Prolog偏好评估程序控制台双输入功能故障排查请求

Let's tackle your problem step by step—your core issue is that the start/0 predicate isn't handling user input correctly, and there's also a subtle bug in your reverse preference rule that could cause unexpected behavior or infinite recursion.

First, Fix the Reverse Preference Rule #

Your current reverse rule has two critical issues:

preferencia(X,Y,D):-preferencia(Y,X,D),!. % reverse preferenciaance

1. Logical mistake: If Ana prefers Maria with a value of 1, Maria should logically prefer Ana with -1, not the same value.
2. Infinite recursion risk: If no direct preference exists between two people, this rule will keep calling itself back and forth forever.

Replace it with this corrected version:

preferencia(X, Y, D) :-
    preferencia(Y, X, RevD),
    D is -RevD,
    !.

This ensures reverse preferences are the negative of the original, and the cut (!) prevents unnecessary backtracking that could lead to loops.

Fix the start/0 Predicate #

Your original start only reads one input value and passes an invalid list structure to eval/2. We need to capture two names from the user and feed them correctly to your evaluation function. Here are two working options:

Option 1: Single Comma-Separated Input

Let users enter both names in one line (e.g., rui, maria):

start :-
    write('Pick 2 Persons to make a group (e.g., rui, maria): '),
    read((Name1, Name2)), % Reads the comma-separated pair as a compound term
    eval([Name1, Name2], DS),
    write('Preference value: '),
    write(DS),
    nl. % Adds a newline for readability

Option 2: Two Separate Inputs

Let users enter one name, press enter, then enter the second:

start :-
    write('Enter first person: '),
    read(Name1),
    write('Enter second person: '),
    read(Name2),
    eval([Name1, Name2], DS),
    write('Preference value: '),
    write(DS),
    nl.

Why This Works #

• Both versions correctly capture two names from the user, no unbound variables left hanging.
• We pass the list [Name1, Name2] directly to eval/2, which matches your working static test call (eval([rui,maria],DS)).
• The fixed reverse rule ensures that even if there's no direct preference fact for a pair (like Maria to Ana), the program will compute the inverse correctly instead of looping.

Test It Out #

If you run start and input rui, maria, the program will return Preference value: 1—just like your static test. If you input maria, rui, it will return -1 (thanks to the fixed reverse rule), which aligns with logical preference symmetry.

内容的提问来源于stack exchange，提问作者Macronaldo