Homework 6 Solutions
Give pushdown automata that recognize the following languages. Give both a drawing and 6-tuple specification for each PDA. (a) A = { w ∈ {0 1}
Pushdown Automata Exercises
30-May-2006 Pda and cfg. 5. pushdown automaton A is specified by. A = ({q0q1}
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A DFA with k states can only “count to k”. •. Solution: extend NFAλ by adding memory From CFG to PDA example. S. → aSb
Solutions for CSE303 Homework 5 1. Construct nondeterministic
Construct nondeterministic pushdown automata (npda) that accept the following regular languages. Note: Observe that all the languages are regular languages so
Pushdown Automata - Examples - Lecture 18 Section 2.2
05-Oct-2009 Examples. Example (Pushdown automaton). Note that this solution is inspired by the grammar. S → SS
Two-Way Tree Automata Solving Pushdown Games
01-Mar-2006 For our particular application we simplify the definition of two-way automata ... We present here a simple example of pushdown game to illustrate ...
Pushdown Automata - Examples - Lecture 18 Section 2.2
06-Oct-2008 Examples. Example (Pushdown automaton). Note that this solution is inspired by the grammar. S → SS
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02-Nov-2012 automaton equipped with a stack-based memory. ○ Each transition. ○ is based on the current input symbol and the top of the ...
Homework 6 Solutions
Give pushdown automata that recognize the following languages. Give both a drawing and 6-tuple specification for each PDA. (a) A = { w ? {0 1}
Pushdown Automata (PDA)
Q) Does a PDA that accepts by empty stack need any final state specified in the design? Page 15. Example: L of balanced p parenthesis. PDA that
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We start with standard problems on building pda for a given language Construct pushdown automata for the following languages. ... Solutions. 1a The pda ...
Pushdown Automata - Examples - Lecture 18 Section 2.2
Oct 5 2009 Balanced. Parentheses. Algebraic. Expressions. Assignment. Homework Review. Solution. A PDA for the first of these languages is.
Exercise Sheet 4
Nov 27 2014 Exercise 4.4 (Pushdown Automata). Create a PDA that recognizes the following language. L = {aibjck
i
k = i + j}. Solution: ...
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A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. ? Each transition. ? is based on the current input symbol and the top
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Tutorial 3 Context Free Languages and Pushdown Automata
convert a context free grammar to a (non-deterministic) PDA (Q 3.1); Solution: There are two parse trees for ¬ true ? false for example:.
Solutions for CSE303 Homework 5 1. Construct nondeterministic
Construct nondeterministic pushdown automata (npda) that accept the following regular languages. Note: Observe that all the languages are regular languages so
QUESTION BANK SOLUTION Unit 1 Introduction to Finite Automata
This distinguishes it from the deterministic finite automaton (DFA) all languages that can be recognized by a non-deterministic pushdown automaton.
Pushdown Automata - Stanford University
A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory Each transition is based on the current input symbol and the top of the stack optionally pops the top of the stack and optionally pushes new symbols onto the stack Initially the stack holds a special symbol Z 0that indicates the bottom of the stack
Pushdown Automata - University of North Carolina at Chapel Hill
A pushdown automaton(PDA) is essentially a finite automaton with a stack Example PDA accepting Initially the symbol the stack 0is on Acceptance can be by final state or empty stack = 01 ?0: 0 Stack Input string: 0011 Current input A PDA can be defined by a 7-tuple ?? 0 0 : A finite set of states
18404J F2020 Lecture 4: Pushdown Automata CFG PDA
Pushdown Automata (PDA) Finite control “head” ba ba a input appears on a “tape” c (pushdown)Schematic diagram for DFA or NFA stack Schematic diagram for PDA Operates like an NFA except can write-add or read-remove symbols from the top of stack push pop Example: PDA for ! = 0$1$ & ? 0
Pushdown Automata Exercises - Leiden University
16 A two-way pushdown automaton may move on its input tape in two directions As usual for two-way automata we assume that the begin and end of the input tape is marked by special symbols In this way the automaton can recognize those positions Describe a two-way pda for each of the following languages (a) f anbncn j n 2 N g (easy)
Pushdown Automata (()PDA) - Washington State University
How to convert an final state PDA into an empty stack PDA? P F==> P N construction Main idea: Whenever P F reaches a final state just make an -transition into a new end state clear out the stack and acceptnew end state clear out the stack and accept
Solutions for CSE303 Homework 5 Solution - Stony Brook University
Solutions for CSE303 Homework 5 1 Construct nondeterministic pushdown automata (npda) that accept the following regular languages Note: Observe that all the languages are regular languages so the solutions are essentially NFA’s (ornpda’s with inactive stack) For all the languagesfis the ?nal state (a)L1=L(aaa?b)
Searches related to pushdown automata examples solutions filetype:pdf
• A DPDA is simply a pushdown automata without non-determinism – i e no epsilon transitions or transitions to multiple states on same input – Only one state at a time • DPDA not as powerful a non-deterministic PDA – This machine accepts a class of languages somewhere between regular languages and context-free languages
Solutions for CSE303 Homework 5
1. Construct nondeterministic pushdown automata (npda) that accept the following regular languages.
Note: Observe that all the languages are regular languages, so the solutions are essentially NFA"s (or
npda"s with inactive stack). For all the languages,fis the final state. (a)L1=L(aaa?b) Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={q0,q1,q2,f}, Σ ={a,b},F={f}, andΔ as the set of following rules acceptsL1.
