## finite automata – Using Myhill Nerode to prove a language is not regular

Let $$L$$ be your language and let $$equiv$$ be the equivalence relation defined by $$x equiv y$$ iff $$x in Sigma^*$$ and $$y in Sigma^*$$ have no distinguishing extension (a distinguishing extension is a word $$z in Sigma^*$$ such that exactly one of $$xz$$ and $$yz$$ belongs to $$L$$).

Given $$i,j in mathbb{N}$$ with $$i < j$$, $$0^i$$ and $$0^j$$ cannot belong the same equivalence class since $$1^i$$ is a distinguishing extension for them: $$0^i 1^i in L$$ but $$0^j 1^i notin L$$.

This shows that $$L/equiv$$ is not finite, and hence $$L$$ is not regular.

## automata – DFA to Regular Expression

I have a DFA

And I don’t know how to turn it into regular expression. I think the problem lies with every state going back to q0, so I can’t figure out how what to do with the symbols.

Would greatly appreciate if someone could help or at least start me off in the right path?

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## regex – Expressão regular para validar string e int em php

Preciso validar essa string.

`\$scop = "ES-1236";` chave de acesso valor fixo `ES-` dinamico `1236` sendo o dinamico apenas números inteiros.

para validar estou criando uma variavel com a busca
`\$lets = "ES-";`

e a expressão para validar

``````if(preg_match("/{\$lets}/i", \$let)) {
echo true;
}
``````

porém nesse formato ele está aceitando se passar somente `ES-` mas gostaria que ele obrigasse a ter um número inteiro após o `ES-`

## plotting – What regular cartesian expression \$y=f(x)\$ has a graph of the “number 7-like” shape?

It seems to be a simple question but I’m stuck. I was trying to transform many fuctions that are partly resembling `7` like $$ax^3+bx^2+cx+d$$, `sawtooth wave`, `Heavyside step`, $$|x|$$, a lot of `trigonometry function family` expressions, `elliptic curve` but failed to do what I need exactly. Finally I have to get an `upside-down trapezoid-like` graph like such:

with a single regular cartesian expression $$y=f(x)$$. All I’ve come up with is an `upside-down triange` with poorly drown discontinuities.

However it was made with an implicit formula `0 == ((UnitStep(-y)*(2 + y*(2/3)))/Abs(x)) - 1` and `ContourPlot` after all which’s not acceptable.

``````ContourPlot({0 == ((UnitStep(-y)*(2 + y*(2/3)))/Abs(x)) - 1}, {x, -5,
5}, {y, -5, 5}, Exclusions -> None, WorkingPrecision -> 16)
``````

I know there is a way to do something similar with transformation of a circle on the complex plane but it’s too difficult for me and I expect to produce the plot with reals.

## finite automata – prove that if a language is regular then so is the reverse of that language

Below is a problem from Dexter C Kozen’s Automata and Computability followed by my attempt at a solution. Please provide feedback on my proof. Are there any errors/leaps in logic?


Problem Statement

My attempt:

$$T$$ is regular if and only if there exists a NFA accepting $$T$$. Let $$N=(Q,∑,Δ,S,F)$$ such that $$L(N)=T.$$

I will show that there exists a NFA accepting $$T$$ if and only if there exist a NFA $$N_{R}=(Q,∑,Δ_R,S_R,F_R)$$ such that $$L(N_R)=revT.$$

Define: $$Δ_R(q,a)={p in Q : q in Δ(p,a)}$$, $$S_R=F$$, $$F_R=S$$

Lemma 1: if $$A⊆Q,$$ then $$hat{Δ}_R(hat{Δ}(A,x),rev(x))=A$$ for all $$x in ∑^✱$$.

Base Cases:

$$hat{Δ}_R(hat{Δ}(A,ε),ε)=hat{Δ}_R(A,ε)=A$$ by Kozen (6.1). So, equality holds for the empty string.

$$hat{Δ}_R(hat{Δ}(A,a),rev(a))=$$$$bigcup_{qinhat{Δ}(A,a) } {p in Q : q in Δ(p,a) }=$$

$${p in Q : Δ(p,a) ∩ hat{Δ}(A,a) not =∅ }=A$$
by Kozen (6.2), the fact that $$rev(a)=a$$, and the definition of $$Δ_R$$.

Inductive Step:

Assume $$hat{Δ}_R(hat{Δ}(A,x),rev(x))=A$$.

$$hat{Δ}_R(hat{Δ}(A,xa),rev(xa))=hat{Δ}_R(hat{Δ}(A,xa),arev(x))=$$

by definition of string reversal in problem statement.
$$hat{Δ}_R(bigcup_{qinhat{Δ}(A,x)}Δ(q,a), arev(x))$$

by Kozen definition (6.2) page 33

$$bigcup_{qinhat{Δ}(A,x)}hat{Δ}_R(Δ(q,a), arev(x))$$

by Kozen Lemma 6.2 page 34

$$bigcup_{qinhat{Δ}(A,x)}hat{Δ}_R(hat{Δ}_R(Δ(q,a), a),rev(x))=$$

by Kozen Lemma 6.1

$$hat{Δ}_R(hat{Δ}_R(hat{Δ}(A,xa), a),rev(x))=$$

by Kozen Lemma 6.2

$$hat{Δ}_R(hat{Δ}_R(hat{Δ}(hat{Δ}(A,x),a), rev(a)),rev(x))=$$

by Kozen Lemma 6.1 and by the fact that $$rev(a)=a$$

$$hat{Δ}_R(hat{Δ}(A,x),rev(x))=$$

by the base case for a single character

$$A$$ by assumption.

