text:
Paris – Gare du Nord to Gare de Lyon Transfer time for a family

When I book with the Eurostar site to a destination in the South of France, they only offer a short Paris transfer time (54 minutes) between the Gare du Nord Eurostar and the Gare de Lyon TGV. Is this possible without any problems with 2 adults and 2 traveling children (without buggy)? How would you travel to avoid being in a hurry, and are there any tips to avoid surprises and delays?

Can my wife travel to Ireland with an EUA family visa from the UK?

We have to go to the funeral of my mother. My wife has an EUA family visa valid until September. Can she travel to Ireland with the EUA Family Visa? Many Thanks

Non-liner optimization of a family of functions

I would like to have tips on the following problem (solution methods, problem category, etc.).
The context is about the simultaneous registration of multiple images ($$n$$ images, $$f_ {i, j}$$ the function that maps coordinates $$(x_i, y_i)$$ to $$(x_j, y_j)$$,

To let $$f_ {i, j} ^ k (x, y)$$ a polynomial function of order $$k$$ from $$mathbb {R} ^ 2$$ to $$mathbb {R} ^ 2$$With $$i, j in [1, n]$$,

For example when $$k = 0$$Such functions can be expressed as $$f_ {i, j} ^ 0 (x, y) = (a_ {ij} ^ 0, a_ {ij} ^ 1)$$,
Or when $$k = 1$$: $$f_ {i, j} ^ 1 (x, y) = (a_ {ij} ^ 0 + b_ {ij} ^ 0 x + c_ {ij} ^ 0 y, a_ {ij} ^ 1 + b_ {ij} ^ 1 y + c_ {ij} ^ 1 y)$$,

Accepted $$k$$ is fixed, is waived in the following.

This family of functions should take into account as much as possible in terms of property: $$f_ {i, j} = f_ {i, k} (f_ {k, j})$$,

I want to use this property to reduce the number of unknowns and get robust estimates of the coefficients.
In addition, I have a large number of matches that can be expressed as such: $$v = f_ {i, j} (u)$$,

My problem is this: How can I calculate the functions robustly? $$f_ {1, i}$$ ?

If $$k = 0$$The problem is quite simple: $$f_ {i, j} = f_ {i, k} (f_ {k, j}) = f_ {i, k} + f_ {k, j}$$ so I can write an over-constrained linear system and solve it with a Moore-Penrose inverse or an algorithm like RANSAC.

If k = 1 or 2, I do not know exactly how to proceed. I think I could try to design a custom convergence scheme with a given order of equation resolution and some convergence iterations.

As an example, if I solve $$f_ {1,2}$$to get then $$f_ {1,3}$$ I can use my matches of the form $$v = f_ {1,3} (u)$$ but also the games like $$v = f_ {2,3} (u) => f_ {1,2} (v) = f_ {1,3} (u)$$

Thanks, Thomas

Should I tell my son that he was abused?

Personally, if your positive person does not remember, I would not tell him. Do not tell him. There is a reason why he does not remember – the brain will work out information to self-stabilize when needed. If he does not ask, he does not need to know. Really not, if he does not know, there is no reason to know it, because now everything is fine, but if you tell him he might be upset that you did not do it all, what you told him earlier or Do not tell me, believe me, just leave it in the past. I would not tell him that he is too young, it could destroy his confidence.

https://s.yimg.com/tr/i/5f326b49e7764432b7746a07b9bb6309_A.jpeg

Anonymous
· 9 hours ago

,

8166

BlackHatKings: proxy lists
Posted by: Afterbarbag
Post Time: June 11, 2019 at 16:56.

Visa – 2004/38 / EC: application for a family residence card (review under EU law) in the Netherlands after more than 90 days in the country

TL; DR (because this has become a MUCH longer contribution than I expected – sorry!): Should we be worried that my non-EU friend will file a request for review against EU law (namely Directive 2004/38 / EG and for whom we feel well otherwise he qualifies to the IND here in the Netherlands some time after his initial 90 days in the country (with a visa that is technically valid until 2022 but with an alleged stated stay of 90 days) have already passed? Thank you in advance!

Hello everyone, I have decided to post this question here (and also on the ImmigrationBoards website), as I have seen a number of similar (though not identical) questions here and some of you really seem to know what you are I hope you can calm me down at least a little bit. Here are some similar questions that I have already discussed:

Does a residence permit permit crossing a Schengen visa?

What are the rules for a non-EU family member traveling with an EU national?

I will try to simplify as much as possible, which is a fairly complex situation. I am a British citizen and my friend is Thai (we are neither married nor in any way in a registered partnership, but we are actually living together). We have lived together in Thailand for almost seven years so that the length of our coexistence for the purposes of 2004/38 / EC will be sufficient to qualify him as my non-EU family member and for all other eligibility requirements of 2004/38 / EC are fulfilled – there is no problem.

In September of last year, he successfully received a free, accelerated entry visa to facilitate entry from the Dutch Embassy in Bangkok – everything went according to plan – which, I understand, was issued as a Schengen Type C visa.

The multiple visa is valid until 2022 (that was unexpected!), And we arrived here on the 12th of March this year in the Netherlands. I continue my freelance freelance paperwork that I did for a few years in Bangkok, and my friend has not worked for some time, but intends to seek work here.

We had no problems when we passed the passport control at Schiphol Airport – we were not even asked a single question (unlike in 2017, when we arrived in Schiphol at that time, with a three-month Schengen type C tourist with multiple entries ) The visa was issued by the Dutch Embassy courtesy to save us time after the outsourced application center had screwed up our application for an EU family visa 2004/38 / EC. On this occasion, the immigration officer in Schiphol asked a few friendly questions, but still nothing out of the ordinary.

Since we arrived here in March, we have rented an apartment (which took much longer than expected) and – after my registration at the local government (municipality / town hall) a few weeks ago and a subsequent home visit to two of their employees, the latter due to a now solved problem in the registration of the land in the community by our landlord – only a few days ago I received a postal confirmation that I was successfully registered in the population register.

I have planned the various interdependent phases of our "arrival bureaucracy" – as both of us need to be legally and fully registered here – for some time, and these (renting an apartment and registering with the community) were two of the first points on my list of six. The remaining four are as follows (as far as I know, these are still correct, but I am very happy to be told otherwise if they are addressed incorrectly or in an order that does not make sense):

1. register with the KvK on my own; I have an appointment for this next Wednesday, June 12, and we need the certificate of that registration for the next step.

2. I am registering as an EU citizen with the Dutch Immigration Service (IND) to obtain proof of residency (this would normally not be required as an EU citizen but unfortunately we need this proof of residency permit Friend can request a similar proof of his right of residence (and in any case it is not a bad thing to have my credentials in view of the forthcoming Brexit) Proof of entitlement I must, in particular, submit the self-employment registration certificate and prove that my self-employment includes a real work (business accounts / monthly business records / accountant letter etc) .I hope i can get along with copies of bills sent to my clients);

3. Let my friend apply to the IND for a "review against EU law" (ie his entitlement under 2004/38 / EC) to obtain a certificate of legal residence (which I assume) is or comes next to the actual residence card in credit card format – please correct me if I'm wrong!); In particular, in order to apply, he must present my own proof of residency, as above, together with evidence of my work and income (with luck I may use the same certificate and bills for self-employment registration, etc.) in the previous step, but if not, then in the last few days I have accepted only a full-time job offer. Therefore, at this point, I should have a signed employment contract, which I can use instead, if necessary) and proof of our relationship and our coexistence abroad (all this was previously demonstrated in the original visa application, so we can hopefully reuse the same evidence. However, if the IND insists that our previous joint Thai leases, etc., be legalized, we may have a bit of a problem, as our last landlord was not cooperative in this regard and my friend's birth certificate is the only one we have before ours Legalizing departure – in this situation we may have to rely on the photos of our course relationship that we have also submitted with the visa application and hope for the best of it); and finally

4. Have my friend register with the local community as I have already done. For this he needs the residence card that he should have received in the previous step (therefore he could not register with the community at the same time as I, as an EU citizen, did not need a residency card).

Now come the potential problem and my question and the part to which I am seeking some advice or at least a little confirmation:

In an ideal world (and indeed, as we had originally expected), all of the above steps would have been completed within the first three months of our arrival here, as this is the period for which this is the case under the 2004/38 EC legislation My friend (and I) can stay here without asking any further questions and without fulfilling other criteria (eg the need for me to work). This means that it is generally expected that you have completed at least Step 3 (its registration with the IND) until the end of these three months.

