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Rev

Required fields*

Day of the week of an ambiguous date

Goal

Write a program or function that returns the day of the week for a date, eg.

01/06/2020 -> Mon

However, it's unknown if the date is in the format mm/dd/yyyy or dd/mm/yyyy. If you can be certain of the day of the week, return it. If there is uncertainty, return an error.

02/07/2020 -> Err (Thu? Fri?)

Input

A date in the format #/#/####.

The numbers can optionally have leading zeros.
The year should be read verbatim, assume that 19 = 19 AD, not 19 = 2019 AD.
The separators can be any non-digit character.
Assume the input date is valid in at least one format (ie. not 13/13/2020).
At a minimum, support dates from 2000 through 2030 (which are likely covered by any modern OS and language with a date library.)

Requirements

If the first and second numbers:
- are the same (03/03/2020), the date is unambiguous.
- either is greater than 12 (13/09/2020, 09/13/2020), the date is unambiguous.
- are different and both 12 or less (09/11/2020), the date is ambiguous.

Output

If the date is unambiguous, return the day of the week.
If the date is ambiguous:
- If both versions of the date have the same day of the week, return the day of the week.
- If both versions of the date have a different day of the week, return an error.
A day of the week should be returned as a string of the day's three-letter abbreviation ("Mon"), case-insensitive.
An error should be returned as any other three-character string, not all whitespace, eg. "Err", "NaN", "???".

Test Cases

02/07/2020 -> Err
01/06/2020 -> Mon
05/05/2020 -> Tue
24/06/2020 -> Wed
05/09/2030 -> Thu
03/13/2020 -> Fri
29/02/2020 -> Sat
03/12/2000 -> Sun

Scoring

The code with the least byte count in a week wins the tick.

Answer*

# [05AB1E](https://github.com/Adriandmen/05AB1E/wiki/Commands), <s>88</s> <s>82</s> <s>79</s> <s>76</s> <s>79</s> <s>73</s> 72 [bytes](https://github.com/Adriandmen/05AB1E/wiki/Codepage)

#ÂÀ‚Dε`UÐ3‹12*+>₂*T÷s3‹Xα¬Ésт%D4÷O.•A±₁γβCüIÊä6’C•3ôsè}DËiнës€¨13‹WiтëθÏ

-3 bytes thanks to *@Arnauld* (initially -6, but +3 bytes again, since Jan./Feb. 2000 would still result in the previous century in the formula). 
-7 bytes thanks to *@Grimmy*.

Input is space-separated; and outputs in lowercase with `100` as error.

[Try it online](https://tio.run/##AYoAdf9vc2FiaWX//yPDgsOA4oCaRM61YFXDkDPigLkxMiorPuKCgipUw7dzM@KAuVjOscKsw4lz0YIlRDTDt08u4oCiQcKx4oKBzrPOskPDvEnDisOkNuKAmUPigKIzw7Rzw6h9RMOLadC9w6tz4oKswqgxM@KAuVdp0YLDq864w4///zAxIDA2IDIwMjA) or [verify all test cases](https://tio.run/##yy9OTMpM/V9Waa@k8KhtkoKSfaXSf@XDTYcbHjXMcjm3NSH08ATjRw07DY20tO0eNTVphRzeXgwSiDi38dCaw53FF5tUXUwOb/fXe9SwyPHQxkdNjec2n9vkfHiP5@Guw0vMHjXMdAbKGB/eUnx4Ra3L4e7MC3sPry5@1LTm0ApDkDnhmRebDq8@t@Nw/38lvTCd/9FKBkYKBuYKRgZGBko6SgaGCgZmcI6pAhBBOUYmaDKWQI4xmGOsYGgMV2apADQQpgwoA@IYgDlA3SbI9hjCZGB6gJxYAA).

**Explanation:**

Since 05AB1E has no Date builtins, we'll have to calculate the *Day of the Week* manually.

The general formula to do this is:

$${\displaystyle h=\left(q+\left\lfloor{\frac{13(m+1)}{5}}\right\rfloor+K+\left\lfloor{\frac{K}{4}}\right\rfloor+\left\lfloor{\frac{J}{4}}\right\rfloor-2J\right){\bmod{7}}}$$

Where for the months March through December:

- \$q\$ is the \$day\$ of the month (`[1, 31]`)
- \$m\$ is the 1-indexed \$month\$ (`[3, 12]`)
- \$K\$ is the year of the century (\$year \bmod 100\$)
- \$J\$ is the 0-indexed century (\$\left\lfloor {\frac {year}{100}}\right\rfloor\$)

And for the months January and February:

- \$q\$ is the \$day\$ of the month (`[1, 31]`)
- \$m\$ is the 1-indexed \$month + 12\$ (`[13, 14]`)
- \$K\$ is the year of the century for the previous year (\$(year - 1) \bmod 100\$)
- \$J\$ is the 0-indexed century for the previous year (\$\left\lfloor {\frac {year-1}{100}}\right\rfloor\$)

Resulting in in the day of the week \$h\,ドル where 0 = Saturday, 1 = Sunday, ..., 6 = Friday. 
[Source: Zeller's congruence](https://en.wikipedia.org/wiki/Zeller%27s_congruence)

Since the input is guaranteed to be in the year-range \$[2000, 2030]\,ドル we can modify \$+ \left\lfloor{\frac{J}{4}}\right\rfloor - 2J\$ to a hard-coded \$-35\$ to save some bytes. EXCEPT, when it's Jan./Feb. 2000, in which case it's the previous century. For those, we use `¬És`, to add 1 to our hard-coded replacement of \$J\$. 
In addition, we won't need the \${\bmod {7}}\,ドル since we only use it to index into the string list, and 05AB1E uses modular indexing anyway.

