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CSCI 2670 Introduction to Theory of Computing

CSCI 2670 Introduction to Theory of Computing. September 20, 2005. Agenda. Last week Context-free grammars Examples, definition, strategies for building This week More on CFG’s Chomsky normal form, Pushdown automata, pumping lemma for CFG’s. Announcement.

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CSCI 2670 Introduction to Theory of Computing

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  1. CSCI 2670Introduction to Theory of Computing September 20, 2005

  2. Agenda • Last week • Context-free grammars • Examples, definition, strategies for building • This week • More on CFG’s • Chomsky normal form, Pushdown automata, pumping lemma for CFG’s

  3. Announcement • Recommended problems to do prior to next Tuesday (9/27) • 2.8, 2.9, 2.12, 2.13, 2.15, 2.16, 2.30 d • I will post solutions to these problems on Friday • Midterm next Tuesday • Chapters 1 & 2 • I will hand out practice midterm tomorrow and solutions the next day • I will hold extra office hours on Friday 11:00 – 12:30

  4. Chomsky normal form • Method of simplifying a CFG Definition: A context-free grammar is in Chomsky normal form if every rule is of one of the following forms A  BC A  a where a is any terminal and A is any variable, and B, and C are any variables or terminals other than the start variable if S is the start variable then the rule S  ε is the only permitted  rule

  5. CFG’s and Chomsky normal form Theorem: Any context-free language is generated by a context-free grammar in Chomsky normal form. Proof idea: Convert any CFG to one in Chomsky normal form by removing or replacing all rules in the wrong form • Add a new start symbol • Eliminate ε rules of the form A  ε • Eliminate unit rules of the form A  B • Convert remaining rules into proper form

  6. Convert a CFG to Chomsky normal form • Add a new start symbol • Create the following new rule S0 S where S is the start symbol and S0 is not used in the CFG

  7. Convert a CFG to Chomsky normal form • Eliminate all ε rules A  ε, where A is not the start variable • For each rule with an occurrence of A on the right-hand side, add a new rule with the A deleted R  uAv becomes R  uAv | uv R  uAvAw becomes R  uAvAw | uvAw | uAvw | uvw • If we have R  A, replace it with R  ε unless we had already removed R  ε

  8. Convert a CFG to Chomsky normal form • Eliminate all unit rules of the form A  B • For each rule B  u, add a new rule A  u, where u is a string of terminals and variables, unless this rule had already been removed • Repeat until all unit rules have been replaced

  9. Convert a CFG to Chomsky normal form • Convert remaining rules into proper form • What’s left? • Replace each rule A  u1u2…uk, where k  3 and ui is a variable or a terminal with k-1 rules A  u1A1 A1  u2A2 … Ak-2  uk-1uk

  10. Example S  S1 | S2 S1 S1b | Ab A  aAb | ab | ε S2 S2a | Ba B  bBa | ba| ε Step 1: Add a new start symbol

  11. Example S0  S S  S1 | S2 S1 S1b | Ab A  aAb | ab | ε S2 S2a | Ba B  bBa | ba | ε Step 2: Eliminate ε rules

  12. Example S0  S S  S1 | S2 S1 S1b | Ab | b A  aAb | ab S2 S2a | Ba | a B  bBa | ba Step 3: Eliminate all unit rules

  13. Example S0  S1b | Ab | b | S2a | Ba | a S  S1b | Ab | b | S2a | Ba | a S1 S1b | Ab | b A  aAb | ab S2 S2a | Ba | a B  bBa | ba Step 4: Convert remaining rules to proper form

  14. Example S0  S1b | Ab | b | S2a | Ba | a S  S1b | Ab | b | S2a | Ba | a S1 S1b | Ab | b A  aA1 | ab A1  Ab S2 S2a | Ba | a B  bB1| ba B1  Ba

  15. Pushdown automata • Similar to finite automata, but for CFG’s • Finite automata are not adequate for CFG’s because we cannot keep track of what we’ve done • At any point, we only know the current state, not previous states • Need memory • PDA’s are finite automata with a stack

  16. a a b b a a b b x y z Finite automata and PDA schematics State control FA State control PDA Stack: Infinite LIFO (last in first out) device

  17. Example read ε & pop $ off stack read ε & push $ on stack read ε & push ε on stack read 0 & push 0 on stack read 1 & pop 0 off stack Language accepted: {0n1n | n  0}

  18. Differences between PDA’s and NFA’s • Transitions read a symbol of the string and push a symbol onto or pop a symbol off of the stack • Stack alphabet is not necessarily the same as the alphabet for the language • e.g., $ marks bottom of stack in previous (0n1n) example

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