
(wait a minute, Gregorian calendar or Julian calendar? I suggested [1901-2099] but you decide to use [1900-2100] so they're different for some inputs) — user202729
– user202729, Commented Apr 2, 2018 at 16:14

Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (ES6), 55 bytes

Saved 6 bytes thanks to @Neil

Takes input in currying syntax (year)(month). Returns false for asymmetric, true for centrally symmetric and 0 for completely symmetric.

y=>m=>(n=(g=_=>new Date(y,m--,7).getDay())()+g())&&n==7

Try it online!

How?

We define the function g() which returns the weekday of yyyy/mm/01, as an integer between 0 = Monday and 6 = Sunday.

g = _ => new Date(y, m--, 7).getDay()

Because getDay() natively returns 0 = Sunday to 6 = Saturday, we shift the result to the expected range by querying for the 7th day instead.

Then we define:

n = g() + g()

Because the constructor of Date expects a 0-indexed month and because g() decrements m after passing it to Date, we actually first compute the weekday of the first day of the next month and then add that of the current one.

Completely symmetric months

Completely symmetric months start with a Monday and are followed by a month which also starts with a Monday. This is only possible for February of a non-leap year.

- Feb -------------- - Mar -------------- Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su -------------------- -------------------- 01 02 03 04 05 06 07 01 02 03 04 05 06 07 08 09 10 11 12 13 14 08 09 10 11 12 13 14 15 16 17 18 19 20 21 15 16 17 18 19 20 21 22 23 24 25 26 27 28 22 23 24 25 26 27 28 29 30 31

This leads to n = 0.

Centrally symmetric months

Centrally symmetric months are months for which the sum of the weekday of their first day and that of the next month is 7.

- M ---------------- - M+1 -------------- Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su -------------------- -------------------- 0 1 [2] 3 4 5 6 0 1 2 3 4 [5] 6 -------------------- -------------------- 01 02 03 04 05 01 02 06 07 08 09 10 11 12 03 04 05 06 07 07 09 13 14 15 16 17 18 19 ... 20 21 22 23 24 25 26 27 28 29 30 31

Hence the second test: n == 7.

No built-in, 93 bytes

Uses Zeller's congruence. Same I/O format as the other version.

y=>m=>(n=(g=_=>(Y=y,((m+(m++>2||Y--&&13))*2.6|0)+Y+(Y>>2)-6*~(Y/=100)+(Y>>2))%7)()+g())&&n==7

Try it online!

I thought it was true, false and filenotfound instead of 0… — Angs
– Angs, Commented Apr 1, 2018 at 10:00

Razvan Socol · Accepted Answer · 2018-04-01 12:02:12Z

T-SQL, 213 bytes (strict I/O rules)

SET DATEFIRST 1SELECT CASE WHEN a+b<>8THEN'a'WHEN a=1THEN''ELSE'centrally 'END+'symetric'FROM(SELECT DATEPART(DW,f)a,DATEPART(DW,DATEADD(M,1,f)-1)b FROM (SELECT CONVERT(DATETIME,REPLACE(s,'.','')+'01')f FROM t)y)x

The above query considers the strict input/output formatting rules.

The input is taken from the column s of a table named t:

CREATE TABLE t (s CHAR(7)) INSERT INTO t VALUES ('2018.04'),('2018.03'),('2010.02'),('1996.02')

Ungolfed:

SET DATEFIRST 1 SELECT *, CASE WHEN a+b<>8 THEN 'a' WHEN a=1 AND b=7 THEN '' ELSE 'centrally ' END+'symetric' FROM ( SELECT *,DATEPART(WEEKDAY,f) a, DATEPART(WEEKDAY,DATEADD(MONTH,1,f)-1) b FROM (SELECT *,CONVERT(DATETIME,REPLACE(s,'.','')+'01')f FROM t)y ) x

SQLFiddle 1

T-SQL, 128 bytes (permissive I/O rules)

SET DATEFIRST 1SELECT CASE WHEN a+b<>8THEN 1WHEN a=1THEN\END FROM(SELECT DATEPART(DW,d)a,DATEPART(DW,DATEADD(M,1,d)-1)b FROM t)x

If the format of the input and of the output can be changed, I would choose to input the first day of the month, in a datetime column named d:

CREATE TABLE t (d DATETIME) INSERT INTO t VALUES ('20180401'),('20180301'),('20100201'),('19960201')

The output would be 1 for asymetric, 0 for symetric, NULL for centrally symetric.

