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#Union concatenation kleene star order series#

The deterministic counterpart of such class of automata defines the deterministic union-free (d-union-free, for short) languages. Obviously such an automaton has exactly one accepting state. The language class is also characterised by a special class of finite automata: 1CFPAs have exactly one cycle-free accepting path from each of their states. Consequently, (nondeterministic) union-free languages are described by regular expressions using only concatenation and Kleene star. Union-free expressions are regular expressions without using the union operation. In other words, for any of these three choices of y y y, the string s s s cannot be pumped in a way that results in a string that has an equal number of a a a’s and b b b’s ( ( (the definition of the language L ). In any of these three cases, we can easily verify that the third constraint for the pumping lemma, x y i z ∈ L ∀ i ≥ 0 xy^iz \in L\ \forall i \geq 0 x y i z ∈ L ∀ i ≥ 0, does not hold. This means that the y y y segment is contained somewhere in the a a a’s and b b b’s section.

y = a j b k y = a^jb^k y = a j b k for some j j j and k, k, k, where j + k ≤ p.

This means that y y y is contained purely in the b b b’s section.

y = b j y = b^j y = b j for some j ≤ p j \leq p j ≤ p.

This means that y y y is contained purely in the a a a’s section.

y = a j y = a^j y = a j for some j ≤ p j \leq p j ≤ p.

There are three possible places in the string that we can assign to be y y y: Now by the pumping lemma, ∣ y ∣ ≤ p |y| \leq p ∣ y ∣ ≤ p. This s s s is in L L L since it has p p p a a a’s and p p p b b b’s. Certainly s = a p b p s = a^pb^p s = a p b p is longer than p p p, so we choose this for the s s s string. We need a string s s s that is longer than or equal to the length of p p p. By the pumping lemma, there exists an integer pumping length p p p for L L L.

X ∘ Y X \circ Y X ∘ Y and X Y XY X Y both mean “X concatenated with Y.” Languages can also be concatenated: L 1 ∘ L 2 L_1 \circ L_2 L 1 ∘ L 2 means strings from L 1 L_1 L 1 are written first, and then strings from L 2 L_2 L 2 come after.įormally, concatenation is defined as follows: A ∘ B = is a regular language. There are two main ways to write a concatenation. A language is a regular language if there is a finite automaton that recognizes it.įor example, this machine recognizes the language of strings that have an even number of zeroes since any string that has an even number of zeroes will go from the start state to an accepting state.Ī regular language can be represented by a string of symbols and operations.Ĭoncatenation is an operation that combines two symbols, sets, or languages. We say that the machine M M M recognizes the language L L L if M M M accepts all strings w w w that are in L L L.

#Union concatenation kleene star order series#

M M M is said to accept a string w w w if the machine starts in a start state, undergoes some series of state transitions, and ends up in an accepting state. A finite state machine, M, M, M, describes a given language, L L L. This means that regular languages can be described by a simple state machine diagram.

it doesn't know how many times the switch has been flipped on or off).įinite automata generate or model regular languages. The machine above can know what state it is currently in, on or off, but it doesn't know what the history of flips was to get it to where it is (i.e. For example, a finite automaton can generate a regular language to describe if a light switch is on or off, but it cannot keep track of how many times the light was switched on or off. Regular languages and finite automata can model computational problems that require a very small amount of memory.