Is it a leap year?

Question

This challenge is quite simple. You will take an input which will be a year from 1801 to 2400, and output if it is a leap year or not.

Your input will have no newlines or trailing spaces:

1954

You will output in any way that you like that clearly tells the user if it is or isn't a leap year (I will accept y or n for yes/no)

You can get a list of leap years here: http://kalender-365.de/leap-years.php I would note that leap years are not ever four years always. 1896 is a leap year, but 1900 is not. The years that follow this "skip" are:

1900
2100
2200
2300

Test cases:

1936 ->  y
1805 ->  n
1900 ->  n
2272 ->  y
2400 ->  y 

EDIT: This is based on a standard Gregorian calendar: http://www.epochconverter.com/date-and-time/daynumbers-by-year.php

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Answer 1

APL, 16 14 12 characters

Returns 0 for a leap year, 1 for a non-leap year.

≥/⌽×4 25 4⊤⎕

Try this solution on tryapl.org. Note that I have changed the solution to the dfn {≥/⌽×4 25 4⊤⍵} as tryapl.com does not support ⎕ (take user input). Note that ⎕ is an empty box, not a missing character.

The same solution in J:

4 25 4>:/@|.@:*@#:]

Explanation

Dyadic ⊤ (encode) represents its right argument in the base specified by its left argument. I use base 4 25 4 in this solution. This represents the year y as a polynomial

y mod 400 = 100 a + 4 b + c where b < 100 and c < 4.

Let propositions α, β, and γ represent if a, b, and c are non-zero: Proposition γ is false if y is dividable by 4, β ∧ γ is false if y is dividable by 100 and α ∧ β ∧ γ is false if y is dividable by 400.

A truth table (* representing “don't care”) were proposition Δ represents if y is a leap-year obtains:

α β γ | Δ
0 0 0 | 1
1 0 0 | 0
* 1 0 | 1
* * 1 | 0

The following statement expresses Δ in α, β, and γ:

Δ = ¬((α → β) → γ)).

Due to the structure of this statement, one can express ¬Δ as the reduction ≥/⌽α β γ where &geq; implements ←. This leads to the answer I am explaining right now.

Answer 2

Pyth, 11 bytes

!%|F_jQ*TT4

This full program read from STDIN and prints True for leap years and False otherwise.

Thanks to @Jakube for suggesting Pyth and basically porting my CJam code.

Verify the test cases yourself in the Pyth Compiler/Executor.

How it works

     jQ*TT   Returns the evaluated input in base 10 × 10.
  |F_        Swaps the digit order and reduces using logical OR.
             So far, we've achieved 1954 -> [19, 54] -> 54 || 19.
!%        4  Returns the logical NOT of the result modulo 4.
             This prints True for multiples of 4 and False otherwise.

Answer 3

CJam, 12 bytes

rS+2m<~e|4%!

This full program read from STDIN and prints 1 for leap years and 0 otherwise.

Verify the test cases yourself in the CJam interpreter.

How it works

r   e# Read from STDIN.
S+  e# Append a space.
2m< e# Rotate two characters to the left.
~   e# Evaluate.
    e# So far, we achieved "1954" -> "54 19" -> 54 19.
e|  e# Logical OR; keep the leftmost non-zero integer.
4%! e# Logical NOT of the kept integer modulo 4.
    e# This pushes 1 for multiples of 4 and 0 otherwise.

Answer 4

Javascript (ES6), 21 characters

The standard rule is that y is a leap year if 4 divides y and if either 100 doesn't divide y or 400 does divide y. In code,

y%4 == 0 && (y%100 != 0 || y%400 == 0)

There's no need for that 100 and 400. Instead it suffices to check whether 16 or 4 divides y, with 16 chosen if 25 divides y, 4 otherwise. Golfed, this becomes

!(y%(y%25?4:16))

A javascript function that implements this is 21 characters long:

l=y=>!(y%(y%25?4:16))

Perl, 28 26 characters

Same idea, but in perl.

$_=$_%($_%25?4:16)?"n":"y"

Run using the -lp options. For example,

perl -lpe '$_=$_%($_%25?4:16)?"n":"y"'

With the test set as input, this produces

1936
y
1805
n
1900
n
2272
y
2400
y

Answer 5

Pyth, 19 15 14 bytes

xFm!%Q^d2[2TyT

Way too easy. Try it online: Demonstration or Test harness

edit: Missed, that you can print Truthy/Falsy values, instead of n/y. -4 byte

edit 2: Used the square root idea of Martin. -1 byte

Explanation

                 implicit: Q = input number
         [         generate a list with the numbers:
          2          2
           T         10
            yT       2*10 = 20
  m              map each number d to:
   !%Q^d2          not(Q mod d^2) // True if Q % d^2 == 0 otherwise False
xF               fold by xor

Answer 6

JavaScript (ES6) 27

The rule: (y%4==0) && (y%100!=0 || y%400==0)

