# how to identify leap years

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Anisia a on 18 May 2020
Commented: Steven Lord on 18 May 2020
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

Stijn Haenen on 18 May 2020
Edited: Stijn Haenen on 18 May 2020
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

Show 1 older comment
Stijn Haenen on 18 May 2020
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?
Stephen Cobeldick on 18 May 2020
These are all divisible by four: 1500, 1700, 1800, 1900, 2100, 2200, 2300, 2500, but none of them are leap years.
Steven Lord on 18 May 2020
Yes. The Wikipedia page has the algorithm for how to determine whether a year is a leap year.