Warmup: Design a CFG to generate the language \(\{a^i b^j \mid j \geq i \geq 0\}\)
Sample derivation:
Design a PDA to recognize the language \(\{a^i b^j \mid j \geq i \geq 0\}\)
Theorem 2.20: A language is generated by some context-free grammar if and only if it is recognized by some push-down automaton.
Definition: a language is called context-free if it is the language generated by a context-free grammar. The class of all context-free language over a given alphabet \(\Sigma\) is called CFL.
Consequences:
Quick proof that every regular language is context free
To prove closure of the class of context-free languages under a given operation, we can choose either of two modes of proof (via CFGs or PDAs) depending on which is easier
To fully specify a PDA we could give its \(6\)-tuple formal definition or we could give its input alphabet, stack alphabet, and state diagram. An informal description of a PDA is a step-by-step description of how its computations would process input strings; the reader should be able to reconstruct the state diagram or formal definition precisely from such a descripton. The informal description of a PDA can refer to some common modules or subroutines that are computable by PDAs:
PDAs can “test for emptiness of stack” without providing details. How? We can always push a special end-of-stack symbol, \(\$\), at the start, before processing any input, and then use this symbol as a flag.
PDAs can “test for end of input” without providing details. How? We can transform a PDA to one where accepting states are only those reachable when there are no more input symbols.
Suppose \(L_1\) and \(L_2\) are context-free languages over \(\Sigma\). Goal: \(L_1 \cup L_2\) is also context-free.
Approach 1: with PDAs
Let \(M_1 = ( Q_1, \Sigma, \Gamma_1, \delta_1, q_1, F_1)\) and \(M_2 = ( Q_2, \Sigma, \Gamma_2, \delta_2, q_2, F_2)\) be PDAs with \(L(M_1) = L_1\) and \(L(M_2) = L_2\).
Define \(M =\)
Approach 2: with CFGs
Let \(G_1 = (V_1, \Sigma, R_1, S_1)\) and \(G_2 = (V_2, \Sigma, R_2, S_2)\) be CFGs with \(L(G_1) = L_1\) and \(L(G_2) = L_2\).
Define \(G =\)
Suppose \(L_1\) and \(L_2\) are context-free languages over \(\Sigma\). Goal: \(L_1 \circ L_2\) is also context-free.
Approach 1: with PDAs
Let \(M_1 = ( Q_1, \Sigma, \Gamma_1, \delta_1, q_1, F_1)\) and \(M_2 = ( Q_2, \Sigma, \Gamma_2, \delta_2, q_2, F_2)\) be PDAs with \(L(M_1) = L_1\) and \(L(M_2) = L_2\).
Define \(M =\)
Approach 2: with CFGs
Let \(G_1 = (V_1, \Sigma, R_1, S_1)\) and \(G_2 = (V_2, \Sigma, R_2, S_2)\) be CFGs with \(L(G_1) = L_1\) and \(L(G_2) = L_2\).
Define \(G =\)
Summary
Over a fixed alphabet \(\Sigma\), a language \(L\) is regular
iff it is described by some regular expression
iff it is recognized by some DFA
iff it is recognized by some NFA
Over a fixed alphabet \(\Sigma\), a language \(L\) is context-free
iff it is generated by some CFG
iff it is recognized by some PDA
Fact: Every regular language is a context-free language.
Fact: There are context-free languages that are not nonregular.
Fact: There are countably many regular languages.
Fact: There are countably infinitely many context-free languages.
Consequence: Most languages are not context-free!
Examples of non-context-free languages
\[\begin{aligned} &\{ a^n b^n c^n \mid 0 \leq n , n \in \mathbb{Z}\}\\ &\{ a^i b^j c^k \mid 0 \leq i \leq j \leq k , i \in \mathbb{Z}, j \in \mathbb{Z}, k \in \mathbb{Z}\}\\ &\{ ww \mid w \in \{0,1\}^* \} \end{aligned}\] (Sipser Ex 2.36, Ex 2.37, 2.38)
There is a Pumping Lemma for CFL that can be used to prove a specific language is non-context-free: If \(A\) is a context-free language, there is a number \(p\) where, if \(s\) is any string in \(A\) of length at least \(p\), then \(s\) may be divided into five pieces \(s = uvxyz\) where (1) for each \(i \geq 0\), \(uv^ixy^iz \in A\), (2) \(|uv|>0\), (3) \(|vxy| \leq p\). We will not go into the details of the proof or application of Pumping Lemma for CFLs this quarter.