Golfed: !(y%100<1&&y%400||y%4) (mainly using De Morgans's law)

A function implementing the rule:

l=y=>!(y%100<1&&y%400||y%4)

A test (run in Firefox) just to be sure:

l=y=>!(y%100<1&&y%400||y%4)

for(o=[],i=0;i<700;i++)
  y=i+1800,
  x=l(y),
  o[i/100|0]=(o[i/100|0]||'')+y+(x?' <b>Y</b>':' <i>N</i>')+'\n'
    
R.innerHTML='<td>'+o.join('</td><td>')+'</td>'
console.log(o[1])

td { white-space: pre; font-family: monospace; padding: 8px}

b { color: red; }
i { color: blue; }

<table>
  <tr id=R></tr>
</table>

Answer 7

TI-BASIC, 20 17 16 13

Because it is tokenized, TI-BASIC is often competitive at simple math challenges, but not this one since there is no "divisible" command. Maybe it is after all, but this is still longer than CJam and Pyth.

This uses David Hammond's method.

not(fPart(Ans/4/4^not(fPart(sub(Ans

Old code at 16 bytes:

not(fPart(Ans/16not(fPart(sub(Ansnot(fPart(Ans/4

Ungolfed:

not(fPart(Ans/16) and not(fPart(Ans/100) and not(fPart(Ans/4))))

fPart( is "fractional part"; exponentiation has higher precedence than division. In TI-BASIC, close-parens are optional.

I use undocumented behavior of the sub( command, usually used to get a substring: when its argument is a number instead of a string, it divides the number by 100. It will work on a TI-83 or 84 series calculator.

20 -> 17 by rearranging code to allow removal of close-parens; 17 -> 16 by replacing 400 with 16; 16 -> 13 by using David Hammond's idea.

Answer 8

Pip, 13 bytes

This one was more interesting than it at first appeared. It took some finagling, but I was finally able to replace those lengthy references to 400 with 4 and the h variable (=100).

!(a%h?aa/h)%4

Outputs 1 for leap year, 0 for non-leap year. Explanation:

               a is command-line argument (implicit)
  a%h?aa/h     If a is divisible by 100, divide it by 100; otherwise, leave it alone
 (        )%4  The result mod 4 is 0 if it's a leap year, nonzero otherwise
!              Negate and (implicitly) print

Answer 9

Regex, 83 62 38

Thanks to Toby for tips about combining both halves of the regex.

If we focus on 1801..2400 range only and assume input are integers:

(?!00)([02468][048]|[13579][26])(00)?$

Test in Ruby (^ = \A and $ = \Z because Ruby) for the desired range:

r = /(?!00)([02468][048]|[13579][26])(00)?\Z/
(1801..2401).each do |year|
  leap = year % 4 == 0 && ((year % 100 != 0) || (year % 400 == 0))
  leap_regex = !year.to_s[r].nil?
  if leap != leap_regex
    print 'Assertion broken:', year, " ", leap, " ", leap_regex, "\n"
  end
end

---

(Bonus) for something that should work not only for 1801..2400, but for any non-negative year:

^\d*(0000|(?!00)([13579][26]|(^|[02468])[048])(00)?)$

Test in Ruby (^ = \A and $ = \Z because Ruby) for first 100000 years:

r = /\A\d*(0000|(?!00)([13579][26]|(\A|[02468])[048])(00)?)\Z/
100000.times do |year|
  leap = year % 4 == 0 && ((year % 100 != 0) || (year % 400 == 0))
  leap_regex = !year.to_s[r].nil?
  if leap != leap_regex
    print 'Assertion broken:', year, " ", leap, " ", leap_regex, "\n"
  end
end

Answer 10

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Answer 11

IBM System Z assembler - 56 bytes.

(96 bytes of source. Previously 712 384 202 bytes of source, 168 byte executable).

Smaller version still. No longer saves caller's registers, changes to literal storage, changed addressing mode.

 l        CSECT      
         using l,15 
         l  5,y     
         n 5,f      
         bnz r      
         xr 4,4     
         l 5,y      
         d 4,c      
         ch 4,i     
         bne i      
         n 5,f      
         bnz r      
i        dc h'0'    
r        b  *       
y        dc f'2004' 
f        dc f'3'    
c        dc f'100'  
         end 

New version. This will ABEND with a S0C1 if it's a leap year, and loop if it isn't. Hopefully that fulfils the requirement of indicating the result.

l        CSECT             
         ASMDREG           
         SYSSTATE archlvl=2
         IEABRCX  DEFINE   
         save  (14,12)     
         larl  r9,s        
         using s,r9        
         st 13,w+4         
         la 13,w           
         st 13,w+8         
         la 5,2004         
         st 5,y            
         n 5,=f'3'         
         bnz r             
         xr r4,r4          
         l 5,y             
         d r4,=f'100'      
         ch r4,=h'0'       
         bne i             
         n 5,=f'3'         
         bnz r             
i        dc h'0'           
r        b  0              
s        dc 0d'0'          
y        ds f              
w        ds 18f            
         ltorg             
         end  

