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Hence we store it in a string called 'path'. " then i minusab I; path plusab "N" elif path= " " then stuck := true C entrance is a dead end C else int p =upb path; C now backtrack and prune path C char lastmove = path[p]; if lastmove = "E" then j rninusab I elif lastmove = "S" then i rninusab I elif lastmove = "W' then j plusab l elif lastmove = "N" then i plusab 1 fi; path:= path [I :p- I] fi end' od; if stuck then print("N 0 PATH POSSIBLE") else print (("MAZE SOLVED BY FOLLOWING THE PATH" ,newline,path)) fi There is a problem if the path has length greater than the width of the lineprinter page, since the print routine abandons the output of a string when the end of the line is reached.

1 Perform this algorithm for the date 29 Feb 2000. Apart from the input and output, the algorithm comprises two logically distinct calculations, and their isolation as separate steps makes it easier to follow and check. A self-contained piece of program that takes some data like 'y', or even no data, and defines some operation possibly producing a result, like 'advance days(y)', can be made into a procedure. There are standard procedures like 'sin' or 'sqrt' that can be used in any program by writing 'sin(x)' or 'sqrt(x)' where 'x' is real.

3 Given the declaration 'struct (string nationality, int age, struct(int day,[I :3) char month, int year)birthdate) john', write assignment statements to make John British and born on Christmas day 1940. We now use the procedural approach in writing a program for the calendar problem, including the new language features. The input is a CALENDARS: PROCEDURES 41 date, like 3 MAY 1999 . 'read(date)' will read this row of denotations into their respective fields. The output is to be of the form "3 MAY 1999 IS A MONDAY".

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