Recall: A set \(X\) is said to be closed under an operation \(OP\) if, for any elements in \(X\), applying \(OP\) to them gives an element in \(X\).
| True/False | Closure claim |
|---|---|
| True | The set of integers is closed under multiplication. |
| \(\forall x \forall y \left( ~(x \in \mathbb{Z} \wedge y \in \mathbb{Z})\to xy \in \mathbb{Z}~\right)\) | |
| True | For each set \(A\), the power set of \(A\) is closed under intersection. |
| \(\forall A_1 \forall A_2 \left( ~(A_1 \in \mathcal{P}(A) \wedge A_2 \in \mathcal{P}(A) \in \mathbb{Z}) \to A_1 \cap A_2 \in \mathcal{P}(A)~\right)\) | |
| The class of regular languages over \(\Sigma\) is closed under complementation. | |
| The class of regular languages over \(\Sigma\) is closed under union. | |
| The class of regular languages over \(\Sigma\) is closed under intersection. | |
| The class of regular languages over \(\Sigma\) is closed under concatenation. | |
| The class of regular languages over \(\Sigma\) is closed under Kleene star. | |
| The class of context-free languages over \(\Sigma\) is closed under complementation. | |
| The class of context-free languages over \(\Sigma\) is closed under union. | |
| The class of context-free languages over \(\Sigma\) is closed under intersection. | |
| The class of context-free languages over \(\Sigma\) is closed under concatenation. | |
| The class of context-free languages over \(\Sigma\) is closed under Kleene star. | |
Regular sets are not the end of the story
Many nice / simple / important sets are not regular
Limitation of the finite-state automaton model: Can’t “count", Can only remember finitely far into the past, Can’t backtrack, Must make decisions in “real-time"
We know actual computers are more powerful than this model...
The next model of computation. Idea: allow some memory of unbounded size. How?
To generalize regular expressions: context-free
grammars
To generalize NFA: Pushdown automata, which is like an NFA with access to a stack: Number of states is fixed, number of entries in stack is unbounded. At each step (1) Transition to new state based on current state, letter read, and top letter of stack, then (2) (Possibly) push or pop a letter to (or from) top of stack. Accept a string iff there is some sequence of states and some sequence of stack contents which helps the PDA processes the entire input string and ends in an accepting state.
Is there a PDA that recognizes the nonregular language \(\{0^n1^n \mid n \geq 0 \}\)?
The PDA with state diagram above can be informally described as:
Read symbols from the input. As each 0 is read, push it onto the stack. As soon as 1s are seen, pop a 0 off the stack for each 1 read. If the stack becomes empty and we are at the end of the input string, accept the input. If the stack becomes empty and there are 1s left to read, or if 1s are finished while the stack still contains 0s, or if any 0s appear in the string following 1s, reject the input.
Trace the computation of this PDA on the input string \(01\).
Trace the computation of this PDA on the input string \(011\).
A PDA recognizing the set \(\{ \hspace{1.5 in} \}\) can be informally described as:
Read symbols from the input. As each 0 is read, push it onto the stack. As soon as 1s are seen, pop a 0 off the stack for each 1 read. If the stack becomes empty and there is exactly one 1 left to read, read that 1 and accept the input. If the stack becomes empty and there are either zero or more than one 1s left to read, or if the 1s are finished while the stack still contains 0s, or if any 0s appear in the input following 1s, reject the input.
Modify the state diagram below to get a PDA that implements this description:
Definition A pushdown automaton (PDA) is specified by a \(6\)-tuple \((Q, \Sigma, \Gamma, \delta, q_0, F)\) where \(Q\) is the finite set of states, \(\Sigma\) is the input alphabet, \(\Gamma\) is the stack alphabet, \[\delta: Q \times \Sigma_\varepsilon \times \Gamma_\varepsilon \to \mathcal{P}( Q \times \Gamma_\varepsilon)\] is the transition function, \(q_0 \in Q\) is the start state, \(F \subseteq Q\) is the set of accept states.