OK, so not the shortest (although it might be once we look at the actual executed code plus the interpreter size...)

leapyear CSECT                                                
         ASMDREG                                              
         SYSSTATE archlvl=2                                   
         IEABRCX  DEFINE                                      

         save  (14,12)                                        

         larl  r9,staticArea                                  
         using staticArea,r9                                  
         st r13,w_savea+4       .Save callers savearea        
         la r13,w_savea         .Address my savearea          
         st r13,w_savea+8         . and save it               

         open  (O,OUTPUT)             .open file              

         la r5,1936             .r5 = input year              
         st r5,years            .Save year                    

         cvd r5,double          .Convert year to p-decimal    
         mvc edarea,=xl8'4020202020202120' .Move in edit mask 
         ed edarea,double+4      .Make packed decimal year printable                              
         mvc outrec(4),edarea+4  .Move year string to output area                             
         bas r10,isitleap       .Call leap year routine       

         close (O)              .Close files            
         b return               .Branch to finish

isitleap ds 0h                                                      
         mvi outrec+5,c'N'      .Set default value                                   
         n r5,=f'3'             .Are last 2 bits 0 (Divisible by 4)?
         bnz notleap            .No - not leap                      
         xr r4,r4               .Clear R4                           
         l r5,years             .Reload r5 with year                
         d r4,=f'100'           .divide r4/r5 pair by 100           
         ch r4,=h'0'            .Remainder 0?                       
         bne isleap             .No - leap year                     
         n r5,=f'3'             .Quotient divisible by 4?           
         bnz notleap            .No - not leap                      

isleap   ds    0h                                                   
         mvi outrec+5,c'Y'      .Move in leap year indicator                                    

notleap  ds    0h                                                   
         put O,outrec           .Print output record                                    
         br r10                 .Return to caller                   

* Program termination                                               
return   ds 0h                                                      
         l r13,w_savea+4         .Restore callers savearea          
         return (14,12),,rc=0    .Restore registers and return    
* storage areas                                                     
staticarea  dc 0d'0'                                                
outrec      ds cl10                                                 
years       ds f                                                    
w_savea     ds 18f                save area                         
edarea      ds cl8                    .edit area                    
double      ds d                                                    
* Macros and literals                                               
         print nogen                                                
O        dcb   recfm=F,lrecl=6,dsorg=PS,ddname=O,macrf=PM           
         print gen                                                  
*                                                                   
         ltorg                         literal storage              
         end  

Output:

ABEND S0C1 for a leap year, S222 (when the CPU time has run out) if not.

1936 Y 1805 N 1900 N 2272 Y 2400 Y

(when run multiple times)

Answer 12

CJam, 18 16 bytes

q~[YAK]f{2#%!}:^

Gives 1 (truthy) for leap years and 0 (falsy) otherwise.

Run all test cases here.

Explanation

q~                 e# Read and eval input.
  [YAK]            e# Push an array containing 2, 10, 20 (the square roots of the
                   e# relevant divisors).
       f{    }     e# Map this block onto that array, also passing in the input year.
         2#        e# Square the divisor.
           %!      e# Modulo followed by logical negation. Gives 1 if the year is divisible
                   e# by the given divisor and 0 otherwise.
                   e# At this point we have one of the following arrays:
                   e#   [0 0 0] - not a leap year
                   e#   [1 0 0] - a leap year
                   e#   [1 1 0] - not a leap year
                   e#   [1 1 1] - a leap year
              :^   e# Reduce XOR onto this array, which gives 1 if there is an odd number
                   e# of 1s and 0 if there's an even number.

Answer 13

Mathematica, 40 27 bytes, 17 chars

#∣4∧(#∣100<U+F523>#∣400)

Uses 17 chars, but 27 bytes. Thanks to @alephalpha for the tip. Note that the vertical bars are actually U+2223 for divides. The <U+F523> should be replaced with the corresponding character.

Answer 14

R, 29

!(Y=scan())%%4&Y%%100|!Y%%400

Test run

> !(Y=scan())%%4&Y%%100|!Y%%400
1: 1936
2: 1805
3: 1900
4: 2272
5: 2400
6: 2200
7: 
Read 6 items
[1]  TRUE FALSE FALSE  TRUE  TRUE FALSE

Answer 15

C, 81

I can do shorter, but this one neatly sticks to 'char' types, without parsing the argument (e.g. with atoi):

main(c,v)char**v;{char*p=*v+9;p-=2*(96==*p+p[1]);putchar("ynnn"[(2**p^p[1])&3]);}

It must be invoked with a name 4 characters long, because it makes the standard assumption that arguments immediately follow the program name, separated by NULs. Furthermore, it assumes that the single argument is encoded in ASCII and has no leading space.

