# If else problem for year

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RAHUL ANTIL on 13 Jun 2019
Commented: abdul kabeer on 29 Mar 2023
Write a function called valid_date that takes three positive integer scalar inputs year, month, day. If these three represent a valid date, return a logical true, otherwise false. The name of the output argument is valid. If any of the inputs is not a positive integer scalar, return false as well. Note that every year that is exactly divisible by 4 is a leap year, except for years that are exactly divisible by 100. However, years that are exactly divisible by 400 are also leap years. For example, the year 1900 was not leap year, but the year 2000 was.
valid = valid_date(2018,4,1)
valid = valid_date(2018,4,31)
Matthew Myers on 16 Jan 2021
Edited: DGM on 21 Feb 2023
Hello everyone, I wanted to post my attempt at this and try to understand my mistakes in writing my code.
I notice that during the function, the month does not trigger the "limit" variable that I have set and I cannot understand why. I would appreciate any help with this.
I tried to take your advice Rik in takling each issue one at a time, but I didn't go about it in the right way
function[out] = valid_date(year,month,day)
out = false
if nargin<3
return
elseif ~isscalar(year) || year<1 || year ~= fix(year)
return
elseif ~isscalar(month) || month<1 || month>12 || month ~= fix(month)
return
end
persistent limit;
if isempty(limit), limit = 0;end
if month == (1:2:11)
limit = 31;
elseif month == (4:2:12)
limit = 30;
elseif month == 2
limit = 28;
end
%leap year part. Month must be february
if year == (0:4:3000) && month == 2
limit = 29;
elseif year == (0:100:3000) && month == 2
limit = 28;
else year == (0:400:3000) && month == 2
limit = 29;
end
if ~isscalar(day) || day<1 || day>limit|| day ~= fix(day)
return;
end
if isscalar(day) && day>=1 && day<=limit
out = true;
end
end
Rik on 16 Jan 2021
I don't understand your indentation and why you are using a persistent variable. I would also suggest to always make sure that conditionals are scalar. I would also suggest making sure your code can handle dates beyond the year 3000.
Did you use the debugger to step through your code line by line?

James Tursa on 14 Jun 2019
Edited: James Tursa on 14 Jun 2019
All of those if-elseif blocks make the code difficult to read, and difficult to debug as well. I would advise against that approach, and instead tackle each issue one at a time and code only for that issue. That makes the logic of each test much easier to read and to debug. For example, here is an outline of what the code could do using words:
function valid = valid_date(year,month,day)
valid = false; % Set a default return value
% Check for positive integer scalar inputs
if( the year is not a positive integer scalar )
return;
end
if( the month is not a positive integer scalar )
return;
end
if( the day is not a positive integer scalar )
return;
end
% Check for proper month
if( the month is not between 1 and 12 inclusive )
return;
end
% Construct an array of the number of days in each month
days_in_month = a 12-element array of the number of days in each month;
% Check to see if this is a leap year
is_leap_year = (you put some code here to determine if the year is a leap year)
% If it is a leap year, change February number of days to 29
if( is_leap_year )
Change the days_in_month value for February to 29
end
% Check to see if the day number is valid
if( the day is greater than the number of days in the month )
return;
end
% Passed all of our checks, so the date must be valid
valid = true;
return;
end
Once you are satisfied that the words do what you want (you should check this yourself ... maybe I missed something that needs to be checked), then you need to write actual code for all of the worded places. The code you write for this outline will in many places be pieces of the code you have already written in your if-elseif blocks above.
Rik on 31 Oct 2020
You don't need to, it is just an easy method. If you do assign false as a kind of default value, you can simply use return to exit the function and say the date is invalid.
Rik on 18 Dec 2020
You are actually explicitly writing on the public internet that you want to cheat? I admire your courage.
You are correct: it would be much faster to copy code by someone else. However, you would not learn. Next time you will need someone else to make your homework again. If you are in a hurry, that probably means you delayed doing your homework. Why should others come to your rescue?

Edited: DGM on 21 Feb 2023
I have tried this program and i got all the answers correct. Please check
function valid=valid_date(y,m,d)
valid = false;
if(((isscalar(y) && y>=1 && y==fix(y))&& (isscalar(m) && m>=1 && m==fix(m) && m<=12)...
&& (isscalar(d) && d>=1 && d==fix(d) && d<=31))==1)
c1= (ismember(m,[4,6,9,11]) && ismember(d,[1:30]));
c2=(ismember(m,[1,3,5,7,8,10,12]) && ismember(d,[1:31]));
if ((c1==1 || c2==1)==1)
valid = true;
else
if ((mod(y,4)==0&&mod(y,100)~=0 || mod(y,400)==0&&mod(y,100)==0)==1)
if (ismember(d,[1:29])==1)
valid = true;
end
return;
else
if (ismember(d,[1:28])==1)
valid = true;
end
return;
end
end
end
end
##### 3 CommentsShow 1 older commentHide 1 older comment
Mohammad Aiyoob Rahmani on 18 Jun 2020
this code is right
it is working properly
abdul kabeer on 29 Mar 2023
This is good

Salman P H on 28 Apr 2020
function valid = valid_date(x,y,z)
t = (isscalar(x) && isscalar(y) && isscalar(z));
if t==false
valid = false;
return;
end
if (x<=0 || y<=0 || z<=0)
valid = false;
return;
end
if any(rem([x, y, z], 1))
valid = false;
return;
end
if (((rem(x,4)==0) && (rem(x,100)~=0)) || (rem(x,400)==0))
a=1;
else
a=0;
end
if (x>0) && a==1
if (y==1 || y==3 || y==5 || y==7 || y==8 || y==10 || y==12) && (z>0 && z<=31)
valid = true;
elseif (y==4 || y==6 || y==9 || y==11) && (z>0 && z<=30)
valid = true;
elseif (y==2 && (z>0 && z<=29))
valid = true;
else
valid = false;
end
elseif x>0 && a==0
if (y==1 || y==3 || y==5 || y==7 || y==8 || y==10 || y==12) && (z>0 && z<=31)
valid = true;
elseif (y==4 || y==6 || y==9 || y==11) && (z>0 && z<=30)
valid = true;
elseif (y==2 && (z>0 && z<=28))
valid = true;
else
valid = false;
end
end
arjun gupta on 11 Oct 2022
Perfect :))
khaula on 1 Jan 2023
after trying all other codes, i found this one correct. thanks

Rik on 13 Jun 2019
You can do this two ways:
Option 1 is to do the actual work. Is the input correct? Does the month entered actually have at least as many days as the day input (taking leap years into account)?
Or option 2: cheat by using the builtin functions to convert your input to a datetime scalar, and extract the year,month,day numbers from that. If those match the input, your date is valid. You should still put in a check if the inputs are positive scalars.
Divya Nangaru Sudhakar on 24 Jun 2019
I tried executing that function explained by James's, I dint get the output. Can you please help me with this function.
Thank you
Rik on 24 Jun 2019
James didn't provide a full function, because it is a homework exercise. If you want me to make your homework, first make sure I'll get the points from your teacher. (You can find guidelines for posting homework on this forum here.)
Did you read the documentation for numel? And how did you try to implement either my solution or the solution by James?