We can save 2 more bytes, by just removing the \$-35\$ all together, since it's a multiple of 7 anyway. 
And one more byte, by changing the \$\left\lfloor{\frac{13(m+1)}{5}}\right\rfloor\$ to \$\left\lfloor{\frac{26(m+1)}{10}}\right\rfloor\,ドル since 05AB1E has a single-byte builtin for `26` and `10` (which are `₂` and `T` respectively).

***Code explanation:***

# # Split the (implicit) input-string by spaces
 Â # Bifurcate it (short for Duplicate & Reverse copy)
 À # Rotate the reversed copy once towards the left
 ‚ # Pair the top two values together
 # (we now have a pair [[day,month,year],[month,day,year]])
 D # Duplicate it
 ε # Map over both values in this pair:

Now comes Zeller's congruence into play:

` # Push the day, month, and year separated to the stack
 U # Pop and save the year in variable `X`
 Ð # Triplicate the month
 3‹ # Check if the month is below 3 (Jan. / Feb.),
 # resulting in 1 or 0 for truthy/falsey respectively
 12* # Multiply this by 12 (either 0 or 12)
 + # And add it to the month
 # This first part was to make Jan. / Feb. 13 and 14

> # Month + 1
 ₂* # Multiplied by 26
 T÷ # Integer-divided by 10
 s3‹ # Check if the month is below 3 again (resulting in 1 / 0)
 Xα # Take the absolute difference with the year
 ¬ # Push the first digit (without popping the mYear itself)
 É # Check if its odd (1 if 1; 0 if 2)
 s # Swap to take the mYear again
 т% # mYear modulo-100
 D4÷ # mYear modulo-100, integer-divided by 4
 O # Take the sum of all values on the stack

And then we'll continue:

.•A±₁γβCüIÊä6’C•
 # Push compressed string "satsunmontuewedthufri"
 3ô # Split it into parts of size 3: ["sat","sun","mon","tue","wed","thu","fri"]
 s # Swap to get the earlier calculated number
 è # And index it into this list (0-based and with wrap-around)
 }D # After the map: duplicate the resulting two days
 Ëi # If they are both the same:
 н # Leave just one of them
 ë # Else (they are different):
 s # Swap, to take the initially duplicated pair of 
 # [[day,month,year],[month,day,year]]
 €¨ # Remove the year from each
 13‹ # Check for the remaining values if they are smaller than 13
 W # Push the flattened minimum (without popping)
 i # If it's truthy (so day and month are both below 13):
 т # Push 100 (as error value)
 ë # Else:
 θ # Pop and leave just the last one (either [0,1] or [1,0])
 Ï # And only keep the day at the truthy index
 # (after which the top of the stack is output implicitly)

[See this 05AB1E tip of mine (section *How to compress strings not part of the dictionary?*)](https://codegolf.stackexchange.com/a/166851/52210) to understand why `.•A±₁γβCüIÊä6’C•` is `"satsunmontuewedthufri"`.

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one. Explanations of your answer make it more interesting to read and are very much encouraged.
…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).

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\$\begingroup\$ "Look mah" would look less "mah" as "Looks meh" :D I would edit it in, but I'm not sure if the "mah" is the "Dutch version" of "meh" :) \$\endgroup\$

RGS
– RGS

2020年03月19日 09:59:48 +00:00
Commented Mar 19, 2020 at 9:59
1

\$\begingroup\$ You can probably get rid of \$J\$ since we're only asked to support dates from 2000 through 2030. \$\endgroup\$

Arnauld
– Arnauld

2020年03月19日 10:14:20 +00:00
Commented Mar 19, 2020 at 10:14
\$\begingroup\$ @Arnauld Thanks, -6 bytes by hard-coding \$+\left\lfloor {\frac {J}{4}}\right\rfloor -2J\$ to \$-35\$. :) \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2020年03月19日 11:48:20 +00:00
Commented Mar 19, 2020 at 11:48
1

\$\begingroup\$ My bad, my previous comment was wrong, I deleted it. #Ð2£TSǝ => #ÂÀ for -4. Rotate the list of days back to what it was so you can delete the -36 (-35 is a noop mod 7). \$\endgroup\$

Grimmy
– Grimmy

2020年03月19日 22:55:10 +00:00
Commented Mar 19, 2020 at 22:55
1

\$\begingroup\$ 13*5÷ => 2*T÷ for -1. \$\endgroup\$

Grimmy
– Grimmy

2020年03月20日 11:02:00 +00:00
Commented Mar 20, 2020 at 11:02

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