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Design contex-free grammars that generate: L 1 = { u v : u ∈ {a,b}*, v ∈ {a, c}*, and |u| ≤...
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Design contex-free grammars that generate: L 1 = { u v : u ∈ {a,b}* , v ∈ {a, c}*, and |u| ≤ |v| ≤ 3 |u| }. L 2 = { a p b q c p a r b 2r : p, q,
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Transcript of Design contex-free grammars that generate: L 1 = { u v : u ∈ {a,b}*, v ∈ {a, c}*, and |u| ≤...
Design contex-free grammars that generate:
L1= { u v : u ∈ {a,b}* , v ∈ {a, c}*, and |u| ≤ |v| ≤ 3 |u| }.
L2= { ap bq cp ar b2r : p, q, r ≥ 0}
Midterm tutorial: Monday June 25 at 7:30pm, ECS 116.
Tutorial this week: Bring any questions you have about assignment #3 or old midterms. Material covered if you have no other questions: Context-free grammars and Pushdown Automata.
Assignment #3 is due at the beginning of class on Tuesday June 26.
3
Today:
We will go back over more examples of context-free grammars, pushdown automata and parse trees. I previously went through that material fairly quickly so you would be able to do the assignment last weekend if you wanted to.
L= {w {a,b}* : w has abb and a prefix and bba as a suffix}
1. Create a regular context-free grammar that generates L.
2. Give a context-free grammar for L that is not regular.
Recall: A regular context-free grammar can have at most one non-terminal on the RHS of each rule, if there is a non-terminal, it is the last symbol on the RHS of the rule.
To accept, there must exist a computation which:
1.Starts in the start state with an empty stack.
2. Applies transitions of the PDA which are applicable at each step.
3. Consumes all the input.
4. Terminates in a final state with an empty stack.
L(M)= { w Σ* : (s, w, ε) ├* (f, ε, ε) for some final state f in F}.
L= { w wR : w {a, b}* }
State Input Pop Next state
Push
s a ε s A
s b ε s B
s ε ε t ε
t a A t ε
t b B t ε
Start state: s, Final State: {t}
PDA’s are non-deterministic:
Guessing wrong time to switch from s to t gives non-accepting computations.
Some non-accepting computations on aaaa:1. Transfer to state t too early:(s, aaaa, ε) ⊢ (s, aaa, A)⊢ (t, aaa, A) ⊢ (t, aa, ε) Cannot finish reading input because stack is empty.2. Transfer to state t too late:(s, aaaa, ε) ⊢ (s, aaa, A)⊢(s, aa, AA) ⊢(s, a, AAA) ⊢(t, a, AAA) ⊢ (t, ε , AA) Cannot empty stack.
Accepting computation on aaaa:
(s, aaaa, ε) ⊢ (s, aaa, A)⊢ (s, aa, AA) ⊢ (t, aa, AA) ⊢ (t, a, A) ⊢ (t, ε, ε)
The computation started in the start state with the input string and an empty stack.It terminated in a final state with all the input consumed and an empty stack.
State Input PopNextstate
Push
s a ε s B
s a A s ε
s b ε s A
s b B s ε
L= {w in {a, b}* : w has the same number of a’s and b’s}
Start state: s
Final states: {s}
State Input Pop Next state
Push
s ε ε t X
t a X t BX
t a A t ε
t a B t BB
t b X t AX
t b A t AA
t b B t ε
t ε X f ε
L= {w in {a, b}* : w has the same number of a’s and b’s}
State state: s, Final states: {f}A more deterministic solution:
Stack will never contain both A’s and B’s.
X- bottom of stack marker.
Parse Trees- G = (V, Σ, R S)
Nodes of parse trees are each labelled with one symbol from V ⋃ {ε} .
Node at top- root, labelled with start symbol S.
Leaves- nodes at the bottom.
Yield- concatenate together the labels on the leaves going from left to right.
Find the leftmost derivation for this parse tree.
To pump:
Look for a path from root with a repeated non-terminal.
Path:
S - A - T - A
Non-terminal A is repeated.
Yield from second copy downwards gives x.
Consider tree from first copy downwards.
Yield before x is v. Yield after
x is y.
Yield after y is z.
Yield of whole tree before v is u.
Cut off part of tree from second copy downwards.
Cut off part from first copy downwards.
To pump zero times:
New yield:
u x z = b bc c
= u v0 x y0 z
To pump twice:
Pumping twice: New yield is:
u v v x y y z
= u v2 x y2 z
To pump three times:
New yield is:
u vvv x yyy z
= u v3 x y3 z
To pump n times:
New yield is:
= u vn x yn z
The Pumping Theorem for Context-Free Languages:
Let G be a context-free grammar.
Then there exists some constant k which depends on G such that for any string w which is generated by G with |w| ≥ k,
there exists u, v, x, y, z, such that
1. w= u v x y z,
2. |v| + |y| ≥ 1, and
3. u vn x yn z is in L for all n ≥ 0.
1.Draw a parse tree for the following derivation:
S C A C C A b b b b A b b
b b B b b b b a A a a b b
b b a b a a b b
2. Show on your parse tree u, v, x, y, z as per the pumping theorem.
3. Prove that the language for this question is an infinite language.
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