Review: ShiftReduce Parsing. Bottomup parsing uses two actions: BottomUp Parsing II. Shift ABC xyz ABCx yz. Lecture 8. Reduce Cbxy ijk CbA ijk


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1 Review: ShiftReduce Parsing Bottomup parsing uses two actions: BottomUp Parsing II Lecture 8 Shift ABC xyz ABCx yz Reduce Cbxy ijk CbA ijk Prof. Aiken CS 13 Lecture 8 1 Prof. Aiken CS 13 Lecture 8 2 Recall: he Stack Left string can be implemented by a stack op of the stack is the Shift pushes a terminal on the stack Reduce pops 0 or more symbols off of the stack production rhs pushes a nonterminal on the stack production lhs Key Issue How do we decide when to shift or reduce? xample grammar: * ) Consider step * We could reduce by giving * A fatal mistake! No way to reduce to the start symbol Prof. Aiken CS 13 Lecture 8 3 Prof. Aiken CS 13 Lecture 8 1
2 Handles Intuition: Want to reduce only if the result can still be reduced to the start symbol Assume a rightmost derivation S * αxω αβω hen X β in the position after α is a handle of αβω Handles Cont.) Handles formalize the uition A handle is a string that can be reduced and also allows further reductions back to the start symbol using a particular production at a specific spot) We only want to reduce at handles Note: We have said what a handle is, not how to find handles Prof. Aiken CS 13 Lecture 8 5 Prof. Aiken CS 13 Lecture 8 6 Important Fact #2 Important Fact #2 about bottomup parsing: In shiftreduce parsing, handles appear only at the top of the stack, never inside Why? Informal induction on # of reduce moves: rue initially, stack is empty Immediately after reducing a handle rightmost nonterminal on top of the stack next handle must be to right of rightmost nonterminal, because this is a rightmost derivation Sequence of shift moves reaches next handle Prof. Aiken CS 13 Lecture 8 Prof. Aiken CS 13 Lecture 8 8 2
3 Summary of Handles In shiftreduce parsing, handles always appear at the top of the stack Handles are never to the left of the rightmost nonterminal herefore, shiftreduce moves are sufficient; the need never move left Bottomup parsing algorithms are based on recognizing handles Recognizing Handles here are no known efficient algorithms to recognize handles Solution: use heuristics to guess which stacks are handles On some CFGs, the heuristics always guess correctly For the heuristics we use here, these are the SLR grammars Other heuristics work for other grammars Prof. Aiken CS 13 Lecture 8 9 Prof. Aiken CS 13 Lecture 8 10 Grammars All CFGs Viable Prefixes It is not obvious how to detect handles Unambiguous CFGs SLR CFGs will generate conflicts At each step the parser sees only the stack, not the entire input; start with that... α is a viable prefix if there is an ω such that α ω is a state of a shiftreduce parser Prof. Aiken CS 13 Lecture 8 11 Prof. Aiken CS 13 Lecture
4 Huh? What does this mean? A few things: A viable prefix does not extend past the right end of the handle It s a viable prefix because it is a prefix of the handle As long as a parser has viable prefixes on the stack no parsing error has been detected Important Fact #3 Important Fact #3 about bottomup parsing: For any grammar, the set of viable prefixes is a regular language Prof. Aiken CS 13 Lecture 8 13 Prof. Aiken CS 13 Lecture 8 1 Important Fact #3 Cont.) Important Fact #3 is nonobvious We show how to compute automata that accept viable prefixes Items An item is a production with a. somewhere on the rhs he items for ) are.).).) ). Prof. Aiken CS 13 Lecture 8 15 Prof. Aiken CS 13 Lecture 8 16
