Structural Induction Principles Suppose Context Free Languages - PDF document

Structural Induction • Principles – Suppose Context Free Languages – U is a set, – I is a subset of U (BASIS), – Op is a set of operations on U (INDUCTION). Parse Trees and Ambiguity – L is a subset of U defined recursively as follows: • I ⊆ L • L is closed under each operation in Op • L is the smallest set satisfying 1 & 2 Structural Induction Structural Induction • Then • Show that all strings x generated by • S → ε (1) – To prove that every element of L has some • S → 0S1 (2) property P, it is sufficient to show: • S → 1S0 (3) • 1. Every element of I has property P • S → SS (4) • 2. The set of elements of L having property P is closed under Op – Must have the property n 0 (x) = n 1 (x) – #2: If x ∈ L has property P, Op(x) also must have property P Structural Induction Structural Induction • U = { 0, 1 } * • Show for BASE case: • I = { ε } – ε has equal number of 0s and 1s • Show that P is closed under Op • Ops • S → 0S1 (2) – If x has property P (INDUCTIVE HYPOTHESIS) • S → 1S0 (3) – Then Op (x) also has property P • S → SS (4) • P: n 0 (x) = n 1 (x) – If ( n 0 (x) = n 1 (x)) then n 0 (Op(x)) = n 1 (Op(x)) 1

Plan for 2 nd half Structural Induction • INDUCTION: Show true for each Op • Ambiguous Grammars and Parse Trees • S → 0S1 • S → 1S0 • Questions? • S → SS • DONE! • Questions Parse Trees Another example • Find a CFG to describe: • Graphical means to illustrate a derivation of – L = {a i b j c k | i = k} a string from a grammar • S → B (1) – Root of the tree = start variable • S → aSc (2) – Interior nodes = other variables • B → bB (3) • Children of nodes = application of a production rule • B → ε (4) – Leaf nodes = Terminal symbols – Can also write as • S → B | aSc • B → bB | ε Another example Parse Tree S • Let’s derive a string from L: aabbcc • An inorder traversal of Rule 2 – S ⇒ aSc rule 2 the tree will give the a S c S ⇒ aaScc rule 2 the string derived. Rule 2 a S c S ⇒ aaBcc rule 1 S ⇒ aabBcc rule 3 B Rule 1 S ⇒ aabbBcc rule 3 b B Rule 3 S ⇒ aabb ε cc rule 4 Rule 3 b B = aabbcc Rule 4 ε 2

Recall our example from last One more example time • Defining the grammar for algebraic • Show derivation for a + a * a expressions – Production rules – S ⇒ S + S rule 1 – S → S + S (1) S ⇒ a + S rule 6 S → S – S (2) S ⇒ a + S * S rule 3 S → S * S (3) S ⇒ a + a * S rule 6 S → S / S (4) S ⇒ a + a * a rule 6 S → (S) (5) S → a (6) Parse Tree One more example – S ⇒ S + S • Another derivation for a + a * a S rule 1 – S ⇒ S * S rule 3 S ⇒ a + S S + S S ⇒ S * a rule 6 rule 6 S ⇒ a + S * S S ⇒ S + S * a rule 1 S * S a rule 3 S ⇒ a + S * a rule 6 S ⇒ a + a * S S ⇒ a + a * a rule 6 a a rule 6 S ⇒ a + a * a rule 6 Parse Tree Parse trees – S ⇒ S * S S rule 3 S S S ⇒ S * a S * S S + S S * S rule 6 a S + S S ⇒ S + S * a a S + S rule 1 S * S a S ⇒ a + S * a a a a a rule 6 a a S ⇒ a + a * a Same string, 2 derivations rule 6 3

Ambiguity Famous programming language ambiguity • A CFG is said to be ambiguous if there is at • Dangling else least 1 string in L(G) having two or more – <stmt> → if (<expr>) <stmt> | distinct derivations. if (<expr>) <stmt> else <stmt> | <some_other_stmt> if (expr1) if (expr2) f(); else g(); if (expr1) if (expr2) f(); else g(); To which if does the else belong? Famous programming language ambiguity Famous programming language ambiguity stmt stmt if ( expr ) stmt else stmt if ( expr ) stmt expr1 g(x); expr1 if ( expr ) stmt if ( expr ) stmt else stmt expr2 f(x); expr2 f(x); g(x); In this derivation, the else belongs to the 1 st if if (expr1) if (expr2) f(); else g(); if (expr1) if (expr2) f(); else g(); Famous programming language ambiguity Famous programming language ambiguity stmt s2 • A way to fix this – <stmt> → <s1> | <s2> if ( expr ) s1 <s1> → if (<expr>) <s1> else <s1> | <otherstmt> <s2> → if (<expr>) <stmt> | expr1 if (<expr>) <s1> else <s2> if ( expr ) s1 else s1 <s1> represents if statements with matching else <s2> represent if statements with at least 1 unmatched if expr2 f(x); g(x); The <s1> in the rule for <s2> will assure that all statements between if and else will not have a dangling else. if (expr1) if (expr2) f(); else g(); 4

Ambiguity Ambiguity • Some languages are inherently ambiguous • Showing a grammar is ambiguous is easy – All possible grammars that generate the – Find a string x in the L(G) that has two language are ambiguous derivations • Unfortunately, there is no algorithm that can • Showing a particular grammar is not tells us whether a grammar is ambiguous or ambiguous is usually difficult. not. • Showing that any grammar is not ambiguous is not possible. Derivations Derivations a + a + a • Leftmost derivations S S – A leftmost derivation is one where the leftmost variable in the current string is always the first S + S S + S to get replaced via a production rule. a S + S – A rightmost derivation is one where the S + S a rightmost variable in the current string is a a always the first to get replaced via a production a a rule. rightmost derivation leftmost derivation Ambiguity Removing ambiguities • As it turns out (we won’t prove this) • Since some languages are inherently ambiguous – This cannot always be done – In unambiguous grammars,leftmost derivations • In fact, will always be unique. – We can/will show there is no “algorithm” for – In unambiguous grammars,rightmost determining if a CFG is ambiguous derivations will always be unique. • However, – On a case by case basis, ambiguities can be eliminated 5

Example Example • Abbreviated grammar for algebraic • This grammar has two problems expressions – Production rules 1. Precedence of operators is not respected – S → S + S (1) a * a + a should be interpreted as (a*a) + a S → S * S (2) 2. Sequence of identical operators can be grouped either from the left or the right S → (S) (3) a + a + a can be interpreted as either (a+a)+a or a + (a S → a (4) + a) Example Example • Solution • Our new grammar – S → S + T | T – Introduce some new variables – T → T * F | F • Factor – expression that cannot be broken up by either * or + – a – F → (S) | a – (S) • Note that • Term – expression that cannot be broken up by + – All Factors – all recursion is leftmost – T * F – * has higher precedent than + • Expression – all possible expression – a + a + a + a * a is interpreted as – All Terms • ((a+a) + a) + (a*a) – S + T Example Example S a + a + a + a * a • It can be shown S + T – That every string x, that is generated by this new grammar, has only one leftmost derivation S + T T * F – As such this new grammar is unambiguous a S + T F F – Done using induction on the |x|. a a F a a 6

Summary Summary – Today • Ambiguity • Context Free Grammars – A grammar is ambiguous if there is a string generated • Parse Trees and Ambiguity by the grammar that has two distinct derivations. – Some languages are inherently ambiguous • All grammars that generate the language are ambiguous – There is no algorithm to determine if any given grammar is ambiguous • Proving a grammar to be ambiguous is easy • Proving that a grammar is not is hard. – Questions? Next time • Pushdown Automata • CFG Problem Session 7