Foundations of Computer Science Lecture 25 Context Free Grammars (CFGs) Solving a Problem by “Listing Out” the Language Rules for CFGs Parse Trees Pushdown Automata
Last Time DFAs: State transitioning machines which can be implemented using basic technology. Powerful: can solve any regular expression. (Finite sets, complement, union, intersection, concatenation, Kleene-star). Computing Model Analyze Model Do we need a Rules to: 1. Capabilities: what can be solved? 1. Construct machine; new model? 2. Limitations: what can’t be solved? 2. Solve problems. DFAs fail at so simple a problem as equality. That’s not acceptable. We need a more powerful machine. Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 2 / 16 Adding Memory →
Adding Memory DFAs have no scratch paper. It’s hard to compute entirely in your head. Stack Memory. Think of a file-clerk with a stack of papers. 0 0 0 0 0 1 1 1 1 1 The clerk’s capabilities: see the top sheet; remove the top sheet ( pop ) q 0 q 7 q 1 yes push something new onto the top of the stack. or q 6 q 2 0 no access to inner sheets without removing top. q 5 q 3 no q 4 DFA with a stack is a pushdown automaton (PDA) How does the stack help to solve { 0 • n 1 • n | n ≥ 0 } ? 1: When you read in each 0, write it to the stack. 2: For each 1, pop the stack. At the end if the stack is empty, accept . The memory allows the automaton to “remember” n . Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 3 / 16 Today →
Today: Context Free Grammars (CFGs) Solving a problem by listing out the language. 1 Rules for Context Free Grammars (CFG). 2 Examples of Context Free Grammars. 3 English. Programming. Proving a CFG solves a problem. 4 Parse Trees. 5 Pushdown Automata and non context free languages. 6 Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 4 / 16 Recursively Defined Language →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε → 01 Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε → 01 → 0011 Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε → 01 → 0011 → 000111 no Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε → 01 → 0011 → 000111 no A Context Free Grammar is like a recursive definition. 1: S → ε ε ∈ L 0 n 1 n Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Recursively Defined Language: Listing a Language. L 0 n 1 n = { 0 • n 1 • n | n ≥ 0 } 1 ε ∈ L 0 n 1 n . [basis] 2 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n . [constructor rule] 3 Nothing else is in L 0 n 1 n . [minimality] To test if 0010 ∈ L 0 n 1 n : generate strings in order of length and test each for a match: ε → 01 → 0011 → 000111 no A Context Free Grammar is like a recursive definition. 1: S → ε ε ∈ L 0 n 1 n 2: S → 0 S 1 x ∈ L 0 n 1 n → 0 • x • 1 ∈ L 0 n 1 n Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 5 / 16 Rules for CFGs →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals S 1: Write down the start variable (form the first production rule, typically S ). Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals 2: S 0 S 1 1: ε 1: Write down the start variable (form the first production rule, typically S ). 2: Replace one variable in the current string with the expression from a pro- duction rule that starts with that variable on the left. “Replace variable with expression , no matter where (independent of context)” Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals 2: S 0 S 1 1: ε 1: Write down the start variable (form the first production rule, typically S ). 2: Replace one variable in the current string with the expression from a pro- duction rule that starts with that variable on the left. 3: Repeat step 2 until no variables remain in the string. “Replace variable with expression , no matter where (independent of context)” Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals 2: 2: S 0 S 1 00 S 11 1: 1: ε 01 1: Write down the start variable (form the first production rule, typically S ). 2: Replace one variable in the current string with the expression from a pro- duction rule that starts with that variable on the left. 3: Repeat step 2 until no variables remain in the string. “Replace variable with expression , no matter where (independent of context)” Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals 2: 2: 2: S 0 S 1 00 S 11 000 S 111 1: 1: 1: ε 01 0011 1: Write down the start variable (form the first production rule, typically S ). 2: Replace one variable in the current string with the expression from a pro- duction rule that starts with that variable on the left. 3: Repeat step 2 until no variables remain in the string. “Replace variable with expression , no matter where (independent of context)” Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
Rules for Context Free Grammars (CFGs) Production rules of the CFG: 1: S → ε 2: S → 0 S 1 Each production rule has the form → variable expression ↑ ↑ P, Q, R, S, T, . . . string of variables and terminals 2: 2: 2: 2: S 0 S 1 00 S 11 000 S 111 0000 S 1111 1: 1: 1: 1: ε 01 0011 000111 1: Write down the start variable (form the first production rule, typically S ). 2: Replace one variable in the current string with the expression from a pro- duction rule that starts with that variable on the left. 3: Repeat step 2 until no variables remain in the string. “Replace variable with expression , no matter where (independent of context)” Creator: Malik Magdon-Ismail Context Free Grammars (CFGs): 6 / 16 cfg bal →
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