Explanation:

main(c,v)
char**v;
{
    char *p = *v+9;
    if (p[0] + p[1] == '0'+'0')
        p -= 2;
    putchar("ynnn"[((*p << 1) ^ p[1])&3]);
}

*v+9 is the position of the 'tens' digit in v[1]+2.

If the 'tens' and 'units' characters add to 96, we end in 00, so back up two characters, so that 'tens' and 'units' point to the century number.

Now xor 'units' with twice the 'tens', mod 4. This works because 10==±2 mod 4, so the lower bit of the 'tens' can just toggle bit 1 of the 'units'. We use the result as an index into our remainders table, printing y only if the modular result is zero.

Answer 16

Befunge-98, (41 bytes)

&:4%#v_:aa*%#v_28*%|
"y",;<;@,"n";>;  ;#[

Simplicity is awesome.

Answer 17

sed, 55

s/00$//
y/0123456789/yNnNyNnNyN/
/N.$/y/ny/yn/
s/.\B//g

• First line divides exact centuries by 100.
• Second line gives 'N' to odd digits, 'y' to 4s, and 'n' to non-4s.
• Third line swaps 'y' and 'n' if there's an odd penultimate digit (because 10 is 2 mod 4)
• Final line deletes all but the last character

Note that non-leap years may be printed as n or N depending on whether they are even or odd. I consider this a creative interpretation of the rule which allows alternatives to 'yes' and 'no' without specifying that they have to be consistent.

Answer 18

Python2 - 37

g=lambda x:(x%4or x%400and x%100<1)<1

Note that if a is a nonnegative integer, then a<1 is a short way of writing not bool(a). The last <1 thus effectively converts the expression in the parentheses to a boolean and negates the result.

Applying the function g to an integer n between 1801 and 2400 will return True if n is a leap year, and False otherwise.

Answer 19

Haskell, 19 bytes

f i=gcd 80i>gcd 50i

Try it online!

Answer 20

KDB(Q), 27 bytes

{0=x mod(4 400)0=x mod 100}

Explanation

               0=x mod 100      / boolean of 100 divisibility
        (4 400)                 / 0b -> 4, 1b -> 400
 0=x mod                        / boolean of 4/400 divisibility
{                         }     / lambda

Test

q){0=x mod(4 400)0=x mod 100}1936 1805 1900 2272 2400
10011b

Answer 21

Julia, 30 28 bytes

y->(y%4<1&&y%100>0)||y%400<1

This creates an unnamed function that accepts an integer argument and returns a boolean value. To call it, give it a name, e.g. f=y->....

Ungolfed:

function f(y)
    (y % 4 == 0 && y % 100 != 0) || y % 400 == 0
end

Example:

julia> for y in [1936, 1805, 1900, 2272, 2400] println(f(y)) end
true
false
false
true
true
true

Answer 22

PHP - 45 bytes

$b=$argv[1]%400;echo !$b|!($b%4)&!!($b%100);

Nothing that special really, just abusing type-juggling.

Answer 23

C, 37 34 30 bytes

l(y){y=y%(y%25?4:16)?110:121;}

Wandbox

Answer 24

Java 8, 49 45 22 20 bytes

n->n%(n%25<1?16:4)<1

-2 bytes thanks to @OlivierGrégoire.

Try it online.

Some 22 bytes solutions:

n->n%25<1?n%16<1:n%4<1

Try it online.

n->(n%25<1?n%16:n%4)<1

Try it online.

java.time.Year::isLeap

Try it online.

Explanation:

n->                // Method with integer parameter and boolean return-type
   n%          <1  //  Return whether the integer is divisible by:
     (n%25<1?      //   If the input is divisible by 25:
             16    //    Check if its divisible by 16
            :      //   Else:
             4)    //    Check if its divisible by 4 instead

Answer 25

C#, 23 bytes

y=>y%25<1?y%16<1:y%4<1;

Try it online!

Full source, including test cases:

using System;

namespace CountingLeapYears
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<int,bool>s=y=>y%25<1?y%16<1:y%4<1;
            Console.WriteLine(s(1936)); //y
            Console.WriteLine(s(1805)); //n
            Console.WriteLine(s(1900)); //n
            Console.WriteLine(s(2272)); //y
            Console.WriteLine(s(2400)); //y
        }
    }
}

Answer 26

T-SQL 37 22 bytes

Saved 15 bytes thanks to BradC's comment.

The usual hardcoded variable for lack of a stdin.

e.g.

DECLARE @ NVARCHAR(4) = '2016'

Then the solution is:

PRINT ISDATE(@+'0229')

Answer 27

Mathematica, 15 bytes

LeapYearQ@*List

Mathematica has a built-in for everything. List is there to make the integer input into a compact DateList object. Returns True or False.

Answer 28

PHP, 28 bytes

Why calculate when there´s a builtin?

<?=checkdate(2,29,$argv[1]);

prints 1 for leap year, empty string for no leap year.
Add + after <?= for 1/0.