5 Items Cont.) he only item for X is X. Items are often called LR0) items Intuition he problem in recognizing viable prefixes is that the stack has only bits and pieces of the rhs of productions If it had a complete rhs, we could reduce hese bits and pieces are always prefixes of rhs of productions Prof. Aiken CS 13 Lecture 8 1 Prof. Aiken CS 13 Lecture 8 18 xample Consider the input ) hen ) is a state of a shiftreduce parse is a prefix of the rhs of ) Will be reduced after the next shift Item.) says that so far we have seen of this production and hope to see ) Generalization he stack may have many prefixes of rhs s Prefix 1 Prefix 2... Prefix n1 Prefix n Let Prefix i be a prefix of rhs of X i α i Prefix i will eventually reduce to X i he missing part of α i1 starts with X i i.e. there is a X i1 Prefix i1 X i β for some β Recursively, Prefix k1 Prefix n eventually reduces to the missing part of α k Prof. Aiken CS 13 Lecture 8 19 Prof. Aiken CS 13 Lecture
6 An xample Consider the string * ): * ) is a state of a shiftreduce parse is a prefix of the rhs of ) is a prefix of the rhs of * is a prefix of the rhs of * An xample Cont.) he stack of items.). *. Says We ve seen of ) We ve seen of We ve seen * of * Prof. Aiken CS 13 Lecture 8 21 Prof. Aiken CS 13 Lecture 8 22 Recognizing Viable Prefixes Idea: o recognize viable prefixes, we must Recognize a sequence of partial rhs s of productions, where ach sequence can eventually reduce to part of the missing suffix of its predecessor An NFA Recognizing Viable Prefixes 1. Add a dummy production S to G 2. he NFA states are the items of G Including the extra production 3. For item α.xβ add transition α.xβ X αx.β. For item α.xβ and production X γ add α.xβ X.γ Prof. Aiken CS 13 Lecture 8 23 Prof. Aiken CS 13 Lecture 8 2 6
7 An NFA Recognizing Viable Prefixes Cont.) 5. very state is an accepting state 6. Start state is S NFA for Viable Prefixes of the xample ). ).).) ) Prof. Aiken CS 13 Lecture * * * *.. *. Prof. Aiken CS 13 Lecture 8 26 NFA for Viable Prefixes in Detail 1) NFA for Viable Prefixes in Detail 2)... Prof. Aiken CS 13 Lecture 8 2 Prof. Aiken CS 13 Lecture 8 28
8 NFA for Viable Prefixes in Detail 3) NFA for Viable Prefixes in Detail ). ).... ) * *. Prof. Aiken CS 13 Lecture Prof. Aiken CS 13 Lecture 8 30 NFA for Viable Prefixes in Detail 5) NFA for Viable Prefixes in Detail 6). ).)..... ).)....)... * *. Prof. Aiken CS 13 Lecture Prof. Aiken CS 13 Lecture
9 NFA for Viable Prefixes in Detail ) NFA for Viable Prefixes in Detail 8). ).)....). ) ).. ).)....). ) ).... * *. Prof. Aiken CS 13 Lecture Prof. Aiken CS 13 Lecture 8 3 NFA for Viable Prefixes in Detail 9) NFA for Viable Prefixes in Detail 10). ).)....). ) ).... ).)....). ). )..... * *. Prof. Aiken CS 13 Lecture Prof. Aiken CS 13 Lecture
10 NFA for Viable Prefixes in Detail 11) NFA for Viable Prefixes in Detail 12). ).)....). ). ).... ).)....). ). ).....* *. Prof. Aiken CS 13 Lecture 8 3..* * *. *. Prof. Aiken CS 13 Lecture 8 38 NFA for Viable Prefixes in Detail 13) ). ).).) ) *..* *. Prof. Aiken CS 13 Lecture 8 39 * *. ranslation to the DFA.....).. *.. ). *.....) *. *. *.) *. *.).) * ) ). Prof. Aiken CS 13 Lecture
11 Lingo he states of the DFA are canonical collections of items or canonical collections of LR0) items he Dragon book gives another way of constructing LR0) items Valid Items Item X β.γ is valid for a viable prefix αβ if * αxω αβγω by a rightmost derivation After parsing αβ, the valid items are the possible tops of the stack of items Prof. Aiken CS 13 Lecture 8 1 Prof. Aiken CS 13 Lecture 8 2 Items Valid for a Prefix An item I is valid for a viable prefix α if the DFA recognizing viable prefixes terminates on input α in a state s containing I he items in s describe what the top of the item stack might be after reading input α Valid Items xample An item is often valid for many prefixes xample: he item.) is valid for prefixes... Prof. Aiken CS 13 Lecture 8 3 Prof. Aiken CS 13 Lecture 8 11
