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!

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

[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 6

[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 !!

Answer 7

[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 8

[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 9

[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 10

[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 11

[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 12

[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