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CPSC 325 - Compiler Tutorial 4 Midterm Review

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Deterministic Finite Automata (DFA) Q: finite set of states Σ: finite set of “letters” (input alphabet) δ: Q × Σ → Q (transition function) q 0 : start state (in Q) F: set of accept states (subset of Q) Acceptance: input consumed with the automaton in a final state.

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DFA Example q1 q2 1 0 0 1 δ01 q1 q2 q1q2 Transition function: Start: Final:

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DFA Example S q1 q2 r1 r2 a b a ab b b ab a

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Non-deterministic Finite Automata (NFA) Transition function is different δ: Q × Σ ε → P(Q) P(Q) is the powerset of Q (set of all subsets) Σ ε is the union of Σ and the special symbol ε (denoting empty) String is accepted if there is at least one path leads to an accept state, and input is consumed.

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NFA Example q1q2q3q4 0, 1 1 0, ε1 0, 1 δ01ε q1{q1}{q1, q2} q2{q3} q3{q4} q4{q4} What strings does this NFA accept?

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Converting an NFA to a DFA For set of states S, closure(S) is the set of states that can be reached from S without consuming any input. For a set of states S, DFAedge(S, c) is the set of states that can be reached from S by consuming input symbol c; algorithm: for each state s in S add closure(δ(s,c)) to DFAedge(S,c)

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NFA to DFA conversion (cont ’ d) A state of the DFA after reading a given input letter represents the set of states that the NFA might have reached with the same input letter. Each state of the DFA that contains a final state of the NFA is a final state of the DFA. Number of states of the DFA exponential (in the worst case) in the number of states of the NFA (2 n ).

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NFA to DFA conversion (formally) If Q is set of states for NFA, P(Q) is set of states for DFA. Σ is same for both automata. If q 0 is start state of NFA, closure(q 0 ) is start state of DFA. Each state of the DFA that contains a final state of the NFA is a final state of the DFA. Transition function for DFA is: (S,c)=DFAedge(S, c), which is the set of states that can be reached from S by consuming input symbol c.

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NFA →DFA Example (1) NFA example (reg.ex.: 01(00|11)*10 ) 01 00 1110 ε ε ε ε q0q0 q1q1 q2q2 q6q6 q7q7 q8q8 q3q3 q4q4 q5q5 q9q9 q 10 q 11 ε

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NFA →DFA Example (2) State01 S: q 0 q1q1 q1q1 q 2 q 6 q 3 q 9 D: q 2 q 6 q 3 q 9 q7q7 q 4 q 10 q7q7 q 8 q 2 q 6 q 3 q 9 q 4 q 10 q 11 q 5 q 2 q 6 q 3 q 9 q 8 q 2 q 6 q 3 q 9 q7q7 q 4 q 10 F: q 11 q 5 q 2 q 6 q 3 q 9 q7q7 q 4 q 10

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NFA →DFA Example (3) 0 q0q0 q1q1 1 1 q2q3q6q9q2q3q6q9 q7q7 q 4 q 10 q 11 q2q3q6q8q9q2q3q6q8q9 q2q3q6q5q9q2q3q6q5q9 00 1 0,1 1 1 0 00 0 1 1

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Exercise What language is accepted by following NFA? Convert it to a DFA. 01 0 10 0 1 q0q0 q1q1 q2q2 q5q5 q6q6 q3q3 q4q4 q7q7 q8q8 1 0 1 1 1 0 1

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A solution 01(00|11)*10 DFA: 01 0 1 0 q0q0 q1q1 q2q2 q5q5 q6q6 q 3, q 7 q4q4 q8q8 1 0 1 0 {} 0,1 1 0 1 0 1

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Regular Expression *: denote 0 or more occurrence +: denote 1 or more occurrence | : OR Example: the sentence contains string 0110 (0|1)*0110(0|1)*

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Converting Regular grammars to (N)FA (0|1)*0110(0|1)* q0q1q2 q3q4 01 10 0,1

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Ambiguity Grammar G is unambiguous iff every sentence in L(G) has a unique leftmost (or rightmost) derivation – Fact: unique leftmost or unique rightmost implies the other A grammar without this property is ambiguous – Note that other grammars that generate the same language may be unambigious

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Solutions The grammar has no notion of precedence or associatively Solution: – Create a non-terminal for each level of precedence – Isolate the corresponding part of the grammar – Force the parser to recognize higher precedence sub expressions first

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Classic Expression Grammar expr ::= term + expr | term – expr | term term ::= factor * term | factor / term | factor factor ::= int | ( expr ) int ::= 0 | 1 | 2 | … | 9

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Another Classic example Grammar for conditional statements...... stmt ::= ifStmt | whileStmt ifStmt ::= if ( cond ) stmt | if ( cond ) stmt lese stmt ……

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Solving Ambiguity Fix the grammar to separate if statements with else clause and if statement with no else - add lots of non-terminals Use some ad-hoc rule in parser - “else matches closest unpaired if”

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Reading for Mid-term Converting NFA to DFA Converting Regular grammars to (N)FA Parsing bottom up or top down General phases of a compiler Chomsky Type 0-3 languages Left recursion causing problem for top down parser and solution to it empty rules causing problems for bottom-up parsers … others Input

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