# Could They Be The Same Day Of The Week?

### Challenge

Given a non-negative integer, output whether it is possible for two dates (of the Gregorian calendar) differing by exactly that many years to share a day of the week. A year is assumed to be a leap year either if it is divisible by 4 but not by 100, or if it is divisible by 400.

Output may be:

• falsey/truthy (in either orientation)
• any two distinct values
• one distinct value and one being anything else
• by program return code
• by success/error
• by any other reasonable means - just ask if you suspect it may be controversial

But not by two non-distinct sets of values, except for falsey/truthy (as this would allow a no-op!)

### Detail

This is whether the input is a member of OEIS sequence A230995.

Members:

0, 5, 6, 7, 11, 12, 17, 18, 22, 23, 28, 29, 33, 34, 35, 39, 40, 45, 46, 50, 51, 56, 57, 61, 62, 63, 67, 68, 73, 74, 78, 79, 84, 85, 89, 90, 91, 95, 96, 101, 102, 106, 107, 108, 112, 113, 114, 117, 118, 119, 123, 124, 125, 129, 130, 131, 134, 135, 136, 140, 141, 142, 145, 146, 147, 151, 152, 153, 157, 158, 159, 162, 163, 164, 168, 169, 170, 173, 174, 175, 179, 180, 181, 185, 186, 187, 190, 191, 192, 196, 197, 198, 202, 203, 204, 208, 209, 210, 213, 214, 215, 219, 220, 221, 225, 226, 227, 230, 231, 232, 236, 237, 238, 241, 242, 243, 247, 248, 249, 253, 254, 255, 258, 259, 260, 264, 265, 266, 269, 270, 271, 275, 276, 277, 281, 282, 283, 286, 287, 288, 292, 293, 294, 298, 299, 304, 305, 309, 310, 311, 315, 316, 321, 322, 326, 327, 332, 333, 337, 338, 339, 343, 344, 349, 350, 354, 355, 360, 361, 365, 366, 367, 371, 372, 377, 378, 382, 383, 388, 389, 393, 394, 395
plus
400, 405, 406, 407, 411, ...


Non-members:

1, 2, 3, 4, 8, 9, 10, 13, 14, 15, 16, 19, 20, 21, 24, 25, 26, 27, 30, 31, 32, 36, 37, 38, 41, 42, 43, 44, 47, 48, 49, 52, 53, 54, 55, 58, 59, 60, 64, 65, 66, 69, 70, 71, 72, 75, 76, 77, 80, 81, 82, 83, 86, 87, 88, 92, 93, 94, 97, 98, 99, 100, 103, 104, 105, 109, 110, 111, 115, 116, 120, 121, 122, 126, 127, 128, 132, 133, 137, 138, 139, 143, 144, 148, 149, 150, 154, 155, 156, 160, 161, 165, 166, 167, 171, 172, 176, 177, 178, 182, 183, 184, 188, 189, 193, 194, 195, 199, 200, 201, 205, 206, 207, 211, 212, 216, 217, 218, 222, 223, 224, 228, 229, 233, 234, 235, 239, 240, 244, 245, 246, 250, 251, 252, 256, 257, 261, 262, 263, 267, 268, 272, 273, 274, 278, 279, 280, 284, 285, 289, 290, 291, 295, 296, 297, 300, 301, 302, 303, 306, 307, 308, 312, 313, 314, 317, 318, 319, 320, 323, 324, 325, 328, 329, 330, 331, 334, 335, 336, 340, 341, 342, 345, 346, 347, 348, 351, 352, 353, 356, 357, 358, 359, 362, 363, 364, 368, 369, 370, 373, 374, 375, 376, 379, 380, 381, 384, 385, 386, 387, 390, 391, 392, 396, 397, 398, 399
plus
401, 402, 403, 404, 408, ...


This is so the shortest answer in each language wins!

• Can output be: program ends (within less than 30 seconds) if input belongs to the sequence, or runs indefinitely (infinite loop) otherwise? – Luis Mendo Dec 31 '17 at 4:35
• @LuisMendo I'll allow a program that does that so long as it is accompanied with a program that provides the time limit (so one may acquire it prior for one's hardware). It is indeed controversial though :) – Jonathan Allan Dec 31 '17 at 4:57
• In which situation is a number divisible by 400 but is not divisible by 100? – ATaco Dec 31 '17 at 5:20
• @ATaco In none. The exception to the every fourth year rule are years that are divisible by 4 and 100, but not by 400. – Dennis Dec 31 '17 at 5:27
• @ATaco maybe the wording is clearer now – Jonathan Allan Dec 31 '17 at 5:38

# MATL, 17 bytes

0Gv@+5:YcYO8XOda


The program halts if the input belongs to the sequence, or runs indefinitely (infinite loop) otherwise.

Let n be the input. The code executes a loop that tests years 1 and 1+n; then 2 and 2+n; ... until a matching day of the week is found. If no matching exists the loop runs indefinitely.

The membership function for n is periodic with period 400. Therefore, at most 400 iterations are needed if n belongs to the sequence. This requires less than 20 seconds in Try It Online. As a proof of this upper bound, here's a modified program that limits the number of iterations to 400 (by adding @401<* at the end). Note also that this bound is loose, and a few seconds usually suffice.

Try it online!

