Create function to calculate leap year

Hi I need to create a function that determines a leap year with the conditions

function result = is_leap_year(year)
% IS_LEAP_YEAR Determine if a given year is a leap year
% RESULT = IS_LEAP_YEAR(YEAR) returns TRUE if YEAR is a leap year, and
% FALSE otherwise.
% An error is raised if YEAR is not a positive integer.

a year is a leap year if it is divisible by four, unless it is also divisible by 100, unless it is also divisible by 400. I have been able to create a function however cannot work out how to do the false otherwise statement. Also with true and false I get 1(true) and 0 (false). Is this acceptable? Here's my function

function [ status ] = leapyear( year )
if mod(year, 400) == 0
status = true;
elseif mod(year, 4) == 0 && mod(year, 100) ~= 0
status = true;
else
status = false;
end
end

[Andrei Bobrov]

leapyear=@(yr)rem(yr,400)&rem(yr,100)&~rem(yr,4)

[Walter Roberson]

Andrei, for the year 2000, your code would start with rem(2000,400) which would be 0, so your code would return logical 0 even though 2000 was a leap year.

Jack, your code fails if given an input of 0.

[Andrei Bobrov]

Hi Walter! Yes!

"if it is a multiple of 4 and not divisible by 100, or a multiple of 400"(Wikipedia on russian)

leapyear=@(yr)~rem(yr,400)|rem(yr,100)&~rem(yr,4)

[chantsal bymba]

How to describe it

（Write a script which calls your function with a range from [1952 to 2020] and displays the output.）

[Walter Roberson]

years = 1795:1810;
for yidx = 1 : length(years)
    result(yidx) = leapyear(years(yidx));
end

Answer 1

[Fangjun Jiang]

TF=~(datenum(year,2,29)==datenum(year,3,1))

Using your approach, it should be like below. This is primarily for learning the if-else statement using only the necessary condition checking.

function [ status ] = leapyear( year )
if mod(year, 4) == 0
    status = true;
    if mod(year, 100) == 0
        status=false;
        if mod(year,400)==0
            status = true;
        end
    end
else
    status=false;
end

true and false are displayed as 1 and 0, but they are different than the double precision number 1 and 0. They are logical. Type this to find out.

class(true)
class(false)
class(1)
class(0)

To make it work for vector, I recommend Andrei's approach.

>> ly=@(yr)~rem(yr,400)|rem(yr,100)&~rem(yr,4);
>> ly(1999:2005)

ans =

       0     1     0     0     0     1     0

[Walter Roberson]

Golfing...

TF=datenum(year,2,29)~=datenum(year,3,1);

And beware if your year is a 2-digit number.

[Fangjun Jiang]

Nice, Walter! 2-digit year number should work too, it is default to the current century.

[Fangjun Jiang]

All right, I'll play alone...

T=datenum(y,2,29)~=datenum(y,3,1);

[Walter Roberson]

Renaming variables doesn't count in golfing, not until you need more than 26 variables (in which case it would be fair game to code so that the variables used most frequently had the shorter names.)

2 digit year numbers are considered to be within the current century in string representations, not in numeric representations.

The terms of the original assignment called for the function to error if the year was not a positive integer; your code does not make that test. Your code will return 1 (true) for an input of 0 -- but of course there was no year 0 so it could not have been a leap year.

Answer 2

[Paulo Silva]

Take a look at this ISLEAP Function , also look at Jan Simon comment to that file.

Answer 3

[Rebecca]

Hi - @Jack, I have the same assignment (we might have class together)

I think I must have some settings weird, when I type

function [ status ] = leapyear (year)

Error: Function definitions are not permitted in this context.

...and would always error when year = 23.

Error using ==> year at 23

Please enter D.

I figured it seemed to think year was to do with the date and clock so I substituted my own word for practise (star, beep, whatever), but then it wanted me to define this word before I started my input in the first place. I know you guys are down to the core pieces of the assignment, but I could use some baby step help.

I'm also pretty sure we're meant to use the rem function, but from all the research I've done so far, mod seems to work similarly - if not better, anyone want to explain?

[Walter Roberson]

Move the code from the "function" keyword on to the end of the file, to a file named leapyear.m, deleting it where it is now. The remaining code must *not* be saved in leapyear.m

The year = 23 is referring to line 23 of your file named "year", not to the year numbered 23.

