how to identify leap years

Hi,

I am trying to create a function that takes 3 +ve integer scalar inputs year, month and date. If these 3 represent a valid date return true otherwise return false. My code is running for most of the input. But I am having problem with leap years. The code I wrote is given below. Can anyone please point out my mistake.

function valid= valid_date(year, month, date)
v=1;
a= mode(year,4);
b= mode(year,100);
c= mode(year,400);

if ~isscalar(year) || year<1 || year~=fix(year) || ~isscalar(month) || month<1 || month~=fix(month) ||~isscalar(date)||  date<1 || date~=fix(date)
    v= 0;   
end
if v==0
    valid=false;
    return
end
 if 0>month || month>12
        v= 0;
    elseif (month==1 ||month==3|| month==5 || month==7 || month==8|| month==10|| month==12) 
        if 0<date&& date<=31
            v=1;
        else 
            v=0;
        end
    elseif (month== 4 ||month==6|| month==9 || month==11) 
        if 0<date&& date<31
            v=1;
        else
            v=0;
        end
    elseif month==2 
        if date==29
            if (a==0 && b~=0) || c==0
               v=1;
            else
               v=0;
            end
        elseif 0<date && date<29
                v=1;
        else
                v=0;
        end
 end
if v==0
    valid=false;
else
    valid=true;
end

Answer 1

[Siva Charan]

Use this subfunction and call from the main function.

if the year is divisible by 100 and not divisible by 400, it is not a leap year. millennium years(1600, 1700, 1800...) should be divisible by 400 to be leap years, for others, any year that is divisible by 4 can be a leap year.

function leapyear = checkleap(year)
     if mod(year,100)==0
         if mod(year,400)==0
            leapyear = true;
         else
             leapyear = false;
         end
     elseif mod(year,4) == 0
         leapyear = true;
     else
         leapyear = false;
     end
end

Answer 2

[Stijn Haenen]

There is a leap year every four years, so you can use this:

if mod(year,4)==0
    'leap year'
else
    'not a leap year'
end

[Walter Roberson]

That is not correct. The year 1900 is divisible by 4 but was not a leap year. But 2000 was a leap year.

[Stijn Haenen]

Ah, i didnt know that.

Then it should be somthing like this:

if mod(year/100,4)==0 && mod(year,4)==0
    'leap year'
else
    'not a leap year'
end

Or are there any other exceptions in leap years?

[Stephen23]

These are all divisible by four: 1500, 1700, 1800, 1900, 2100, 2200, 2300, 2500, but none of them are leap years.

[Steven Lord]

Yes. The Wikipedia page has the algorithm for how to determine whether a year is a leap year.