CS 5010: Problem Set 8

Out: Monday, October 31, 2016

Due: Monday, November 7, 2016 at 600pm local time.

The goal of this problem set is to give you practice using everything that you've learned this semester.

NOTE: Both of the problems on this problem set are brand-new for Fall 2016. We do our best to make sure they are bug-free when we release them, but please keep an eye out for updates. We will announce all updates on Piazza.

You must use DrScheme's HtDP Intermediate Student Language with Lambda. Use list abstractions like filter, fold, and map whenever they are helpful. As before, you will be penalized for failing to use these when they are "obviously" applicable.

Remember that you must turn in all the usual design-recipe deliverables, including the following:

You also must turn in a call graph for your solution, as in the last few problem sets. Call this file call-graph-1 or call-graph-2 with an appropriate suffix, and bring a paper copy of each of these to your codewalk.

If you really need to do so, it is ok to write two mutually-recursive functions using general recursion. If you do this, you need make sure that your halting measure decreases on EVERY recursive call to the function, even if that call comes via the other function.

In addition to rackunit and "extras.rkt", you may also require sets.rkt (but be careful-- make sure that the functions of sets.rkt apply correctly to your data!)

Note: not everything on this problem set requires the use of invariants or the use of general recursion. Part of your task is to figure out when you need these things and when you do not. Remember, it is the purpose statement that determines whether or not you need to state an invariant.

Remember that you must follow the design recipe. Your deliverables include the data definitions (including interpretation and templates), contract and purpose header, strategies, code, and tests. Be sure to sync your work and fill out a Work Session Report at the end of every work session. Use the Work Session Report for PS08 .

As usual, the rubric for grading is here.

  1. Your employers at NextPython, Inc., the makers of GarterSnake, have introduced a new product, which they call Simplified Garter Snake (SGS). Simplified GarterSnake is just like GarterSnake, except for the following:

    Your task is to check SGS programs for potential infinite recursion. Consider the following SGS program:

    (define some-loops
         (make-def 'f1 (list 'x) (make-appexp 'no-loop (list (make-varexp 'x))))
         (make-def 'f2 (list 'u 'y) (make-appexp 'f1 (list (make-varexp 'y))))
         (make-def 'f3 (list 'x 'u)
                   (make-appexp 'f1 (list (make-appexp 'f4
                                                 (list (make-varexp 'u)
                                                       (make-varexp 'w)))
                                          (make-varexp 'z))))
         (make-def 'f4 (list 'x 'y)
                   (make-appexp 'f5
                                (list (make-varexp 'y)
                                      (make-varexp 'u))))
         (make-def 'f5 (list 'u)
                   (make-appexp 'f2
                                (list (make-appexp 'f3 empty))))
         (make-def 'no-loop (list 'x) (make-varexp 'x))))

    In this program, neither f1 nor f2 can lead to an infinite loop. However, f3 calls f4, which calls f5, which calls f3 again. So any call to f3, f4, or f5 may lead to an infinite recursion.

    Note that this program takes advantage of the global scope of function names. For example, f1 calls no-loop, even though the definition of no-loop comes after the definition of f1.

    Recall that the syntax-tree representation of GarterSnake is defined as follows:

    ;; A Program is a ListOfDefinition.
    (define-struct def (name args body))
    ;; A Definition is a (make-def Variable ListOfVariable Exp)
    ;; name is the name of the function being defined
    ;; args is the list of arguments of the function
    ;; body is the body of the function.
    (define-struct varexp (name))
    (define-struct appexp (fn args))
    ;; An Exp is one of
    ;; -- (make-varexp Variable)
    ;; -- (make-appexp Variable ListOfExp)
    ;; (make-varexp v)                   represents a use of the variable v
    ;; (make-appexp f (list e1 ... en))  represents a call to the function
    ;;                                   named f, with arguments e1,..,en
    ;; A Variable is a Symbol.

    You are to turn in a file named q1.rkt that provides the following functions:

    any-loops? : Program -> Boolean
    GIVEN: a valid SGS program p (that is, a GS program that obeys the
    restrictions listed above).
    RETURNS: true iff there is some function f in p that calls itself
    either directly or indirectly, as in the example above.
  2. (Resolution Theorem-Proving) The motivation for this problem is something called the satisfiability problem and resolution theorem-proving.

    Here's the set-up:

    Imagine that we have a number of boolean variables a, b, c, d, etc., which can either be true or false. You probably know that any boolean formula can be expressed in conjunctive normal form, that is, as a formula that looks like

    (a ∨ ¬b ∨ c) & (d ∨ b) & (¬a ∨ c)

    We will use conventional terminology for formulas in conjunctive normal form, in which a disjunction like (a ∨ ¬b ∨ c) is called a clause, and each of its disjuncts, which are either variables or their negations, are called literals. For example, the clause (a ∨ ¬b ∨ c) contains the literals a, ¬b, and c.