[Rebecca]

Ah - I'm guessing DON'T delete 'year' but it's not a file I want to be using?

so save [status] = leapyear (year) in leapyear.m but nothing else? (if that's not right I'm really sorry for misunderstanding)

[Walter Roberson]

function [ status ] = leapyear (year)

and the rest of your code that implements the test for any one year should go in leapyear.m

The parts of the file that you had _before_ the "function" line should stay where they were... unless, that is, your original file happened to be named leapyear.m in which case you need to choose a different name for the parts before the "function" line.

[Rebecca]

fantastic! - I had a friend help and learnt the difference between the scrip screen etc. (thank you!)

I added an extra line in to the above to make sure that year was an integer:

if year <=0 || rem(year,1) ~=0

error('either not positive or non integer')

end

(next I'll learn how to make it look light grey and code for this site)

Answer 4

[Walter Roberson]

http://www.nmm.ac.uk/explore/astronomy-and-time/time-facts/leap-years and http://en.wikipedia.org/wiki/Gregorian_calendar

Julian Calendar established 46 BC, but leap years not properly added up through 7 CE

8 CE to 1581 CE - every year divisible by 4 was a leap year; the Julian Calendar.

1582 CE - Gregorian Calendar decreed, with the 4 except 100 except-except 400 structure. Some countries adopt this in October 1582 CE

1582 CE onward - various countries adopt at different times. Some of the changes were strange; some areas went back to the Julian calendar

1752 - Britain and colonies adopt Gregorian Calendar in September 1752 CE

1752 CE onward - more changes in various countries

1912 CE - 1928 CE: Republic of China is officially Gregorian but in practice local rules are followed

1918 CE Russia becomes second last major country to change. The "October Revolution" of 1917 was October in Julian calendar but November in Gregorian

1926 CE Turkey becomes last western country to change.

1929 CE Republic of China goes officially Gregorian again

The next 700 years: arguments from the Orthodox Christian churches as they have not adopted the Gregorian Calendar for some purposes; it will next make a difference in 2800 CE...

Answer 5

[Fahad alshammari]

function year
year = input (' Enter the year : ')
if mod(year,4) == 0 disp ('true')
elseif mod(year,100) == 0 disp ('falus')
elseif mod(year,400) == 0 disp ('true')
else disp ('falus')
end end

[Walter Roberson]

No this would say true for 1900. Also it is not possible for a number to be divisible by 100 without being divisible by 4 so your elseif clauses can never be reached.

[Fahad alshammari]

can u help me fix it plz ?

[Walter Roberson]

Reverse the order of your tests. Multiple of 400 is a leap year, multiple of 100 is not a leap year unless you already passed the 400 test, multiple of 4 is a leap year unless you already failed the 100 test.

Answer 6

[Shubham Agrawal]

a=input('Enter the year');
if (((a/400-floor(a/400))==0) | (((a/4-floor(a/4))==0)&(a/100-floor(a/100)~=0)))==1 
    disp('This is a leap year');
else
    disp('This is not a leap year');
end

Answer 7

[Shanjul Shrivastava]

function result = is_leap_year(year)
if year<=0; error("year is not a positive integer");
 elseif(mod(year,4))~=0 | (mod(year,4)==0 & mod(year,100)==0 & mod(year,400)~=0)
  result="false";
else 
  result="true"; 
 end
end

[Walter Roberson]

This does not appear to return true or false, as required by the assignment.

[Shanjul Shrivastava]

Thanx. I hope its fine now

[madhan ravi]

There’s an inbuilt function for true & false

[Walter Roberson]

The string() object "true" is not the same thing as TRUE, which is a logical value.

Also, there are no circumstances under which the code is expected to error(). Years before 46 BCE are not leap years. See the list at https://www.mathworks.com/matlabcentral/answers/14859-create-function-to-calculate-leap-year#answer_20236

Answer 8

[Dongchi Jiang]

clc;
year=input('Enter a year: \n');
if (rem(year,400)==0 | rem(year,4)==0 & rem(year,100)~=0)
    disp('The year is a leap year')
else
    disp('The year is not a leap year')
end

[Hardik Senta]

thank you for your great answer.