This also corresponds to the maximum period of 90 days (in a 180-day period) during which non-EU / EEA / Swiss nationals with a Schengen short-stay visa should generally remain in the Schengen area without applying for a longer-term visa , Residency permit. This period of 90 days is, of course, generally applicable to short stays, even if the visa, as in the case of my friend, actually expires in 2022 (which is approximately four years since it was issued in 2022) 2018); Like most of this length, his visa still has a "length of stay" field, which is "90". [days]and usually that would be true.

Obviously, all the other bureaucratic procedures that led to Step 3 took longer than expected, meaning that none of the four steps above are yet complete. Importantly, my friend is not yet registered with the IND in step 3 – and here's the big, his first 90 consecutive days in the Schengen area (and in his case in the Netherlands, since he was not) abroad since Arrival) expire today (Sunday, June 09). (Last minute to ask advice, I know, I know – doh!)

As already mentioned, my appointment as a self-employed person on Wednesday, 12 June (since I had now accepted a full-time job, I could only have waited for this to come through as proof of my work, but that would have been the case) Delayed things further – and I will definitely be on my own). Fortunately, I can pay a little more on that day to get an extract of the certificate there, so there should be no more waiting time.

Given the fact that the individual steps are dependent on the previous steps, I have so far planned to agree the dates with the IND only when the previous steps were completed and I received the certificate of self-employment registration and other documents in the Hand had to get started. But since we're here, I plan to call the IND on Tuesday (closed on Monday for a holiday, and an appointment for this type of application seems to be one of the few you can). t make it online!) to make the appointments now, although I still do not know how long it could take. (At least I can talk to someone there, and we should have an appointment to submit our applications in the diary.)

My main question is: Do you think that your knowledge and / or experience may cause problems in filing my friend 's application for review under EU law (2004/38 / EC) and issuing a residence permit (and)? therefore his residence card) because he has passed the last 90 days when he personally visits the office of the IND to file his application?

Let's just wait and go ahead as planned, or – and I think there's no other alternative here to apply for an (emergency) extension of his existing stay or to give him an extra sticker in his passport to express him Instead, we should choose the ultimate alternative and have him return to Thailand the next day or so (before we are over the apparent three-day deadline). & # 39; Period at the end of a 90-day stay, in apparently no enforcement action was taken) and then presumably because of the 90/180 day rule and the need to wait for a new one, another full 90 days wait 180-day period, and then just come and start over to complete the whole process?

Of course, this last option is all the riskier and less attractive than being a British citizen and none of us knows exactly what's going to happen in terms of Brexit over the next 90 days. That is, it is quite conceivable that Brexit might have already arrived by the time he has to return here. In that case, I would no longer be an EU citizen and he would no longer have the right to settle here in the Netherlands as my (non-) EU family member as he is now.

Of course my strong preference would be for us to stay here now until everything is settled, but of course I do not want him encountering immigration problems now or in the future when leaving the Schengen area, or worse, when we come back in.

(Second thought: In view of the 2004/38 / EC, the family member seems to be entitled to spend together with the EU citizen up to three months in any Schengen state, before a request for a residence permit, etc. must be made the time that they can spend throughout Schengen space (at least based on the interpretation in SuperRoo's answer here and in phoog's answer here), provided they move enough (ie they do not stay longer than three months) and are when traveling with the EU citizen, could it also be worth traveling to another Schengen state for a short time, crossing the border into Belgium – or, theoretically, saying that we did? – and then to the Netherlands to return to where we left off (does it make sense to technically meet the requirements of the rules, is it worth considering, unnecessary or unhelpful?)

Here is my personal point of view: I increasingly doubt this, given the gravity of the situation, but I still do not believe that the 90/180 day rule applies to family members outside the EU (assuming they are traveling) or resident in the EU EU). Here and here @phoog explains in detail and with respect to certain parts of Directive 2004/38 / EC that the 90-day deadline for (non-) EU family members does not apply because they are not considered as third-country nationals. Nationals of the country for this purpose. That makes perfect sense to me (and I also noticed while writing that if the 90-day deadline were indeed to apply, there would certainly be a gap between family members who have three months to ask no questions at all a specific Schengen country (as well as their EU relative) and allegedly only 90 days in the Schengen area (or entry stamps for non-visa nationals) – provided that 90 days are usually less than three months, it seems just not fit.

@phoog also points out that the free movement of the family member together with the EU citizen results from Directive 2004/38 / EC and not from a visa (which merely provides proof of entry) or a residence card (which in turn is easier) the existing right to free movement, as it actually does, and without which the right still exists in full).

However, this separate answer is one of a number that takes a slightly different view, suggesting that the 90/180 day rule will still apply unless and until the family member has received a residence card.

I have sent a request to the Consular Contact Center of the Dutch Government, which I have found very helpful in the past (and which, above all, reacts rather quickly), and they have advised:

"If I understand that correctly, you would like to know if your partner with a Schengen short-stay visa can stay in the Schengen area for more than 90 days.

I'm afraid your friend can not stay longer than 90 days in the Schengen area with a visa. Please note that this type of visa is only valid for short stays (maximum of 90 days within a period of 180 days). I would advise him to leave the Schengen area before the 90-day deadline. If this is not the case, it is overwhelmed, which can lead to a warning of the Schengen Information System. "

However, in light of the above, I find it increasingly unlikely that what you have told me is true (I wonder if the consular officer of the consular center may not fully acknowledge the rights granted to the family member until 2004). / 38 / EC, with or without visa OR residence card, and the effects that may actually result in "crossing" in this case).

In addition, numerous contributions such as this have highlighted that Directive 2004/38 / EC also states that family members who fail to meet their obligations, for example, as in this case, should register with the IND on time, should only produce reasonable results and non-discriminatory Sanctions ". (@phoog points out here as well.) The IND's own website even states (albeit in connection with a slightly different topic, but still related to "overcrowding" or "illegal" residence in the Netherlands) that the Application of The ban on returning to the Netherlands, for example, does not apply to EU / EEA citizens and their family members.

(Nevertheless, the person in the consular center points to the risk of a Schengen Information System alert if my friend remains "overly".)

Even if my friend exceeds the original 90-day period by a few days (or even longer) (that is, even if they interpret him as I do not)) before he finally submits his application to the IND can submit, it sounds like the worst he / she gets in the application is a slap on the wrist and maybe a fine – or maybe, if we're lucky, not even that.

I think writing everything down in such a long, long post (sorry!) Has helped me to calm down a bit – but I'm still a bit torn, which I can do best, especially if someone in the past has had experiences with it or something similar. Should I be worried? Or am I not worried about anything? What are your thoughts? Thank you in advance – and thank you so much for reading!

(Whew!)

Right or wrong, tree family is balanced? (Data structures)

A family of trees is said to be balanced if, for each tree in the family, the height is O (log (n)).

For a tree family where there is a constant c, the height difference between the two subtrees of the node for each node in the tree is max c.

I understand that this statement is true. This is the general case for AVL trees. I just do not know how to prove it formally.

Many Thanks.

[ Politics ] Open question: Why do conservatives talk about "Christian and family values" when Trump's behavior deviates from both?

[ Politics ] Open question: Why do conservatives talk about "Christian and family values" when Trump's behavior deviates from both? ,


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### A Note on Bounding Regret of the C$^2$UCB Contextual Combinatorial Bandit

We revisit the proof by Qin et al. (2014) of bounded regret of the C$^2$UCB contextual combinatorial bandit. We demonstrate an error in the proof of volumetric expansion of the moment matrix, used in upper bounding a function of context vector norms. We prove a relaxed inequality that yields the originally-stated regret bound.

### Sparse Dueling Bandits

The dueling bandit problem is a variation of the classical multi-armed bandit in which the allowable actions are noisy comparisons between pairs of arms. This paper focuses on a new approach for finding the "best" arm according to the Borda criterion using noisy comparisons. We prove that in the absence of structural assumptions, the sample complexity of this problem is proportional to the sum of the inverse squared gaps between the Borda scores of each suboptimal arm and the best arm. We explore this dependence further and consider structural constraints on the pairwise comparison matrix (a particular form of sparsity natural to this problem) that can significantly reduce the sample complexity. This motivates a new algorithm called Successive Elimination with Comparison Sparsity (SECS) that exploits sparsity to find the Borda winner using fewer samples than standard algorithms. We also evaluate the new algorithm experimentally with synthetic and real data. The results show that the sparsity model and the new algorithm can provide significant improvements over standard approaches.

