Feb 8 2011 Exercise 2.1.19 (formulation). Show that the language. L = {an
Note that M? accepts the string 100 ? C = { w
Regular Expressions. • Nonregular Languages. CS 341: Chapter 1. 1-3. Introduction L(M3) is the language of strings over ? that do not have length 1.
different order from what is done below may result in a different (but also correct) regular 3. Prove that the following languages are not regular.
Context-free languages are not closed under intersection or complement. This will be shown later. 2. Page 3. 1.5 Intersection with a regular language.
Answer: A CFG is in Chomsky normal form if each of its rules has one of 3 Consider the language L = {?M?
Note that if we start a derivation it never finishes
Chapter 3: Fluency 5. Does comprehension strategy instruction improve reading? ... effect on reading development than did multiple-skill instruction.
1.3.3 Quitting MATLAB . 1.4.9 Entering multiple statements per line . ... The seven lab sessions include not only the basic concepts of MATLAB but also ...
CHAPTER 3 • N-GRAM LANGUAGE MODELS. When we use a bigram model to predict the conditional probability of the next word we are thus making the following
theorem: if L1andL2 regular thenL1L2regularFor each of the statements below decide whether it is true orfalse If it is true prove it If not give a counterexample All parts refer to languages over the alphabetfabg 1 If L1 µL2 andL1is not regular thenL2is not regular
language of size one is regular when break up L into two or more smaller languages Also many groups that did include the correct base case neglected to give a justification as to why a language of size 1 is regular or proved it only for the language {?} A valid justification
Show that the language L = {anbm: n ? m} is not regular 5 Prove or disprove the following statement: If L1 and L2 are not regular languages then L1 ? L2 is not regular 6 Show that the language L = {x ? {a b}* : x = anbambamax(mn)} is not regular 7 Show that the language L = {x ? {a b}* : x contains exactly two more b's than a
(h) If L? = L1 ? L2 is a regular language and L1 is a regular language, then L2 is a regular language. (i) Every regular language has a regular proper subset. (j) If L1 and L2 are nonregular languages, then L1 ? L2 is also not regular. 4. Show that the language L = {anbm : n ? m} is not regular. 5.
Show that for each n >= 1, the language B nis regular. For each n>=1, we built a DFA with the n states q 0 , q 1, …, q n-1to count the number of consecutive a’s modulo n read so far. For each character a that is input, the counter increments by 1 and jumps to the next state in M. It accept the string if and only if the machine stops at q 0.
If L is a regular language, then so is L? = {w : w ? and wR ? L}. If C is any set of regular languages, ?C (the union of all the elements of C) is a regular language. (g) L = {xyxR : x, y ? ?*} is regular. (h) If L? = L1 ? L2 is a regular language and L1 is a regular language, then L2 is a regular language.
7. First, let L' = L ? a*b*, which must be regular if L is. We observe that L' = anbn+2 : n ? 0. Now use the pumping lemma to show that L' is not regular in the same way we used it to show that anbn is not regular. 8. We use the pumping lemma.