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

  • 8
    You should be more clear: A given year is a leap year if and only if it is (divisible by 4)∧((divisible by 100)→(divisible by 400)). – LegionMammal978 May 26 '15 at 11:34
  • Your input will have no newlines or trailing spaces. Dang it, that would have saved me 2 bytes... – Dennis May 26 '15 at 17:10
  • 2
    You should extend the accepted input range to AD 1601 thru 2400. This covers two 400-year Gregorian cycles (which proleptically start on Monday). – David R Tribble May 26 '15 at 18:17
  • 2
    Does falsy if leap year and truthy if not a leap year count as "clearly tells the user if it is or isn't"? – lirtosiast May 28 '15 at 21:27
  • 1
    why did you delete your account? – noɥʇʎԀʎzɐɹƆ Jul 18 '16 at 20:28

50 Answers 50

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.com. 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, βγ is false if y is dividable by 100 and αβγ 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 = ¬((αβ) → γ)).

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

  • 1
    ≠/×4 25 4⊤⎕ with 1 for leap, 0 for non-leap. – Adám Jan 16 at 10:38

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.

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.
  • I've got a few more 12 byte alternatives. Maybe you can find something in them to bring it down to 11? r2/~~\e|i4%!, r2/~~\~e|4%!, r2/:~~\e|4%!, r2/S*~\e|4%! and the 13 byte r2/:~W%:e|4%! – Martin Ender May 26 '15 at 18:40
  • @MartinBüttner: There's also r2/:i:\e|4%! (12) and r2/:i(fe|~4%! (13). I've even tried GolfScript (which doesn't require r), but or4 is interpreted as a single token. If only the input had a trailing newline... – Dennis May 26 '15 at 19:00

Javascript (ES5), 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
  • I had used your suggestion in my answer, did not see yours. Now I have rolled back. Note: You should specify EcmaScript 6, or else someone will complain 'not working in Chrome' – edc65 May 28 '15 at 21:59
  • @edc65: Well, he should specify EcmaScript 6 because it is EcmaScript 6. Arrow function notation (y=>...) is an ES6 feature. – Tim Čas Sep 5 '16 at 11:50

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

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>

  • 2
    You can reduce this by six characters if you use !(y%(y%25?4:16)) instead of !(y%100<1&&y%400||y%4). For those bothered by the ternary operator, you could use !(y%(4<<2*!(y%25))) and still save three characters over !(y%100<1&&y%400||y%4). – David Hammen May 28 '15 at 20:48
  • 1
    David Hammen's suggestion is identical to his answer, so I think you should keep the length as 27. – lirtosiast May 28 '15 at 21:52

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

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
  • 1
    If you have (?!) you can combine the two halves: (?!00)([02468][048]|[13579][26])(00)?$ - for 38. That won't work for one-digit years, though. – Toby Speight May 27 '15 at 16:45

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)

  • Down to 376 bytes by making storage areas minimum size (13 bytes), removing staging area 'leapflag' and only including a single year (rather than 5) in the program. – Steve Ives May 27 '15 at 10:08
  • 384 bytes by providing a slightly formatted output: – Steve Ives May 27 '15 at 10:29
  • 1
    +1 for interesting and educational choice of language. :-) – Toby Speight May 27 '15 at 17:02
  • I could save a few bytes by abandoning convention and not bothering to save the callers registers at the start, seeing as the program never returns to the caller. This is Very Bad Form. – Steve Ives May 29 '15 at 12:47

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

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.

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.

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.

  • 2
    This is one of those puzzles where Mathematica offers a solution that feels kind of cheaty: LeapYearQ[#]& – zeldredge May 26 '15 at 13:23
  • 1
    You can use to represent Divisible : #∣4&&(!#∣100||#∣400)&, 21 characters, 27 UTF-8 bytes. – alephalpha May 26 '15 at 13:54
  • @zeldredge Still, that's not shorter than the APL solution. – FUZxxl May 26 '15 at 14:10
  • @alephalpha Alternatively, you could use U+F523 (\[Implies]) to make it #∣4&&(#∣100<U+F523>#∣400)& for 19 chars (but still 27 bytes). – LegionMammal978 May 26 '15 at 15:15
  • This is a standard loophole; you're using a function that does exactly the required functionality. This is verboten. – FUZxxl May 26 '15 at 15:33

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

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.

Befunge-98, (41 bytes)

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

Simplicity is awesome.

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.

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

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.

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

PHP - 45 bytes

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

Nothing that special really, just abusing type-juggling.

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
        }
    }
}

C, 37 34 30 bytes

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

Wandbox

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)

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).

  • 1
    "You will output in any way that you like that clearly tells the user if it is or isn't a leap year." Can you save a couple bytes by returning 1 or 0 rather than 'Y' or 'N'? (I don't really know C at all, just guessing.) – Alex A. May 28 '15 at 18:54
  • @AlexA. Thanks for the edit - now I know how to highlight syntax :) I considered it. The ASCII codes are both two digits, so no gain from that (by the way, I'm using uppercase Y and N to save 2 bytes, since lowercases have 3 digits). They are sequential, so that could be useful. Unfortunately, due to operator precedence, I get the same byte count: main(y){scanf("%d",&y);y=!(y%(y%100?4:400))+48;puts(&y);}. I can go down to 48 bytes if I can output an empty line for leap years and any character (ASCII 1-99) otherwise, but I feel like it's a bit bending the rules. What do you think? – Andrea Biondo May 28 '15 at 20:19
  • I must have done something wrong when counting chars. It's 57, not 59 :) – Andrea Biondo May 28 '15 at 20:21
  • 1
    Yeah, I'd say that's bending the rules, but you could comment on the question and ask the OP for confirmation. A good tool for counting bytes is this--I think a lot of the folks here use it. – Alex A. May 28 '15 at 20:22
  • Nah, I'll leave it as it is :) – Andrea Biondo May 28 '15 at 20:36

Haskell, 43 41 bytes

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

  • I don't know Haskell, but can ==0 be <1? – lirtosiast May 28 '15 at 21:13
  • You are right, it can. – nfs May 29 '15 at 9:04

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)
  • Would it be shorter to define a function? – Leaky Nun Jul 22 '16 at 16:55
  • probably, but I will edit it later. – OldBunny2800 Jul 22 '16 at 16:59
  • You have updated the main code to be a function, but not the ungolfed code? – Destructible Lemon Jul 23 '16 at 2:25
  • Oh oops. Thanks – OldBunny2800 Jul 23 '16 at 2:29
  • I thought LOLCODE has automatic type coercion, meaning that any non-zero value is equivalent to WIN.. – Leaky Nun Jul 23 '16 at 2:38

Java 8, 17 bytes

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

It almost looks like a golfing language lol

T-SQL 37 bytes

The usual hardcoded variable for lack of a stdin.

e.g.

DECLARE @ NVARCHAR(4) = '2016'

Then the solution is:

PRINT IIF(ISDATE(@+'0229')=1,'y','n')

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.

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