### Bilinear Bandits with Low-rank Structure

We introduce the bilinear bandit problem with low-rank structure where an action is a pair of arms from two different entity types, and the reward is a bilinear function of the known feature vectors of the arms. The problem is motivated by numerous applications in which the learner must recommend two different entity types as one action, such as a male / female pair in an online dating service. The unknown in the problem is a $d_1$ by $d_2$ matrix $\mathbf{\Theta}^*$ with rank $r \ll \min\{d_1,d_2\}$ governing the reward generation. Determination of $\mathbf{\Theta}^*$ with low-rank structure poses a significant challenge in finding the right exploration-exploitation tradeoff. In this work, we propose a new two-stage algorithm called "Explore-Subspace-Then-Refine" (ESTR). The first stage is an explicit subspace exploration, while the second stage is a linear bandit algorithm called "almost-low-dimensional OFUL" (LowOFUL) that exploits and further refines the estimated subspace via a regularization technique. We show that the regret of ESTR is $\tilde{O}((d_1+d_2)^{3/2} \sqrt{r T})$ (where $\tilde{O}$ hides logarithmic factors), which improves upon the regret of $\tilde{O}(d_1d_2\sqrt{T})$ of a naive linear bandit reduction. We conjecture that the regret bound of ESTR is unimprovable up to polylogarithmic factors.

### Alternating Linear Bandits for Online Matrix-Factorization Recommendation

We consider the problem of online collaborative filtering in the online setting, where items are recommended to the users over time. At each time step, the user (selected by the environment) consumes an item (selected by the agent) and provides a rating of the selected item. In this paper, we propose a novel algorithm for online matrix factorization recommendation that combines linear bandits and alternating least squares. In this formulation, the bandit feedback is equal to the difference between the ratings of the best and selected items. We evaluate the performance of the proposed algorithm over time using both cumulative regret and average cumulative NDCG. Simulation results over three synthetic datasets as well as three real-world datasets for online collaborative filtering indicate the superior performance of the proposed algorithm over two state-of-the-art online algorithms.

### Gaussian Process Bandits for Tree Search: Theory and Application to Planning in Discounted MDPs

We motivate and analyse a new Tree Search algorithm, GPTS, based on recent theoretical advances in the use of Gaussian Processes for Bandit problems. We consider tree paths as arms and we assume the target/reward function is drawn from a GP distribution. The posterior mean and variance, after observing data, are used to define confidence intervals for the function values, and we sequentially play arms with highest upper confidence bounds. We give an efficient implementation of GPTS and we adapt previous regret bounds by determining the decay rate of the eigenvalues of the kernel matrix on the whole set of tree paths. We consider two kernels in the feature space of binary vectors indexed by the nodes of the tree: linear and Gaussian. The regret grows in square root of the number of iterations T, up to a logarithmic factor, with a constant that improves with bigger Gaussian kernel widths. We focus on practical values of T, smaller than the number of arms. Finally, we apply GPTS to Open Loop Planning in discounted Markov Decision Processes by modelling the reward as a discounted sum of independent Gaussian Processes. We report similar regret bounds to those of the OLOP algorithm.

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Finite-time attitude synchronization with distributed discontinuous protocols

# Finite-time attitude synchronization with distributed discontinuous protocols

Jieqiang Wei, Silun Zhang, Antonio Adaldo, Johan Thunberg, Xiaoming Hu, Karl H. Johansson *This work is supported by Knut and Alice Wallenberg Foundation, Swedish Research Council, and Swedish Foundation for Strategic Research.J. Wei, A. Adaldo and K.H. Johansson are with the ACCESS Linnaeus Centre, School of Electrical Engineering. S. Zhang and X. Hu are with School of engineering sciences. KTH Royal Institute of Technology, SE-100 44 Stockholm, Sweden. {jieqiang, silunz, adaldo, hu, kallej}@kth.se. Johan Thunberg is with Luxembourg Centre for Systems Biomedicine, UniversitÃ© du Luxembourg,6, avenue du Swing, L-4367 Belvaux. {johan.thunberg}@uni.lu
###### Abstract

The finite-time attitude synchronization problem is considered in this paper, where the rotation of each rigid body is expressed using the axis-angle representation. Two discontinuous and distributed controllers using the vectorized signum function are proposed, which guarantee almost global and local convergence, respectively. Filippov solutions and non-smooth analysis techniques are adopted to handle the discontinuities. Sufficient conditions are provided to guarantee finite-time convergence and boundedness of the solutions. Simulation examples are provided to verify the performances of the control protocols designed in this paper.

Agents and autonomous systems, Finite-time attitude synchronization, Network Analysis and Control, Nonlinear systems

## I Introduction

Originally motivated by aerospace developments in the middle of the last century [5, 15], the rigid body attitude control problem has continued to attract attention with many applications such as aircraft attitude control [2, 32], spacial grabbing technology of manipulators [21], target surveillance by unmanned vehicles [24], and camera calibration in computer vision [20]. Furthermore, the configuration space of rigid-body attitudes is the compact non-Euclidean manifold , which poses theoretical challenges for attitude control [3].

Here we review some related existing work. As attitude systems evolves on —a compact manifold without a boundary—there exists no continuous control law that achieves global asymptotic stability [6]. Hence one has to resort to some hybrid or discontinuous approaches. In [17], exponential stability is guaranteed for the tracking problem for a single attitude. The coordination of multiple attitudes is of high interest both in academic and industrial research, e.g., [11, 26, 29]. In [18] the authors considered the synchronization problem of attitudes under a leader-follower architecture. In [23], the authors provided a local result on attitude synchronization. Based on a passivity approach, [25] proposed a consensus control protocol for multiple rigid bodies with attitudes represented by modified Rodrigues parameters. In [31], the authors provided a control protocol in discrete time that achieves almost global synchronization, but it requires global knowledge of the graph topology. Although there exists no continuous control law that achieves global asymptotic stability, a methodology based on the axis-angle representation obtains almost global stability for attitude synchronization under directed and switching interconnection topologies is proposed in [30]. These control laws were later generalized to include various types of vector representations including the Rodrigues Parameters and Unit Quaternions [29]. Besides these agreement results, [16, 28, 34] provided distributed schemes for more general formation control of attitude in space .

Among all the properties of attitude synchronization schemes, the finite-time convergence is an important one, because in practice it is desired that the system reaches the target configuration within a certain time-interval; consider, for instance, satellites in space that shall face a certain direction as they move in their orbits, or cameras that shall reach a certain formation to quickly follow an object. So far, finite-time attitude control problems are studied in different settings, e.g., [12, 35]. In [12], finite-time attitude synchronization was investigated in a leader-follower architecture, namely all the followers tracking the attitude of the leader. In [35], quaternion representation was employed for finite-time attitude synchronization. Both works used continuous control protocols with high-gain.

In this paper, we shall focus on the finite-time attitude synchronization problem, based on the axis-angle representations of the rotations without a leader-follower architecture, using discontinuous control laws. Two intuitive control schemes are proposed. The first scheme employs a direction-preserving sign function to guarantee finite-time synchronization almost globally, namely, the convergence holds for almost all the initial conditions. The other scheme, motivated by binary controllers for scalar multi-agent systems, e.g., [7, 19, 9, 14], uses the component-wise sign function. Compared to the first scheme, the second one is more coarse, in the sense that only finite number of control outputs are employed, and guarantees finite-time convergence locally. Since these control schemes are discontinuous, nonsmooth analysis is employed throughout the paper.

The structure of the paper is as follows. In Section II, we review some results for the axis-angle representation for rotations in and introduce some terminologies and notations in the context of graph theory and discontinuous dynamical systems. Section III presents the problem formulation of the finite-time attitude synchronization problem. The main results of the stability analysis of the finite-time convergence are presented in Section IV, where an almost global and local stability are given in Subsection IV-A and IV-B respectively. Then, in Section V, the paper is concluded.

Notation. With and we denote the sets of negative, positive, non-negative, non-positive real numbers, respectively. The rotation group The vector space of real by skew symmetric matrices is denoted as . denotes the -norm and the -norm is denoted simply as without a subscript.

