I was given a regular expression, and I am suppose to covert it to NFA and then DFA. Here's the regular expression:
a ( b | c )* a | a a c* b
Then I coverted this to NFA using Thomson's algorithm:
and here's the DFA:
Can someone please take a quick look at let me know if I am wrong or right?
Since this is very likely homework, I'm hesitant to just give you the complete correct solution.
Your NFA appears correct, but has a lot of superfluous states that aren't necessary but do not adversely affect its correctness. (At first glance it looks like you could remove 11 states.)
Your DFA is incorrect, though. This is because when you branch off to begin handling one condition of the string or the other, you later rejoin them together. This allows it to take the path from an accepted string matching a(b|c)*a and take in another b or c by travelling to nodes 15,17 or 11. It then accepts this string even though it doesn't match your expression.
What you need to do is basically stop this from happening. If you have additional questions feel free to ask.
I highly recommend making a list of test strings that you know should be and shouldn't be accepted, and then trace them through, making sure your automata ends in the correct (accept or reject) state.
Related
As above, I have a transition graph, but I'm not sure how to find the language of it, seems to me that there are a lot of possibilities, but I must be misunderstanding somehow. My understanding is that any word that leads from the initial to the final state is accepted. Surely there are a lot of different ways to achieve this. aab, ab, abb, abbaab. As I understand it, a language is a set of all the possible words, but if there are a vast amount of possible words, how can you find the language? Im a first year University student, this is part of my homework, but I'm not just trying to get you to do it for me, I want to understand it - if this doesn't make sense/ is in the wrong place I apologise in advance, thanks. Here's my graph
Some regular languages are finite - they contain a finite number of strings. When you have a finite number of things, that means you can count them all and get to the end eventually; you can write them all down if they're words in a language. Writing down all the words in a language is a way of giving an extensive definition of a language.
However - there are languages which are not finite. They do not contain any number of words you can count from beginning to end, or ever completely write down. If you think of all natural numbers (1, 2, …, 100, …) as strings in the language of decimal representations, clearly there are not finitely many of them. There are infinitely many. You cannot give extensive definitions of infinite languages (except, possibly, by suggestive use of the ellipsis, as I have done in my example). In these cases, you must describe the strings which are included and/or excluded and rely on the reader to deduce membership for any particular case.
Giving a finite automaton is one perfectly reasonable way of giving a criterion according to which membership in the language can be determined: run the string through the automaton and see if it is accepted. In this sense, asking what the language of a finite automaton is can be viewed as trivial: it accepts the language of strings that leave the finite automaton in an accepting state.
Another way of describing the language is to give a grammar or, for regular languages, a regular expression. These are not necessarily more or less helpful ways of describing a language than the finite automaton you already have is.
Typically, in coursework, when you are asked to describe the language of a finite automaton, you are probably being asked to give a plain, English-language definition - a simple one - of the strings, and possibly provide some set-builder notation. That's what we'll try to do here.
Your NFA loops in q0, accepting any number of a, until it sees at least one a. If it sees a b before an a, it crashes. After that, if it sees at least one b, it can accept; it can see any number of b, but after the initial run of a, it can never again see two a in a row. The machine accepts only if it ends with a b.
Taken together, this might be a description in plain English that is good for this language:
All strings that begin with at least one a, which end with b and which do not contain the substring aa after the first instance of b.
A regular expression for this language is aa*(bb*a)*.
given a grammar like
<term>::= x[i]+exp(x[i]) | x[i]
<i>::= 1|2|3
Does a way exist to force the use of the same "i" in one solution of non terminal symbol ? So, I want to avoid solutions like x[1]+exp(2) or x[3]+exp(1)
Does a way exist to avoid that the same "i" is used in one solution of non terminal symbol ?So, I want to avoid solutions like x[1]+exp(1)
No, that's not possible with a context-free grammar.
This is essentially what "context-free" means. Every non-terminal in a production can be expanded independently without regard to the context in which it appears.
Of course, if i really only has three possible values, you can enumerate the finite number of legal productions, according to any definition of "legal" which you find convenient. But that gets really messy when the number of possibilities increases.
The most convenient solution is generally to accept the base syntax and check for concordance (or difference) in the associated semantic rule. That also allows for better error messages.
So I am going to find the occurrence of s in d. s = "infinite" and d = "ininfinintefinfinite " using finite automaton. The first step I need is to construct a state diagram of s. But I think I have problems on identifying the occurrence of the string patterns. I am so confused about this. If someone could explain a little bit on this topic to me, it'll be really helpful.
You should be able to use regular expressions to accomplish your goal. Every programming language I've seen has support for regular expressions and should make your task much easier. You can test your regexs here: https://regexr.com/ and find a reference sheet for different things. For your example, the regex /(infinite)/ should do the trick.
I'm tasked with writing a regular grammar based on a regular expression.
Given the regular expression a*b can be written as S -> b | aS
Is it incorrect that ba* as a regular grammar is S -> b | Sa?
I'm told the correct answer is in fact S -> bA, A -> ^| aA but I don't see the difference myself.
An explanation would be greatly appreciated!
IIRC, both your answer and the one being called "correct" are correct. See this. What you have constructed is a "left regular grammar", while the proponent(s) of the "correct" answer obviously prefer a "right regular grammar". There are other arbitrary rules that may be held more or less pedantically, like the "no empty productions" rule, but they don't really affect the class of regular languages, just the compactness of the grammar you use for a particular language, as your example highlights - a single production with two alternatives vs. two productions, one with a single clause, and one with two alternatives, one of which is empty.
Can someone give me a brief description of how to determine the precedence of one thing with respect to another in a grammar? I can't seem to find a good answer in my books or online even though i'm pretty sure its very simple. Example
S-> B
B-> B and T | T
T-> T or F | F
F-> not F| (B)|true|false
The question was "what is the precedence of "and" with respect to "or" in the above grammar?".
"and" has lower precedence than "or", but I'm not sure why.
The symbol B represents the and statement, and its definition contains T which is the or. You can get from and to or, but not the other way around. So if you think of the language statement as a stack, you'll keep doing replacements and adding to the top of the stack until you wind up evaluating the whole statement, and as you then pop items off you'll reverse direction and the or statement will always be above (and therefore evaluate before) the and. Hope that makes sense.