Answer 29

SmileBASIC, 40 bytes

INPUT Y?0DTREAD STR$(Y)+"/02/29"OUT,,,?1

Prints 0 1 if it's a leap year, otherwise prints 0 and errors.

DTREAD parses a date string in the form of YYYY/MM/DD and gives the year, month, and day. It can also return the day of the week, and trying to get the day of the week of a day that doesn't exist (Feb. 29 of a non-leap year) will cause an error.

This is shorter than the obvious answer using mod, because SB uses MOD instead of %, which takes up 3-5 characters.

Answer 30

Jelly, 7 bytes

ọ4,25>/

Try it online!

How it works

ọ4,25>/  Main link. Argument: n (1801 - 2400)

ọ4,25    Test how many times n is divisible by 4 and 25.
     >/  Verify that the order of 4 is higher than the order of 25.

Answer 31

x86 machine code, 14 bytes

00000000: adfd ad2d 3030 74fa 86c4 d50a a803       ...-00t.......

Unassembled

AD         LODSW           ; SI += 2 to start with last two digits
FD         STD             ; set direction for LODSW to decrement
AD         LODSW           ; load two ASCII digit chars into AX
2D 3030    SUB  AX, '00'   ; convert to decimal, test if year ends in 00?
74 FA      JZ   -6         ; if so, look at the first two digits instead
86 C4      XCHG AL, AH     ; endian convert
D5 0A      AAD             ; base convert from 10 to binary
A8 03      TEST AL, 3      ; mod 4 = 0?

Input string in SI, output is ZF if is a leap year.

Explanation

If the year does not end in 00, it is a leap year if last two digits mod 4 is 0. If it does end in 00, it will be a leap year if the first two digits mod 4 is 0. Example:

Leap year:

• 1936: 36 mod 4 == 0
• 2400: 24 mod 4 == 0

NOT leap year:

• 1805: 05 mod 4 == 1
• 1900: 19 mod 4 == 3

Test Program Output

Download and test LEAP.COM.

Answer 32

Excel, 30 bytes

David Hammen's method (A1 serves as input):

=MOD(A1,IF(MOD(A1,25),4,16))=0

Built-in functions can get you down to 23 bytes. However, this approach doesn't work on 1900 because of bug compatibility with Lotus 1-2-3:

=MONTH(DATE(A1,2,29))=2

You could even save two more bytes by using Portugese locale:

=MÊS(DATA(A1;2;29))=2

Answer 33

Javascript ES6, 32, 29, 26

Any of following lines works:

f=y=>new Date(y,2,0).getDate()&1
g=y=>!(y&3)^!(y%100)>!(y%400)
h=y=>!(y&3|y%100<1&&y%400)

Answer 34

C, 57 bytes

Takes the input from stdin, with or without trailing spaces/newline. Only works on little endian machines (yeah, like everyone is on BE these days). Outputs Y or N.

main(y){scanf("%d",&y);y=y%(y%100?4:400)?78:89;puts(&y);}

Explanation

Ungolfed:

int main(int y) {
   scanf("%d", &y);
   y = y % (y % 100 ? 4 : 400) ? 'N' : 'Y';
   puts(&y);
}

First, scanf reads the year as an integer in y. Then, y is modulo'ed with 4 or 400 depending on whether the year is divisible by 100. If the remainder is zero, the ASCII code for Y is assigned to y, otherwise it gets the ASCII code for N. The value of y is now 0x000000??, where 0x?? is the assigned character. Being on a little-endian machine, in memory this is stored as ?? 00 00 00. This is a NULL-terminated C string, containing only the assigned characters. The address of y is passed to puts and the char is printed (with a trailing newline).

Answer 35

Haskell, 43 41 39 bytes

l x=mod x(if mod x 25<1then 16else 4)<1

If x is divisible by 25, check if it is also divisible by 16, making it divisible by the least common multiple of 25 and 16, which is 400.
If x is not divisible by 25, check if it is divisible by 4.

Inspired by @David Hammen's answer in JavaScript.

Answer 36

PHP, 34 30 bytes

echo+!($argn%($argn%25?4:16));

I'm using the + sign to convert false (which would normally be converted to an empty string) to 0, because that seems to comply closer to the "clearness" requirement (maybe not?). With standard truthy/falsy rules, 1 byte can be saved.

Note: I shamelessly implemented the divisibility by 25 trick from @David Hammen. Without his algorithm it would be 37 bytes:

<?=+!(($b=$argv[1])%400^$b%100^$b%4);

Run like this:

echo 1900 | php -nR 'echo+!($argn%($argn%25?4:16));';echo

Tweaks

• Saved 4 bytes by using $argn

Answer 37

PowerShell, 31 bytes

I am excited to say that I golfed this shorter than the builtin!

param($a)!($a%(4,16)[!($a%25)])

Outputs true for leap years and false otherwise.

