1 Lecture 27 Parse/Derivation Trees –Leftmost derivations, rightmost derivations Ambiguous Grammars –Examples Arithmetic expressions If-then-else Statements –Inherently ambiguous CFL’s

2 Context-Free Grammars Parse Trees Leftmost/rightmost derivations Ambiguous grammars

3 Parse Tree Parse/derivation trees are structured derivations –The structure graphically illustrates semantic information about the string Formalization of concept we encountered in regular languages unit –Note, what we saw before were not exactly parse trees as we define them now, but they were close

4 Parse Tree Example Parse tree for string ( )(( )) and grammar BALG –BALG = (V, , S, P) V = {S},  = {(, )}, S = S P = S --> SS | (S) | –One derivation of ( )(( )) S ==> SS ==> (S)S ==> ( )S ==> ( )(S) ==> ( )((S)) ==> ( )(( )) –Parse tree S SS SS(()) S()

5 Comments about Example Syntax: –draw a unique arrow from each variable to each character that is a direct child of that variable –The derived string can be read in a left to right traversal of the leaves Semantics –The tree graphically illustrates the nesting structure of the string of parentheses S SS SS(()) S()

6 Leftmost/Rightmost Derivations There is more than one derivation of the string ( )(( )). –S ==> SS ==> (S)S ==>( )S ==> ( )(S) ==> ( )((S)) ==> ( )(( )) –S ==> SS ==> (S)S ==> (S)(S) ==> ( )(S) ==> ( )((S)) ==> ( )(( )) –S ==> SS ==> S(S) ==> S((S)) ==> S(( )) ==> (S)(( )) ==>( )(( )) Leftmost derivation –Leftmost variable is always expanded –first one above is a leftmost derivation Rightmost derivation –Rightmost variable is always expanded –third one above is a rightmost derivation S SS SS(()) S()

7 Comments Fix a string and a grammar –Any derivation corresponds to a unique parse tree –Any parse tree can correspond to many different derivations Example –The one parse tree corresponds to all three derivations Unique mappings –For any parse tree, there is a unique leftmost/rightmost derivation that it corresponds to S SS SS(()) S() –S ==> SS ==> (S)S ==>( )S ==> ( )(S) ==> ( )((S)) ==> ( )(( )) –S ==> SS ==> (S)S ==> (S)(S) ==> ( )(S) ==> ( )((S)) ==> ( )(( )) –S ==> SS ==> S(S) ==> S((S)) ==> S(( )) ==> (S)(( )) ==>( )(( ))

8 Example S ==> SS ==> SSS ==> (S)SS ==> ( )SS ==> ( )S ==> ( ) –The above is a leftmost derivation of the string ( ) from the grammar BALG –Draw the corresponding parse tree –Draw the corresponding rightmost derivation S ==> SS ==> S ==> SS ==> S ==> (S) ==> ( ) S ==> (S) ==> (SS) ==> (S(S)) ==> (S( )) ==> (( )) –The above is a rightmost derivation of the string (( )) from the grammar BALG –Draw the corresponding parse tree –Draw the corresponding leftmost derivation S ==> (S) ==>(SS) ==> (S) ==> ((S)) ==> (( ))

9 Ambiguous Grammars Examples: Arithmetic Expressions If-then-else statements Inherently ambiguous grammars

10 Ambiguous Grammars A grammar G is ambiguous if there exists a string with two or more distinct parse trees –(2 or more distinct leftmost/rightmost derivations) Example –Grammar AG is ambiguous String aaa has 2 rightmost derivations –S ==> SS ==> SSS ==> SSa ==> Saa ==> aaa –S ==> SS ==> Sa ==> SSa ==> Saa ==> aaa »Draw the corresponding parse trees »Draw the corresponding leftmost derivations

11 2 Simple Examples Grammar BALG is ambiguous –String ( ) has >2 leftmost derivations S ==> (S) ==> ( ) S ==> (S) ==> (SS) ==>(S) ==>( ) S ==> SS ==> (S)S ==> ( )S ==> ( ) –Draw the corresponding parse trees –Draw the corresponding rightmost derivations Grammar ABG is NOT ambiguous –Consider any string x in {a i b i | i > 0} There is a unique parse tree for x

12 Legal Arithmetic Expressions Develop a grammar MATHG = (V, , S, P) for the language of legal arithmetic expressions –  = {0, 1, +, *, -, /, (, )} –Strings in the language include * *( ) –Strings not in the language include )(

13 Grammar MATHG 1 V = {E, N}  = {0, 1, +, *, -, /, (, )} S = E P: –E --> N | E+E | E*E | E/E | E-E | (E) –N --> N0 | N1 | 0 | 1

