'Prolog : append to list

I have facts like that : student(stud01, 'Programming 1', 90). student(stud01, 'Math 1', 78). student(stud01, 'Statistics 1', 94). student(stud01, 'Electronics 1', 81). student(stud01, 'Management', 66). student(stud01, 'English', 83).

i want to build a predict that take the student id and and put all its grade in list and after that take the max grade

i try this : stuGrade(ID,List):- stuGrade([],ID,List).

stuGrade(TmpList,ID,List):- student(ID,Course,Grade), not(mem([ID],TmpList)), !, append_List([Grade],TmpList,NewTmpList), stuGrade(List,ID,NewTmpList).

stuGrade(List,ID,List).

maxStudentGrade(ID,Max):- stuGrade(ID,L), max(L,Max).

but the problem that after using ?-trace. i see that the output list containing only the first grade (90) how can i slove this problem

listprologappendmaxpredicate


Solution 1:[1]

Suppose that you begin with ?- stuGrade(stu01, List).

Then you will get the first student entry at least. And you say there is some issue after that.

So, a simple way to check is to test by using the first part of the query, such that,

?- ID = stu01, TmpList = [],
   student(ID,Course,Grade),
   not(mem([ID],TmpList)), !.

When you find its result is false, stop and check the last predicate.

Using break points

trace/0 and nodebug/0 are nice tools.

You may want to check issues by like the following,

stuGrade(TmpList, ID, List) :-
  student(ID, Course, Grade),
  trace,  % <------------------ Put a break point here.
  not(mem([ID], TmpList)),
  nodebug, % <----------------- Here, stop tracing.
  !,
  append_List([Grade], TmpList, NewTmpList),
  stuGrade(List, ID, NewTmpList).

Good luck.

Solution 2:[2]

And you code...

stuGrade(ID,List):- stuGrade([],ID,List).

stuGrade(TmpList, ID, List):-
  student(ID, Course, Grade),
  not(mem([ID], TmpList)),
  !, 
  append_List([Grade], TmpList, NewTmpList),
  stuGrade(List, ID, NewTmpList).

stuGrade(List,ID,List).

Not bad. But, consider the first clause of stuGrade/3.

When you get a Grade, it's prepended to TmpList and you get NewTmpList. But at the next line, the newly built list is put as the third argument, as you know that it's for the final returned List.

It cannot be that. The first parameter of stuGrade/3, TmpList, is the accumulating parameter. Then NewTmpList might be placed as the first argument of the recursive query, such that stuGrade(NewTmpList, ID, List).

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Yau-Hsien Huang
Solution 2 Yau-Hsien Huang

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