If we can run it on a server (or with a login) configured for BRITISH language, we can remove SET DATEFIRST 1 saving 15 more bytes.

SQLFiddle 2

Nice work. Not sure if it'll work in all versions, but on SQL 2012 I was able to save 15 bytes by using CONVERT(DATETIME,s+'.01') instead of REPLACE. You can also drop the space in FROM (SELECT — BradC
– BradC, Commented Apr 3, 2018 at 14:58
It works, but it's dependant on the DATEFORMAT setting. For example, if we use SET LANGUAGE BRITISH, then CONVERT(DATETIME,'2018.02.01') would be January 2nd, instead of February 1st. — Razvan Socol
– Razvan Socol, Commented Apr 3, 2018 at 16:46

Angs · Accepted Answer · 2018-04-01 09:20:38Z

Haskell, 170 bytes

import Data.Time.Calendar import Data.Time.Calendar.WeekDate a%b=((\(_,_,a)->a).toWeekDate.fromGregorian a b$1)!gregorianMonthLength a b 1!28=2 4!29=1 7!30=1 3!31=1 _!_=0

Returns 2 for centrally symmetric, 1 for symmetric and 0 for asymmetric

@TuukkaX Sorry for confusion - this is my first challenge, I've changed the rules so they also allow permissive output formats so it can be more "in spirit" of the code-golf. — mkierc
– mkierc, Commented Apr 1, 2018 at 9:38

pwn'cat · Accepted Answer · 2018-04-03 11:23:58Z

5

Python 2, 118 104 bytes

Thanks to Jonathan Allan and Dead Possum for the improvements!

from calendar import* def f(*d):_=monthcalendar(*d);print all(sum(_,[]))+(_[0].count(0)==_[-1].count(0))

Python 3, 122 105 bytes

from calendar import* def f(*d):_=monthcalendar(*d);print(all(sum(_,[]))+(_[0].count(0)==_[-1].count(0)))

Input

First is the year
Second is the month

Output

0 = no symmetry
1 = central symmetry
2 = complete symmetry

edited Apr 3, 2018 at 11:23

answered Apr 1, 2018 at 17:30

pwn'cat

1511 silver badge5 bronze badges

3

\$\begingroup\$ Welcome to the site! You may not assume that input is stored in a variable (such as Y or M), so this is currently a snippet and invalid. If you change the variables to calls to input() however, this will be perfectly fine. \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2018-04-01 17:45:57 +00:00
Commented Apr 1, 2018 at 17:45
1

\$\begingroup\$ @cairdcoinheringaahing Thanks for the welcome! Fixed user input :) \$\endgroup\$

pwn'cat
– pwn'cat

2018-04-01 18:11:39 +00:00
Commented Apr 1, 2018 at 18:11
\$\begingroup\$ Welcome! Tweaks for -9 bytes here - all import, unpacked input, _[0]+_[-1]->sum(..) \$\endgroup\$

Dead Possum
– Dead Possum

2018-04-02 12:37:20 +00:00
Commented Apr 2, 2018 at 12:37
1

\$\begingroup\$ Some tricks to get it down 13 bytes here \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2018-04-02 13:04:35 +00:00
Commented Apr 2, 2018 at 13:04
1

\$\begingroup\$ ...and another byte using Dead Possum's sum trick - here \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2018-04-02 13:19:27 +00:00
Commented Apr 2, 2018 at 13:19

Add a comment |

Galen Ivanov · Accepted Answer · 2018-04-03 08:06:17Z

Red, 199, 168 161 bytes

func[d][t: split d"."y: do t/1 m: do t/2 a: to-date[1 m y]b: a + 31 b/day: 1 b: b - 1 if(1 = s: a/weekday)and(7 = e: b/weekday)[return 1]if 8 - e = s[return 2]0]

Try it online!

