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.

Hi Folks,

I have tried and ended up in errors. For example(valid_date(2018, 4, 1) failed,The last day of every month

Variable valid has an incorrect value. valid_date(2000, 1, 31) failed..., Random leap years

Variable valid has an incorrect value. valid_date(1624, 2, 29) failed..., Random dates

Variable valid has an incorrect value. valid_date(1433, 6, 28) failed...)

Kindly point out the errors I have made!

function valid = valid_date(year,month,day)
if (nargin==3)
    if (fix(valid_date)&&isscalar(valid_date))
        if((rem(year,4)==0||rem(year,400)==0) && rem(year,100)~=0)
            if((month==1||3||5||7||9||11) && 0<day<32)
            valid=true;
            elseif(month==2 && 0<day<30)
            valid=true;
            elseif((month==4||6||8||10||12) && 0<day<31)
            valid=true;
            else
            valid=false;    
            end
        else
            if((month==1||3||5||7||9||11) && 0<day<32)
            valid=true;
            elseif((month==4||6||8||10||12) && 0<day<31)
            valid=true;
            elseif(month==2 && 0<day<29)
            valid=true;
            else
            valid=false;
            end
        end
        else
        valid=false;
        end
    else
        valid=false;
    end

[Stephen23]

Note that this syntax

month==1||3||5||7||9||11

is equivalent to:

(((((month==1)||3)||5)||7)||9)||11

which, because 11 is non-zero, will always be true. You should start thinking about using vectors, which would make that comparison easy (hint: use any)

[Guillaume]

As Stephen points out, inventing your own syntax for the if tests is not going to work very well. Note that to check if a number is a member of the set {a,b,c}, you'd use ismember(x, [a, b, c]), not some invented x==a||b||c syntax.

In this case, it didn't even matter that the comparison didn't make sense because of:

if fix(valid_date)&&isscalar(valid_date)  %removed unnecessary parentheses
    %...
end

which recurse into valid_date, this time with no argument. Since valid_date with 0 inputs always return false (thankfully, breaking the recursion), and since fix(false) is always false, the above if is always false.

As coded, the function can be simply reduced to

function valid = valid_date(~, ~, ~)
    valid = false;
end

[Rakeshwar Elango]

Thank you Stephen and Guillaume as the newly invented statement of me didn't through any error so I was wandering wheather where it went wrong. Thank you for your suggesion!

[DGM]

See also:

https://www.mathworks.com/matlabcentral/answers/498288-how-to-write-a-function-to-determine-if-a-date-entered-is-valid

https://www.mathworks.com/matlabcentral/answers/503149-write-a-function-three-positive-integer-scalar-inputs-year-month-day

https://www.mathworks.com/matlabcentral/answers/1618405-need-help-on-this-program

Answer 1

[James Tursa]

Looks like you are taking the same class as Rahul. Rather than repeat my answer here, I will simply direct you to the link:

https://www.mathworks.com/matlabcentral/answers/466971-if-else-problem-for-year

[Gloria Paul]

Been stuck on this assignment for awhile now. Very close to giving up

[Karina Medina Barzola]

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

[Divyanshu Gupta]

Hello I tried floowing code but its not working can anyone tell why?

function valid = valid_date(year,month,day)

if isinteger(year) && year>0 && isinteger(month) && month==1||month==3||month==5||month==7||month==8||month==10||month==12 && day < 32 && isinteger(day) && day>0

valid = true;

elseif isinteger(year) && year>0 && isinteger(month) && month==4||month==6||month==9||month==11 && day <31 && isinteger(day) && day>0

valid = true;

elseif isinteger(year) && year>0 && isinteger(year/4) && isinteger(year/400) && isinteger(month) && month == 2 && isinteger(day) && day>0 && day < 30

valid = true;

elseif isinteger(year) && year>0 && ~isinteger(year/4) && isinteger(month) && month == 2 && isinteger(day) && day>0 && day < 29

valid = true;

else

valid = false;

end

Answer 2

[Martín Avalos Postigo]

function valid = valid_date(year,month,day)
if sum(rem([year,month,day],1))==0 && sum([year,month,day]>0)==3
    if ismember(month,[1,3,5,7,8,10,12]) && day<32
        valid=true;
    elseif ismember(month,[4,6,9,11]) && day<31
        valid=true;
    elseif month==2 && ismember(sum(rem(year,[4,100,400])==0),[1,3]) && day<30
        valid=true;
    elseif month==2 && ismember(sum(rem(year,[4,100,400])==0),[0,2]) && day<29
        valid=true;
    else
        valid=false;
    end
else
    valid=false;
end

[Jaime Joaquin Moulic]

Thank you!!

