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41 lines
1.9 KiB
Prolog
41 lines
1.9 KiB
Prolog
% month_days(Month, DaysInMonth, DaysBeforeMonth)
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% returns the number of days in Month and the number of days it takes to reach that date in year 2023
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% prolog works by defining facts and rules and when queried about them returns values
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% it works different from functional programming in this aspect since it
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% doesn't just follow instruction it returns output based on world state
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month_days('01', 31, 0).
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month_days('02', 28, 31).
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month_days('03', 31, 59).
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month_days('04', 30, 90).
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month_days('05', 31, 120).
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month_days('06', 30, 151).
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month_days('07', 31, 181).
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month_days('08', 31, 212).
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month_days('09', 30, 243).
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month_days('10', 31, 273).
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month_days('11', 30, 304).
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month_days('12', 31, 334).
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% day_of_year(Date, DayOfYear) converts a date to day of year number later used to calculate interval
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% It also checks if number of days given by user is smaller or equal to number of days in the given month and if no then it gives fail
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day_of_year(Date, DayOfYear) :-
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atom_chars(Date, Chars),
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append(DayChars, MonthChars, Chars),
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atom_chars(Day, DayChars),
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atom_chars(Month, MonthChars),
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month_days(Month, DaysInMonth, MonthDays),
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atom_number(Day, DayNumber),
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((DayNumber =< DaysInMonth, DayNumber > 0) -> DayOfYear is MonthDays + DayNumber) ; fail.
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% interval(Date1, Date2) prints the number of days between Date1 and Date2
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% We always expect date to be in format ddmm where 'd' stands for day and 'm' stands for month
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% if the month or day is just a single digit we expecte there to be zero in front
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% (like 0505 for 5th of may or 1105 for 11th of may or 0511 for 5th of november)
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% write(Interval) prints out result, nl writes newline to make output conform to project requirements as much as possible
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interval(Date1, Date2) :-
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(day_of_year(Date1, DayOfYear1), day_of_year(Date2, DayOfYear2) ->
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Interval is abs(DayOfYear2 - DayOfYear1),
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write(Interval), nl
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;
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write('Invalid input.'), nl, fail
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). |