## Ii Preliminaries

In this section, we briefly review some essentials about rigid body attitudes [27], graph theory [4], and give some definitions for Filippov solutions [13].

Next lemma follows from Euler’s Rotation Theorem.

###### Lemma 1.

The exponential map

 exp:so(3)→SO(3) (1)

is surjective.

For any and given as

 ^p:=⎛⎜⎝0−p3p2p30−p1−p2p10⎞⎟⎠, (2)

Rodrigues’ formula is the right-hand side of

 exp(^p)=I3+sin(θ)θ^p+1−cos(θ)θ2(^p)2, (3)

where and is the rotation matrix through an angle anticlockwise about the axis . For where is not symmetric, we define the inverse of as

 log(R)=θ2sin(θ)(R−R⊤), (4)

where . We define as the zero matrix in . Note that (4) is not defined for . The Riemannian metric for is defined as where is the Frobenius norm.

One important relation between and is that the open ball in with radius around the identity, which is almost the whole , is diffeomorphic to the open ball in via the logarithmic and the exponential map defined in (4) and (3).

An undirected graph consists of a finite set of nodes and a set of edges of unordered pairs of . To each edge , we associate a weight . The weighted adjacency matrix is defined by if and otherwise. Note that and that as no self-loops are allowed. For each node , its degree is defined as . The graph Laplacian is defined as where is a diagonal matrix such that . As a result, . We denote the set of neighbors of node as . If the edges are ordered pairs of , the graph is called a directed graph, or digraph for short. An edge of a digraph is denoted by (with ) representing the tail vertex and the head vertex of this edge. A digraph with unit weights is completely specified by its incidence matrix , where , with equal to if the th edge is towards vertex , and equal to if the th edge is originating from vertex , and otherwise. The incidence matrix for undirected graphs is defined by adding arbitrary orientations to the edges of the graph. Finally, we say that a graph is connected if, for any two nodes and , there exists a sequence of edges that connects them. In order to simplify the notation in the proofs, we set the weight to be one. All the results in this paper hold for the general case where the ’s are elements in .

In the remainder of this section, we discuss Filippov solutions. Let be a map from to and let denote the collection of all subsets of . The Filippov set-valued map of , denoted , is defined as

 F[f](x):=⋂δ>0⋂μ(S)=0¯¯¯¯¯co{f(B(x,δ)∖S)},

where is a subset of , denotes the Lebesgue measure, is the ball centered at with radius and denotes the convex closure of a set . If is continuous at , then contains only the point .

A Filippov solution of the differential equation on is an absolutely continuous function that satisfies the differential inclusion

 ˙x(t)∈F[f](x(t)) (5)

for almost all . A Filippov solution is maximal if it cannot be extended forward in time, that is, if it is not the result of the truncation of another solution with a larger interval of definition. Next, we introduce invariant sets, which will play a key part further on. Since Filippov solutions are not necessarily unique, we need to specify two types of invariant sets. A set is called weakly invariant if, for each , at least one maximal solution of (5) with initial condition is contained in . Similarly, is called strongly invariant if, for each , every maximal solution of (5) with initial condition is contained in . For more details, see [10, 13]. We use the same definition of regular function as in [8] and recall that any convex function is regular.

For locally Lipschitz, the generalized gradient is defined by

 ∂V(x):=co{limi→∞∇V(xi)∣xi→x,xi∉S∪Ωf}, (6)

where is the gradient operator, is the set of points where fails to be differentiable and is a set of measure zero that can be arbitrarily chosen to simplify the computation, since the resulting set is independent of the choice of [8].

Given a set-valued map , the set-valued Lie derivative of a locally Lipschitz function with respect to at is defined as

 LTV(x):= {a∈R∣∃ν∈T(x)% such that (7) ζTν=a,∀ζ∈∂V(x)}.

If is convex and compact , then is a compact interval in , possibly empty.

The th row of a matrix is denoted as . For any matrix , we denote as and as . A positive definite and semidefinite (symmetric) matrix is denoted as and , respectively. The vectors denote the canonical basis of . The vectors and represents a -dimensional column vector with each entry being and , respectively. In this paper, we define the direction-preserving signum as

 sign(w)={w∥w∥ if w≠0,0 if w=0, (8)

for . The component-wise signum is denoted as

 signc(w)=[sign(w1),…,sign(wk)]⊤, (9)

where . Notice that for scalars, these two signum functions coincide. Furthermore, component-wise signum is coarser than direction-preserving in the sense that there is only a finite number of elements in the range for a fixed dimension .

## Iii Problem formulation

We consider a system of agents (rigid bodies). We denote the world frame as and the instantaneous body frame of agent as where . Let be the attitude of relative to at time , and, when , the corresponding axis-angle representation be given by

 ^xi(t)=log(Ri(t)). (10)

The kinematics of is given by [27]

 ˙xi=Lxiωi,i∈I (11)

where is the control signal corresponding to the instantaneous angular velocity of relative to expressed in the frame , and the transition matrix is defined as

 Lxi= I3+^xi2+(1−sinc(∥xi∥)sinc2(∥xi∥2))(^xi∥xi∥)2 = sinc(∥xi∥)sinc2(∥xi∥2)I3+(1−sinc(∥xi∥)sinc2(∥xi∥2))xix⊤i∥xi∥2+^xi2 =: L1xi+^xi2, (12)

where is defined as for all and , see [27]. Note that for , the function is concave and belongs to . Then the symmetric part of , denoted by , is positive semidefinite. More precisely, if . Moreover, is Lipschitz on for any (see [30]).

The system (11) can be written in a compact form as

 ˙x=Lxω (13)

where

 x =[x⊤1,…,x⊤n]⊤, (14) Lx =blockdiag(Lx1,…,Lxn), ω =[ω⊤1,…,ω⊤n]⊤.

For the multi-agent system (13), we assume that the agents can communicate with each other about the state variables via an undirected connected graph . The aim is to design control protocols for such that the absolute rotations of all agents converge to a common rotation in the world frame in finite time, i.e.,

 ∃T>0,¯R∈SO(3) s.t. Ri→¯R,∀i∈I, as t→T. (15)

This is equivalent to that converges to the consensus space

 C={x∈R3n∣∃¯x∈R3 such that x=\mathds1n⊗¯x} (16)

in finite time. We shall propose two distributed controllers that achieve this goal.

## Iv Main result

In this section, we shall first present a control law that guarantees that the rotations of all the rigid bodies converge to a common rotation for any initial condition for all . Note that this initial condition in is equivalent to under the axis-angle representation. In order to avoid the singularity of the logarithmic map (4), the control law makes sure that the constraint is met for all and for all time . We consider controllers of the following form

 ωi=fi (∑j∈Ni(xj−xi)),i∈I, (17)

with maps and the elements in the set are the neighbors of agent . Now the closed-loop system can be written in a compact form as

 ˙x=Lxf(−^Lx) (18)

where , and is the Laplacian of the graph. Our control design is based on the signum function. More precisely, we consider the case when some of the functions are or , while the others satisfy certain continuity assumptions to be defined in the following subsections. We propose two control protocols which guarantee almost global, in the sense of , and local convergence, respectively. As discontinuities are introduced into (18) by the signum functions, we shall understand the trajectories in the sense of Filippov, namely an absolutely continuous function satisfying the differential inclusion

 ˙x ∈F[Lxf(−^Lx)](x) (19) =LxF[f(−^Lx)](x) =:F1(x)

for almost all time, where we used Theorem 1(5) in [22].

### Iv-a Control law for global convergence

In this subsection, we shall design a controller such that finite-time synchronization is achieved for any initial condition by using the direction preserving defined in (8). It might seem natural to let for all . However, the following example shows that this simple controller does not guarantee for all Filippov solutions.

###### Example 1.

Consider the system

 ˙x1 =Lx1sign(x2+x3−2x1) ˙x2 =Lx2sign(x1+x3−2x2) ˙x3 =Lx3sign(x2+x1−2x3)

defined on a complete graph. We show that for such that , the trajectories can violate the constraints , for some .