Builtin:

[datetime]::IsLeapYear($args[0])

Though, if I wanted to stretch the statement 'clearly tells the user if it is or isn't a leap year' and do something non-standard, I could save 3 bytes and use:

param($a)$a%(4,16)[!($a%25)]

This outputs 0 for leap years and 1 or higher for non-leap years, which I don't like since I'd prefer to return a more standard truthy value for leap years.

Answer 38

LOLCODE, 228 202 159 bytes

HOW IZ I f YR a
MOD OF a AN 100
O RLY?
YA RLY
MOD OF a AN 4
O RLY?
YA RLY
b R 1
OIC
NO WAI
MOD OF a AN 400
O RLY?
YA RLY
b R 0
NO WAI
b R 1
OIC
OIC
IF U SAY SO

Ungolfed:

HAI 1.3 BTW "HAI" does nothing functionally in current versions and does not throw an error if you omit it.
HOW IZ I leap YR input
    I HAS A output
    DIFFRINT MOD OF input AN 100 AN 0 BTW Thanks @LeakyNun, In LOLCODE any non-empty values, i.e. 0, "", etc. default to WIN.
    O RLY?
        YA RLY
            BOTH SAEM MOD OF a AN 4 AN 0
            O RLY?
                YA RLY
                    output R WIN BTW "WIN" is true, but in the actual program I used 1 as a truthy value because it's shorter.
            OIC
        NO WAI
            DIFFRINT MOD OF a AN 400 AN 0
            O RLY?
                YA RLY
                    output R FAIL BTW "Fail" is false, but in the actual program I used 0 as a falsy value.
                NO WAI
                    output R WIN
            OIC
    OIC
    FOUND YR output BTW This statement is implied in the golfed version.
IF U SAY SO BTW "KTHXBYE", just like "HAI" has no functional purpose and throws no error on omission.
KTHXBYE

In Python ungolfed, because LOLCODE is confusing:

def leap:
    if(year % 100 != 0):
        if(year % 4 == 0):
            output = true
    else:
        if(year % 400 != 0):
            output = false
        else:
            output = true
    return(output)

Answer 39

Retina, 24 bytes

[^04]00$

.+
$*
....

^$

Try it online!

Answer 40

PHP, 45 bytes

<?=date_create($argv[1].'-1-1')->format('L');

Not the shortest, but using built in functions. Totally not code-golf.

Answer 41

Jelly, 8 bytes

ȷ2*ḍ¥×4ḍ

Try it online!

Answer 42

Bash, 14 bytes

date -d$1/2/29

Try it online!

Outputs via exit code, 0 for leap year and 1 otherwise.

date -d STRING will display the date indicated by STRING. $1/2/29 represents February 29th of $1, the argument. If STRING is not valid, i.e. Feb 29 does not exist, then date errors out.

Answer 43

C++, 50 43 bytes

-7 bytes thanks to Zacharý

int l(int y){return!(y%4)&&y%100+!(y%400);}

And the test code is :

auto t = {
    1936,1805,1900,2272,2400
};

for (auto&a : t) {
    std::cout << "Year : " << a << " is " << (l(a) ? "" : "NOT") << " a leap year\n";
}

Answer 44

Ruby, 22 bytes

->n{1>n%(n%25<1?16:4)}

Try it online!

The challenge is quite old, but I noticed there was no valid Ruby answer yet. Nothing particularly original, anyway.

Answer 45

D, 43 42 bytes

T l(T)(T y){return!(y%4)&&y%100+!(y%400);}

Try it online!

Another port of @HatsuPointerKun's C++ answer.

Answer 46

Kotlin, 30 bytes

{it%4<1&&(it%100>0||it%400>0)}

Try it online!

Answer 47

Common Lisp, 46 bytes

(=(mod(setq x(read))(if(>(mod x 25)0)4 16))0)

Try it online!

Based on the David Hammen’s method.

Answer 48

Clam, 33 31 bytes

=a*rp|e%a*"400"0&%a*"100"e%a*40

Try it online!

-2 bytes thanks to ASCII-only

Outputs true for leap years. Outputs false for other years

Explanation

=a*rp|e%a*"400"0&%a*"100"e%a*40
=a*r                             read input and store in a*
    p                            Print..
       %a*"400"                    a* % 400
      e                            ==
               0                   0
     |                             OR
                 %a*"100"            a* % 100
                &                    AND
                          %a*4       a* % 4
                         e           ==
                              0      0

Very awkward explanation, but basically what it does is this:

print(year % 400 == 0 || (year % 100 && year % 4 == 0))

Clam follows JS rules for truthy and falsey values, meaning year % 100 is true if it does not equal 0 (and it's shorter than adding an e before it and an 0 after it)

Resulting JS:

myVar = read();
console.log(year % 400 == 0 || (year % 100 && year % 4 == 0));

Answer 49

MathGolf, 9 bytes

4♀`*]÷~≤*

Try it online!

