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ITEC 380
2016fall
ibarland

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hw09
R5: Adding environments
prolog lists

Due Dec.09 (Fri) 23:59 Dec.11 (SatSun.) 17:00
Submit: a hardcopy with your prolog code. If doing R6 for extra credit, submit those files as well, with a “>>>R5” comment by your modified lines.

We continue to build on the R language implementation from previous homeworks (R2 sol'n posted; R4 sol'n discussed, and full source by Mon 17:00; note that R5 can be based off of the R2 solution for full credit). (R2 specs R4 specs You may implement this homework in either Java or Racket (or another language, if you clear it with me). Label each section of lines you change with a comment “;>>>R5”. You don't need to turn in any hardcopy of unchanged-code (but do submit a fully-working copy in the drop-box).

  1. (10pts) Prolog list

    Write the following Prolog predicates. Do not use append.

    1. (3pts) last(List, Item), which succeeds exactly when Item is the last item in List. This rule should fail if List is empty, of course. (This happens to be the book's Chpt.16, programming exercise #6.)
    2. (2pts) nextToLast(List, Item) which succeeds exactly when Item is the next-to-last item in List. (This rule should fail if List has fewer than two items, of course.)
    3. (2pts) lastTwoReversed(List, ListOf2) which succeeds exactly when ListOf2 contains the last and the second-to-last item of List (in that order, and nothing else).
    4. (3pts) reverseLastTwo(List, NewList) succeeds exactly when NewList is like List except that the last two items have been reversed. (This rule will fail if List has fewer than two items.)
    All of the predicates fail if the first argument is not a list. Some examples (test cases) are provided, below.

    Note that xsb Prolog contains several list functions which you are NOT to use for this assignment (e.g. append and reverse). Also, for full credit, don't merely reverse the input list and operate on that result.

    As ever, your program should follow good style, including appropriate white space, meaningful variable names, and as well as a header comment with your name, the name of the assignment, etc.. (You can have comments describing how a predicate works if you want; however, you can also assume your program is being read by somebody who understands the fundamentals of Prolog.)

  2. R5 (10pts; This problem and the next are really the crux of the project.)
    Deferred evaluation: R5 doesn't add any new syntax, but it is a different evaluation model which will give us more expressive semantics.

    There are two problems1 with the substitution approach used above: we can't make recursive functions, and it doesn't generalize to assigning to variables. We solve these problems with deferred substitution:

    Your test cases should include a recursive function, as well as the example below. Also, since eval now takes an extra argument, that suggests three to four check-expects with various environments (lists of bindings):

    A step sideways: This algorithm as described lets us add recursive functions, but it also doesn't handle some expressions that R4 did! For example, :o make-adder :B m -> :B n -> |n m ;)| :U ! !make-adder 3! 4! gives an error "unbound identifier: m" if no substitution has been done, The problem will be fixed in R6: in the first example, !make-adder 3! returns a function whose body involves m and n, but not the binding of m to 3. (To get this approach to work we'd need to return the function and its bindings; however, R6's static scoping is even better.)

    Note that this gives us dynamic scoping (which we'll mention in class):

    :o m 100 
    :U :o addM :B x -> |x m ;)|
       :U | :o m 5 :U !addM 3!
            :o m 4 :U !addM 3! 
            ;)
          |
    
    evaluates to 15, not 206.

  3. (extra-credit — 15pts total) R6: Implement static scope (closures). Copy your R0-R4 file/project to a new R54. You shouldn't need any additional test cases for R6; the tests for R0-R5 should suffice, although anyone or two R5 examples depending on dynamic binding should now have a new expected-result.
  4. Further extra-credit options (of varying difficulty):
  5. Extra credit (5pts)6 (Scope)
    In Advanced Student, the form set! changes the binding of a variable:

    (define x 5)
    (set! x (* 2 x))
    ; the variable x is now bound to 10.