12 . Valid Items for..... *.. *.) * *..) *...) * *..) ) )... )...) * LR0) Parsing Idea: Assume stack contains α next input is t DFA on input α terminates in state s Reduce by X β if s contains item X β. Shift if s contains item X β.tω equivalent to saying s has a transition labeled t Prof. Aiken CS 13 Lecture 8 5 Prof. Aiken CS 13 Lecture 8 6. LR0) Conflicts LR0) has a reduce/reduce conflict if: Any state has two reduce items: X β. and Y ω. LR0) has a shift/reduce conflict if: Any state has a reduce item and a shift item: X β. and Y ω.tδ LR0) Conflicts..... *.. *.) * *..) *....) *. )...) wo *. shift/reduce * conflicts with.) LR0) rules ) ). Prof. Aiken CS 13 Lecture 8 Prof. Aiken CS 13 Lecture
13 SLR LR = Lefttoright scan SLR = Simple LR SLR improves on LR0) shift/reduce heuristics Fewer states have conflicts SLR Parsing Idea: Assume stack contains α next input is t DFA on input α terminates in state s Reduce by X β if s contains item X β. t FollowX) Shift if s contains item X β.tω Prof. Aiken CS 13 Lecture 8 9 Prof. Aiken CS 13 Lecture SLR Parsing Cont.) If there are conflicts under these rules, the grammar is not SLR he rules amount to a heuristic for detecting handles he SLR grammars are those where the heuristics detect exactly the handles SLR Conflicts..... *.. *.) * *..) *....) *. ).. Follow) = { ), $ }.) Follow) *. = {, ), $ } * No.) conflicts with ) SLR rules! ). Prof. Aiken CS 13 Lecture 8 51 Prof. Aiken CS 13 Lecture
14 Precedence Declarations Digression Lots of grammars aren t SLR including all ambiguous grammars We can parse more grammars by using precedence declarations Instructions for resolving conflicts Precedence Declarations Cont.) Consider our favorite ambiguous grammar: * ) he DFA for this grammar contains a state with the following items: *.. shift/reduce conflict! Declaring * has higher precedence than resolves this conflict in favor of reducing Prof. Aiken CS 13 Lecture 8 53 Prof. Aiken CS 13 Lecture 8 5 Precedence Declarations Cont.) he term precedence declaration is misleading hese declarations do not define precedence; they define conflict resolutions Not quite the same thing! Naïve SLR Parsing Algorithm 1. Let M be DFA for viable prefixes of G 2. Let x 1 x n $ be initial configuration 3. Repeat until configuration is S $ Let α ω be current configuration Run M on current stack α If M rejects α, report parsing error Stack α is not a viable prefix If M accepts α with items I, let a be next input Shift if X β. a γ I Reduce if X β. I and a FollowX) Report parsing error if neither applies Prof. Aiken CS 13 Lecture 8 55 Prof. Aiken CS 13 Lecture
15 Notes If there is a conflict in the last step, grammar is not SLRk) SLR xample Configuration DFA Halt State Action * $ 1 shift k is the amount of lookahead In practice k = 1 Prof. Aiken CS 13 Lecture 8 5 Prof. Aiken CS 13 Lecture Configuration * $..) *.. * *. *..) *. *.).) 11 * ) ). 8..) * SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift Prof. Aiken CS 13 Lecture 8 59 Prof. Aiken CS 13 Lecture
16 Configuration * $....) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Configuration * $....) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 61 Prof. Aiken CS 13 Lecture 8 62 SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift * $ 11 shift Configuration * $....) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 63 Prof. Aiken CS 13 Lecture
