Answered: Question 5: Use the pumping lemma to…

Question 5: Use the pumping lemma to show that the following languages are not regular. 1. A₁ = {012 | n >= 0} 2. A3{an | n >= 0} (Here, a 2^n means a string of 2n a's.)

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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None of the answers can be two choices; there can only be one answer.
helo by Peter Linz, "Introduction to Formal Languages and Automaton", 6th edition, 2017. The cut is from chapter 2, section 1. thanks :)
∑ = {C,A,G,T}, L = { w : w = CAjGnTmC, m = j + n }. For example, CAGTTC ∈ L; CTAGTC ∉ L because the symbols are not in the order specified by the characteristic function; CAGTT ∉ L because it does not end with C; and CAGGTTC ∉ L because the number of T's do not equal the number of A's plus the number of G's. Prove that L ∉ RLs using the RL pumping theorem.
5. a. Write a lambda expression for a Comparator that compares strings in increasing order of their length. Comparator c = b. Write a lambda expression for a Comparator that compares strings in increasing order of their last characters. Comparator c =
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-] Use the CFL Pumping Lemma to show each of the following language not to be context-free: a) {a^n b^n c^i | i < n} b) {www w is a binary string over {0,1}}
Using C++ Please add the specific explanation. Thanks. Huffman coding is used to compress data. The idea is straightforward: represent more commonlonger strings with shorter ones via a basic translation matrix. The translation matrix is easilycomputed from the data itself by counting and sorting by frequency.For example, in a well-known corpus used in Natural Language Processing called the "Brown"corpus (see nltk.org), the top-20 most frequent tokens, which are words or punctuation marksare listed below associated with frequency and code. The word "and" for example requireswriting three characters. However, if I encoded it differently, say, using the word "5" (yes, Icalled "5" a word on purpose), then I save having to write two extra characters! Note, the word"and" is so frequent, I save those two extra characters many times over!Token Frequency Codethe 62713 1, 58334 2. 49346 3of 36080 4and 27932 5to 25732 6a 21881 7in 19536 8that 10237 9is 10011 10was 9777 11for 8841 12`` 8837 13''…
4. For E = {a, b}, construct dfa's that accept the sets consisting of (d) all strings with at least one b and exactly two a's. (e) all the strings with exactly two a's and more than three b's.
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QIX. Construct a TM that accepts the following languages. 12. L12={a?rb" | n 21} 13. L5 = {02^n |n 21 }. i.e., the occurrences of 0 is always powers of 2, means, every time, you can divide the resultant to 2, recursively.
1. The homomorphism h is defined by h(a) = 01 and h(b) = 10. What is h(aaba)? a) aaba b) 01010101 = c) 01011001 d) 010101 2. Let h be the homomorphism defined by h(a) = 01, h(b) = 10, h(c) = 0, and h(d) : 1. If we take any string w in (0+1)*, h¨¹ (w) contains some number of strings, N(w). For example, h¹ (1100) = {ddcc, dbc}, i.e., N(1100) = 2. We can calculate the number of strings in h¨¹ (w) by a recursion on the length of w. For example, if w = 00x for some string x, then N(w) = N(0x), since the first 0 in w can only be produced from c, not from a. Complete the reasoning necessary to compute N(w) for any string w in (0+1)*. Then, choose the correct value of N(01101). a) 8 b) 6 16 d) 4 3. The operation DM(L) is defined as follows: 1. Throw away every even-length string from L. 2. For each odd-length string, remove the middle character. For example, if L = {001, 1100, 10101}, then DM(L) = {01, 1001}. That is, even-length string 1100 is deleted, the middle character of 001 is removed…