Draw the state diagram and give the formal definition of a PDA with \(\Sigma = \Gamma\).
Draw the state diagram and give the formal definition of a PDA with \(\Sigma \cap \Gamma = \emptyset\).
For the PDA state diagrams below, \(\Sigma = \{0,1\}\).
| Mathematical description of language | State diagram of PDA recognizing language |
|---|---|
| \(\Gamma = \{ \$, \#\}\) | |
| \(\Gamma = \{ \sun, 1\}\) | |
| \(\{ 0^i 1^j 0^k \mid i,j,k \geq 0 \}\) | |
Note: alternate notation is to replace \(;\) with \(\to\)
Big picture: PDAs were motivated by wanting to add some memory of unbounded size to NFA. How do we accomplish a similar enhancement of regular expressions to get a syntactic model that is more expressive?
DFA, NFA, PDA: Machines process one input string at a time; the computation of a machine on its input string reads the input from left to right.
Regular expressions: Syntactic descriptions of all strings that match a particular pattern; the language described by a regular expression is built up recursively according to the expression’s syntax
Context-free grammars: Rules to produce
one string at a time, adding characters from the middle, beginning, or
end of the final string as the derivation proceeds.
Definitions below are on pages 101-102.
| Term | Typical symbol | Meaning |
| or Notation | ||
| Context-free grammar (CFG) | \(G\) | \(G = (V, \Sigma, R, S)\) |
| The set of variables | \(V\) | Finite set of symbols that represent phases in production pattern |
| The set of terminals | \(\Sigma\) | Alphabet of symbols of strings generated by CFG |
| \(V \cap \Sigma = \emptyset\) | ||
| The set of rules | \(R\) | Each rule is \(A \to u\) with \(A \in V\) and \(u \in (V \cup \Sigma)^*\) |
| The start variable | \(S\) | Usually on left-hand-side of first/ topmost rule |
| Derivation | \(S \Rightarrow \cdots \Rightarrow w\) | Sequence of substitutions in a CFG (also written \(S \Rightarrow^* w\)). At each step, we can apply one rule to one occurrence of a variable in the current string by substituting that occurrence of the variable with the right-hand-side of the rule. The derivation must end when the current string has only terminals (no variables) because then there are no instances of variables to apply a rule to. |
| Language generated by the context-free grammar \(G\) | \(L(G)\) | The set of strings for which there is a derivation in \(G\). Symbolically: \(\{ w \in \Sigma^* \mid S \Rightarrow^* w \}\) i.e. \[\{ w \in \Sigma^* \mid \text{there is derivation in $G$ that ends in $w$} \}\] |
| Context-free language | A language that is the language generated by some context-free grammar |
Examples of context-free grammars, derivations in those grammars, and the languages generated by those grammars
\(G_1 = (\{S\}, \{0\}, R, S)\) with rules \[\begin{aligned} &S \to 0S\\ &S \to 0\\ \end{aligned}\] In \(L(G_1)\) …
Not in \(L(G_1)\) …
\(G_2 = (\{S\}, \{0,1\}, R, S)\) \[S \to 0S \mid 1S \mid \varepsilon\] In \(L(G_2)\) …
Not in \(L(G_2)\) …
\((\{S, T\}, \{0, 1\}, R, S)\) with rules \[\begin{aligned} &S \to T1T1T1T \\ &T \to 0T \mid 1T \mid \varepsilon \end{aligned}\]
In \(L(G_3)\) …
Not in \(L(G_3)\) …
\(G_4 = (\{A, B\}, \{0, 1\}, R, A)\) with rules \[A \to 0A0 \mid 0A1 \mid 1A0 \mid 1A1 \mid 1\] In \(L(G_4)\) …
Not in \(L(G_4)\) …
Design a CFG to generate the language \(\{a^n b^n \mid n \geq 0\}\)
Sample derivation: