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Homework 6: Symbolic Evaluation of Boolean Expressions

Due: Wednesday, Mar. 9, 2011

Extra Credit (100 pts.)

Overview

Write a Scheme function reduce that reduces boolean expressions (represented in Scheme notation) to simplified form. For the purposes of this assignment, boolean expressions are Scheme expressions constructed from:

  • the boolean constants true and false ;
  • boolean variables (represented by symbols other than true, false, not, and, or, implies, and if) that can be bound to either true or false.
  • the unary operator not .
  • the binary operators and, or, and implies, and
  • the ternary operator if.

The course staff is providing functions parse and unparse in the file parse.ss\ that convert boolean expressions in Scheme notation to a simple inductively defined type called boolExp and vice-versa. The coding of parse and unparse is not difficult, but it is tedious (like most parsing) so the course staff is providing this code rather than asking students to write it. The Scheme primitive read: -> SchemeExp is a procedure of no arguments that reads a Scheme expression from the console. DrScheme pops up an input box to serve as the console when (read) is executed.

These parsing functions rely on the following Scheme data definitions. Given

a boolExp is either:

  • a boolean constant true and false ;
  • a symbol S representing a boolean variable;
  • (make-Not X) where X is a boolExp ;
  • (make-And X Y) where X and Y are boolExps;
  • (make-Or X Y) where X and Y are boolExps;
  • (make-Implies X Y) where X and Y are boolExps; or
  • (make-If X Y Z) where X , Y , and Z are boolExps.

A bool-SchemeExp is either:

  • a boolean constant true or false;
  • a symbol S;
  • (list 'not X) where X is a bool-SchemeExp ;
  • (list op X Y) where op is 'and , 'or , or 'implies where X and Y are bool-SchemeExps;
  • (list 'if X Y Z) where X, Y, and Z are bool-SchemeExps.

The provided functions parse and unparse have the following signatures.

The course staff is also providing a very simple test file\ for the eval and reduce functions and a file\ containing a sequence of raw input formulas (to be parsed by parse function in parse.ss\). A good solution to this problem will include much more comprehensive test data for all functions, including some much larger test cases for reduce. The normalize function is difficult to test on large data because the printed output for some important normalized trees (represented as DAGs (Directed Acyclic Graphs) in memory) is so large.

Given a parsed input of type boolExp , the simplification process consists of following four phases:

  • Conversion to if form implemented by the function convert-to-if.
  • Normalization implemented by the function normalize.
  • Symbolic evaluation implemented by the function eval.
  • Conversion back to conventional boolean form implemented by the function convert-to-bool.

A description of each of these phases follows. The reduce function has type bool-SchemeExp -> bool-SchemeExp.

Conversion to if form

A boolean expression (boolExp) can be converted to if form by repeatedly applying the following rewrite rules in any order until no rule is applicable.

In these rules, X and Y denote arbitrary boolExps}. The conversion process always terminates (since each rewrite strictly reduces the number of logical connectives excluding {{make-If) and yields a unique answer independent of the order in which the rewrites are performed. This property is called the Church-Rosser property, after the logicians (Alonzo Church and Barkley Rosser) who invented the concept.

Since the reduction rules for this phase are Church-Rosser, you can write the function convert-to-if using simple structural recursion. For each of the boolean operators And, Or, Not, Implies, and if, reduce the component expressions first and then applying the matching reduction (except for if for which there is no top-level reduction).

The following examples illustrate the conversion process:

We suggest simply traversing the tree using the structural recursion template for type boolExp and converting all structures (other than if) to the corresponding if structures.

Write an inductive data definition and template for boolean formulas in if form, naming this type ifExp. (Note: make-If is the only constructor, other than variables and constants, for ifExp.

The provided function parse: input -> boolExp takes a Scheme expression and returns the corresponding boolExp.

Normalization

An ifExp is normalized iff every sub-expression in test position is either a variable (symbol) or a constant (true or false). We call this type norm-ifExp .

For example, the ifExp (make-If (make-If X Y Z) U V)) is not a norm-ifExp because it has an If construction in test position. In contrast, the equivalent ifExp (make-If X (make-If Y U V) (make-If Z U V)) is normalized and hence is an norm-ifExp.

The normalization process, implemented by the function normalize: ifExp -> norm-ifExp eliminates all if constructions that appear in test positions inside if constructions. We perform this transformation by repeatedly applying the following rewrite rule (to any portion of the expression) until it is inapplicable:

This transformation always terminates and yields a unique answer independent of the order in which rewrites are performed. The proof of this fact is left as an optional exercise.

In the normalize function, it is critically important not to duplicate any work, so the order in which reductions are made really matters. Do NOT apply the normalization rule above unless U and V are already normalized, because the rule duplicates both U and V. If you reduce the consequent and the alternative (U and V in the left hand side of the rule above) before reducing the test, normalize runs in linear time (in the number of nodes in the input); if done in the wrong order it runs in exponential time in the worst case. (And some of our test cases will exhibit this worst case behavior.)

Hint: define a sub-function head-normalize that takes three norm-ifExps X, Y, and Z and constructs a norm-ifExp equivalent to (makeIf X Y Z). This help function processes X because the test position must be a variable or a constant, yet X can be an arbitrary norm-ifExp. In contrast, (head-normalize X Y Z) never even inspects Y and Z because they are already normalized and the normalizing transformations performed in head-normalize never place these expressions in test position.

The following examples illustrate how the normalize and head-normalize functions behave:

Once a large formula has been normalized, do not try to print it unless you know that the formula is small! The printed form can be exponentially larger than the internal representation (because the internal representation can share subtrees).

Before you start writing normalize, write the template corresponding to the inductive data definition of norm-ifExp.

Symbolic Evaluation

The symbolic evaluation process, implemented by the function eval: norm-if-form environment -> norm-if-form, reduces a norm-if-form to simple form. In particular, it reduces all tautologies (expressions that are always true) to true and all contradictions (expressions that are always false) to false.

Symbolic evaluation applies the following rewrite rules to an expression until none is applicable (with one exception discussed below):

The notation M[X <- N] means M with all occurrences of the symbol X replaced by the expression N. It is very costly to actually perform these subtitutions on =norm-if-form= data. To avoid this computational expense, we simply maintain a list of bindings which are pairs consisting of symbols (variable names) and boolean values {true, false. The following data definition definition formally defines the binding type.

A binding is a pair (make-binding s v) where s is a symbol (a variable) and v is a boolean value (an element of { true, false }.

An environment is a (list-of binding).

When the eval function encounters a variable (symbol), it looks up the symbol in the environment and replaces the symbol it's boolean value if it exists.

These rewrite rules do not have the Church-Rosser property. The last two rewrite rules are the spoilers; the relative order in which they are applied can affect the result in some cases. However, the rewrite rules do have the Church-Rosser property on expressions which are tautologies or contradictions.

If the final rule is applied only when X actually occurs in either Y or Z, then the symbolic evaluation process is guaranteed to terminate. In this case, every rule either reduces the size of the expression or the number of variable occurrences in it.

We recommend applying the rules in the order shown from the top down until no more reductions are possible (using the constraint on the final rule). Note that the last rule should only be applied once to a given sub-expression.

Conversion to Boolean Form

The final phase converts an expression in (not necessarily reduced or normalized) If form to an equivalent expression constructed from variables and { true, false, And, Or, Not, Implies, If. This process eliminates every expression of the form

where one of the arguments {X, Y, Z is a constant { true, false }.

Use the following set of reduction rules to perform this conversion

where X , Y , and Z are arbitrary If forms. This set of rules is Church-Rosser, so the rules can safely be applied using simple structural recursion.

Points Dsitribution

  • convert-to-if (10%)
  • normalize (20%)
  • eval (20%)
  • convert-to-bool (10%)
  • reduce (40%)
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