I'm beginner for Automata and JFLAP. I have action Finite Automata which tells that on any character in the language, it must go to another state. How am I implement this on JLAP.Thanks
Related
As we know, the definition of 'finite state automata' is:
Then we have this finite state automat described as:
Then we have the conclusion:
Question is : Instead of accept an empty string, what if the automat first read string is '2', which is not belong to the alphabet(0,1) of this automata. will this automata still go to accept state?
pics quoted from book <Introduction to the Theory of Computation>
When a machine is defined, one of its items is the alphabet (second one in your definition). We are not allowed to test our machine outside the scope of the alphabet. Therefore, all strings of L (accepted strings) and L-bar (rejected strings) should come out of Sigma-start.
In one of my cs classes they mentioned that the difference between context-free grammar and context-sensitive grammar is that in CSG, then the left side of the production rule has to be less or equal than the right side.
So, one example they gave was that context-sensitive grammars can't have an empty string because then the first rule wouldn't be satisfied.
However, I have understood that regular grammars are contained in context-free, context-free are contained in context-sensitive, and context-sensitive are contained in recursive enumerable grammars.
So, for example if a grammar is recursive enumerable then is also of the type context-sensitive, context-free and regular.
The problem is that if this happens, then if I have a context-free grammar that contains an empty string then it would not satisfy the rule to be counted as a context-sensitive, but then a contradiction would occur, because each context-sensitive is context-free.
Empty productions ("lambda productions", so-called because λ is often used to refer to the empty string) can be mechanically eliminated from any context-free grammar, except for the possible top-level production S → λ. The algorithm to do so is presented in pretty well every text on formal language theory.
So for any CFG with lambda productions, there is an equivalent CFG without lambda productions which generates the same language, and which is also a context-sensitive grammar. Thus, the prohibition on contracting rules in CSGs does not affect the hierarchy of languages: any context-free language is a context-sensitive language.
Chomsky's original definition of context-sensitive grammars did not specify the non-contracting property, but rather an even more restrictive one: every production had to be of the form αAβ→αγβ where A is a single symbol and γ is not empty. This set of grammars generates the same set of languages as non-contracting grammars (that was also proven by Chomsky), but it is not the same set. Also, his context-free grammars were indeed a subset of context-sensitive grammars because by his original definition of CFGs, lambda productions were prohibited. (The 1959 paper is available online; see the Wikipedia article on the Chomsky hierarchy for a reference link.)
It is precisely the existence of a non-empty context -- α and β -- which leads to the names "context-sensitive" and "context-free"; it is much less clear what "context-sensitive" might mean with respect to an arbitrary non-contracting rule such as AB→BA . (Note 1)
In short, the claim that "every CFG is a CSG" is not technically correct given the common modern usage of CFG and CSG, as cited in your question. But it is only a technicality: the CFG with lambda productions can be mechanically transformed, just as a non-contracting grammar can be mechanically transformed into a grammar fitting Chomsky's definition of context-sensitive (see the Wikipedia article on non-contracting grammars).
(It is also quite common to allow both context-sensitive and context-free languages to include the empty string, by adding an exception for the rule S→λ to both CFG and CSG definitions.)
Notes
In Chomsky's formulation of context-free and -sensitive grammars, it was unambiguous what was meant by a parse tree, even for a CSG; since Chomsky is a linguist and was seeking a framework to explain the structure of natural language, the notion of a parse tree mattered. It is not at all obvious how you might describe the result of AB → BA as a parse tree, although it is quite clear in the case of, for example, a A b → B.
There is a DFA.
But I don't know this DFA means.
I want to know this DFA's Regular expression and description
(Description like 'this automata accept suffix is 011)
Blue state is start state and red states are accept states.
see the link below....it will tell about DFA.
http://en.wikipedia.org/wiki/Deterministic_finite_automaton
Does exists one algorithmic to convert a Linear Grammar Right to the equal Linear Grammar Left?
For every right-linear grammar, there exists an equivalent left-linear grammar that generates the same language, and vice-versa.
Use the grammar to build the FSA that recognizes the language generated by the original grammar.
Swap initial states with final states.
Invert arrows orientation.
If multiple initial states are present, set them as not initial, create a dummy initial state and link it with them using spontaneous moves.
From the modified FSA, obtain another right-linear grammar, using the "standard" approach.
Reverse the right side of every production of the grammar.
You should get an equivalent left-linear grammar.
I am preparing contex free grammar for an exam. I couldn't understand why the language
{ a^n b^n | n>=0}
is context free but not regular. Why is it not regular?
When can we say that an expression is not regular?
Thanks
An expression is not regular if it cannot be matched (exactly) by a regular expression or (equivalently) a finite state machine. See also context free language and regular language.
Like said in previous answers, its context free because you can express it with context free grammar.
For example: S -> aSb | ε
Its not regular because you cannot express it with finite state machine nor regular expressions. You should be able to count the number of As and check that number of Bs match. This cannot be done with finite states as n could be anything
The standard approach is to use the Pumping Lemma
Well you can say it is context free because you can express it using a Context-Free Grammar. It is not regular however because a regular expression (and finite automata) can not represent that language.