First, by Theorem 1(1) in [22], we have for any , there exists an , independent of , such that the ball . Second, suppose for some . Then there exists such that the vector . Hence,

 x(t)=\mathds1⊗((t−t0)ε1¯x∥¯x∥+¯x), t⩾t0

is a Filippov solution. Indeed,

 ˙x(t) =ε1\mathds1⊗¯x∥¯x∥ =Lx(t)ε1\mathds1⊗¯x∥¯x∥ ∈Lx(t)F[sign(−^Lx)](x(t))

where the second equality follows from

 Lx(t)ε1\mathds1⊗¯x∥¯x∥ =Lx(t)x(t)ε1∥¯x∥+(t−t0)ε1 =x(t)ε1∥¯x∥+(t−t0)ε1 =ε1\mathds1⊗¯x∥¯x∥.

Here we used the fact that . Then for large enough , can be larger than . The solutions of the type with a non-constant function is called sliding consensus.

The previous example motivates us to consider the following assumption.

###### Assumption 1.

For some set , the function in (18) satisfies the following conditions:

1. For , is locally Lipschitz continuous and satisfies and for all ;

2. For , the function .

Note that Condition (i) in Assumption 1 corresponds to that is direction preserving.

Before showing the result for finite-time convergence, we formulate a condition for the controller (17) satisfying Assumption 1 such that the set is strongly invariant for the dynamics (18).

###### Lemma 2.

Consider the differential inclusion (19) satisfying Assumption 1. If one of the following two conditions is satisfied

1. and ;

2. and ,

then the set , where is a constant, is strongly invariant. This implies that is strongly invariant.

###### Proof.

We use a Lyapunov-like argument to prove that for any initial condition in , all the solutions of (19) will remain within the set.

Consider the Lyapunov function candidate . Notice that is convex, hence regular. Let

 (20)

The generalized gradient of is given as

 ∂V(x)=co{ei⊗xi∣i∈α(x)}. (21)

Next, let be defined as

 Ψ={t⩾0∣both ˙x(t) and ddtV(x(t)) exist}. (22)

Since is absolutely continuous (by definition of Filippov solutions) and is locally Lipschitz, by Lemma 1 in [1] it follows that for a set of measure zero and that

 ddtV(x(t))∈LF1V(x(t)) (23)

for all , so that the set is nonempty. For , we have that is empty, and hence by definition. Therefore, we only consider in the rest of the proof.

By using Theorem 1(4) and (5) in [22], the differential inclusion can be enlarged as follows

 ˙x ∈F1(x) (24) ⊂n\raisebox-1.29pt\Large×i=1F[Lxifi(−^Lix)](x) =n\raisebox-1.29pt\Large×i=1LxiF[fi(−^Lix)](x) =n\raisebox-1.29pt\Large×i=1LxiF[fi](−^Lix) =:F2(x),

where the first equality follows from Assumption 1 and the fact that is continuous for with , thus we can use Theorem 1(5) in [22]. Moreover, we obtain that for all . For the rest of the proof, we shall show by considering two cases.

Case 1: For , i.e., , the following two subcases can be distinguished.

1. and . There is such that is locally Lipschitz and direction preserving. Then, by using the definition of the Filippov set-valued map, one can show that for all (recall that ). As is nonempty (by considering ), there exists such that for all , see the definition (7). By choosing , it follows that , which implies that .

2. and . In the following, we consider the Filippov solution of system (19), which can be written as

 ˙x1 =Lx1x2−x1∥x2−x1∥, (25) ˙x2 =Lx2x1−x2∥x1−x2∥.

Then it can be shown that, for (i.e., ), any element in the Filippov set-valued map of (25) satisfies . Stated differently, the following implication holds for :

 ν∈F[h](x),x∈C⇒ν1=−ν2. (26)

Next, by recalling that , it follows that

 ∂V(x)=co{e1⊗x1,e2⊗x2} (27)

with . Now, following a similar reasoning as in item (i) on the basis of the definition of the set-valued Lie derivative in (7), it can be concluded that , so that for all .

Case 2: For , take an index such that . Note that such indeed exists. Namely, assume in order to establish a contradiction that for all . Then, it holds that

 0=x⊤i^Lix=∑j∈Nix⊤i(xi−xj). (28)

Since , it follows from (28) that for all . By repeating this argument and recalling that the interconnection topology is connected, it follows that for all , i.e., . This is a contradiction to .

For the index satisfying , it follows from Assumption 1 that there exists such that

 F[fi](−^Lix)=−γi^Lix, (29)

i.e., for any it holds that . Note that this is a result of the direction-preserving property of either the vectorized signum function (for nonzero argument, then ) or the Lipschitz continuous function (by Assumption 1). Then, choosing as (recall that ), it follows that for any we have

 ζ⊤ν= −γix⊤iLxi^Lix = −γix⊤i^Lix < 0, (30)

where the second equality is based on .

Summarizing the results of the two cases leads to

 maxLF2V(x)⩽0 (31)

if for all . Since the trajectory is absolutely continuous, we have that if for all , all the trajectories remain within the set . ∎

###### Remark 1.

As indicated in Example 1, sliding consensus can happen when and . This will violate the strong invariance of the set with , which will introduce singularity for the axis-angle representation for rotations.

Before we prove the finite-time convergence, we provide a sufficient condition for that all Filippov solutions of (19) converge to consensus asymptotically.

###### Lemma 3.

Under the same assumptions of Lemma 2, all Filippov solutions of (19) asymptotically converge to static consensus.

###### Proof.

Similar to the proof of Lemma 2, we shall prove that the conclusion holds for the bigger inclusion given by (24). In this proof cases (i) and (ii) can be handled with the same arguments.

Consider the Lyapunov function candidate , which is convex, hence regular. The generalized gradient of is given as follows:

 ∂V(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩^Lx√x⊤^Lx % if x∉C,co{limy→x^Ly√y⊤^Ly:y∉C} if x∈C. (32)

Next we shall calculate the Lie derivative of by considering two cases.

1. If , the Lie derivative is given as

 LF2V(x) =x⊤^L√x⊤^LxF2 (33) =∑i∈Ix⊤^L⊤iLxiF[fi](−^Lix)√x⊤^Lx.

Here we have for . Indeed, it is because: (1), the conditions that the matrix for satisfying ; (2), Assumption 1 about direction preservation, and (3), the set is strongly invariant for . Moreover, if , the set . Hence .

2. If , it can be seen that implies . Hence if , it has to be . In fact, by taking , we have that .

Next, by Theorem 3 in [9], it holds that all Filippov solutions of (24) converge to the set asymptotically. The remaining task is to characterize the set . So far we have shown that , which implies that . By the fact that is closed, we have . Moreover, when , where , which implies that remains constant. In conclusion, asymptotic convergence to static consensus is guaranteed.

Now we are ready for the main result of this section.

###### Theorem 4.

Assume that for all and that the graph is connected. Consider the multi-agent system (18) satisfying Assumption 1 and the corresponding differential inclusion (19). Then, all Filippov solutions converge to consensus in finite time if one of the following conditions holds:

• and ;

• and .

###### Proof.

The proof is separated into two parts, one for each of the conditions.

(i) Without loss of generality, we assume that satisfies the condition (i) in Assumption 1 while are . Similar to Lemma 2, instead of proving the conclusion for the differential inclusion (19), we shall show that it holds for the bigger inclusion given by (24).

Consider the Lyapunov function candidate . We shall show that there exists such that for any initial condition with .

In the proof of Lemma 3, we have shown that for , the Lie derivative is given by (33). By the fact that with and is direction preserving, we have

 LF2V⩽∑ni=2x⊤^L⊤iLxiF[sign](−^Lix)√x⊤^Lx. (34)

Furthermore, the Filippov set-valued map

 F[sign](−^Lix)=⎧⎨⎩{−^Lix∥^Lix∥} if ∥^Lix∥≠0,{v∣∥v∥⩽1} if ∥^Lix∥=0, (35)

which implies that

 x⊤^L⊤iLxiF[sign](−^Lix) (36) = ⎧⎪⎨⎪⎩{−x⊤^L⊤iLxi^Lix∥^Lix∥} if ∥^Lix∥≠0,{∥^Lix∥} if ∥^Lix∥=0.

Note that, for any satisfying for all , there exists , which only depends on , such that for all . This implies that

 x⊤^L⊤iLxiF[sign](−^Lix)⊂(−∞,−c1∥^Lix∥]. (37)

So far we have shown that, for any , it holds that

 a⩽−c1∑ni=2∥^Lix∥√x⊤^Lx. (38)

Furthermore, by using that , which is based on the connectivity of the graph , we have

 ∥^L1x∥=∥∥ ∥∥n∑i=2^Lix∥∥ ∥∥⩽n∑i=2∥∥^Lix∥∥, (39)

where the triangle inequality is used. Then, the use of (39) in (38) yields

 (40)

By exploiting the observation that is a graph Laplacian, it holds that

 L=U⊤ΛU,L⊤L=U⊤Λ2U, (41)

where is a diagonal matrix with real-valued eigenvalues satisfying and for . The matrix collects the corresponding right-eigenvectors. From (41), it can be seen that

 L⊤L−cL⩾0 (42)

for any . Consequently, using , it follows that

 (n∑i=1∥^Lix∥)2=x⊤^L⊤^Lx⩾cx⊤^Lx. (43)

After taking the square root (note that for all ) in (43) and substituting the result in (40), the result

 a⩽−c1√c2√x⊤^Lx√x⊤^Lx=−c1√c2 (44)

follows, which proves finite-time convergence to consensus by Proposition 4 in [9] and Lemma 3.

(ii) By using a similar reasoning, we have that for any , it satisfies that where satisfying for all . This again implies finite time convergence.

###### Remark 2.

Theorem 4 provides sufficient conditions for finite-time convergence of the protocol (17) satisfying Assumption 1. However, we conjecture these sufficient conditions to be necessary as well. Namely, for the case , we expect that all the Filippov solutions of (19) converge to consensus in finite time if and only if ; and for the case , we expect that the finite-time synchronization is achieved if and only if . We show that the latter statement holds according to the following argument.

If and , then . In this case, we can only have asymptotic convergence if . Indeed, the right-hand side of (18) is Lipschitz; therefore, finite-time convergence to an equilibrium can not occur.

Unfortunately, for the case , we can not prove the necessity, which leaves it as an open question.

We close this subsection by demonstrating the result in Theorem 4 and conjecture in Remark 2 by an example.

###### Example 2.

Consider the three-agent system

 ˙x1 =Lx1(x2−x1) ˙x2 =Lx2sign(x1+x3−2x2) (45) ˙x3 =Lx3sign(x2−x3),

defined on a line graph with. Notice that this system meets condition (i) in Theorem 4. A phase portrait and trajectory of this system are depicted in Fig. 1. There, we can see that finite-time consensus is achieved.

Next, modify the system to

 ˙x1 =Lx1(x2−x1) ˙x2 =


text:
# Root over NFS - Another Approach

## George Gousios, cs98011@icsd.aegean.gr

v1.0, 2001-09-12

This HOWTO does not intend to replace the existing Root over NFS Howto's. It is just another approach, particularly useful in large system installations. It is the result of many days of trying to setup a system for the University of the Aegean computer labs. The installation method described here is up and running. The HOWTO is dedicated to all of those guys who programmed these exceptionally good OS and tools. Also dedicated to all people that encouraged me to write it.

## 1.Introduction

This document does not resemble a common HOWTO, meaning referencing to general principles, but it is rather an on-hand approach to a by nature complex matter. It borrows the structure of the current Root over NFS , but differs from it in the following points:

• It provides a working solution fom the distribution used. The distribution specific points should be applicable to all major distributions (RedHat,SuSE,Debian).
• It uses more up to date tools, ex NFS v3.0, kernel 2.4.0, dhcp instead of bootparamd.
• All steps are described in detail, letting the reader to adapt them to his own system. No scripts!

This HOWTO expects that you have a general knowledge of what you are up to, so first read the Diskless Nodes HOW-TO.

## 1.1 The setting

It is a common case a University computer lab to have a lot PC's running Windows 98 or/and NT and a powerful UNIX server to satisfy the need of an alternative operating environment. This UNIX server is most of times idle or meerly accessed by telnet and running stupid tasks. On the other hand, students, especially those attending a computer science department, feel like taking full advantage of it, just for fun or for "educational purposes" (breaking in, hacking it...). The restrictive environment of telnet does not allow us to enjoy the use of a power server.There are 2 alternatives to that:

• Try to persuade the department' s headmaster to approve of the purchase of a bunch of new Unix workstations.
• Try to persuade the same guy to approve of transforming the server to a diskless node server.

The network at the computer lab consists of the following.

• UNIX server: SUN Enterprise 3500 with 2 64 bit SPARC@366 Mhz processors and 512 MB of memory. A real monster, isn't it?
• "Dumb" target workstations: 60-70 PC's with variable configurations, ranging from PII 266 to PIII 450 with 64-128 MB RAM.

The task I had to accomplish was the following: Provide a complete working solution without new expenses and without modifying anything but the necessary on the server.

## 1.2 The alternatives

Being the responsible for the project, I had to choose between a variety of solutions about it. I choose the following, for the reasons illustrated:

• The new 2.4 kernel: It provides a robust and fast solution, using less memory than the old 2.2 series. If it is important for your users to attach devices to their PC's then it is the only solution. Also provides NFS v3, and more efficient memory management.
• The KDE 2.1.1 desktop environment: VERY stable, easy to use, Internet enabled, makes the transition from Windows to Linux desktop almost effortless. GNOME + Afterstep is another option, but not as mature as a solution as KDE.
• SuSE 7.0 distribution: My favorite one, IMHO the most balanced between ease of use and understanding of a Linux system structure.

## 1.3 General Principles

To be able to boot a Linux system, you have to provide it with the following:

• The  /sbin directory. There exists the  init programm, which is responsible for starting other programms and start up scripts during the boot process. Also, the  /sbin directory contains the startup scripts in the case of SuSE, some useful programms like the portmap programm and many other programms that are needed before you mount the /usr directory.
• The /lib directory. It contains the libc libraries that are absolutely necessary if your init is dynamically linked.
• The /bin directory. It contains file commands and shells for running startup scripts.
• The  /etc directory. It contains configuration files for most programms and the rc.d directories that is the default for startup scripts.
• The /var directory. It is a spool area for programms that want to write somewhere. It is divided into many subdirectories with alternate usability.
• The /dev directory. It contains character and block special devices that allow programms to communicate with the computers devices via the kernel.
You should notice that after a clean install, the total size of these directories is not that big, ranging from 30 to 40 MB. The main load of files exists in the /usr and /opt directories. So, it is possible to create a directory for every diskless client containing the above listed directories and mount points for directories like /usr that will be exported by the server. The boot process, as assumed by this document, is the following:
1. The user reboots the computer, and using a diskette boots the Linux kernel.
2. The kernel takes control of the system, identifies the system devices, and uses BOOTP to obtain the IP address matching the NIC 's hardware address.
3. The init programm is started. Before switching to a run level, it calls a script described in the /etc/inittab file. This script is responsible for building the library cache, initialise and mount a swap file, load some system specific kernel modules and set the hostname.
4. The boot script finishes and the init programm switches to the specified runlevel. It starts to execute the scripts located into the /etc/rc.d/rcX directory where 'X' is the name of the runlevel. These scripts are responsible for starting the portmapper and mounting the NFS exported /usr, /home  and /opt directories.
5. The user is able to login.
To sum up, the system administrator has to do the following tasks:
• Prepare a clean install of the system to be exported to the diskless hosts.
• Create the host specific directories
• Control what is going to be started during the diskless clients' boot proces
• Prepare the server to export some directories and start a bootp service.

## 2.Setting up the server

The first, and less tricky, thing to do is to setup the server. The server must be prepared to run these services:

• NFS, preferably version 3, for exporting the following directories: /usr, /lib/modules, /opt (at least at SuSE) and /home (unless you have a dedicated file server).
• DHCP server (in bootp mode), for matching the clients' MAC addresses to IP addresses.
Also, the administrator has to create directories for each client, containing nessesary startup files and programs. The directory scheme created for the installation described was like this one:

 /usr/local/linux- |-/base- | |-/bin | |-/sbin | |-/etc | |-/workstations- | | | |-195.251.160.100 | | |-/bin | | |-/sbin | | |-/etc | | | |-195.251.160.101 | |-195.251.160.102 | |-base(symbolic link to ../base)

The /base directory contains the whole file system you want to export to your clients. The per IP directories contain files that are needed before mounting the /usr or /lib/modules directories, like the /etc folder. This is a confortable directory structure for 2 purposes: i) You can easily create a basic system at the base directory and copy the per workstation files at the workstation directories easily, with an entry level bash script ii) You can easily add or delete or update workstations by modifying the directories under /workstations. A script for copying the appropriate files (which will be discused later) can be found in Appendix A.