    Given a choice of truth values for each variable, a clause is true if and only if at least one of its literals is true. Thus, the empty clause, which contains no literals, is always false. Similarly, a a choice of truth values for each variable, a formula is true if and only if all of its clauses are true. So the empty formula, which contains no clauses, is always true.

    The question we want to answer is: given a formula in conjunctive normal form, is there a choice of truth values for its variables that makes the formula true? For any choice of truth values for the variables, This is called thesatisfiability problem. For our example, the assignment

    a = T, b = F, c = T, d = T

    makes the formula true, and we say that the formula is satisfiable.

    We say that a formula F implies a formula G iff every choice of truth values that makes F true also makes G true. (If you know about truth tables, this is saying that G is true on every line of the truth table that makes F true.)

    So, if a formula F implies a contradiction, which is never true, then the original formula F must also never be true. In this case, we say that F is unsatisfiable.

    As it turns out, there is a simple method for deriving a contradiction from a formula, if one exists. This is called the rule of resolution. It says:

    (a1 ∨ a2 ∨ ... b)         (¬b ∨ c1 ∨ c2 ∨ ...)
          (a1 ∨ a2 ∨ ... ∨ c1 ∨ c2 ∨ ...)

    This rule says that if we have the two formulas above the line, then we can derive the formula below the line. Both the b and the ¬b are deleted in the conclusion.

    In our example clauses above, (a ∨ ¬b ∨ c) & (d ∨ b) & (¬a ∨ c) , the first two clauses resolve on the b to get (a ∨ c ∨ d), and the first and third clauses resolve on a to get (¬b ∨ c).

    We call b and ¬b complementary literals, Disjunctions like (a1 ∨ a2 ∨ ... b) are treated as sets, so the b and the ¬b may appear anywhere in their clauses. It is possible for a pair of clauses to have more than one pair of complementary literals, in which case the rule applies in two ways, yielding more than one conclusion.

    Observe that the conclusion of the rule of resolution, as written in the box above, is implied by its hypotheses: if both of the original clauses are true and b is true, then one of the c's must be true in order to make the second clause true. Similarly, if b is false, then one of the a's must be true in order to make the first clause true. So if both clauses are true, then either one of the a's is true or one of the c's is true and this is what the conclusion says.

    So, if you can derive the empty clause (which is always false) from some set of clauses F, then you've shown that your set of clauses F implies the empty clause, and therefore that it's unsatisfiable.

    Further, there is a theorem that says that this is an "if and only if": a formula form is unsatisfiable if and only if the empty clause is derivable from the formula.

    So to solve the satisfiability problem, we start with a set of clauses, and repeatedly apply the resolution rule until we either derive the empty clause or run out of new clauses to generate. (This always halts. Why? One of your deliverables is a halting measure for this algorithm.) If we derive the empty clause, we know our original set of clauses was unsatisfiable. If we derive all the clauses it is possible to derive, then we know (by the theorem above) that it is satisfiable.

    For example, consider the following set of clauses:

    (a ∨ ¬b ∨ c)
    (d ∨ b)
    (¬a ∨ c)

    From these clauses we can derive the empty clause, as follows:

    (a \/ ~b \/ c)    (~a \/ c)
           (~b \/ c)                (b)
                        (c)                    (~c)

    In general, you will have to consider all generated clauses at every step, not just ones from the original set.

    Most of the preceding discussion is the motivation for this problem. We've presented the problem in terms of for propositional (boolean) logic, but the technique extends to first-order logic. This is all standard material dating back to at least 1965; if you want to understand it better, go look it up on the internet. However, if you rely on anything you found on the internet, or elsewhere, you must cite it in a footnote, just as you would for any other term paper.

    Now we can get to specifics:

    Here are the specified pieces of the data definitions. You may not need all of these.

    A Variable is a Racket Symbol.
    A Literal is one of
    -- (make-pos Variable)  Interp: a literal containing the variable
    -- (make-neg Variable)  Interp: a literal containing the negation of
                                    the variable
    A Clause is a SetOfLiteral

    (Note that make-pos is analogous to make-varexp in GarterSnake: it provides a way of injecting a variable into a boolean expression. make-neg injects the variable and puts a negation on it at the same time.)

    You are to turn in a file named q2.rkt that provides the following function:

    is-null-derivable? : ListOfClause -> Boolean
    GIVEN: a list of clauses
    RETURNS: true iff the empty clause is derivable from the given
    clauses using the rule of resolution as given above.

    In addition, you must provide the following functions that we need in order to test is-null-derivable? :

    make-pos : Symbol -> Literal
    make-neg : Symbol -> Literal
    GIVEN: A Variable v
    RETURNS: The literal v or ¬v
    make-clause : ListOfLiteral -> Clause
    GIVEN: a list of literals, possibly with duplications
    RETURNS: a clause containing exactly those literals

    You get to design the representation of literals and clauses. In particular, it is possible but not required that you have structs for both make-pos and make-neg. You need only provide the constructors above, but you will need to document your data definitions in the usual way.

Last modified: Tue Nov 1 17:09:12 Eastern Daylight Time 2016