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# 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

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unction valid=valid_date(year,month,date)

if nargin==3

valid1=true;

else valid=false;

return

end

v1=[year]; v2=[month]; v3=[date];

if isscalar(v1)==true && isscalar(v2)==true && isscalar(v3)==true

valid2=true;

else valid=false;

return

end

if year>0 && 0<month && month<=12 && 0<date && date<=31

valid3=true;

else valid=false;

return

end

a=year/4; b=year/100; c=year/400;

if rem(year,4)==0 && rem(year,100)~=0

valid4=true;

else valid4=false;

end

if rem(year,100)==0 && rem(year,400)~=0

valid5=true;

else valid5=false;

end

if rem(year,400)==0

valid4=true;

else valid4=false;

end

if (month==1||3||5||7||8||10||12 && date<=31) || (month==2 && date<=29) || (month==4||6||9||11 && date<=30) && valid4==true && valid5==false

valid6=true;

else valid6=false;

end

if (month==1||3||5||7||8||10||12 && date<=31) || (month==2 && date<=28) || (month==4||6||9||11 && date<=30) && valid5==true && valid4==false

valid7=true;

else

valid7=false;

end

if valid1==true && valid2==true && valid3==true && valid6==true

valid=true;

elseif valid1==true &&valid2==true && valid3==true && valid7==true

valid=true;

else

valid=false;

return

end

why this code is not working for 2018/4/31 and 2003/2/29? and other random dates. but is works for non scalar and random leap years

##### 8 Comments

Panagiotis Papias
on 8 Feb 2021

### Answers (11)

SANTOSH KAMBLE
on 3 May 2020

Edited: SANTOSH KAMBLE
on 4 May 2020

function [valid]=valid_date(year,month,date)

y=year;m=month;d=date;

if ~isscalar(y)|| ~isscalar(m) || ~isscalar(d)

valid=false;

return

end

if y>=1 && m==1 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==2 && d>=1 && d<=28 && mod(y,4)~=0

valid=true;

elseif y>=1 && m==2 && d>=1 && d<=29 && ((mod(y,4)==0 && mod(y,100)~=0 )|| mod(y,400)==0)

valid=true;

elseif y>=1 && m==3 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==4 && d>=1 && d<=30

valid=true;

elseif y>=1 && m==5 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==6 && d>=1 && d<=30

valid=true;

elseif y>=1 && m==7 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==8 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==9 && d>=1 && d<=30

valid=true;

elseif y>=1 && m==10 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==11 && d>=1 && d<=30

valid=true;

elseif y>=1 && m==12 && d>=1 && d<=31

valid=true;

else

valid=false;

end

end

##### 15 Comments

anil Reddy
on 21 Oct 2020

if i take (0<month<=12) in place of ((12>=month) && (month>0)) why am i getting mistake

Rik
on 21 Oct 2020

Because that is not how Matlab operations work. The explanation provided by mlint explains it:

per isakson
on 29 Jul 2020

Edited: per isakson
on 29 Jul 2020

Would this function pass?

function [ valid, dt ] = valid_date( y, m, d )

% The names of the input arguments, (year,month,day), are they

% mandatory? Are these names chosen to shadow functions in the

% finance toolbox?

try

% datetime is "smart". Doc says: "Each element of DateVector

% should be a positive or negative integer value [...]. If an

% element falls outside the conventional range, datetime adjusts

% both that date vector element and the previous element."

dt = datetime( y, m, d );

% If datetime didn't adjust any element the input is a

% valid date.

% valid = all( [ year(dt)==y, month(dt)==m, day(dt)==d ] );

vec = datevec( dt );

valid = all( [ vec(1)==y, vec(2)==m, vec(3)==d ] );

catch

% datetime throws an exception when not all input values are

% integers

dt = datetime.empty;

valid = false;

end

end

##### 1 Comment

Rik
on 13 Aug 2020

I would hope so. As an instructor I would applaud lateral thinking like this. Just as I would accept code where someone did this:

for n=1%use a loop because it is a requirement

output=sum(data,2);

end

Iccu OUMOUACHA
on 25 May 2020

function valid=valid_date(year, month, day)

if year<=0 || month<=0 || day<=0 || mod( year , 1 )~=0 || mod( month , 1 )~=0 || mod( day , 1 )~=0