## 2.1 Setting up the NFS server

An NFS server can be set up in two ways:

• Using the /etc/exports file at BSD-compliant Unices like Linux of FreeBSD.
• Using the /etc/dfs/dfstab at SysV Unices like Solaris.
/etc/exports: The /etc/exports file controls the directories to be exported and the export options per workstation. It has a structure like the following (Linux):

/path/to/dir1   ws1(options) ws2(options)....
/path/to/dir2   ws3(options) ws1(options)....

Options include ro or rw, root_squash, wsize, tcp, version.

Have a look at the nfs or the exports man page and the NFS Howto for a more detailed description of what these options mean.

/etc/dfs/dfstab:A typical dfstab file on Solaris should look like the following:

share -F nfs -o rw=193.250.160@,ro=193.250.161@ /export/home
share -F nfs -o ro=193.250.160@,root=193.250.161.132 /export/engineering

Of course, these options are discused in detail at the dfstab man page.

The directories we want to export are /usr/local/linux/base/usr, /usr/local/linux/base/opt, /usr/local/linux/base/lib/modules and /home, assuming that you 've followed the suggested structure.

Optimising NFS

Of course, this is none of our business but here are some general principles:

• Reduce the TCP window size (parameter wsize for Linux) to whatever is closest to the MTU of your network type. For Ethernet, a good value of wsize is 2048 bytes as long as the MTU is 1536 bytes. This is generally a good idea because the main traffic load between the clients and the server consists of little packets and only in the case of starting large programms like X or StarOffice there is a big number of fragmented packets. Of course this may vary in your case, according to the needs of your users.
• If you plan to have a large installation, break the space for your workstations into 2 or more SCSI disks. This will allow consequent writes and reads on both disks, increasing responce and reducing latency before a request completes
• Always use NFS v3 over TCP. The main reason for migrating from v2 to v3 is the writeback case it offers on both the workstation and the server. Also, mounting NFS over TCP lets you use the first recomentation. \end{itemize} For further optimising use a packet analyzer like Ethereal or tcpdump and dicide your needs.By the way, Sun has written an excellent guide to optimizing NFS performance which, although emphasised on Solaris, is applicable to every modern Unix and is accessible online at http://docs.sun.com.

## 2.2 Setting up the DHCP/BOOTP server

Although there are many DHCP or BOOTP servers 'out there', some of which are proprietary, the best option is to use the reference IETF DHCP server. It is the least vulnerable and the most extensible DHCP available. The main server configuration is done through the /etc/dhcpd.conf file. This file is divided into two sections, the general server configuration and the host specific configuration. A typical dhcpd.conf file looks like this, in case that the DHCP/BOOTP server is used in BOOTP mode:

 subnet 193.250.160.0 netmask 255.255.255.0 { range 193.250.160.10 193.250.160.12; } host george{ hardware ethernet 00:60:08:2C:22:20; fixed-address 193.250.160.10; } host earth{ hardware ethernet 00:A0:24:A5:FD:E0; fixed-address 193.250.160.12; }

This structure is fairly easy to be understood by everyone. For every diskless client we have to supply the programm with a 'host' declaration providing a pair of hardware and IP adresses. The host name provided in the 'host' statement can be everything, but there is a conversion to use the real host name of the client having the specific IP. The range statement in the subnet declaration is not necessary to be the range that you want your clients to have. In fact, if these clients are normal workstations with an operating system that during its boot uses DHCP to obtain an IP address it is not recommended to have the same IP for their operation as diskless clients. If you have specific needs, have a look at dhcpd.conf man page.

Another difficulty is how to obtain the IP - MAC address pairs for a large network. The solution is a nice little programm called arpwatch. This programm runs at the background and keeps track of the IP - MAC address pairs of the computers that your computer has contacted in a file that you have specified. The only thing you have to do is to ping the computers you want. At Appendix B there is a script that starts arpwatch, pings a range of subsequent IP's and creates the dhcpd.conf file. If you want to do it manually, start arpwatch when your network is at its peak of usage and wait for some time. On a shared medium network (Ethernet, Tokenring) arpwatch will track down all different IP 's and hardware addresses.

## 2.3 Preparing the base system

To prepare the base system just install your favorite distribution to a mountable partition on a hard disk with a Unix like operating system already installed. Install all the programms you want to be available to your users. Then you have to transfer the whole partition preserving the links and the character or block devices. This is best done using the tar programm. Boot the previously installed system and execute the following command, assuming that you have mounted the new partition at /mnt:

 tar cpvf system.tar /mnt/.

This command will create a tar archive at the current directory with the whole system to be served to the diskless clients. Then just copy the tar archive to the server using a CDROM or through the network and extract it at the base directory. The command to do this is:

 tar xvf system.tar /usr/local/linux/base

## 3.1 Errata

In order to setup the clients, we have to work on the base system. First, we will make some modifications to the startup scripts by hand and second we will boot a workstation with the base system to make sure it works and to polish some details. Note that this part is very distribution specific and perhaps some of those described here are not applicable to your case. I can only guarantee that this works for SuSE 7.0. Please, feel free to send me distribution specific copies of this page!

## 3.2 Fiddling with scripts and files!

After init is started, it executes a script described in /etc/inittab. This script has a very spesific job to do: Bring the system in a state that other programms can be started. In most distributions I can think of this script does the following:

1. Mounts the /proc, /dev/pts and swap filesystems.
2. Activates raid arrays and fscks the root filesystem.
5. Executes user defined client scripts.
6. Set some kernel parameters.
On most distributions I have checked this script is very well commented and it is possible for an experienced user to remove some lines that are not wanted or not applicable during a network boot. I 've also noticed that all programms started do not require the /usr directory to be mounted. If you are trying to netboot a host you must do the following modifications to this script:

• Remove all entries that do fsck or initialise raid arrays, and add to the top of the script this command : mount -o remount,rw / because the client has to have rw access to the root directory when it boots.
• Do not let the kernel deamon start until all partitions are mounted
• Mount a swap partition. This is described later.
• Start the portmapper. If your system has a specific directory for starting bootup scripts, place the portmapper startup script there giving it the highest priority possible, for example: ln -s /etc/rc.d/portmap /etc/rc.d/boot/S01portmap if you are using SuSE.
• Place the NFS filesystem mounting script in the system specific directory for boot scripts with priority lower than the portmapper, for example ln -s /etc/rc.d/nfs /etc/rc.d/boot/S02nfs for SuSE.
• Remove all entries that automount local partitions, and all entries that start an automounter deamon for RedHat.

### How to setup a swap partition

This is tricky business! Swapping over NFS is not allowed by the kernel and not functioning either. You cannot use swapon on files that are on an NFS mounted filesystem. We have to do some tricks to enable it:

1. Create the swap file. Its size can be variable but for a machine with 128 MB of RAM a swap size of 40-50 MB seems reasonable. The command to create the swap file is: dd if=/dev/zero of=/var/swap bs=1k count=Xk where X stands for the number of MB your swap should be. It is also a necessity to put the swap file under /var as long as it is mounted at boot.
2. Format the swap file using the mkswapfs command.
3. Initialise a loopback device using the swap file. The command is losetup /dev/loop0 /var/swap.
4. Mount the loopback device with the command mount /dev/loop0 swap.
You have to initialise a swap partition at the very beginning of the boot process. So place commands 2-4 somewhere near to the top of the startup script. The first command is very time consuming,especially in the case of a loaded network so just copy a swap file in the base system and do not delete it when you create directories for each host.

### Modifying /etc/fstab

The /etc/fstabfile contains entries for automounting file systems at boot. In our case, we have to place the following lines at the end of it:

 server_IP:/usr/local/linux/base/usr /usr nfs nfsvers=3,wsize=2048,tcp 0 0 server_IP:/usr/local/linux/base/opt /opt nfs nfsvers=3,wsize=2048,tcp 0 0 server_IP:/usr/local/linux/base/lib/modules /lib/modules nfs nfsvers=3 wsize=2048,tcp 0 0 fileserver_IP:/home /home nfs nfsvers=3,wsize=2048,tcp 0 0

Also, do not forget to comment out lines that mount local partitions. Save this file as /etc/fstab.new because it should not be activated yet, as long as we have to boot the base system first.

