Basic logic programming in Scala

Question

I have a fun little problem where I'm presented with a logic clause like so:

Rule 1. A, B and C are unique, and are numbers from 1 to 3 (so every number is used).
Rule 2. B < 2
Rule 3. C > 2

Now (assuming this quick example I just came up with actually validates :P), it's easy to see

A = 2
B = 1
C = 3

But that's a very contrived example.

What about if you have 20 (or ten thousand) rules, and they overlapped. Assume there is valid single answer, and you have access to the rules somehow (be it a list of predicates say).

Is there a nice, general, intuitive solution? I know Prolog could solve it with certain libraries, but my attempts proved futile. I know I could just brute force permutate all numbers in the range, and then manually check they conform to the rules. But that seems pretty lame.

Are there any constraints on the form of the predicates, or are they completely arbitrary? — Miles Sabin, May 14 '12 at 08:01
I guess they are arbitrary. They must be useful though - i.e. no constraint should constrain where another also already does so. — Dominic Bou-Samra, May 14 '12 at 09:26
You may be interested in Styla, a lightweight Scala-based Prolog interpreter available from: http://code.google.com/p/styla/ — Paulo Moura, May 14 '12 at 15:45

Tomasz Nurkiewicz · Accepted Answer · 2012-05-14T07:18:55.780

Brute force solution

First I create all possible inputs based on rule 1 which happens to be all permutations:

val all = (1 to 3).permutations

Then I define remaining rules:

val rule2 = (a: Int, b: Int, c: Int) => b < 2
val rule3 = (a: Int, b: Int, c: Int) => c > 2

And I simply filter out solutions that do not match all the rules:

val solutions = all.
  filter(i => rule2(i(0), i(1), i(2))).
  filter(i => rule3(i(0), i(1), i(2))).
  toList

solutions.toList prints one and only valid solution:

List(Vector(2, 1, 3))

Note that because permutations are generated lazily, this implementation isn't really that ineffective (e.g. rule3 is only applied on solutions that passed rule2).

You can use foldLeft() to apply arbitrary number of predicates:

val rules = List(rule2, rule3)

rules.
  foldLeft(all)((solutionsSoFar, rule) => 
    solutionsSoFar.filter(s => 
      rule(s(0), s(1), s(2))
    )
  )

Drools

Have a look at Drools business rule engine. Just put all possible solutions as knowledge base and define rules using nice DSL.

This is what I have, almost word for word (seriously, our variables have almost the same names :P I was How do you handle it for 10000 rules? I don't think a chained filter works well there. — Dominic Bou-Samra, May 14 '12 at 06:59
@DominicBou-Samra: see my update and [my article about `foldLeft()`](http://nurkiewicz.blogspot.com/2012/04/secret-powers-of-foldleft-in-scala.html) if you are not feeling comfortable with it — Tomasz Nurkiewicz, May 14 '12 at 07:07

score 3 · Answer 2 · answered May 14 '12 at 07:14

This is a job for constraints. There are several constraint solvers for java. My favourite is JaCoP. There is a nice scala DSL for it, so it almost looks like Prolog constraints. Here is the typical SENDMOREMONEY example: http://www.hakank.org/jacop/SendMoreMoney.scala

score 2 · Answer 3 · answered May 14 '12 at 19:26

2

You could be interested by this blog where the author shows in a three-parts article how to do Logic Programming in Scala :

http://ambassadortothecomputers.blogspot.com/feeds/posts/default?alt=rss

answered May 14 '12 at 19:26

bhericher

667
4
13

Here's a better link, starting with part one of that three-part article: http://ambassadortothecomputers.blogspot.com/2011/04/logic-programming-in-scala-part-1.html – owensmartin Jan 28 '18 at 06:49

Basic logic programming in Scala

3 Answers3

Brute force solution

Drools