Ask HN: What's Prolog like in 2024?

    % Facts
    parent(john, mary).
    parent(mary, ann).
    parent(mary, tom).

    % Rules
    ancestor(X, Y) :- parent(X, Y).
    ancestor(X, Z) :- parent(X, Y), ancestor(Y, Z).

    % Query
    ?- ancestor(john, X).

    -- Create and populate the table
    CREATE TABLE Parent (
        parent VARCHAR(50),
        child VARCHAR(50)
    );

    INSERT INTO Parent VALUES ('john', 'mary');
    INSERT INTO Parent VALUES ('mary', 'ann');
    INSERT INTO Parent VALUES ('mary', 'tom');

    -- Recursive query to find ancestors
    WITH RECURSIVE Ancestor AS (
        SELECT parent, child
        FROM Parent
        UNION ALL
        SELECT a.parent, p.child
        FROM Ancestor a
        JOIN Parent p ON a.child = p.parent
    )
    SELECT DISTINCT parent AS ancestor
    FROM Ancestor
    WHERE child IN ('ann', 'tom');

    % Define the base relation
    friend(Person1, Person2).

    % Define friend-of-friend relation
    friend_of_friend(X, Z) :- friend(X, Y), friend(Y, Z), X != Z.

    % Define potential friend recommendation
    % (friend of friend who is not already a friend)
    recommend_friend(X, Z) :- friend_of_friend(X, Z), not friend(X, Z).

    % Count mutual friends for recommendations
    mutual_friend_count(X, Z, Count) :- 
        recommend_friend(X, Z),
        Count = count{Y : friend(X, Y), friend(Y, Z)}.

    % Query to get top friend recommendations for a person
    top_recommendations(Person, RecommendedFriend, MutualCount) :-
        mutual_friend_count(Person, RecommendedFriend, MutualCount),
        MutualCount >= 5,
        MutualCount = max{C : mutual_friend_count(Person, _, C)}.

    WITH RECURSIVE
    -- Base friend relation
    friends AS (
        SELECT DISTINCT person1, person2
        FROM friendship
        UNION
        SELECT person2, person1
        FROM friendship
    ),

    -- Friend of friend relation
    friend_of_friend AS (
        SELECT f1.person1 AS person, f2.person2 AS friend_of_friend
        FROM friends f1
        JOIN friends f2 ON f1.person2 = f2.person1
        WHERE f1.person1 <> f2.person2
    ),

    -- Potential friend recommendations
    potential_recommendations AS (
        SELECT fof.person, fof.friend_of_friend, 
            COUNT(*) AS mutual_friend_count
        FROM friend_of_friend fof
        LEFT JOIN friends f ON fof.person = f.person1 AND fof.friend_of_friend = f.person2
        WHERE f.person1 IS NULL  -- Ensure they're not already friends
        GROUP BY fof.person, fof.friend_of_friend
        HAVING COUNT(*) >= 5  -- Minimum mutual friends threshold
    ),

    -- Rank recommendations
    ranked_recommendations AS (
        SELECT person, friend_of_friend, mutual_friend_count,
            RANK() OVER (PARTITION BY person ORDER BY mutual_friend_count DESC) as rank
        FROM potential_recommendations
    )

    -- Get top recommendations
    SELECT person, friend_of_friend, mutual_friend_count
    FROM ranked_recommendations
    WHERE rank = 1;