     1. (define a 10)
     2. (define b 20)
     3. (define make-foo
     4.   (let {[b 30]}
     5.      (lambda ()            ; ← make-foo is bound to this function.
     6.         (let {[c 40]}
     7.            (lambda (cmd)   ; ← make-foo returns this function as its answer.
     8.              (let {[d 50]}
     9.                (cond [(symbol=? cmd 'incA) (set! a (+ a 1))]
    10.                      [(symbol=? cmd 'incB) (set! b (+ b 1))]
    11.                      [(symbol=? cmd 'incC) (set! c (+ c 1))]
    12.                      [(symbol=? cmd 'incD) (set! d (+ d 1))]
    13.                      [(symbol=? cmd 'get-all) (list a b c d)])))))))
          

    1. The scope of the a declared in line 1 is lines      through     .

    2. The scope of the b declared in line 2 is lines      through     .

    3. The scope of the b declared in line 4 is lines      through     .

    4. The scope of the c declared in line 6 is lines      through     .

    5. The scope of the d declared in line 8 is lines      through     .
    Suppose that make-foo is called exactly three times (but that a function returned by make-foo is not called).

    1. How many variables named a are created?     

    2. How many variables named b are created?     

    3. How many variables named c are created?     

    4. How many variables named d are created?     
    (Hint: Each of the above four answers are different.)
    (0pts) To help you figure out the above, try filling out the following (before you try running it):
    (set! a 500)
    (set! b 600)
    (define counter1 (make-foo))
    (define counter2 (make-foo))
    (define counter3 (make-foo))
    
    (counter1 'get-all)  ; >>> TODO:   (list                    )
    (counter1 'incA)
    (counter1 'incB)
    (counter1 'incC)
    (counter1 'incD)
    (counter1 'get-all)  ; >>> TODO:   (list                    )
    
    (counter2 'get-all)  ; >>> TODO:   (list                    )
    


Here are some examples of the list predicates, for the prolog list questions:

last([1,2,3], 3).
Yes

last([1,2,3], 4).
No

last([1,2,3], Y).
Y=3

last([], Y).
No

last(Y, 3).
Y=[3].

nextToLast([1,2,3], 2).
Yes

nextToLast([1,2,3], 3).
No

nextToLast([1,2,3], Y).
Y=2

nextToLast([1], Y).
No

nextToLast(Y, 3).
Y=[3, _h114],         % does not have to be 114, 'course.
Y=[_h116, 3, _h114].

lastTwoReversed([1,2,3], Y).
Y=[3,2]

lastTwoReversed([1], Y).
No

reverseLastTwo([1,2,3,4], Y).
Y=[1,2,4,3]

reverseLastTwo([1,2], Y).
Y=[2,1]

reverseLastTwo([1], Y).
No


Mutation in Racket

If you want to use mutation in your racket-implementation, use Advanced Student language. This language level includes set! (to assign to variables), and set-struct-field! (to assign to struct fields). Note that these two actions are very different, which is why racket gives them separate names; in Jave assigning-to-field and assigning-to-local-variable look alike, despite giving rise to very different behavior.

Since the mutators return #void, and we want to return a (useful) value from every expression, we will use mutation inside a begin expression:

(define times-called 0)
(define (triplify-and-print n)
   (begin (printf "We use `begin` to call a function for a side-effect, but return a value too.\n")
          (set! times-called (add1 times-called))
          (printf "This function has been called ~a time~a.\n"
                  times-called 
                  (if (= times-called 1) "" "s"))
          (* 3 n)))

(triplify-and-print 5)
(triplify-and-print 2)
     

Btw, it's worth noting that in full-racket, begin is implicit in almost all forms (e.g. function-bodies and cond-branches).


1 A third issue, pointed out in Programming Languages and Interpretation, is that evaluating deeply nested lets is an O(n) algorithm.      
2 Note that the list/map you recur with has the additional binding, but that recursive call shouldn't add/modify the list/map used at the top level. Since java.util.Map is inherently mutable, you'll want to make a new copy of that map recurring.      
3“What, racket uses references-to-structs instead of actual structs? Why didn't you tell us that earlier, Barland?” Because in the absence of mutation, no program's results can distinguish between having an object vs a reference-to-object-which-is-always-dereferenced. So it was like a religious-opinion difference, until getting mutation in Advanced Student.      
4 Presumably you do this for each new homework, but it's a particularly good idea for this one, since R5 is not just R4 plus new code, but instead replaces some of the R4 code (subst) with a different approach.      
5 You can think about which implementation you prefer, when using a language: when seeing the program's internal representation of your code (when debugging, or seeing the compiled result of your code).      
6We will talk about this in lecture (let-over-lambda, to implement O.O.), but only briefly.      

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©2016, Ian Barland, Radford University
Last modified 2016.Dec.11 (Sun)
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