[Alexandar]

Can you please explain what

sum(rem([year,month,day],1))==0 

does? Is it trying to say if there is a remainder of 1 then it is false?

[Walter Roberson]

@Alexandar

rem() of something with 1 is 0 if the item is integer, and is non-zero if the item has a fraction. So the rem([year,month,day],1) is creating a vector of fractions (possibly 0) for year, month, day. Those fractions are summed, and the sum is compared to 0. The hypothesis is that if the fractions are all 0 then the sum of fractions are 0, and if any of the fractions are non-zero, then the sum of the fractions would be non-zero.

However, the code has a bug:

sum(rem([-3.2,4.2,5],1))

ans = 0

compare to

sum(mod([-3.2,4.2,5],1))

ans = 1

Answer 3

[Jatan Shah]

function valid = valid_date (year, month, day)

if nargin<3
    valid = false;
    
elseif ~isscalar(year) || year<1 || year~=fix(year)
    valid = false;
    
elseif ~isscalar(month) || month<1 || month~=fix(month) || month>12
    valid = false;
    
elseif ~isscalar(day) || 1>day || day~=fix(day) || day>31
    valid = false;
    
elseif (month==4 || month==6 || month==9 || month==11) && day>30
    valid = false;
    
elseif month==2 && day>29
    valid = false;
    
elseif month==2 && day>28 && (year/4) ~= fix(year/4) 
    valid = false;
    
elseif   (year/100)==fix(year/100) && (year/400)~=fix(year/400)
    valid = false;
else
    
    valid = true;
    
end

[Aaliyah Gonzales]

this is the perfect code

Answer 4

[Abhinav Gadge]

%works correctly for all cases

function valid=valid_date(y,m,d)
days=[31,28,31,30,31,30,31,31,30,31,30,31]
if (mod(y,4)==0 & mod(y,100) ~=0) | mod(y,400)==0
        days(2)=29
end
if (isscalar(y) & isscalar(m) & isscalar(d)) & y>1  & (m>=1 & m<=12) & (d>=1 & d<=days(m))
    valid=true
else
    valid=false
end
        
end

Answer 5

[VIGNESH B S]

function [valid] = valid_date(year,month,day)
Entered = 0;

if isscalar(year) == 0
    valid = logical(0);
    Entered = 1;
elseif isscalar(month) == 0
    valid = logical(0);
    Entered = 1;
elseif isscalar(day) == 0
    valid = logical(0);
    Entered = 1;
end



if year <= 0
    valid = logical(0);
    Entered = 1;
elseif month <= 0
    valid = logical(0);
    Entered = 1;
elseif day <= 0
    valid = logical(0);
    Entered = 1;
end



leap_year = 0;
if(mod(year,4) == 0)
    leap_year = 1;
    if mod(year,100) == 0
        if mod(year,400) == 0
            leap_year = 1;
        else
            leap_year = 0;
        end
    end
end
if [leap_year,month,day] == [0,2,29]
    valid = logical(0);
    Entered = 1;
end
if [leap_year,month,day] == [1,2,29]
    valid = logical(1);
end
if month == 1
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 10
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 8
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 7
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 5
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 3
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
elseif month == 12
    if day >= 32
        valid = (0>1);
        Entered = 1;
    end
    
elseif month == 4
    if day >=  31
        valid = logical(0);
        Entered = 1;
    end
elseif month == 9
    if day >=  31
        valid = logical(0);
        Entered = 1;
    end
elseif month == 6
    if day >=  31
        valid = logical(0);
        Entered = 1;
    end
elseif month == 11
    if day >=  31
        valid = logical(0);
        Entered = 1;
    end
end
if month == 2
    if day >= 30
        Entered = 1;
        valid = logical(0);
    end
end

if month >= 13
    Entered = 1;
    valid = logical(0);
end


if Entered == 0
    valid = logical(1);
end

end
%A HELL LINES OF CODES!!!!!!! BUT IT WORKED OUT !!

