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ITEC 380
2019fall
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V4 shadowing, function-application; prolog

Due Dec.02 (Mon). Submit all files on D2L, plus hardcopy of any (parts of) files you added for V3/V4. Your prolog queries can be inside a comment of a racket file, at the top of your submitted hardcopy, thanks!

Point values will be changed somewhat, for some problems.

We continue to build on the language implementation of V1/V2. You can implement this homework in either Java or Racket. You may use the provided V2 solution if you want. Please add a comment “>>> V3” or “>>> V4” next to places you add/update for this homework. You don’t need to turn in any hardcopy of unchanged-code (but submit a fully-working copy in the drop-box). Please put new tests into the list “tests” in the V4-test file, as possible/appropriate.

    V3 is just like V2, except we now allow one variable to shadow another.

    For example, we might have a let x := … {: … :} which in turn contains another let x := … {: … :} inside of it. In that case the inner x should shadow the outer one:
    let x := 3 {: let x := 5 {: #yasss x 3# :} :} ⇒ let x := 5 {: #yasss x 3# :} :}1 ⇒ #yasss 5 3# ⇒ 8. And of course, shadowing may occur between non-adjacent scopes: let x := 3 {: let y := 4 {: let x := 5 {: … :} :} :}.

    In technical terms: when substituting, only substitute “free occurrences” of an Id in E1, not any “bound occurrences”2.

    As an example, for the expression let x := 3 {: (:let y := 4 {: #yasss x (:let x := 5 {: #yasss x y# :}:)# :} :):}, you’d give the tree drawn at abstract syntax tree, with bindings identified the right. This corresponds to some runnable test-cases: 

    (check-expect (eval (parse! "let x := 3 {: (:let y := 4 {: #yasss x (:let x := 5 {: #yasss x y# :}:)# :} :):}"))
              (+ 3 (+ 5 4)))
(check-expect (subst "x" 3 (parse! "(:let y := 4 {: #yasss x (:let x := 5 {: #yasss x y# :}:)# :} :)")) 
              (parse! "(:let y := 4 {: #yasss 3 (:let x := 5 {: #yasss x y# :}:)# :} :)") )
    Our goal in doing this is to understand: when substituting free variables, exactly when do we stop and not substitute it in various sub-trees? For example, in the provided image: Why does the top-level "x" bind to the "x" in the middle, but not the "x" in the bottom-right? What rule can you use, when substituting, about precisely where to stop substituting? You might find it helpful to try to explain (in English) to a friend, exactly when you do and don’t substitute.

    1. (0pts) Change the purpose-statement of subst to be “substitute any free occurrences of …”. Note that you will not substitute other binding occurrences, either.
    2. For each of the following, fill in the blanks by replacing the outermost LetExpr with just its body, but substituting its variable respecting shadowing.
      For example (using the program/image above): let x := 3 {: (:let y := 4 {: #yasss x (:let x := 5 {: #yasss x y# :}:)# :} :):}
      ⇒ (:let y := 4 {: #yasss 3 (:let x := 5 {: #yasss x y# :}:)# :} :)
      ⇒ #yasss 3 (:let x := 5 {: #yasss x 4# :}:)#
      ⇒ #yasss 3 (:#yasss 5 4#:)#
      ⇒ 12.
      Complete the blanks in a similar way, below:
      1. let y := 3 {: let x := 5 {: #yasss x y# :} :} ⇒                                              ⇒                                  ⇒ 8
      2. let y := 3 {: let x := y {: #yasss x y#:} :} ⇒                                                                                      ⇒                                  ⇒ 6
      3. let x := 5 {: let y := 3 {: #yasss let x := y {: #yasss x y# :} x #:}:} ⇒                                                                                                                                                                      ⇒                                                                                                                      ⇒                                                                                                                                            ⇒                                  ⇒ 11
      4. let x := 5 {: let x := #yasss x 1# {: #yasss x 2#):} :} ⇒ let      := #yasss      1# {: #yasss      2# :} ⇒ let      :=      {: #yasss      2# :} ⇒ #yasss      2# ⇒ (    )

    3. (5pts) Update V2 to V3, by the necessary changes to enable shadowing.
      You are encouraged to build on your own previous solution, but you may also use the V2 solution (.rkt files posted).

      • The change should be quite small, but is surgically precise.
      • Recall that V2’s subst is essentially the same code as change-name in AncTrees.
        Hint: What is needed for V3’s (and V4’s) subst is essentially the same as hw07's change-eyes-to-until.
      • Please label each section of lines you change with a comment “;>>>V3”.

    V4 adds (non-recursive) functions and function-application to our language: 
    Expr ::= … | FuncExpr | FuncApplyExpr

FuncExpr ::= faq Id mek Expr          “frequent-query”
FuncApplyExpr ::= mod Expr wtf Expr   “modify arg with the function”
    Be sure not to confuse functions with function-application (calling a function) — it’s the difference between square-root (as a function), and the square-root-function-applied-to-4 (or put differently: it’s the difference between a hammer, and hitting something with a hammer).

    Here is the function (λ (x) (+ (* 3 x) 1)) written in V4: faq x mek #yasss #yeet x 3# 1#.
    And, here is the (uniterated) collatz function in V4: 

    faq n mek ?| 2 n  thn  #yeet n 0.5#  els  #yasss #yeet n 3# 1# :|
faq n mek ?| n 2  thn  #yasss #yeet x 3# 1#  els  #yeet n 0.5# :|

    Just as numbers are self-evaluating, so are FuncExprs. Evaluating (an internal representation of) a function results in that same (internal representation of the) function. We won’t actually evaluate the body until the function is applied. (This is exactly how Java, racket, python, javascript, etc. treat functions.)

    A FuncApplyExpr represents calling a function. Here are two expressions, both evaluating to 5·3+1 = 16: 

    let tripleAndInc 
:=  faq x mek #yasss #yeet x 3# 1#
{:  mod 5 wtf tripleAndInc  :}

mod 5 wtf faq x mek #yasss #yeet x 3# 1#
    In FuncApplyExpr, the first Expr had better evaluate to a function. (That is, it might be a FuncExpr, or an Id which gets substitued to a function value. It could also be (say) an ParityExpr or LetExpr which evaluates to a function.)

    1. First, write the following four functions as V4 programs. (You can then modify them as part of your tests.)
      1. A constant function that always returns (say) 17.
      2. the function sqr, which squares its input; give it a name with a V4 LetExpr whose body we’ll just leave as “…”. I.e. write the V4 equivalent of racket (let {[sqr (lambda (x) (* x ))]} …).
      3. the factorial function, written in V4 using a LetExpr with body “…” as in the previous example.
        Note: You won’t able to evaluate function-applications for recursive functions yet (see V5), but we can still write the test cases! (You can comment out that one test case for now, since it’ll trigger a run-time exception otherwise.)
        4. and
      4. The V4 equivalent of the following racket definition make-adder: 
        (define (make-adder n)
  (lambda (m) (+ n m)))

; Two examples of applying this function:
;
(make-adder 3)     ; evals to (lambda (m) (+ 3 m))
((make-adder 3) 4) ; evals to 7
      Then, upgrade V3 so that it allows functions to be represented; label each section of lines you change with a comment “;>>>V4”.
      1. (2pts) Add a struct/class for representing FuncExprs internally.
      2. (2pts) parse! (and tests).
      3. (2pts) expr->string (and tests)
      4. (2pts) eval (and tests)
    2. Implement function-application.
      1. (2pts) Add a struct/class for representing FuncApplyExprs internally.
      2. (2pts) parse! (and tests)
      3. (2pts) expr->string (and tests)
      4. (9pts) eval (and tests). Here, more than half the points are for tests, since you want to try several situations involving shadowing variables. (You don’t need to test eval’ing your factorial function, though.)

        The semantics of eval’ing the function-application mod Expr0 wtf Expr1:

        1. Evaluate Expr0; let’s call the result actual-arg.
        2. Evaluate Expr1; let’s call the result f. (f had better be a function-value!)
        3. Substitute f’s parameter with actual-arg in f’s body; call this new expression E′.
        4. Evaluate E′ and return that value.
        Hey, those semantics are practically the same as LetExpr’s! Indeed, it’s not very different; the function holds the identifier and body; when you eval a function-application then we do the same substitution.

Prolog

As mentioned: Your prolog queries can be inside a comment of a (racket) file, at the top of your submitted hardcopy, thanks! Only include your added rules/facts, not the rest of the provided .pl file.

  1. (1pt) Add another super-person (hero, villain, or neither) and at least three more color-or-weapon preferences to the Prolog superhero knowledge base from lecture.
  2. (3pts) Two super-people are compatible if they share a fighting style, or a color-preference (and they aren’t the same person). Moreover, team(A,B,C) is true if A and B are compatible, and B and C are compatible, but A and C aren’t the same.
    1. define a rule compatible(A,B).
    2. What query will solve for every super-person compatible with bubbles?
    3. define team(A,B,C).
    4. What query will solve for every team starting with bubbles and ending with rafael?

  3. (5pts) Write friendly: We say two super-people are friendly if they are connected through some chain of compatible people. For example, bubbles and rafael would be friendly if compatible(bubbles,magikarp), compatible(magikarp,spongebob), and compatible(spongebob,rafael). (That is: friendly is the transitive closure of the compatible relation.) Everybody, of course, is trivially friendly with themselves (since they’re connected by a chain of 0 compatible others).

The interpreter project is based on the first chapters of Programming Languages and Interpretation, by Shriram Krishnamurthi. As a result, this homework assignment is covered by the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. Although we’re using a different dialect of racket than that book, you might find it helpful to skim it.

1 The notation “let x := 5 {: #yasss x 3# :} ⇒ 5 plus 3 ⇒ 8” is shorthand for 
  eval(parse("let x := 5 {: #yasss x 3# :}"))
= eval(parse("#yasss 5 3#"))
= eval(parse("8"))
Observe how we definitely don’t write “"let x := 5 {: #yasss x 3# :}" = "#yasss 5 3#" = 8” since the two strings are not .equals(·) to each other, and besides strings are never ints. More specifically: we distinguish between “⇒” (“code evaluates to”) and “=” (“equals”, just as “=” has meant since kindergarten).      ↩
2 nor any “binding occurrences”: The first x in let x := 5 {: #yasss x 3# :} is a binding occurrence, and the second x is a bound occurrence. (We say that “a variable is bound inside the scope of its binding occurrence”.)      ↩

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