Right now I'm learning for a test in cs and I have my problems with regular expressions. Here is example of a question I don’t understand.
We have a given alphabet Σ = {0, 1} and L1. L1 represent any word with odd number of 0(zeros) and exactly twice 1 (ones).
The prof. showed us an example how the solution should look like:
Σ = {a, b} with L1.
L1 represents any Word in which the sub-words aa or bb occurs.
L1 = (a ∪ b)* (aa ∪ bb)* (a ∪ b)*
thank you for your help
Try this:
(?mx)
^
(?=0*10*10*$) # check if the word consists of only 0s and exactly two 1s
(?:..){1,}. # check if the word length is odd
$
Demo on RegEx101. There you can find an exhaustive explanation of this regular expression.
Related
I have understood that L={wxw^r|w,x belongs to {a,b}^* } is regular because it turns out to be the pattern of starting and ending with same symbol but I am not getting the proper explanation that how to say L={ww^rx|w,x belongs to {a,b}*} is regular language using DFA design.
Please help me in understanding this!
This is a trick question. The language L as you have specified is the language of the regular expression (a + b)*, that is, any string of a's and b's. The trick is that for any string y = s1.s2.s3...sk where si in {a, b}, we can write y = wxw^R where w is the empty string and x = y. Basically, the trick is that we can always choose w to be the empty string, and in that case we are left with L = {x | x in {a, b}^*}, clearly regular. Another way of thinking about it is this: can you find any string of a's and b's that is not in L? Is it not in L even if you take w to be the empty string?
let L = {wwRu | w, u ∈ {0,1}+}. Is L regular language ? Note that w, u cannot be empty.
I've tried to prove it is not regular language by the pumping lemma, but I failed when w = 0^p1^p, 01^p, (01)^p. Once I take y = 0^p or 1^p, xyyz will be 00.../11.../01^n0... etc.
And I cannot draw its DFA/NFA or write its regular expression to prove it is regular language.
So is L regular or not ? How can I prove it ?
The language is not regular, and we can prove it using the Myhill-Nerode theorem.
Consider the sequence of strings 01, 0101, ..., (01)^n, ...
First, notice that none of these strings are in the language. Any prefix of any of these strings which has even length is of the form (01)^2m for some m, and therefore just a shorter string in the sequence; splitting such a prefix in two either has both substrings start with 0 and end with 1, or else it has the first substring start and end with 0 and the second start and end with 1. In either case, these strings are not of the form w(w^R)u for any w or u.
Next, notice that the shortest possible string which we can append to any of these strings, to produce a string in the language, is always the reverse of itself followed by either 0 or 1. That is, to turn 01 into a string in the language, we must append 100 or 101; there are no shorter strings we can append to 01 to get a string in the language. The same holds true for 0101: 10100 and 10101 are the shortest possible strings that take 0101 to a string in L. And so on for each string of the form (01)^n.
This means that each string of the form (01)^n is distinguishable with respect to the target language w(w^R)u. The Myhill-Nerode theorem tells us that a minimal DFA for a regular language has exactly as many states as there are equivalence classes under the indistinguishability relation. Because we have infinitely many distinguishable strings with respect to our language, a minimal DFA for this language must have infinitely many states. But, a DFA cannot have infinitely many states; this is a contradiction. This means that our language cannot be regular.
The language is REGULAR:
L = 00(0+1)+ + 11(0+1)+ + 0(11)+0(0+1)+ + 1(00)+1(0+1)+
How to construct a NFA which accepts set of all strings w such that n1(w) mod 3 > 1, where n1(w) is the number of 1s in w over the alphabet = {0,1}?
So, basically a NFA which accepts the set of all strings with 2, 5, 8,... number of 1s.
I guess the regular expression for the language would be (0*10*10*)(0*10*10*10*)*
I can create a NFA for the above regular exp, but I'm not sure if this can be reduced further or if it's even correct in the first place. I'm new to regular languages, DFA, NFA stuff. Please help me!
Let me propose this NFA, over alphabet {0,1}:
Q = {A, B, C}
q0 = {A}
F = {C}
d = {(A,0,A)
(A,1,B)
(B,0,B)
(B,1,C)
(C,0,C)
(C,1,A)}
Please check whether it satisfies the requirement.
Im looking forward to design a FA which accepts some kind of string that accept some A and B.
First a string which the number of A is five more times higher than B.
i mean L={w∈{A,B}* and (nA(W)-nB(W) mod 5=0)
And also a FA which accept different number of character in a string:
L={A^n B^m C^k | n,k>0 and m>3}
I design some FAs But they did not work perfectly on this complicated strings.
Any help on how should i design this ?
Unfortunately, your questions are confusing as the english text doesn't agree with the mathematical formula. I will try to answer to these four questions then:
A language which consists of string over {a,b} that the number of a (= #a(w))
is five times as the number of b ( #b(w)),
L = { w in {a,b}* : #a(w)>#b(w) and #a(w)=#b(w)mod5 }
This cannot be done by an NFA. The proof is simple by using the pumping lemma (P.L) with the string a^pb^5p, where p is the constant of P.L.
For the language: L={w∈{A,B}* : (nA(W)-nB(W)) mod 5=0} that you wrote,
you can do it with an DFA that consists of a cycle of 5 states.
The transitions are, if you read a go clockwise if you read b go counter-clocwise. Choose at random one state to be initial state and the same state will be the final state.
For the language L={A^n B^m C^k | n,k>0 and m>3}, it should be easy to find out
if you read L as L=A(A)* B(B)* c^4(C)*
For the language that accepts different number of character in the string (let's say over a,b). The language should be R={ w in {a,b}* : #a(w) not equal #b(w)}
This language again it cannot be recognized by an NFA. If this language was regular (recognzied by an NFA) so would be this language:
L=a*b* intersection (R complement). The language L is {a^n b^n/ n non-negative integer}.
The language L is the first example of most books when they speak about languages that are non-regular.
Hopefully, you will find this answer helpful.
CS student slogging through a logic class. This question has me befuddled
Inductively Defining Sets of Strings
Find an inductive definition for the following set of strings:
S = {apbcr | p is a natural number, and r is a natural number greater than 0} *the p and r are superscripts*
I'd suggest the following definition: