algorithm analysis – Space complexity of using a pairwise independent hash family

I’m trying to analyze the space complexity of using the coloring function $f$ which appears in “Colorful Triangle Counting and a MapReduce Implementation”, Pagh and Tsourakakis, 2011, https://arxiv.org/abs/1103.6073.

As far as I understand, $f:(n) rightarrow (N)$ is a hash function, that should be picked uniformly at random out of a pairwise independent hash functions family $H$. I have a few general questions:

  1. Does the space complexity required by $f$ is affected by the fact that $H$ is $k$-wise independent? Why? (If it does, then also- how?)
  2. What do we know about $|H|$? What if $H$ is $k$-wise independent?
  3. Is there a more space-efficient way to store $f$ than storing an $N times m$ matrix that maps each vertex to its color, using O($N m$) storage words?
  4. Does the total space complexity which is required in order to use $f$ as described in the paper is $|H| cdot O(text{space complexity of } f)$?

Best regards

css – Caracteres especiais em negrito no Safari – font family Gordita

Os caracteres especiais como ã ou ç estão ficando em negrito apenas no Safari versão 13.

Por exemplo:
Escolha as informações… Selecione as opções…

Como já citei, isso acontece apenas no Safari, pelo chrome e IE está tudo ok.

Registering non-EU family member of EU citizen (non-French) in France

But my husband needs to register, is that right?

That is correct.

If he needs to register, will we need to show some rent agreement?

Possibly. See Carte de séjour de membre de la famille d’un citoyen de l’Union/EEE/Suisse. As far as I can tell, though, temporary accommodations should be sufficient, including those obtained through AirBnb. Specifically, the page says that you must provide

Indication relative au domicile : apportée par tout moyen

This means that you can use “any means” to show where you will be living.

can we just show we have 10K euros in cash or this needs to be deposited to some French bank account?

The money does not need to be in a French bank account. See Titre de séjour d’un retraité (ou inactif) citoyen UE/EEE/Suisse:

tout document permettant de justifier de la réalité de vos ressources et la durée pendant laquelle vous en disposerez comme des relevés de compte bancaire, des bulletins de pension, etc.

This indicates that you can use bank statements among other documents, but says nothing about where the banks are located. Any bank should be fine, therefore.

However, be careful about the amount. See Séjour de longue durée d’un Européen en France. You may be asked to show slightly more than €10,000 for a year, and rather more indeed if you are over 65, in which case the amount is nearly €17,000 a year for a couple.

Also note that you will have to provide evidence of sufficient health insurance.

st.statistics – Is linear combination of dependent exponential family random variables(multinomial, normal) in exponential family?

As well known to us, multinomial distribution and normal distribution belong to exponential family. Define
$$Z = sum_{i=1}^{n}a_{i}(X_{i} – b_{i}) + sum_{j=1}^{m}c_{j}Y_{j},$$
where $X_{i}$ is multinomial random variable, $Y_{j}$ is normal random variable, $a_{i}, b_{i}, c_{j}$ are constants. Additionally, $X_{i}, Y_{j} (i = 1,2,cdots,n, j = 1,2,cdots,m)$ are dependent.

Question: does the distribution of random variable $Z$ belong to exponential family?

Furthermore, in general, does the distribution of linear combination of exponential family random variables belong to exponential family?

If yes, can you provide a rigorous proof or a reference? Otherwise, can you provide a counterexample?

Thank you in advance!

customs and immigration – Practical implication of Judgment in Case C-754/18 to family members of EU citizens according to EU 2004/38/EC

CJEU recently made a Judgment in Case C-754/18.

http://curia.europa.eu/juris/documents.jsf?num=C-754/18

What practical implications it may have to family members of EU citizens according to EU 2004/38/EC.

The first point made by CJEU is that a permanent residence card according to Article 20 is at least equivalent to residence card according to Article 10.

Honestly, to claim otherwise is very stupid and I remember that European Commision already said it years before. Finally, there is a final argument to this topic.

The second point, for me much more interesting, is that CJEU decided that once a status as a family member of EU citizen according to EU 2004/38/EC is proven in one member state then a border guard is not allowed to question the status (unless there is an misuse or fraud).

From my experience, the police is very restrictive regarding Article 3/1 Beneficiaries of EU 2004/38/EC “a family member who is accompanying or joining the EU citizen”.

