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

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

[Stephen23]

Most of the bugs in your code have been thoroughly discussed before:

https://www.mathworks.com/matlabcentral/answers/453107-how-can-i-solve-this-can-anyone-provide-me-with-code-whats-wrong-with-my-code

https://www.mathworks.com/matlabcentral/answers/469201-write-a-function-called-valid_date-that-takes-three-positive-integer-scalar-inputs-year-month-day

What is the point in renaming the data?:

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

[Julian Veran]

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

[Walter Roberson]

Julian,

before posting, please use the MATLAB editor to "smart indent". Then when you post, first click on the '>' button in the CODE toolbar section of the Answers editor, and then paste the code into the highlight box that appears: that will cause the code to be posted in formatted form.

[Amrut Umrankar]

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

[Panagiotis Papias]

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

leap_year = false; % We set the leap year as false 

if nargin == 3   

    % Year prerequisites:
    if ~isscalar(year) || year ~= fix(year) || year < 0, valid=false;  %error('Invalid input as a year');
    else
        % leap year prerequisites:
        if rem(year,4) == 0 
            if rem(year,400) == 0, leap_year=true;
            else 
                if rem(year,100) == 0, leap_year=false;
                else leap_year= true;
                end
            end
        end
        
        % Month prerequisites:
        if ~isscalar(month) || month ~= fix(month) || month < 0 || month > 12, valid = false;  %error('Invalid input as a month');
        else
            % Day prerequisites:
            if ~isscalar(day) || day ~= fix(day) || day < 0 || day > 31, valid = false; %error('Invalid input as a day');
            else
                
                % months with odd (i.e.31) number of days:
                if month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month ==12 && day <= 31, valid = true;
                else,valid = false; end
                
                % months with even (i.e.30)number of days:
                if month == 4 || month == 6 || month == 9 || month == 11 && day <= 30; valid = true;
                else, valid = false; end
                    
                % February -special- Prerequisites:
                if month == 2 && day > 28 && leap_year == false, valid = false;  %error('February has maximum 28 days'); 
                elseif month == 2 && day > 29 && leap_year == true, valid = false; %error('February, for a leap year has at maximum 29 days');
                else,valid = true; end
                
            end
        end
    end
    
else
    valid = false; %error('The number of input arguement has to be 3 and be with this order: "year","month","day"');
end

end

[Panagiotis Papias]

The code is running ok, but when it comes to test:

valid_date(2007, 4, 31) fails to comply :/ i am stuck for hours :(((

[Rik]

Did you use the debugger to run your code line by line for this input?

[Panagiotis Papias]

Tbh, no you are right i have to get more familiar with the debugging. However, i followed sb's else instructions and i managed to solve the assignment 1 hr later. I think i got it what i did wrong

Answer 1

[SANTOSH KAMBLE]

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
    

[Walter Roberson]

when y mod 4 is nonzero then y mod 100 cannot be 0, since 100 is 4*25

[SANTOSH KAMBLE]

Thanks for bringing it to my notice,now it's been resolved.

[Walter Roberson]

I would point out to you that if you were to use

if y < 1 | m < 1 | d < 1 | m > 12 | d > 31
    valid = false;
elseif .....
end

then you would not have to keep rechecking y>=1 & d>=1 and that you would be able to skip the d<=31 on months that have 31 days

I would also point out that

if m == 1 || m == 3 || m == 5 (and so on)

handles a lot of cases all in one place without needing to test month by month.

[Santosh Kamble]

That's a very useful suggestion. Thanks a lot.

[Fakhrul Alam Anik]

You didn't write code for "integer" input. It will show any fraction date as valid.

[JIGNESH PATIL]

I am getting incorrect answer only for the last day of every month. What changes needs to be done in the above syntax? Anyone?

[Walter Roberson]

JIGNESH PATIL: what code are you using?

[Saurov Baidya]

nice ....this was simple of all... but lengthy

[Saurov Baidya]

function [valid] = valid_date(y, m, d)
if ~isscalar(y) || ~isscalar(m) || ~isscalar(d)
    valid = false
return
end
if y>=1 && ismember(m,[1 3 5 7 8 10 12]) && d>=1 && d<=31
    valid=true;
elseif y>=1 && ismember(m,[4 6 9 11]) && d>=1 && d<=30
    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;
else valid = false
    return
end
end

[Walter Roberson]

Notice that if you were to use

if ~isscalar(y) || ~isscalar(m) || ~isscalar(d) || y < 1

then you would not have to keep checking y>=1

[Khaled Bin Easin]

function valid = valid_date(year,month,day)
if isscalar(year) && isscalar(month) && isscalar(day)
    valid = year>0 && ((12>=month) && (month>0)) && ((31>=day) && (day>0)) && year==fix(year) && month==fix(month) && day==fix(day) && (((month==1 || month==3 || month==5 || month==7 || month==8 || month==10 || month==12) && day<=31) || ((month==4 || month==6 || month==9 || month==11) && day<=30) || (((((rem(year,4)==0 && rem(year,100)~=0) || (rem(year,400)==0)) && day<=29) || (day<=28)) && month==2));
else 
    valid=false;
end

What about this code!!

[Gourav Soni]

This is good and simple

[Shaked Baharav]

Hello, Can you explain to me what you did in the code?

[anil Reddy]

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

[Rik]

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

Answer 2

[per isakson]

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

[Rik]

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

Answer 3

[Iccu OUMOUACHA]

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?

[Rik]

It doesn't return false, it returns an error:

valid_date([2020 2020], 11, 5)

[Iccu OUMOUACHA]

the problem remained

[Rik]

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.

[Chandan Kumar]

looks like you didnt verifyng the input is scalar or not

Answer 4

[Capulus_love]

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? :(

[Rik]

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.

Answer 5

[Adonis Thirafi Hugo Mafaza]

Why is my code not working?

[Rik]

Because you didn't use the debugging tools to run your code line by line.

Answer 6

[QueenX]

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

[Rik]

It is a good exercise to write a function like this, but why did you post this? What does it add to the previous answers?

[QueenX]

I simply think that there are a lot of ways to solve this code. Just want to share my view. Is there any problem? btw

Answer 7

[Amrut Umrankar]

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

[Rik]

Why did you post this? What does it teach? Why should it not be deleted?

[Amrut Umrankar]

It is just Another way to solve same problem, As I have used pretty basic coding that's why it is big. It can be optimized.

[Rik]

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.

Answer 8

[ABHIJIT BISWAS]

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

Answer 9

[Chandan Kumar]

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

[Rik]

Shorter code is possible with a different strategy, see e.g. this answer.

My opinion would be that the shortest code is not always the best code. The best code in my opinion would be maintainable (so well-commented), as well as fast.

Answer 10

[Shun Yan]

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

[Walter Roberson]

if d==fix(d)&&e~=fix(e)

Suppose d is a vector. Then d==fix(d) would result in a vector the same size. Then you would have the && operation. But the && operation can only be used when both sides evaluate to scalars (exception: if the left side evalutes to scalar false, then with && the right side is ignored.)

[Shun Yan]

oh so tthe isscalar can't be used in tthat whole bunch of &&? So can i just write a seperate if statements with all tthe ~isscalars att the front, I think that'll work right?

[Walter Roberson]

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

Answer 11

[freddy alexander rodriguez torres]

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