17 Configuration * $....) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Configuration * $....) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 65 Prof. Aiken CS 13 Lecture 8 66 SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift * $ 11 shift * $ 3 $ Follow) red... Configuration * $..) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 6 Prof. Aiken CS 13 Lecture
18 .. Configuration * $..) *.. * *. *..) *. *.).) 11 * ) ). 8..) *.. Configuration * $..) *.. * *. *..) *. *.).) 11 * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 69 Prof. Aiken CS 13 Lecture Configuration * $..) *.. * *. *..) *. *.).) 11 * ) ). 8..) * SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift * $ 11 shift * $ 3 $ Follow) red. * $ $ Follow) red. * Prof. Aiken CS 13 Lecture 8 1 Prof. Aiken CS 13 Lecture
19 . Configuration * $...) *.. * * *..) *. *.) 11.) * ) ). 8..) *. Configuration * $...) *.. * * *..) *. *.) 11.) * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 3 Prof. Aiken CS 13 Lecture 8. Configuration * $...) *.. * * *..) *. *.) 11.) * ) ). 8..) *. Configuration * $...) *.. * * *..) *. *.) 11.) * ) ). 8..) * Prof. Aiken CS 13 Lecture 8 5 Prof. Aiken CS 13 Lecture
20 SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift * $ 11 shift * $ 3 $ Follow) red. * $ $ Follow) red. * $ 5 $ Follow) red. Configuration $ * *.) * *..) 11 *....) 6 * *..) ) ). 8..) * Prof. Aiken CS 13 Lecture 8 Prof. Aiken CS 13 Lecture 8 8 Configuration $ * *.) * *..) 11 *....) 6 * *..) ) ). 8..) * SLR xample Configuration DFA Halt State Action * $ 1 shift * $ 3 * not in Follow) shift * $ 11 shift * $ 3 $ Follow) red. * $ $ Follow) red. * $ 5 $ Follow) red. $ accept Prof. Aiken CS 13 Lecture 8 9 Prof. Aiken CS 13 Lecture
21 Notes Skipped using extra start state S in this example to save space on slides Rerunning the automaton at each step is wasteful Most of the work is repeated An Improvement Remember the state of the automaton on each prefix of the stack Change stack to contain pairs Symbol, DFA State Prof. Aiken CS 13 Lecture 8 81 Prof. Aiken CS 13 Lecture 8 82 An Improvement Cont.) For a stack sym 1, state 1... sym n, state n state n is the final state of the DFA on sym 1 sym n Goto able Define goto[i,a] = j if state i A state j goto is just the transition function of the DFA One of two parsing tables Detail: he bottom of the stack is any,start where any is any dummy symbol start is the start state of the DFA Prof. Aiken CS 13 Lecture 8 83 Prof. Aiken CS 13 Lecture
22 Refined Parser Moves Shift x Push a, x on the stack a is current input x is a DFA state Reduce X α As before Accept rror Action able For each state s i and terminal a If s i has item X α.aβ and goto[i,a] = j then action[i,a] = shift j If s i has item X α. and a FollowX) and X S then action[i,a] = reduce X α If s i has item S. then action[i,$] = accept Otherwise, action[i,a] = error Prof. Aiken CS 13 Lecture 8 85 Prof. Aiken CS 13 Lecture 8 86 SLR Parsing Algorithm Notes on SLR Parsing Algorithm Let I = w$ be initial input Let j = 0 Let DFA state 1 have item S Let stack = dummy, 1 repeat case action[top_statestack),i[j]] of shift k: push I[j], k reduce X A: pop A pairs, push X, goto[top_statestack),x] accept: halt normally error: halt and report error Prof. Aiken CS 13 Lecture 8 8 Note that the algorithm uses only the DFA states and the input he stack symbols are never used! However, we still need the symbols for semantic actions Prof. Aiken CS 13 Lecture
23 More Notes Some common constructs are not SLR1) LR1) is more powerful Build lookahead o the items An LR1) item is a pair: LR0) item x lookahead [. *, $] means After seeing * reduce if lookahead is $ More accurate than just using follow sets ake a look at the LR1) automaton for your parser! Prof. Aiken CS 13 Lecture
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