Explanation

4         Push 4
 ♀        Push 100
  `*      Duplicate top two elements of stack and multiply (400)
    ]     Wrap stack in array ([4, 100, 400])
     ÷    Check input for divisibility with all array items
      ~   Dump array onto stack
       ≤  Check if divisibility by 100 is <= than divisibility with 400
        * Multiply with the divisibility bool for 4, works like logical and

The trick is that ≤ ensures that if a number is divisible by 100, it must also be divisible by 400, but if it's not divisible by 100, anything goes.

Answer 50

Japt -!, 8 5 bytes

Locale dependent.

ÐUT k

Try it

Answer 51

Japt -h!, 8 bytes

Input is taken as a string.

k0 ò o%4

Try it online!

Answer 52

Haskell, 35 bytes

l x|x!25<1=x!16<1
l x=x!4<1
(!)=mod

Try it online!

Answer 53

PHP, 35 34 bytes

-1 byte thanks to 640KB

<?=date(L,strtotime($argv[1].-1));

Try it online!

Came across the L option in the PHP date formatter, which returns if the year of the date is a leap year. After that, it was just a matter of trying to get the timestamp of the year in the shortest way (though I'm not too confident this is the shortest)

Answer 54

Reg, 14 bytes

1¿:%[4|]%[;]

Try it online!

How it works (because it contains unprintable characters)

Answer 55

W d, 10 8 bytes

Ö♦Ç⌂╬`§Γ

Explanation:

Uncompressed:

25m4&16|m!

25m        % If the input is modulo-able by 25:
   4&      % Return 4
     16|   % Otherwise, return 16
        m! % Modulo input by the value. Negate the value.

Answer 56

SpecBAS - 77

Not the shortest (but not the longest either).

1 INPUT y: PRINT y;"->";"ny"((y MOD 4=0 AND (y MOD 100<>0) OR y MOD 400=0)+1)

Uses the standard formula, then prints "n" or "y" based on return value (the final +1 at the end is due to strings being 1-based).

Answer 57

Python, 37 bytes

def c(s):return s%16*(s%25<1)<(s%4<1)

Answer 58

Python, 50 48 bytes

import calendar as y
def a(b):print(y.isleap(b))

Answer 59

C#, 26 bytes

y=>DateTime.IsLeapYear(y);

C# lambda (Predicate) where the input is a int and the output is a bool. I use a builtin.

Answer 60

Java 8, 17 bytes

y->y%4<1&y%100>0;

It almost looks like a golfing language lol

Answer 61

BASH, 68 bytes

((!(y % 4) && ( y % 100 || !(y % 400)))) &&  echo "leap" || echo "no"

Answer 62

MATLAB + Aerospace Toolbox, 9 bytes

@leapyear

Anonymous function that determines if the passed year is a leap year, returning 1 for leap and 0 for not.

Ok so MATLAB has a built-in... But there we are.

---

Octave, 13 bytes

@is_leap_year

Try it online!

Octave also has a built-in... cost's 4 more bytes than MATLAB though.

Answer 63

Whitespace, 103 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S N
S _Duplicate_input][S S S T T   S S T   N
_Push_25][T S T T   _Modulo][N
T   S T N
_If_0_jump_to_Label_TRUE][S S S T   S S N
_Push_4][T  S T T   _Modulo][N
T   S S N
_If_0_jump_to_Label_LEAP][N
S N
N
_Jump_to_Label_PRINT][N
S S T   N
_Create_Label_TRUE][S S S T S S S S N
_Push_16][T S T T   _Modulo][N
T   S S N
_If_0_jump_to_Label_LEAP][N
S N
N
_Jump_to_Label_PRINT][N
S S S N
_Create_Label_LEAP][S S S T N
_Push_1][N
S S N
_Create_Label_PRINT][T  N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Explanation in pseudo-code:

Integer i = STDIN-input as integer
If i modulo-25 is 0:
  If i modulo-16 is 0:
    Print 1
  Else:
    Print 0
Else-if i modulo-4 is 0:
  Print 1
Else:
  Print 0

Example runs:

Input: 1936

Command     Explanation                Stack            Heap       STDIN    STDOUT   STDERR

SSSN        Push 0                     [0]              {}
SNS         Duplicate top (0)          [0,0]            {}
SNS         Duplicate top (0)          [0,0,0]          {}
TNTT        Read STDIN as integer      [0,0]            {0:1936}   1936
TTT         Retrieve at heap 0         [0,1936]         {0:1936}
SNS         Duplicate top (1936)       [1936,1936]      {0:1936}
SSSTTSSTN   Push 25                    [1936,1936,25]   {0:1936}
TSTT        Modulo                     [1936,11]        {0:1936}
NTSTN       If 0: Jump to Label_TRUE   [1936]           {0:1936}
SSSTSSN     Push 4                     [1936,4]         {0:1936}
TSTT        Modulo                     [0]              {0:1936}
NTSSN       if 0: Jump to Label_LEAP   []               {0:1936}
NSSSN       Create Label_LEAP          []               {0:1936}
SSSTN       Push 1                     [1]              {0:1936}
NSSN        Create Label_PRINT         [1]              {0:1936}
TNST        Print top as integer       []               {0:1936}            1
                                                                                     error

Program stops with an error: No exit defined.
Try it online (with raw spaces, tabs and new-lines only).