((q0,a,e),(q1,e)) ((q1,a,e),(q2,e)) ((q2,a,e),(q2,e)) ((q2,b,e),(f,e)) (b)L2=L(abb?aba?) Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={r0,r1,r2,r3,f}, Σ ={a,b},F={f}, and Δ as the set of following rules acceptsL2. ((r0,a,e),(r1,e)) ((r1,b,e),(r2,e)) ((r2,b,e),(r2,e)) ((r2,a,e),(r3,e)) ((r3,b,e),(f,e)) ((f,a,e),(f,e)) (c) the union ofL1andL2 Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={s,q0,q1,q2,r0,r1,r2,r3,f}, Σ ={a,b}, F={f}, and Δ as the set of following rules acceptsL1?L2 ((s,e,e),(q0,e)) ((s,e,e),(r0,e)) whereq0andr0are as defined above.2. Construct npda"s that accept the following languages on Σ ={a,b,c}.
(a)L={anb2n:n≥0} Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={s,f}, Σ ={a,b},F={f}, and Δ as the set of following rules acceptsL. ((s,e,e),(f,e)) ((s,a,e),(s,aa)) ((s,b,a),(f,e)) ((f,b,a),(f,e)) (b)L={anbmcn+m:n≥0,m≥0} Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={s0,s1,f}, Σ ={a,b},F={s1}, and Δ as the set of following rules acceptsL. ((s0,e,e),(f,e)) ((s0,a,e),(s0,a)) ((s0,c,a),(f,e)) ((s0,b,e),(s1,b)) ((s1,b,e),(s1,b)) ((s1,c,b),(f,e)) ((f,c,b),(f,e)) ((f,c,a),(f,e)) 13. Find an npda on Σ ={a,b,c}that accepts the languageL={w1cw2:w1,w2? {a,b}?,w1?=wR2}
Solution: Idea: Observe that ifw1=wR2, then the npda should not accept the input. Butw1=wR2is equivalent towR1=w2. Hence, we first, pushw1onto stack, then while matchingw2, if a mismatch isfound (thenwR1?=w2), consume all the input and empty the stack then go to final state; otherwise (then
w R1=w2which meansw1=wR2), the input is not accepted. Thus the npdaM= (K,Σ,Γ,Δ,s,F), whereK={s0,s1,d,f}, Σ ={a,b},F={f}, and Δ as the set of following rules acceptsL. ((s0,a,e),(s0,a)) ((s0,b,e),(s0,b)) ((s0,c,e),(s1,e)) ((s1,a,a),(s1,e)) ((s1,b,b),(s1,e)) ((s1,a,b),(d,e)) ((s1,b,a),(d,e)) ((d,e,a),(d,e)) ((d,e,b),(d,e)) ((d,a,e),(d,e)) ((d,b,e),(d,e)) ((d,e,e),(f,e))4. Construct an npda corresponding to the grammar
S→aABB|aAA,
A→aBB|a,
B→bBB|A
Solution: The npdaM= (K,Σ,Γ,Δ,s,F), whereK={s0,s1}, Σ ={a,b},F={s1}, and Δ as the set of following rules implements the grammar above. ((s0,e,e),(s1,S) ((s1,a,S),(s1,ABB) ((s1,a,S),(s1,AA) ((s1,a,A),(s1,BB) ((s1,a,A),(s1,e) ((s1,b,B),(s1,BB) ((s1,a,B),(s1,BB) ((s1,a,B),(s1,e)Another solution:((s0,e,e),(s1,S)
((s1,e,S),(s1,aABB)) ((s1,e,S),(s1,aAA)) ((s1,e,A),(s1,aBB)) ((s1,e,B),(s1,bBB)) ((s1,e,B),(s1,A)) ((s1,a,a),(s1,e)) ((s1,b,b),(s1,e))5. Show that the languageL={ww:w? {a,b}?}is not context-free.
Solution: Consider the stringambmambm. Now, letvandycontain only the firsta"s and letv=ap andy=aq. Then consideruxz=akbmambm,k < m, which is not inL. For other choices ofvandy, we can make similar arguments (since each symbol occurs exactlymtimes). Thus,Lis not context-free. 26. Show that the languageL={an!:n≥0}is not context-free.
Solution: In this case, this is same as showing thatLis not regular (since the language consists entirely of alphabet over single symbol). Letw=uvxyzsuch thatuv=?. Then we need to show that we havem!-k >(m-1)!. ThereforeLis not context-free (or regular, either).7. Construct Turing machines that will accept the following languages on{a,b}.
(a)L={w:|w|is even} Solution: Here we keep checking off two input symbols and if ultimately we encounter end of input in the start state, we accept the input. The Turing machineM= (K,Σ,δ,s,H), whereK={q0,q1,h},Σ ={a,b,?,?},s=q0,H={h}andδas follows acceptsL.q σδ(q,σ)q0a(q1,→)q0b(q1,→)q1a(q0,→)q1b(q0,→)q0?(h,?)(b)L={w:na(w) =nb(w)}
wherena(w) is number ofa"s inwandnb(w) is number ofb"s inw. Solution: First find leftmostaorb. Ifa(orb) is found, then replace it withxand find a matchingb (ora), replace it withxand then go back to the left end of the tape (so as to find next leftmostaor b). Repeat. The Turing machineM= (K,Σ,δ,s,H), whereK={q0,q1,q2,q3,h},Σ ={a,b,?,?},s=q0,H={h}andδas given below acceptsL.q σδ(q,σ)q0x(q0,→)q0a(q1,x)q0b(q2,x)q1x(q1,→)q1a(q1,→)q1b(q3,x)q2x(q2,→)q2a(q3,x)q2b(q2,→)q3x(q3,←)q3a(q3,←)q3b(q3,←)q3?(q0,→)q0?(h,?)3
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