Proof that language is not regular. $L=\{w\bar{w}|w\in \{0,1\}^* and\ \bar{w}\ is\ one’s\ complement\ of\ w\}$

I’m trying to proof that the following language is not regular using pumping lemma. $ L=\{w\bar{w}|w\in \{0,1\}^* and\ \bar{w}\ is\ one’s\ complement\ of\ w\}$

I started by stating that:

$ |w\bar{w}| = 2p = |xyz|$

Because of Pumping Lemma the following has to be true:

$ 1 \leq |y| \leq |xy| \leq p$

Because $ |xy| \leq p$ and $ |w|=|\bar{w}|$ have to be true, $ \bar{w}$ has to be completly in $ z$ . I now tried somehow to manipulate the first half, so that it always evaluates to a contradiction, but I am stuck.

Draw an NFA over {0,1}

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if I want to design a NFA (that’s NOT A DFA) that accepts the set of all strings that do not contain the substring 1010, is this correct? because I can just accept 1010 by capturing it in the starting state itself, right?

the starting state accepts 0,1 so I can essentially take the string 1010 and self loop it in the starting state itself… is that correct?

A deterministic FA for $0^*1^*$ is required

A deterministic finite automaton without $ \epsilon$ steps for the language $ 0^*1^*$ is required. Any nice picture ? I have created a NFA for this language which has 2 states $ Q_1,Q_2$ , both are accepting, $ Q_1$ is initial, there is a move under $ 0$ from $ Q_1$ to $ Q_1$ , a move under $ 1$ from $ Q_1$ to $ Q_2$ and finally a move from $ Q_2$ to $ Q_2$ under $ 1$ .

Find a function $F$ on $[0,1]$ with moments decaying as $(\ln n)^{-n}$

Let $ F:[0,1]\to\mathbb{R}$ be a measurable function such that $ $ \mu_n(F)=\int_0^1F(t)t^ndt\sim\frac1{(\ln n)^n}\quad\mbox{as}\quad n\to\infty. $ $ More precisely, $ $ 0<c<|\mu_n(F)|(\ln n)^n<C<\infty,\quad\forall n\in\mathbb{N}. $ $ Note that in this case the series $ $ \sum_{n=0}^\infty\mu_n(F)z^n $ $ represents an entire function. The generating function $ $ \sum_{n=0}^\infty\frac{\mu_n(F)z^n}{n!} $ $ is an entire function of minimal exponential type. I am not aware of any explicit description of such a function, so this is already a sign.

Question: Can we find an explicitly given such function $ F$ ? Here by “explicitly given” I mean in a “closed form”, e.g., no series.

I think that by convexity arguments it can be shown that $ F$ cannot be non-negative (non-positive). In fact, I do not need $ F$ be a function; $ F(x)dx$ may be a signed measure on $ [0,1]$ , just given explicitly.

Thank you.

Show if all continuous functions f from X to {0,1} with the discrete topology are constant, then X is connect.

This problem was similar but I couldn’t make it work with the direction I want to show: f is a continuous function from (X,$ \tau$ ) to {0,1} with discrete topology, if f non constant then (X,$ \tau$ ) disconnected

I tried showing the contrapositive (the inverse of the linked problem), which is showing “if X is disconnected, then all the continuous functions f are surjective” but I’m very stuck.

My attempt was: X disconnected –> X = A∪B where A and B are disjoint open sets in X. And since f is continuous, we know f-1({0}), f-1({1}) and f-1({0, 1}) are all open in X. My gut tells me because A and B are open, disjoint and cover X, that I could conclude, without loss of generality, that f(A) = {0} and f(B)={1}, which shows surjectivity. But I can’t figure out if this idea is correct and – if it is – how to formalize it.

Continuous function on $[0,1]$ such that its zeros form a nowhere dense set of positive measure?

I know few facts,

  • if $ f : [0,1] \to \mathbb{R}$ is continuous, $ Z(f) \triangleq f^{-1}(\{0\})$ is closed,
  • there are continuous functions whose zeros are nowhere dense,
  • there are nowhere dense sets of positive measure.

From these facts, I cannot conclude that if $ f:[0,1] \to \mathbb{R}$ is continuous and its zeros $ Z(f)$ form a nowhere dense set, this set $ Z(f)$ is of null measure. Can we prove that it is indeed the case, or exhibit a counter-example?

The language $\{ww | w \in \{0,1\}^{*} \}$ is not a CFL

We have proved that the language $ L = \{\omega\omega | \omega \in \{0,1\}^{*} \} $ is not a CFL, and we did so by using pumping lemma. And the proof is clear to me. But I thought of the following CFG:

$ G = (\{S, S_{1} \},\{0,1\},R,S) $ where R has the following rules:

$ S\rightarrow S_{1}S_{1} | \epsilon $

$ S_{1} \rightarrow 0S_{1} | 1S_{1} | \epsilon $

It feels like that this CFG’s language should be the language $ L$ that I have defined above since each substitution adds the same letter on both sides. But it can’t be since we can use pumping lemma on the word $ 0^{l}1^{l}0^{l}1^{l}$ (where $ l$ is the pumping length). So either I’m not doing the substitution incorrectly, or the CFG’s language contains $ L$ and has more words that I’m not seeing currently…

Can someone help me out and point out where my mistake is?

Pushdown Automata: { x#y | x,y in {0,1}* such that x != y and xi = yi for some i, 1

I was instructed to create the pushdown automata described in the title. Basically, there is a string x#y where x and y are strings of 1s and 0s and there must be at least one difference and at least one similarity between x and y for the string to be accepted. I understand that I must have two separate cases, with one case assuming that |x| = |y| and one case where |x| != |y|. Right now I am only working on the case where |x| = |y|. So far, I am using nondeterminism to find the ith value of x and matching it against the ith value of y. However, this only determines if there is at least one difference between x and y. I am failing to see a solution where I can also determine that there is at least one similarity. If anybody can point me in the right direction, it would be much appreciated.

Monge-Kantorovich duality with a $\{0,1\}$ cost function

Consider the usual Monge-Kantorovich transportation problem where $ X$ and $ Y$ are Polish spaces, $ \mu$ and $ \nu$ are probability measures on $ X$ and $ Y$ , and $ c:X\times Y \to \mathbb{R}^+ \cup \{+\infty \}$ is a lower semi-continuous cost function. The Kantorovich duality theorem states that the transportation cost between $ \mu$ and $ \nu$ is equal to the supremum of $ $ \int_X \varphi~ d\mu +\int_Y \psi~ d\nu $ $ over all $ L_1$ functions $ \varphi(x)$ and $ \psi(y)$ such that $ \varphi(x)+\psi(y)\leq c(x,y)$ for almost all $ x\in X$ and $ y\in Y$ .

My question is: if $ c(x,y)\in \{0,1\}$ for all $ x$ and $ y$ , does it follow that there exists a solution (or “almost exists” a solution) where $ \varphi(x)$ and $ \psi(x)$ only take values in the set $ \{-1,0,1\}$ ? Finite dimensional experiments with linear programs suggest that the answer is “yes” but I cannot tell if they extend to the general setting.