0 - asymmetric

1 - symmetric

2 - centrally symmetric

Mathematica, 137 bytes

a=#~DateValue~"DayName"&;b=a/@{2}~DateRange~{3};Which[{##}=={0,6},1,+##>5,0,1>0,-1]&@@(Position[b,a@#][[1,1]]~Mod~7&)/@{{##},{#,#2+1,0}}&

Pure function. Takes the year and the month as input and returns -1 for asymmetric months, 0 for centrally symmetric months, and 1 for fully symmetric months. Not sure why this language can't convert from a weekday to a number by default...

Digital Trauma · Accepted Answer · 2018-04-03 00:31:08Z

Bash + GNU utilities, 70

date -f- +%u<<<"$1/1+1month-1day $1/1"|dc -e??sad8*la-55-rla+8-d*64*+p

Input is formatted as YYYY/MM.

Output is numeric, as follows:

less than 0: centrally symmetric
exactly 0: symmetric
greater than 0: asymmetric

I assume this output format is acceptable for this question.

Try it online!

tsh · Accepted Answer · 2018-04-03 06:20:45Z

C, 111 bytes

a;g(y,m){y-=a=m<3;return(y/100*21/4+y%100*5/4+(13*m+16*a+8)/5)%7;}f(y,m){a=g(y,m)+g(y,m+1);return(a>0)+(a==7);}

Invoke f(year, month), 0 for completely symmetric, 1 for asymmetric, 2 for centrally symmetric.

IIRC you can abuse UB on GCC by replacing return with y= (the first parameter) and falling out of the function. — Quentin
– Quentin, Commented Apr 3, 2018 at 10:48

Phil H · Accepted Answer · 2018-04-03 14:15:04Z

Perl 6, 74 bytes

{{{$_==30??2!!$_%7==2}(2*.day-of-week+.days-in-month)}(Date.new("$_-01"))}

Bare block, implicitly a function of 1 argument, a string like "2012-02". Returns:

2 # Fully symmetric True # Centrally symmetric False # Asymmetric

When the pattern is symmetric, as .day-of-week increases by 1, .days-in-month would need to move by 2 in order to still match (the month would be starting a day later but need to end a day earlier), so 2 * .day-of-week + .days-in-month gives us a measure of that gap. Modulo 7 it should be 1 to get symmetry, but we can first cheaply check for the non-leap February by checking that total before modulo (Monday and 28 days per month is the minimum possible combination).

I'm surprised that this takes so many bytes, but fully 36 bytes are needed for just making a date and getting the day of the week and days in that month.

Stack Exchange Network

The symmetry of months

Introduction

Task

Rules

9 Answers 9

JavaScript (ES6), 55 bytes

How?

Completely symmetric months

Centrally symmetric months

No built-in, 93 bytes

T-SQL, 213 bytes (strict I/O rules)

T-SQL, 128 bytes (permissive I/O rules)

Haskell, 170 bytes

Python 2, 118 104 bytes

Python 3, 122 105 bytes

Input

Output

Red, 199, 168 161 bytes

More readable:

Mathematica, 137 bytes

Bash + GNU utilities, 70

C, 111 bytes

Perl 6, 74 bytes

Hot Network Questions

The symmetry of months

Introduction

Task

Rules

9 Answers 9

JavaScript (ES6), 55 bytes

How?

Completely symmetric months

Centrally symmetric months

No built-in, 93 bytes

T-SQL, 213 bytes (strict I/O rules)

T-SQL, 128 bytes (permissive I/O rules)

Haskell, 170 bytes

Python 2, 118 104 bytes

Python 3, 122 105 bytes

Input

Output

Red, 199, 168 161 bytes

More readable:

Mathematica, 137 bytes

Bash + GNU utilities, 70

C, 111 bytes

Perl 6, 74 bytes

Related

Hot Network Questions