valid=false;

return

end

if month<=12 && (ismember(month, [4 6 9 11]) && ismember(day, [1:30]))

valid=true;

elseif month<=12 && (ismember(month, [1 3 5 7 8 10 12]) && ismember(day, 1:31))

valid=true;

elseif month==2 && (mod(year,4)==0 && mod(year,100)~=0 || mod(year,400)==0 && mod(year,100)==0) && ismember(day, 1:29)

valid=true;

elseif month==2 && ismember(day, 1:28)

valid=true;

else

valid=false;

return

end

end

it works in Matlab, when I test a non-scalar inputs, it return false. But here it doesn't work!!!!! ((Assessment 2))

What is the problem?

##### 4 Comments

Rik
on 25 May 2020

Then you should change your code.

If you think this comment is unhelpful: show the code you used to ensure the output is false for non-scalar inputs. That is the only way someone will be able to help you.

Capulus_love
on 11 Aug 2020

Edited: per isakson
on 11 Aug 2020

function x = valid_date(year,month,day)

if nargin == 3

if isscalar(year) && isscalar(month) && isscalar(day)

if month > 0 && month <= 12 && day >= 0 && year > 0

if (month == 1 || 3 || 5 || 7 || 8 || 10 || 12 && day <= 31)...

|| (month == 2 && day <= 29) && (month == 4 || 6 || 9 || 11 && day <= 30)

c0 = mod(year,4)

c1 = mod(year,100)

c2 = mod(year,400)

if (c0 == 0 && c1 ~=0) || (c1 == 0 && c2 ~= 0) || c2 == 0

x = 'true'

else

x = 'false'

end

else

x = 'false'

end

else

x = 'false'

end

else

x = 'false'

end

else

x = 'false'

end

end

why this code not working? :(

##### 1 Comment

Rik
on 11 Aug 2020

Let's try some cases that will likely point us to an issue:

valid_date(2020,2,29) %returns true

valid_date(2021,2,29) %returns false

valid_date(2021,2,28) %returns false

That last one is a problem. Can you follow the flow of your code where it should mark Feb 29 as valid only in leap years, and the rest of Feb as valid every year?

Also, your code returns the value as a char array, not a logical.

QueenX
on 31 Oct 2020

Edited: QueenX
on 31 Oct 2020

I did spend whole noon to make it work. Even though it is quite long, hope it is useful for someone.

function valid = valid_date(year, month, day)

if ~isscalar(year) || year ~= fix(year)||year<=0

valid = false;

elseif ~isscalar(month) || month >12 || month<=0 || month ~= fix(month)

valid = false;

elseif ~isscalar(day) || day >31 || day ~= fix(day)|| day<=0

valid = false;

elseif day >29 && month == 2 && rem(year,4) == 0 && rem(year,100)~=0 %leap year

valid = false;

elseif day>29 && month == 2 && rem(year,400) == 0 %leap year

valid =false;

elseif day>30 && (month == 4 ||month == 6||month == 9 ||month == 11)

valid =false;

elseif day>28 && month==2 && rem(year,100) == 0 && rem(year,4)==0 && rem(year,400)~=0 %non leap year

valid =false;

elseif day>28 && month==2 && rem(year,4)~=0 && rem(year,400)~=0 %non leap year

valid =false;

else

valid = true;

end

end

Amrut Umrankar
on 15 Nov 2020

Edited: Amrut Umrankar
on 15 Nov 2020

function valid = valid_date(year, month, day)

if ~isscalar(year) || ~isscalar(month) || ~isscalar(day) || year ~=fix(year) || month ~=fix(month) || day ~=fix(day)

valid = false;

return;

end

if month <= 0 || month >= 13 || day <= 0 || year <=0

valid = false;

return;

end

ma =0;

if month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12

ma = 1;

elseif month == 2

ma = 2;

elseif month == 4 || month == 6 || month == 9 || month == 11

ma= 3;

end

if ma == 1

if day >= 32

valid = false;

else

valid = true;

end

end

if ma == 2

if mod( year , 4 )

if day >= 29

valid = false;

else

valid = true;

end

end

if ~mod( year , 4 )

if ~mod(year , 100)

if ~mod(year, 400)

if day >= 30

valid = false;

else

valid = true;

