4 ms·

How are the two any different? Can you come up with any differences in expressive power between the first-order logic/Horn clauses that underpin Prolog and the

How are the two any different? Can you come up with any differences in expressive power between the first-order logic/Horn clauses that underpin Prolog and the relational algebra that underpins SQL? I wonder whether you're accurately portraying the structural foundations of either Prolog or SQL.

nevertoolate 3y ago
You want to learn or to debate :)
- yowlingcat 3y ago
  I get where you're coming from but your point seems flippant. My answer is "debate" -- and looking at the rest of my subthread, some commenters have raised some great concrete areas around Prolog's unique strengths, which I've found helpful!
hyperpape 3y ago
Relational algebra is not Turing complete. SQL gets Turing completeness by means of various additions/alterations (sometimes pretty weird ones about that). So the foundations are pretty different, and I think this is obvious if you tried to start writing typical programs in SQL/Prolog. Now, some people argue you might want a less expressive logic programming language, especially datalog, so this isn’t a knock-down argument, but it’s a pretty big difference.
- _a_a_a_ 3y ago
  > Relational algebra is not Turing complete. yes it is > SQL gets Turing completeness by means of various additions/alterations. Sigh. No. SQL was a subset of relational until recursive CTEs were included to round it off properly.
  - hyperpape 3y ago
    > yes it is Can you double check and/or give references? Because I specifically doubted my memory here, and everything I found online before posting matches what I thought—-standard relational algebra is not Turing-complete (hence relational queries are guaranteed to terminate).
    - _a_a_a_ 3y ago
      Fair Q, I may be confusing relational and predicate calculus. Will check and get back. Apologies in advance if I'm wrong.
    - _a_a_a_ 3y ago
      I need to do some more checking, but you are right. The original relational algebra was a subset of first-order predicate logic. Adding recursion I believe just gave it parity with 1st order PL, so fulsome apologies and thanks for pointing out my error.

nevertoolate 3y ago

You want to learn or to debate :)

yowlingcat 3y ago
I get where you're coming from but your point seems flippant. My answer is "debate" -- and looking at the rest of my subthread, some commenters have raised some great concrete areas around Prolog's unique strengths, which I've found helpful!

hyperpape 3y ago

Relational algebra is not Turing complete. SQL gets Turing completeness by means of various additions/alterations (sometimes pretty weird ones about that). So the foundations are pretty different, and I think this is obvious if you tried to start writing typical programs in SQL/Prolog. Now, some people argue you might want a less expressive logic programming language, especially datalog, so this isn’t a knock-down argument, but it’s a pretty big difference.

_a_a_a_ 3y ago
> Relational algebra is not Turing complete. yes it is > SQL gets Turing completeness by means of various additions/alterations. Sigh. No. SQL was a subset of relational until recursive CTEs were included to round it off properly.
- hyperpape 3y ago
  > yes it is Can you double check and/or give references? Because I specifically doubted my memory here, and everything I found online before posting matches what I thought—-standard relational algebra is not Turing-complete (hence relational queries are guaranteed to terminate).
  - _a_a_a_ 3y ago
    Fair Q, I may be confusing relational and predicate calculus. Will check and get back. Apologies in advance if I'm wrong.
  - _a_a_a_ 3y ago
    I need to do some more checking, but you are right. The original relational algebra was a subset of first-order predicate logic. Adding recursion I believe just gave it parity with 1st order PL, so fulsome apologies and thanks for pointing out my error.