[Sri]

workd in all inputs that i tried.

Answer 6

[Aramis]

The best answer

function valid = valid_date(year,month,day)
    if not(isscalar(year))|| not(isscalar(month)) || not(isscalar(day))|| min([year month day]) <1 || month>12 || day>31 || nargin<3
        valid = false;
    elseif month == 2
        if mod(year,4) == 0 && not(mod(year,100) == 0) || mod(year,400) == 0
            valid = day<30;
        else
            valid = day<29;
        end
    elseif max(month == [4 6 9 11])
        valid = day<31;
    else
        valid = true;
    end
end

Answer 7

[S HARIPRAKASH]

valid date or not

function valid=valid_date(year,month,day)
%check whether the given input is scalar positive integer or not
%1<=day<=31
%1<=month<=12
if((~isscalar(year))||(~isscalar(month))||(~isscalar(day))||(year<=0)||(month<=0)||(day<=0)||(month>12)||(day>31)||(~(year==fix(year)))||(~(month==fix(month)))||(~(day==fix(day))))
    valid=false;
    return
end
%february has 29 days in leap year and 28 days in non leap year
if (month==2)
    if((mod(year,4)==0 && mod(year,100)~=0)||(mod(year,400)==0))
        if day<30
            valid=true;
            return
        else
            valid=false;
            return
        end
    else
        if day<29
            valid=true;return
        else
            valid=false;return
        end
    end
%check whether the number of given days is within the limit of days that particular month has.
else
    month_days=[31,28,31,30,31,30,31,31,30,31,30,31];
    if day<=month_days(month)
        valid=true;
        return
    else
        valid=false;
        return
    end
end
end

Answer 8

[Fazal Hussain]

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

[Erikc Anderson Cortez Benites]

function valid = valid_date(year, month, day)
if isscalar(year) && isscalar(month) && isscalar(day)
    if month < 1 ||  month > 12
        valid = false;
    else
        if month == 2 && isleapyear(year)
           valid = day <= 29 && day>0 ;  
        elseif sum(month == [1,3,5,7,8,10,12])
            valid = day <= 31 && day>0;  
        else
            valid = day <= 30 && day>0; 
        end
    end
else 
    valid = false;
end 


function re = isleapyear(year)
if mod(year,400) == 0
    re = 1;
elseif (mod(year,4) == 0) && (mod(year,100)) ~= 0 
    re = 1;
else
    re = 0;
end

Answer 10

[Ravleen Kaur]

function valid = valid_date(year,month,day)
if (isscalar(year) && (year>=1) && year==fix(year)) && (isscalar(month) && ((1<=month)&&(month<=12)) && (month==fix(month))) && (isscalar(day) && ((1<=day)) && (day==fix(day)))
    if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12) && day<32 
      valid = true;
    elseif ( month == 4 || month == 6 || month == 7 || month == 9 || month == 11)  && day<31
      valid = true;
    else 
      valid = false;
    end
    if (month == 2)
        if day<29
        valid = true;  return
        elseif day<30 && rem(year,4)==0 && rem(year,100)~=0 || rem(year,400)==0
        valid = true;  return
        else
        valid = false; return
        end
    end
else
    valid = false; return
end

Answer 11

[David Goh]

function valid = valid_date(year,month,day)

if isscalar(year) == 1 && isscalar(month) == 1 && isscalar(day) == 1 && year > 0 && month > 0 && day > 0
    if month <= 12
        if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)
            if day <= 31
                valid = true;
            else 
                valid = false;
            end
        elseif month == 2
            if (rem(year,4)==0 || rem(year,400)==0) && (rem(year,400)~=100 && rem(year,400)~=200 && rem(year,400)~=300)
                if day <= 29
                    valid = true;
                else 
                    valid = false;
                end
            else 
                if day <= 28
                    valid = true;
                else 
                    valid = false;
                end
            end
        else 
            if day <= 30
                valid = true;
            else 
                valid = false;
            end
        end
    else 
        valid = false;
    end
else 
    valid = false;
end

Answer 12

[Muhammad]

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

[Walter Roberson]

It is not efficient to keep checking isscalar()