end

else

if day >= 29

valid = false;

else

valid = true;

end

end

else

if day >= 30

valid = false;

else

valid = true;

end

end

end

end

if ma == 3

if day >= 31

valid = false;

else

valid = true;

end

end

##### 3 Comments

Amrut Umrankar
on 15 Nov 2020

Rik
on 15 Nov 2020

Big code is not a problem. I think undocumented code is a problem if you want to teach people something. You don't explain what your code is doing. Your choice of variable names also doesn't help: why use ma if you can use something more descriptive like MonthType? I know you are determining if a year is a leap year, but you don't explain that. If someone is inexperienced enough to scroll down all the way to your answer, you can't assume that would be clear to them.

If you are posting a complete solution to a homework question, at least try to teach something, instead of only providing the opportunity for cheating.

ABHIJIT BISWAS
on 29 Nov 2020

function isvalid = valid_date(y, m, d)

% Check if the inputs are valid

% Check that they are scalars

if ~(isscalar(y) && isscalar(m) && isscalar(d))

isvalid = false;

% Check that inputs are positive

elseif ~all([y, m, d] > 0)

isvalid = false;

% Check that inputs are integers (not the data type)

elseif any(rem([y, m, d], 1))

isvalid = false;

% Check that m and d are below the max possible

elseif (m > 12) || (d > 31)

isvalid = false;

% The inputs could be a valid date, let's see if they actually are

else

% Vector of the number of days for each month

daysInMonth = [31 28 31 30 31 30 31 31 30 31 30 31];

% If leap year, change days in Feb

if isequal(rem(y, 4), 0) && (~isequal(rem(y, 100), 0) || isequal(rem(y, 400), 0))

daysInMonth(2) = 29;

end

maxDay = daysInMonth(m);

if d > maxDay

isvalid = false;

else

isvalid = true;

end

end

end

##### 0 Comments

Chandan Kumar
on 3 Mar 2021

function valid = valid_date(year,month,date)

y=year;m=month;d=date;

if ~isscalar(y)|| ~isscalar(m) || ~isscalar(d)

valid=false;

return

end

if y>=1 && m==1 && d>=1 && d<=31

valid=true;

elseif y>=1 && m==2 && d>=1 && d<=28 && mod(y,4)~=0

valid=true;

elseif y>=1 && m==2 && d>=1 && d<=29 && ((mod(y,4)==0 && mod(y,100)~=0 )|| mod(y,400)==0)

valid=true;

elseif m<=12 && (ismember(m, [4 6 9 11]) && ismember(d, [1:30]))

valid=true;

elseif m<=12 && (ismember(m, [1 3 5 7 8 10 12]) && ismember(d, 1:31))

valid=true;

else

valid = false

end

% This is the shortest and oring code for all the valid date i could write

% date formaat should be in lie vald_date(year,month,date) to mae the function work

##### 1 Comment

Rik
on 3 Mar 2021

Shun Yan
on 5 Apr 2021

Why would this code not pass the scalar test? %Return false if the input is not scalar

##### 3 Comments

Shun Yan
on 5 Apr 2021

Walter Roberson
on 5 Apr 2021

You need to have the isscalar() check before the other checks.

if ~all(iscalar(a) && isscalar(b) && isscalar(c))

do whatever appropriate for error

end

freddy alexander rodriguez torres
on 12 Apr 2021

Edited: freddy alexander rodriguez torres
on 12 Apr 2021

function valid=valid_date(y,m,d)

k=y/4;

j=y/400;

i=y/100;

if ~isscalar(y) || ~isscalar(m) || ~isscalar(d) || y~=fix(y) || m~=fix(m) || d~=fix(d)

valid=false;

elseif (k==fix(k) || j==fix(j)) && m==2 && d<=29 && i~=fix(i) && d>0

valid=true;

elseif j==fix(j) && m==2 && d<=29 && d>0

valid=true;

elseif (ismember(m, [1 3 5 7 8 10 12])) && d<=31 && d>0

valid=true;

elseif (ismember(m, [4 6 9 11])) && d<=30 && d>0

valid=true;

elseif m==2 && d<=28 && i==fix(i) && d>0

valid=true;

elseif m==2 && d<=28 && i~=fix(i) && d>0

valid=true;

else

valid=false;

end

##### 0 Comments

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