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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Lecture 6: Pushdown automata
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
Pushdown Automata Exercises
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: ...
Pushdown Automata
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
CS 341: Foundations of Computer Science II
Prof. Marvin Nakayama
Homework 6 Solutions
1. Give pushdown automata that recognize the following languages.Give both a drawing
and6-tuple specification for each PDA. (a)A={w? {0,1}?|wcontains at least three1s}Answer:
q1q2q3q40, ε→ε
1, ε→ε
0, ε→ε
1, ε→ε
0, ε→ε
1, ε→ε
0, ε→ε
1, ε→ε
We formally express the PDA as a 6-tuple(Q,Σ,Γ,δ,q1,F), where •Q={q1,q2,q3,q4} •Σ ={0,1} •Γ ={0,1} •transition functionδ:Q×Σε×Γε→ P(Q×Γε)is defined byInput:01ε
Stack:01ε01ε01ε
q1{(q1,ε)}{(q2,ε)} q2{(q2,ε)}{(q3,ε)} q3{(q3,ε)}{(q4,ε)} q4{(q4,ε)}{(q4,ε)}Blank entries are∅.
•q1is the start state •F={q4} Note thatAis a regular language, so the language has a DFA. We can easily convert the DFA into a PDA by using the same states and transitionsand never push nor pop anything to/from the stack. 1 (b)B={w? {0,1}?|w=wRand the length ofwis odd}Answer:
q1q2q3q4ε, ε→$0, ε→0
1, ε→1
0, ε→ε
1, ε→ε
0,0→ε
1,1→ε
Since the length of any stringw?Bis odd,wmust have a symbol exactly in the middle position; i.e.,|w|= 2n+ 1for somen≥0, and the(n+ 1)th symbol inwis the middle one. If a stringwof length2n+ 1satisfiesw=wR, the firstnsymbols must match (in reverse order) the lastnsymbols, and the middle symbol doesn"t have to match anything. Thus, in the above PDA, the transition fromq2to itself reads the firstnsymbols and pushes these on the stack. The transition fromq2toq3nondeterministically identifies the middle symbol ofw, which doesn"t need to match any symbol, so the stack is unaltered. The transition fromq3to itself then reads the lastnsymbols ofw, popping the stack at each step to make sure the symbols after the middle match (in reverse order)the symbols before the middle. We formally express the PDA as a 6-tuple(Q,Σ,Γ,δ,q1,F), where •Q={q1,q2,q3,q4} •Σ ={0,1} •Γ ={0,1,$}(use$to mark bottom of stack) •transition functionδ:Q×Σε×Γε→ P(Q×Γε)is defined byInput:01ε
Stack:01$ε01$ε01$ε
q1{(q2,$)} q2{(q2,0),(q3,ε)}{(q2,1),(q3,ε)} q3{(q3,ε)}{(q3,ε)}{(q4,ε)} q4Blank entries are∅.
•q1is the start state •F={q4} (c)C={w? {0,1}?|w=wR}Answer:
2 q1q2q3q4ε, ε→$0, ε→0
1, ε→1
0, ε→ε
1, ε→ε
0,0→ε
1,1→ε
The length of a stringw?Ccan be either even or odd. If it"s even, then there is no middle symbol inw, so the first half ofwis pushed on the stack, we move fromq2toq3without reading, pushing, or popping anything, and then match the second half ofwto the first half in reverse order by popping the stack. If the length ofwis odd, then there is a middle symbol inw, and the description of thePDA in part (b) applies.
We formally express the PDA as a 6-tuple(Q,Σ,Γ,δ,q1,F), where •Q={q1,q2,q3,q4} •Σ ={0,1} •Γ ={0,1,$}(use$to mark bottom of stack) •transition functionδ:Q×Σε×Γε→ P(Q×Γε)is defined byInput:01ε
Stack:01$ε01$ε01$ε
q1{(q2,$)} q2{(q2,0),(q3,ε)}{(q2,1),(q3,ε)}{(q3,ε)} q3{(q3,ε)}{(q3,ε)}{(q4,ε)} q4Blank entries are∅.
•q1is the start state •F={q1,q4} (d)D={aibjck|i,j,k≥0, andi=jorj=k}Answer:
q1 q2q3q4 q5q6q7q8 a, ε→ab, a→ε c, ε→ε a, ε→ε b, ε→bc, b→ε 3 The PDA has a nondeterministic branch atq1. If the string isaibjckwithi=j, then the PDA takes the branch fromq1toq2. If the string isaibjckwithj=k, then the PDA takes the branch fromq1toq5. We formally express the PDA as a 6-tuple(Q,Σ,Γ,δ,q1,F), where •Q={q1,q2,...,q8} •Σ ={a,b,c} •Γ ={a,b,$}(use$to mark bottom of stack) •transition functionδ:Q×Σε×Γε→ P(Q×Γε)is defined byInput:abcε
Stack:abc$εabc$εabc$εabc$ε
q1{(q2,$),(q5,$)} q2{(q2,a)}{(q3,ε)} q3{(q3,ε)}{(q4,ε)} q4{(q4,ε)} q5{(q5,ε)}{(q6,ε)} q6{(q6,b)}{(q7,ε)} q7{(q7,ε)}{(q8,ε)} q8Blank entries are∅.
•q1is the start state •F={q4,q8} (e)E={aibjck|i,j,k≥0andi+j=k}Answer:
q1q2q3q4q5ε, ε→$ε, ε→ε a, ε→xb, ε→x c, x→ε For everyaandbread in the first part of the string, the PDA pushes anxonto the stack. Then it must read acfor eachxpopped off the stack. We formally express the PDA as a 6-tuple(Q,Σ,Γ,δ,q1,F), where •Q={q1,q2,...,q5} •Σ ={a,b,c} •Γ ={x,$}(use$to mark bottom of stack) •transition functionδ:Q×Σε×Γε→ P(Q×Γε)is defined byInput:abcε
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