Lemma 2: $$hat{Δ}(A ∩ B,x)= hat{Δ}(A,x) ∩ hat{Δ}(B,x)$$

Base Case:

$$hat{Δ}(A ∩ B,ε)=A ∩ B= hat{Δ}(A,ε) ∩ hat{Δ}(B,ε)$$ by definition (6.1) kozen

Inductive step:

Assume $$hat{Δ}(A ∩ B,x)= hat{Δ}(A,x) ∩ hat{Δ}(B,x)$$

$$hat{Δ}(A ∩ B,xa)=bigcup_{qinhat{Δ}(A∩ B,x)}Δ(q,a)=$$
by definition (6.2) kozen
$$bigcup_{qin(hat{Δ}(A,x)∩ hat{Δ}( B,x))}Δ(q,a)=$$
by assumption
$$bigcup_{qinhat{Δ}(A,x)}Δ(q,a)∩bigcup_{qinhat{Δ}(A,x)}Δ(q,a)=hat{Δ}(A,xa)∩hat{Δ}(B,xa)$$
by definition (6.2) kozen and basic set theory

Now I will use lemma 1 and lemma 2 to show $$x in L(N)$$ IFF $$rev(x) in L(N_R)$$

$$x in L(N)$$ IFF $$hat{Δ}(S,x)∩F not = ∅$$ IFF

$$hat{Δ_R}(hat{Δ}(S,x)∩F,rev(x)) not = hat{Δ_R}(∅,rev(x))$$ IFF

$$hat{Δ_R}(hat{Δ}(S,x),rev(x)) ∩hat{Δ_R}(F,rev(x)) not = ∅$$, by lemma 6.2, IFF

$$S ∩hat{Δ_R}(F,rev(x)) not = ∅$$, by lemma 6.1, IFF

$$F_R ∩hat{Δ_R}(S_R,rev(x)) not = ∅$$, by definition of $$F_R,S_R$$, IFF

$$rev(x) in L(N_R)$$ $$∎$$

## finite automata – prove that if a language is regular then so is the reverse of that language

Below is a problem from Dexter C Kozen’s Automata and Computability followed by my attempt at a solution. Please provide feedback on my proof. Are there any errors/leaps in logic? If having access to the text would be helpful, then I am happy to post a link, assuming that doing so is permissible on this form.



My attempt:

$$T$$ is regular if and only if there exists a NFA accepting $$T$$. Let $$N=(Q,∑,Δ,S,F)$$ such that $$L(N)=T.$$

I will show that there exists a NFA accepting $$T$$ if and only if there exist a NFA $$N_{R}=(Q,∑,Δ_R,S_R,F_R)$$ such that $$L(N_R)=revT.$$

Define: $$Δ_R(q,a)={p in Q : q in Δ(p,a)}$$, $$S_R=F$$, $$F_R=S$$

Lemma 1: if $$A⊆Q,$$ then $$hat{Δ}_R(hat{Δ}(A,x),rev(x))=A$$ for all $$x in ∑^✱$$.

Base Cases:

$$hat{Δ}_R(hat{Δ}(A,ε),ε)=hat{Δ}_R(A,ε)=A$$ by Kozen (6.1). So, equality holds for the empty string.

$$hat{Δ}_R(hat{Δ}(A,a),rev(a))=$$$$bigcup_{qinhat{Δ}(A,a) } {p in Q : q in Δ(p,a) }=$$

$${p in Q : Δ(p,a) ∩ hat{Δ}(A,a) not =∅ }=A$$
by Kozen (6.2), the fact that $$rev(a)=a$$, and the definition of $$Δ_R$$.