14 MATHG 1 is ambiguous Come up with two distinct leftmost derivations of the string 11+0*11 –E ==> E+E ==> N+E ==> N1+E ==> 11+E ==> 11+E*E ==> 11+N*E ==> 11+0*E ==> 11+0*N ==> 11+0*N1 ==> 11+0*11 –E ==> E*E ==> E+E*E ==> N+E*E ==> N1+E*E ==> 11+E*E ==> 11+N*E ==> 11+0*E ==> 11+0*N ==> 11+0*N1 ==>11+0*11 Draw the corresponding parse trees E --> N | E+E | E*E | E/E | E-E | (E) N --> N0 | N1 | 0 | 1

15 Corresponding Parse Trees E ==> E+E ==> N+E ==> N1+E ==> 11+E ==> 11+E*E ==> 11+N*E ==> 11+0*E ==> 11+0*N ==> 11+0*N1 ==> 11+0*11 E ==> E*E ==> E+E*E ==> N+E*E ==> N1+E*E ==> 11+E*E ==> 11+N*E ==> 11+0*E ==> 11+0*N ==> 11+0*N1 ==>11+0*11 E E E+ E E* N N 1 1 N 0 N N1 1 E N N 1 1 * E E N 0 + E E N N1 1

16 Parse Tree Meanings E E E+ E E*N N 1 1 N 0 N N1 1 E N N 1 1 * E E N 0 + E E N N1 1 Note how the parse trees captures the semantic meaning of string 11+0*11. More specifically, what number does the first parse tree represent? 11+(0*11) = 11 What number does the second parse tree represent? (11+0)*11 = 1001 (in binary notation)

17 Implications Two interpretations of string 11+0*11 –11+(0*11) = 11 –(11+0)*11 = 1001 What if a line in a program is –MSU_Tuition = 11+0*11; –What is MSU_Tuition? Depends on how the expression 11+0*11 is parsed. This is not good. Ambiguity in grammars is undesirable. In this case, there is an unambiguous grammar for the language of arithmetic expressions

18 If-Then-Else Statements A grammar ITEG = (V, , S, P) for the language of legal If-Then-Else statements –V = (S, BOOL) –  = {adv 75, grade=3.5, grade=3.0, if, then, else} –S = S –P: S --> if BOOL then S else S | if BOOL then S |grade=3.5 | grade=3.0 BOOL --> adv 75

19 ITEG is ambiguous Come up with two distinct leftmost derivations of the string –if adv 75 then grade=3.5 else grade=3.0 –S ==>if BOOL then S else S ==> if adv if adv if adv 75 then S else S ==> if adv 75 then grade=3.5 else S ==> if adv 75 then grade=3.5 else grade=3.0 –S ==>if BOOL then S ==> if adv if adv if adv 75 then S else S ==> if adv 75 then grade=3.5 else S ==> if adv 75 then grade=3.5 else grade=3.0 Draw the corresponding parse trees S --> if BOOL then S |grade=3.5 | grade=3.0 | if BOOL then S else S BOOL --> adv 75

20 Corresponding Parse Trees S ==>if BOOL then S else S ==> if adv if adv if adv 75 then S else S ==> if adv 75 then grade=3.5 else S ==> if adv 75 then grade=3.5 else grade=3.0 S ==>if BOOL then S ==> if adv if adv if adv 75 then S else S ==> if adv 75 then grade=3.5 else S ==> if adv 75 then grade=3.5 else grade=3.0 SS if BthenSelse S adv<90if BthenSgrade=3.0 adv>75 grade=3.5 if BthenS adv<90else S if BthenS adv>75 grade=3.5 grade=3.0

21 Parse Tree Meanings S if BthenSelse S if BthenS adv<90 adv>75 grade=3.5 grade=3.0 S if BthenS else S if BthenS adv<90 adv>75 grade=3.5grade=3.0 If you receive a 90 on advanced points, what is your grade? By parse tree 1 grade = 3.0 By parse tree 2 grade is undefined (as it should be)

22 Implications Two interpretations of string –if adv 75 then grade=3.5 else grade=3.0 –Issue is which if-then does the last ELSE attach to? –This phenomenon is known as the “dangling else” –Answer: Typically, else binds to NEAREST if-then In this case, there is an unambiguous grammar for handling if-then’s as well as if-then-else’s

23 Inherently ambiguous CFL’s A CFL L is inherently ambiguous iff for all CFG’s G such that L(G) = L, G is ambiguous Examples so far –None of the CFL’s we’ve seen so far are inherently ambiguous –While the CFG’s we’ve seen ambiguous, there do exist unambiguous CFG’s for those CFL’s. Later result –There exist inherently ambiguous CFL’s –Example: {a i b j c k | i=j or j=k or i=j=k} Note i=j=k is unnecessary, but I added it here for clarity

24 Summary Parse trees illustrate “semantic” information about strings Ambiguous grammars are undesirable –This means there are multiple parse trees for some string –These strings can be interpreted in multiple ways There are some heuristics people use for taking an ambiguous grammar and making it unambiguous, but this is not the focus of this course There are some inherently ambiguous grammars –Thus, the above heuristics do not always work