Convert the given CFG to CNF: a) E-> E+T | T, T-> T*F | F, F -> 0 | 1 | (E) o where the set of terminals is {+, *, (, ), 0, 1} b) s-> i(B)SK | p, K -> os | ɛ, B -> t | f o where the set of terminals is {(, ), i, o, t, f, p} c) S-> -s | S=S | (S) | p | q o where the set of terminals is {-, =, (, ), p, q}

Convert the given CFG to CNF: a) E-> E+T | T, T-> TF | F, F -> 0 | 1 | (E) o where the set of terminals is {+, , (, ), 0, 1} b) s-> i(B)SK | p, K -> os | ɛ, B -> t | f o where the set of terminals is {(, ), i, o, t, f, p} c) S-> -s | S=S | (S) | p | q o where the set of terminals is {-, =, (, ), p, q}

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Similar questions

Construct PDA for the language L=\ a^ m b^ m cA n | m, n>=1) .?
Present a coding technnique that accepts 2 numbers x,y that will result into a transformation L: R^2 -> R^2 Example: L(x,y) = (x-y,x+y) .
Consider the following BNF: < expr > < expr > + < expr > | < expr > * < expr > | (<expr>) | <digit> <digit> 0| 1|2|3…….|9 What are the terminals and non-terminals?
kindly provide clear (step-by-step) explanations to help me understand the topic.
When we convert CFG: A->AYA le, Y-->if le into CNF, the step two is............ A->A | e, A-->AYA|AY YA| Y, Y-->if | e O Skip A->A | e, A-->AYAJAY YA| Y | e, Y->if A'-->A | e, A-->AYA|AY YA| Y | e, Y->if | e OA->Ale, A-->AYA|AY IYA| Y, Y->if
consider a PDA M = ({g-91+42),la,c).{a.Z,},8.40 Zo.lg.}) shown in below figure. Check the acceptability of string aacaa. a, Za, aza a, a, e C, a, a E, Zg, Ze a, a, aa FIGURE : PDA accepting {a"ca" : n2 1}
please explain step by step
For a Turing machine M, (M) refers to the binary representation of M. For a Turing machine M, L(M) contains the set of all strings accepted by M. For a Turing machine M and an input x € {0,1}*, Steps(M, x) refers to the number of steps taken by M to execute on x before it halts. Here, one step of execution of M on x = one movement (left or right) of the tape head. For a Turing machine M and an input x = {0,1}*, we define the following: ReachCells(M,x) = {i : M reaches ith tape cell when M is executed on x} Informally, it contains all locations on the tape that are visited when M is ecuted on x. The leftmost location on the tape is the first tape cell, the location next to it is the second tape cell, and so on. A string w₁ is an anagram of w2 if w₁ can be obtained by rearranging the alphabets of w2. Formally, if w₁ is an n length string, wê is called an anagram of w₁ if there exists a permutation à on n elements such that π(w₁) = W2.
Consider the languages in Table 1. Match them with the suitable PDA from Table 2. Table 1. Table 2. No. Push-down Automata (a,E/A) (b,e/B) (c,B/E) (d,A/c) (6,E/8) {amb^cdm | m, n > O}. (c,X/8) (d,X/8) {ambmc^dn | n > 0, m >= 0} 8 (d,X/8) {a'b/ckd \ i, j, k, I≥ 0; i = T and j = k}. {amb^c?d9 | m + n = p + q}. {ambmtnc¢ [ m,n >= 0} {amb^c+1 | m ≥0, n >0}. {am+1b^cm | m>0, n ≥ 0}. 5. No. Language 1. 2. 3. 4. 5. 6. 7. (6,8/8) (a,8/X) (b,8/X) (b,e/x) c,X/8) 31 (d,X/e) 9⁰ 91 (a,8/X) (b,E/Y) (c,Y/8) €o (b,e/Y) WA (d,X/8) (b,8/A) (c,A/8) (c,e/8) (a,c/A) (b,E/E) (c,A/8) (a,e/8) (e,e/6) (c,A/E) qo 91 q2 (a,&/X) (b,X/8) (c,&/Y) (d,Y/8) (6,8/8) (c,&/Y) (d,Y/e) ஏo 91 92 (a,e/X) (b,X/E) (b,e/Y) (c,Y/E) (E,6,8) (E,E,E) (a,8/8) .8. (b,E/A) (E,E/E) (E,E,E)
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Subject Name Theory
Let = {0, 1} and consider the following languages 1. LBOUNDED = {(M, x) : M is a TM, x € Σ*, M never moves its pointer more than |x| tape symbols away from the left endmarker |} 2. LDEC = {(M): M is a TM, L(M) is decidable} For each language, determine whether it is decidable or undecidable. If it is decidable, give a decision procedure which decides it. If it is undecidable, you must use Rice's theorem, if applicable, or a mapping reduction, otherwise. [Hint: Given some input string x, you may assume that the number of possible instantaneous configurations of a TM where the pointer never moves more than |x| tape symbols away from the left endmarker is at most |T||*|. (²[+¹). (¹).]