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Code Reviews Solving logical puzzle with negation and undefined aspects in Prolog

Assume this trivial logic puzzle which I have made up: There are three boys, Fred, John and Max. No two of the boys have the same age. Max is older than John. Fred is not the oldest one. Quest...

0 answers · posted 2y ago by Dirk Herrmann‭ · edited 2y ago by Dirk Herrmann‭

Question prolog

#2: Post edited by

Dirk Herrmann‭ · 2022-05-11T23:05:32Z (over 2 years ago)

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Assume this trivial logic puzzle which I have made up:
There are three boys, Fred, John and Max. No two of the boys have the same age. Max is older than John. Fred is not the oldest one. Question 1: Who is the oldest one? Question 2: What could be possible orderings by age?
Obviously, Max is the oldest one. And, from the given information the following two orderings are possible: Max,Fred,John and Max,John,Fred.
But, formulating this trivial logic puzzle in Prolog appeared to be surprisingly difficult (at least for me, as I am still learning Prolog). Particularly the statement "Fred is not the oldest one" caused me a lot of headache. While this restricts the solution space, it still leaves possibilities open with respect to whether Fred is older than John or not.
~~In the end, I came up with the following solution, which works with SWI Prolog:~~
``` prolog
boys(Boys) :- Boys = [fred,john,max].
older(A,B,List) :-
nth0(A_idx,List,A),
nth0(B_idx,List,B),
A_idx < B_idx.
fromoldest(A,B,C) :-
boys(Boys), % there are three boys
permutation(Boys,[A,B,C]), % all of different age
older(max,john,[A,B,C]), % max is older than john
dif(A,fred). % fred is not the oldest
```
I would be glad to get feedback on this approach:
* Is this idiomatic Prolog or what constructs should be preferred?
* Are there more straightforward ways of solving this?
* What about performance (like, use of `permutation`)?

Assume this trivial logic puzzle which I have made up:
There are three boys, Fred, John and Max. No two of the boys have the same age. Max is older than John. Fred is not the oldest one. Question 1: Who is the oldest one? Question 2: What could be possible orderings by age?
Obviously, Max is the oldest one. And, from the given information the following two orderings are possible: Max,Fred,John and Max,John,Fred.
But, formulating this trivial logic puzzle in Prolog appeared to be surprisingly difficult (at least for me, as I am still learning Prolog). Particularly the statement "Fred is not the oldest one" caused me a lot of headache. While this restricts the solution space, it still leaves possibilities open with respect to whether Fred is older than John or not.
In the end, I came up with the following solution, which works with SWI Prolog if you issue the query `?- fromoldest(A,B,C).`:
``` prolog
boys(Boys) :- Boys = [fred,john,max].
older(A,B,List) :-
nth0(A_idx,List,A),
nth0(B_idx,List,B),
A_idx < B_idx.
fromoldest(A,B,C) :-
boys(Boys), % there are three boys
permutation(Boys,[A,B,C]), % all of different age
older(max,john,[A,B,C]), % max is older than john
dif(A,fred). % fred is not the oldest
```
I would be glad to get feedback on this approach:
* Is this idiomatic Prolog or what constructs should be preferred?
* Are there more straightforward ways of solving this?
* What about performance (like, use of `permutation`)?

#1: Initial revision by

Dirk Herrmann‭ · 2022-05-11T23:00:21Z (over 2 years ago)

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Solving logical puzzle with negation and undefined aspects in Prolog

Assume this trivial logic puzzle which I have made up:

There are three boys, Fred, John and Max.  No two of the boys have the same age.  Max is older than John.  Fred is not the oldest one.  Question 1: Who is the oldest one?  Question 2: What could be possible orderings by age?

Obviously, Max is the oldest one.  And, from the given information the following two orderings are possible: Max,Fred,John and Max,John,Fred.

But, formulating this trivial logic puzzle in Prolog appeared to be surprisingly difficult (at least for me, as I am still learning Prolog).  Particularly the statement "Fred is not the oldest one" caused me a lot of headache.  While this restricts the solution space, it still leaves possibilities open with respect to whether Fred is older than John or not.

In the end, I came up with the following solution, which works with SWI Prolog:

``` prolog
boys(Boys) :- Boys = [fred,john,max].

older(A,B,List) :-
    nth0(A_idx,List,A),
    nth0(B_idx,List,B),
    A_idx < B_idx.

fromoldest(A,B,C) :-
    boys(Boys),                 % there are three boys
    permutation(Boys,[A,B,C]),  % all of different age
    older(max,john,[A,B,C]),    % max is older than john
    dif(A,fred).                % fred is not the oldest
```

I would be glad to get feedback on this approach:
* Is this idiomatic Prolog or what constructs should be preferred?
* Are there more straightforward ways of solving this?
* What about performance (like, use of `permutation`)?

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