Input: 2400

Command     Explanation                Stack            Heap       STDIN    STDOUT   STDERR

SSSN        Push 0                     [0]              {}
SNS         Duplicate top (0)          [0,0]            {}
SNS         Duplicate top (0)          [0,0,0]          {}
TNTT        Read STDIN as integer      [0,0]            {0:2400}   2400
TTT         Retrieve at heap 0         [0,2400]         {0:2400}
SNS         Duplicate top (2400)       [2400,2400]      {0:2400}
SSSTTSSTN   Push 25                    [2400,2400,25]   {0:2400}
TSTT        Modulo                     [2400,0]         {0:2400}
NTSTN       If 0: Jump to Label_TRUE   [2400]           {0:2400}
NSSTN       Create Label_TRUE          [2400]           {0:2400}
SSSTSSSSN   Push 16                    [2400,16]        {0:2400}
TSTT        Modulo                     [0]              {0:2400}
NTSSN       If 0: Jump to Label_LEAP   []               {0:2400}
NSSSN       Create Label_LEAP          []               {0:2400}
SSSTN       Push 1                     [1]              {0:2400}
NSSN        Create Label_PRINT         [1]              {0:2400}
TNST        Print top as integer       []               {0:2400}            1
                                                                                     error

Program stops with an error: No exit defined.
Try it online (with raw spaces, tabs and new-lines only).

Input: 1991

Command     Explanation                Stack            Heap       STDIN    STDOUT   STDERR

SSSN        Push 0                     [0]              {}
SNS         Duplicate top (0)          [0,0]            {}
SNS         Duplicate top (0)          [0,0,0]          {}
TNTT        Read STDIN as integer      [0,0]            {0:1991}   1991
TTT         Retrieve at heap 0         [0,1991]         {0:1991}
SNS         Duplicate top (1991)       [0,1991,1991]    {0:1991}
SSSTTSSTN   Push 25                    [0,1991,1991,25] {0:1991}
TSTT        Modulo                     [0,1991,16]      {0:1991}
NTSTN       If 0: Jump to Label_TRUE   [0,1991]         {0:1991}
SSSTSSN     Push 4                     [0,1991,4]       {0:1991}
TSTT        Modulo                     [0,3]            {0:1991}
NTSSN       if 0: Jump to Label_LEAP   [0]              {0:1991}
NSSN        Create Label_PRINT         [0]              {0:1991}
TNST        Print top as integer       []               {0:1991}            0
                                                                                     error

Program stops with an error: No exit defined.
Try it online (with raw spaces, tabs and new-lines only).

Answer 64

ActionScript 2.0, 49 bytes

function a(b){trace(!(b%4)&&b%100+!(b%400)?1:0);}
function a(b){                                  } - define a function
              trace(                          );  - that outputs
                                          ?       - if
                    !(b%4)                        - the input divides by 4
                          &&                      - and
                            b%100                 - doesn't divide by 100
                                 +                - plus
                                  !(b%400)        - does divide by 400
                                           1:0    - 1, else 0

Not a very good explanation, and I might have swapped some stuff, but this still confuses me, so this is the best you're going to get.

Answer 65

C (gcc), 33 bytes

f(i,a){a=!(i%4|!(i%100)&&i%400);}

Try it online!

Answer 66

05AB1E, 9 7 bytes

т‰0Kθ4Ö

Try it online or verify all test cases.

Explanation:

т‰         # Divmod the (implicit) input by 100
           #  i.e. 1900 → [19,00]
           #  i.e. 1936 → [19,36]
           #  i.e. 1991 → [19,91]
           #  i.e. 2000 → [20,00]
  0K       # Remove all 0s
           #  i.e. [19,00] → [19]
           #  i.e. [19,36] → [19,36]
           #  i.e. [19,91] → [19,91]
           #  i.e. [20,00] → [20]
    θ      # Pop and get the last item
           #  i.e. [19] → 19
           #  i.e. [19,36] → 36
           #  i.e. [19,91] → 91
           #  i.e. [20] → 20
     4Ö    # Check if it's divisible by 4 (and output the result implicitly)
           #  i.e. 19 → 0 (falsey)
           #  i.e. 36 → 1 (truthy)
           #  i.e. 91 → 0 (falsey)
           #  i.e. 20 → 1 (truthy)

Answer 67

JavaScript, 19 bytes

y=>!(y%25?y%4:y%16)

Try it online!

Answer 68

Swift, 82 bytes

func l(y:Int){y%4==0 && (y%100 != 0 || y%400==0) ?print("Y"):print("N")}
l(y:1936)

Try it online!

Answer 69

Reality, 1 byte

L

input via stdinput

---

The language was made after the challenge