7

This is NOT homework -- the solution is already in the text. I just failed to understand the solution.

Problem

(run* (q)
  (let [a (== true q)
        b (== false q)]
    b))

Correct Solution

(false)

My believed solution

()

My Confusion

Apparently the "a (== true q)" line is NOT executed, since only b is the goal. This confuses me. My mental model so far for logic programming has been:

  • consider all possible assignemnts to q

  • output the ones that manages to pass through the entire program

    Thus, the "a (== true q)" forces q = true, which makes it impossible to satisfy the "b (== false q)" line.

    However, apparently only "thigns needed to compute the goal" are executed. What's going on? What's the right mental execution model for core.logic / mini-kanren?

Thanks

(BTW, I'm clearly in the wrong, since mini-karen + core.logic agre with each other -- I just want to understand what I'm doing wrong.)

Michel Schinz
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    When your objective is to reach the goal, it is better to leave something that isn't related to your objective to reach the goal – Ankur May 31 '12 at 05:27
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    @Ankur: you're probably right. However, I don't understand it. Can you provide some insight on how "let" fits into the execution model of mini-kanren / core.logic? It's clearly that I do not understand what "let" means -- and it's clear that "let" means something else than in pure scheme/clojure. –  May 31 '12 at 05:29

1 Answers1

9

== produces a goal. But you don't pass the a goal to run. So run doesn't know about it. A comparable situation is this:

(defn call [f] (f))

(call
  (let [a #(println "a")
        b #(println "b")]
    b))

The a function is created but not passed to call. So it is never executed.

kotarak
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  • This makes sense now, despite fooling around with the repl, I didn't realize that == returned functions. [I thought it was just some weird type/macro trick to make core.logic work.] –  May 31 '12 at 10:37