I think you should check the documentation about the relative priority of the two short-circuit logic operators. I do not recommend chaining them together without using () to make the interpretation clear to people reading the code

[Muhammad]

Thanks for pointing that out. I'll definitely revisit the relative priority of the two short-circuit logic operators

Answer 13

[ARYA]

function valid= valid_date(year,month,day)
flag=0;
if(month<=0||month>12||day<=0||day>31)
    flag=0;
else
    if(month==4||month==6||month==9||month==11)
        if day<=30
            flag=1;
        else
            flag=0;
        end
    elseif month==2
        if (mod(year,4)==0&&mod(year,100)~=0)||(mod(year,400)==0)&& day<=29
            flag=1;
        elseif (day<=28)
            flag=1;
        else 
            flag=0;
        end
    else
        if  day<=31
            flag=1;
        else
            flag=0;
        end
    end
end 
if(~isscalar(year)||~isscalar(month)||~isscalar(day))
    flag=0;
    
end

if(flag==0)
    valid=false;
else
    valid=true;
end

[Samiullah]

this is the perfect code

[Samiullah]

thank you

Answer 14

[Paul]

function valid = valid_date(y,m,d);
if ~isscalar(y)||~isscalar(m)||~isscalar(d)||y~=fix(y)||m~=fix(m)||d~=fix(d);
    valid = false;
else
    if y > 0 && (m == 12 || m == 1 || m == 3 || m == 5|| m == 7 || m == 8 || m == 10) && d >= 1 && d <= 31;
        valid = true;
    elseif (m== 4 || m == 6 || m ==9 || m == 11) && d >=1 && d <= 30;
        valid = true;
    elseif m==2 && d>=1 && d<=28;
        valid = true;
    elseif ((mod(y,4)==0 && mod(y,100)~=0)||mod(y,400)==0) && m==2 && d>=1 && d<=29;
        valid = true;
    else 
        valid = false;
    end
end
end

[DGM]

This fails for negative years. When you have a control structure with a single output and you have multiple cases where the output is identical, that's a good sign that you need to simplify things. Similarly, see how many times you're checking that d>=1.

I can't fault you too much, since your example is at least relatively compact already, but it still squarely fits the theme of giant tangled salads of undocumented inline logic in nested control structures. It's easy to miss certain cases when you can't really even read the logic or understand how it's responding to the inputs. You can make it easier on yourself if you use logical variables to describe the aspects of the input, and then simply calculate the output based on a logical combination of those variables. That way you can easily see whether things are being evaluated correctly.

Answer 15

[DGM]

Instead of a giant undocumented salad of inline logic, use logical variables to clearly describe the input conditions as they pertain to the relevant concepts.

function valid = valid_date(y,m,d)
% VALID = VALID_DATE(Y,M,D)
% Synopsis goes here blah blah blah 

    % basic input validation
    scalarinput = isscalar(y) && isscalar(m) && isscalar(d);
    integerinput = ~mod(y,1) && ~mod(m,1) && ~mod(d,1);
    nonsenseinput = y<1 || m<1 || m>12 || d<1 || d>31;
    acceptableinput = scalarinput && integerinput && ~nonsenseinput;
    
    % three classes of month length
    isleapyear = ~mod(y,4) && (mod(y,100) || ~mod(y,400));
    validmonthA = ismember(m,[1 3 5 7 8 10 12]) && d<=31;
    validmonthB = ismember(m,[4 6 9 11]) && d<=30;
    validmonthC = m==2 && d<=(28 + isleapyear); % implicit casting of logicals to numeric!
    
    % assemble the output based on those conditions
    valid = acceptableinput && (validmonthA || validmonthB || validmonthC)
end

The variable names are self-explanatory, and tell you exactly what's being tested on each line. At a glance, you know that the output is only true when the given inputs are scalar, integer, within sensible ranges, and together describe one of three classes of month (31 days, 30 days, and February).

This example leans heavily on implicit casting behaviors in MATLAB. The use of comparision or negation operators will make sure the numeric output of mod() is always logical. Similarly, adding a logical to a numeric value treats the logical as either zero or one.

Could this be made more efficient using an if-else structure? Yes, though for scalar inputs, I doubt that the benefit of early returns are even measurable. The reason I did it this way is simply for emphasis.

Did I miss something? I don't know. I'm not the one taking the online course.