Inductive Step:

Assume $$hat{Δ}_R(hat{Δ}(A,x),rev(x))=A$$.

$$hat{Δ}_R(hat{Δ}(A,xa),rev(xa))=hat{Δ}_R(hat{Δ}(A,xa),arev(x))=$$

by definition of string reversal in problem statement.
$$hat{Δ}_R(bigcup_{qinhat{Δ}(A,x)}Δ(q,a), arev(x))$$

by Kozen definition (6.2) page 33

$$bigcup_{qinhat{Δ}(A,x)}hat{Δ}_R(Δ(q,a), arev(x))$$

by Kozen Lemma 6.2 page 34

$$bigcup_{qinhat{Δ}(A,x)}hat{Δ}_R(hat{Δ}_R(Δ(q,a), a),rev(x))=$$

by Kozen Lemma 6.1

$$hat{Δ}_R(hat{Δ}_R(hat{Δ}(A,xa), a),rev(x))=$$

by Kozen Lemma 6.2

$$hat{Δ}_R(hat{Δ}_R(hat{Δ}(hat{Δ}(A,x),a), rev(a)),rev(x))=$$

by Kozen Lemma 6.1 and by the fact that $$rev(a)=a$$

$$hat{Δ}_R(hat{Δ}(A,x),rev(x))=$$

by the base case for a single character

$$A$$ by assumption.

Lemma 2: $$hat{Δ}(A ∩ B,x)= hat{Δ}(A,x) ∩ hat{Δ}(B,x)$$

Base Case:

$$hat{Δ}(A ∩ B,ε)=A ∩ B= hat{Δ}(A,ε) ∩ hat{Δ}(B,ε)$$ by definition (6.1) kozen

Inductive step:

Assume $$hat{Δ}(A ∩ B,x)= hat{Δ}(A,x) ∩ hat{Δ}(B,x)$$

$$hat{Δ}(A ∩ B,xa)=bigcup_{qinhat{Δ}(A∩ B,x)}Δ(q,a)=$$
by definition (6.2) kozen
$$bigcup_{qin(hat{Δ}(A,x)∩ hat{Δ}( B,x))}Δ(q,a)=$$
by assumption
$$bigcup_{qinhat{Δ}(A,x)}Δ(q,a)∩bigcup_{qinhat{Δ}(A,x)}Δ(q,a)=hat{Δ}(A,xa)∩hat{Δ}(B,xa)$$
by definition (6.2) kozen and basic set theory

Now I will use lemma 1 and lemma 2 to show $$x in L(N)$$ IFF $$rev(x) in L(N_R)$$

$$x in L(N)$$ IFF $$hat{Δ}(S,x)∩F not = ∅$$ IFF

$$hat{Δ_R}(hat{Δ}(S,x)∩F,rev(x)) not = hat{Δ_R}(∅,rev(x))$$ IFF

$$hat{Δ_R}(hat{Δ}(S,x),rev(x)) ∩hat{Δ_R}(F,rev(x)) not = ∅$$, by lemma 6.2, IFF

$$S ∩hat{Δ_R}(F,rev(x)) not = ∅$$, by lemma 6.1, IFF

$$F_R ∩hat{Δ_R}(S_R,rev(x))$$, by definition of $$F_R,S_R$$, IFF

$$rev(x) in L(N_R)$$ $$∎$$

## Expressão Regular Python no Javascript?

Separar os caracteres de uma string com hífen em Python funciona perfeitamente desse jeito

``````import re

regex = r"B(?=(.{1}))"

test_str = "Pêssego"

subst = "-"

result = re.sub(regex, subst, test_str, 0)

if result:
print (result) // P-ê-s-s-e-g-o
``````

Em javascript

``````const regex = /B(?=(.{1}))/g;
const str = `Pêssego`;
const subst = `-`;

const result = str.replace(regex, subst);

console.log(result); // Pês-s-e-g-o
``````

Devo acrescentar algo? Tirar algo? Qual a diferença entre dos dois? Existe alguma outra forma de separar uma string com hifens em em tempo real em um campo de entrada?

## differential geometry – Proof of smoothness of the Exponential map at a point of a complete regular surface in Euclidean space

Suppose S is a regular, connected surface in Euclidean space and $$d_{s}$$ is the intrinsic metric on S. When (S,d) is complete, we know that for each geodesic $$gamma:Jrightarrow S$$ (where J is any interval) there exists a unique geodesic $$eta:mathbb{R}rightarrow S$$ which is an extension of $$gamma$$.

If $$p$$ is any point of S, it is shown in do Carmo (page 284) that if a sufficiently small $$win T_{p}S$$ is chosen, the geodesic $$gamma$$ with initial state $$gamma(0)=p, gamma'(0)=w$$ is well defined at $$t=1$$. The symbol exp$$_{p}(w)$$ is made to denote the point $$gamma(1)in S$$. It is then shown (page 285) that exp$$_{p}$$ can be made into a smooth map on a sufficiently small neighbourhood of $$0in T_{p}S$$. This truth depends on the theorem of solutions of systems of ODE’s which says that the solution depends smoothly on initial conditions.

By what I wrote above, when S is complete, $$p$$ is any point and $$w$$ any tangent vector to S at $$p$$, the symbol exp$$_{p}(w)$$ is well defined, but how can we show smoothness of exp$$_{p}$$ on all of $$T_{p}S$$? Am I missing something elementary here? We know it is smooth around $$0$$, but the proof depended on covering $$p$$ with a chart and applying the theorem from analysis. It seems we cannot argue this way in the general case.

In do Carmo I did not see this technical question discussed, and neither in Tapp.