### Explanation

           % Do...while
0Gv       %   Push column vector [0; n], where n is the input number
@+        %   Add k, element-wise. Gives [k; k+n]
5:        %   Push row vector [1, 2, 3, 4, 5]
Yc        %   Horizontal "string" concatenation: gives the 2×6 matrix
%   [k, 1, 2, 3, 4, 5; k+n, 1, 2, 3, 4, 5]. The 6 columns
%   represent year, month, day, hour, minute, second
YO        %   Convert each row to serial date number. Gives a column
%   vector of length 2
8XO       %   Convert each date number to date string with format 8,
%   which is weekday in three letters ('Mon', 'Tue', etc).
%   This gives a 2×3 char matrix such as ['Wed';'Fri']
d         %   Difference (of codepoints) along each column. Gives a
%   row vector of length 3
a         %   True if some element is nonzero, or false otherwise
% End (implicit). The loop proceeds with the next iteration
% if the top of the stack is true


## Old version, 24 bytes

400:"0G&v@+5:YcYO8XOdavA


Output is 0 if the input belongs to the sequence, or 1 otherwise.

Try it online!

### Explanation

400         % Push row vector [1, 2, ..., 400]
"           % For each k in that array
0G&v      %   Push column vector [0; n], where n is the input number
@+        %   Add k, element-wise. Gives [k; k+n]
5:        %   Push row vector [1, 2, 3, 4, 5]
Yc        %   Horizontal "string" concatenation: gives the 2×6 matrix
%   [k, 1, 2, 3, 4, 5; k+n, 1, 2, 3, 4, 5]. The 6 columns
%   represent year, month, day, hour, minute, second
YO        %   Convert each row to serial date number. Gives a column
%   vector of length 2
8XO       %   Convert each date number to date string with format 8,
%   which is weekday in three letters ('Mon', 'Tue', etc).
%   This gives a 2×3 char matrix such as ['Wed';'Fri']
d         %   Difference (of codepoints) along each column. Gives a
%   row vector of length 3
a         %   True if some element is nonzero, or false otherwise
v         %   Concatenate vertically with previous results
A         %   True if all results so far are true
% End (implicit). Display (implicit)

• Seems good, my request was really that I'd like to know the worst case input, or have a program which forces the 400 iterations -- that way one could get an upper bound wherever one chooses to run it. (BTW I think the infinite loop is, in practice, ended by an out of bounds error.) – Jonathan Allan Dec 31 '17 at 18:35
• @JonathanAllan Thanks. I see. I've added a modified program that limits the number of iterations to 400. It takes about 14 seconds, so I'm using 20 seconds as upper bound – Luis Mendo Dec 31 '17 at 18:59

# Python 2, 58 bytes

u=-abs(200-input()%400)-4
print u/100+5>(u-8)*5/4%7>u%4/-3


Try it online!

A direct formula.

• This is nice. I believe you can save 2 bytes with 5*u/4%7-3 instead of (u-8)*5/4%7. – Jonathan Allan Dec 31 '17 at 3:35
• Save 2 more by using the success/error option with 1/(...) instead of print .... – Jonathan Allan Dec 31 '17 at 4:01

# Jelly, 20 18 bytes

99R4ḍj‘ṡ%4ȷ$S€P7ḍ  Outputs 1 for members, 0 for non-members. Try it online! ### How it works 99R4ḍj‘ṡ%4ȷ$S€P7ḍ  Main link. Argument: n

99                  Set the return value to 99.
R                 Range; yield [01, .., 99].
4ḍ               Test each element in the range for divisibility by 4.
j             Join the resulting array, using itself as separator.
The result is an array of 9801 Booleans indicating whether the
years they represent have leap days.
‘            Increment the results, yielding 1 = 365 (mod 7) for non-leap
years, 2 = 366 (mod 7) for leap years.
%4ȷ$Compute n % 4000. ṡ Take all slices of length n % 4000 of the result to the left. S€ Take the sum of each slice. P Take the product of the sums. 7ḍ Test for divisibility by 7.  # Python 2, 83 bytes lambda i:all(c(y)-c(y-i)for y in range(401)) c=lambda n:(5*(n/4)+n%4-n/100+n/400)%7  Try it online! Direct port of my Haskell answer. # Haskell, 76 bytes -35 bytes thanks to Jonathan Allan. -2 bytes thanks to Lynn. f i=or[c y==c$y+i|y<-[0..400]]
c n=(5*n#4+n%4-n#100+n#400)%7
(%)=mod
(#)=div


Try it online!

Using the OEIS PARI program's algorithm.

• 5*(n#4) can be 5*n#4 also! – Lynn Dec 30 '17 at 22:35

# Pyth, 32 bytes

iI7*FsM.:hMs.iKiIL4S99*98]K%Q400


Try it here! (Click "Switch to Test Suite" to verify more test cases at once)

### How?

Uses a cool trick I just added to the "Tips for golfing in Pyth" thread.

iI7*FsM.:hMs.iKiIL4S99*98]K%Q400 | Full program. Reads from STDIN, outputs to STDOUT.

S99           | Generate the integers in 1 ... 99.
L               | For each integer N in that list...
iI 4              | Check if 4 is invariant over applying GCD with N.
| This is equivalent to checking if 4 | N.
K                  | Store the result in a variable K.
.i         *98]K     | And interleave K with the elements of K wrapped
| into a list and repeated 98 times.
s                     | Flatten.
hM                      | Increment.
.:                        | And generate all the substrings...
%Q400 | Of length  % 400.
sM                          | Sum each.
*F                            | And apply folded product.
iI7                              | Check if 7 is invariant when applied GCD with the
| product (basically check whether 7 | product).
| Implicitly output the appropriate boolean value.


# Python 3, 110 107 bytes

from datetime import*
lambda n,d=date:0in[d(i,1,1).weekday()-d(i+n%400,1,1).weekday()for i in range(1,999)]
`

Try it online!

-3 bytes thanks to Mr. Xcoder.