Firstly, the police does not even know the EU law. Then, they demand an absolute proof of accompanying or joining according to Article 3/1.

Do I understand that this police practise is over? The CJEU said clearly that a residence card according to Article 10 or 20 proves the status and police cannot further question it.

I am currently forcing criminal charges against German police as they completely failed to recognise my wife as a family member of EU citizen according to EU 2004/38/EC making all kind of stupid excuses. My wife clearly identified herself with residence cards according to Article 10 and 20.

What is your experience?

ag.algebraic geometry – A complex analytic interpretation of multiplicity on the special fiber of a flat family

Let $X$ be a variety over $mathbb C$ and $pi: Xto Delta$ be a flat morphism over the unit disk $Delta={z:|z|<1}$. Let $Z$ be a component of $X_0=pi^{-1}(0)$. The multiplicity of $Z$ is defined as the order of vanishing of $pi^*(t)$ on $Z$, where $t$ is the local coordinate of $0inDelta$.

I’m trying to understand the multiplicity through a complex analytic point of view, via the following way.

Let $pin Z$ be a general point (in particular, $p$ is a smooth point of $Z$ and does not meet other components of $X_0$). Let $Delta_p$ is a holomorphic disk in $X$ which is transversal to $Z$ at $p$. Then the restriction

$$pi|_{Delta_p}:Delta_pto Delta$$

is identified with $zmapsto z^m$ for some $mge 1$.

Question: Is it true that $m$ coincides with multiplicity of $Z$ in general?

My motivating example is the family of plane curves $X={y^2-tx=0}subset mathbb C^2_{(x,y)}times Delta$, take $p=(1,0,0)$ and $Delta_p={(1,t,t^2):tin Delta}$ which is 2-to-1 to $Delta$, so the degree coincides with the multiplicity of ${y^2=0}$. Also the degree is not dependent on the choice of $p$ (except at $p=(0,0,0)$ where the total space is singular) and choice of normal disk.

This leads to the construction in my question as above, and I think the answer should be true. However, I haven’t seen such a complex analytic description in any reference before. Is that true? I would appreciate it if anyone can show me proof or reference.

p.s., My original question was posted on MSE here but I received no comment/answer so far.

Residency card of a family member of EEA national

I am married to an Irish wife. I am living in the UK for almost 5 and a half years and my 5 year visa is about to expire. I have a visa says residence card of a family member of a EA national. Am I supposed to apply for a settlement visa? How much does it cost and how much do they extend my visa after?

pr.probability – Constructing a family of random variables for a given dependency graph

To make sure that we are all on the same page, I will state the following two definitions:

Definition A: Two families of random variables $(X_a)_{ain A}$ and $(X_b)_{bin B}$ are said to be independent if and only if the two sigma fields $sigma({X_a:ain A})$ and $sigma({X_b:bin B})$ are independent. Ref.

Definition B: Consider any undirected, simple graph $G=(V,E)$, $V$ are the vertices and $E$ the edges, and a family of random variables $(X_a)_{ain V}$. Then $G$ is a dependency graph for the $X_a$ if the following is true: For any disjoint subsets $I,Jsubset V$ such that there is no edge between any vertex of $I$ and any vertex of $J$ (i.e. for all $iin I$ and $jin J$ we have ${i,j}notin E$), we have that $(X_i)_{iin I}$ and $(X_j)_{jin J}$ are independent families of random variables.

Example: If you consider the graph $G$ with vertices $V={1,2,3}$ and no edges ($E=emptyset$), then it is a dependency graph for $(X_1,X_2,X_3)$ if and only if the random variables are independent.

My question: Suppose I am given a finite, undirected, simple graph $G=(V,E)$. I want to construct a family of random variables $(X_a)_{ain V}$ such that $G$ is a dependency graph for the $X_a$. Note that there are many such families of random variables (for example if the $X_a$ are an independent family then $G$ automatically is a dependency graph for them), so I would like to have one where we have “as few independencies as possible”. I.e. if there is an edge between $i,jin V$, then I would like $X_i$ and $X_j$ to not be independent. Is there any known result on how to get such random variables? Is it known if such random variables always have to exist? (Seeing for instance exercise 18 from here, it seems such random variables should always exist.)


Cross-posted on Mathematics Stackexchange: https://math.stackexchange.com/questions/3724271.