You must provide the system with to files to let the users perform a login. To do this just copy the files /etc/passwd and /etc/shadow from your file server to the base system. Notice that you have to do it every time you add a user to the system, or a user changes his/her password, so can best be done by creating a cron job.

## 3.3 Booting the base system

To boot the base system we have to create a boot disk first. Go to the next section and create a boot disk as recommended. Please, change the 'append' line to this one:

 append init=/sbin/init root=/dev/nfs ip=X:Y:195.251.160.254:255.255.255.0:::'off' nfsroot=Y:/usr/local/linux/base vga=0x318 (Of course, in a sigle line)
where X stands for an unused IP address in your network and Y for the IP address of the NFS server. Of course, you have to export the /usr/local/linux/base directory from the NFS server with the rw,no_root_squash options. Now boot the base system. Everything should work OK, but I don' t think that there is a possibility that you succeeded from the first boot! There are many obscure points, that you have forgotten to edit or I have forgotten to mention.

This is the standard method to boot the base system and to add programms or a new kernel to your installation. So backup the files you have edited as well as the boot disk image.

After succeeding to boot the system, you are in a complete linux enviroment. Login as root and enjoy a first ride in your newly created system! Now comes the hard time... You have to disable some services that startup automatically and remove some programms not needed by the users.

## 3.4 Configuring the system

Nearly all distributions start these services:

• inetd, the Internet superdeamon responsible for starting other deamons like telnet, ftp etc.
• syslogd, the logging deamon. Not needed on a diskless client not needed because all the modifications are done to files easyly replacable.
• httpd, the apache webserver. Not needed for obvious reasons.
• dhcpclient. Needed for automatic aquisition of an IP address. At out case, this is done by the kernel.
• lpd, the line printer deamon. This is needed only when you have a printer connected to a host. In most cases, this is not needed.
Also, according to your installation, there may be started sshd, nscd, cupsd and other network services not needed on clients. To disable these services, remove their entries from the runtime directory under /etc/rc.d/X. There is a more elegant way to do this under SuSE or RedHat, using Yast or Linuxconfig. For Yast, go to System administration ---> Change configuration file and using search locate the entries for every service you want to stop.

Then, uninstall all these services from the base system. The only service that seems reasonable to me to be left running is the NameServer caching deamon, which is able to reduce network traffic a lot.

Now, you have to edit some files:

• /etc/resolv.conf Used to provide a nameserver. Add these entries: nameserver xxx.xxx.xxx.xxx and domain xxxxx , replacing x with the correct values.
• /etc/hosts Used to match IP addresses to host names localy. Provide the basic servers' names of your network.
• /etc/nntpserver Used to provide a news server. Just append the nameserver 's hostname.
• /etc/fstab Restore the fstab.new file we have created earlier.

### Configuring the language

Perhaps, you do not leave in the US or the UK, like me, so you have to configure the language. This is simply done through the .profile file. Just add the following: export LANG="X"where X is your natural language. Then, download a console font which supports your codepage and set, with the help of Yast, the keyboard keymap. Copy .profile to /etc/skel of the file server or to all the users' home directories.

### The X window system

If you want to provide a working X enviroment for clients with different graphics hardware, you have to use the XFBDev server. If you followed the instructions on howto create a boot disk, you would now be in framebufer mode at 1024x768@16M colors, which is sufficient for use with X windows. Now, you have to configure the X server to load the framebuffer driver. SuSE provides an exellent tool for configuring X wherher it might be version 3 or 4. It is called sax for X 3.3.x and sax2 for X 4.x. To use XFBDev driver start sax with the -s XF86_FBDev option and configure the server according to your hardware. In case you do not use SuSE, most of the work must be done by hand. Create a basic /etc/X11/XF86Config file using xf86config4. Please choose entries that are as much as possible closer to your needs. Then edit the /etc/X11/XF86Config. This file is devided into sections that start with the keyword 'Section' and end with 'EndSection'. Do the following modifications:

• Section "Files": Add the path to the direcory where you 've put your fonts.
• Section "Module": Load the GLX module if you want REALLY SLOW Open GL graphics (Load glx)!
• Section "InputDevice, Driver="mouse"": Add the following lines if you want to use a wheel mouse:
 Option "Buttons" "5" Option "ZAxisMapping" "4 5"
• Section "Device": Replace everything with the following:
 BoardName "AutoDetected" Driver "fb" Identifier "Device[0]" VendorName "AutoDetected
• Section "Modes": Replace everything with the following:
 Identifier "Modes[0]" Modeline "1024x768" 71.39 1024 1040 1216 1 400 768 768 776 802
• Section "Screen": Replace everything with the following
 DefaultDepth 16 SubSection "Display" Depth 16 Modes "1024x768" EndSubSection Device "Device[0]" Identifier "Screen[0]" Monitor "Monitor[0]"
• Section "ServerLayout": Replace everything with the following:
 Identifier "Layout[all]" InputDevice "Keyboard[0]" "CoreKeyboard" InputDevice "Mouse[1]" "CorePointer" Screen "Screen[0]"
and then replace the first argument of the InputDevice directives with the identifiers which can be found earlier in the file.
I thing that it should be a working configuration for framebuffer systems. For further reference take a look at the XF86Config and the xf86cfg4 man pages. You will find a working XF86Config file at Appendix C.

### Configuring network access for KDE2

KDE is the most extensible, configurable and internet enabled window manager available, even if we count some commercial ones that are proud of it! To download KDE, ftp to ftp.kde.org and get the rpms for your distribution. There, you can also find vanilla sources and other related projects.

## 7.2 Appendix B - A script to create the dhcpd.conf file using arpwatch

 #!/bin/bash #A script that starts arpwatch, pings a range of addresses and creates an #/etc/dhcpd.conf file from the output of arpwatch. #The arp.dat2dhcpd.conf programm is described later. #Do not forget to edit the i variable and the while statement to specify #the range of the addresses you want to ping i=128; echo "Starting arpwatch";echo arpwatch while [ "$i" -lt 253 ] do addr=195.251.160.$i echo "Now pinging $addr" ping -c 5$addr >/dev/null i=$(($i+1)) done echo exit killproc arpwatch echo "Creating /etc/dhcpd.conf" cat /var/lib/arpwatch/arp.dat |arp.dat2dhcpd.conf >/etc/dhcpd.conf

### The arp.dat2dhcpd.conf script

 #!/usr/bin/perl -n ($ether,$ip,$stup1,$name) = split; if ($name eq "") { print " host host$i { hardware ethernet $ether; fixed-address$ip; } "; $i++;} else{ print " host$name { hardware ethernet $ether; fixed-address$ip; } "}

## 7.3 Appendix C - A sample XF86Config file

 #This file should let X 4.0.1 work in 1024x768@16M colors #with the fbdev driver using the linux's framebuffer Section "Files" RgbPath "/usr/X11R6/lib/X11/rgb" FontPath "/usr/X11R6/lib/X11/fonts/75dpi:unscaled" FontPath "/usr/X11R6/lib/X11/fonts/local" FontPath "/usr/X11R6/lib/X11/fonts/misc:unscaled" FontPath "/usr/X11R6/lib/X11/fonts/100dpi:unscaled" FontPath "/usr/X11R6/lib/X11/fonts/Type1" FontPath "/usr/X11R6/lib/X11/fonts/URW" FontPath "/usr/X11R6/lib/X11/fonts/Speedo" FontPath "/usr/X11R6/lib/X11/fonts/misc" FontPath "/usr/X11R6/lib/X11/fonts/75dpi" FontPath "/usr/X11R6/lib/X11/fonts/100dpi" FontPath "/usr/X11R6/lib/X11/fonts/PEX" FontPath "/usr/X11R6/lib/X11/fonts/cyrillic" FontPath "/usr/X11R6/lib/X11/fonts/latin2/misc" FontPath "/usr/X11R6/lib/X11/fonts/latin2/75dpi" FontPath "/usr/X11R6/lib/X11/fonts/latin2/100dpi" FontPath "/usr/X11R6/lib/X11/fonts/latin7/75dpi" FontPath "/usr/X11R6/lib/X11/fonts/kwintv" FontPath "/usr/X11R6/lib/X11/fonts/truetype" FontPath "/usr/X11R6/lib/X11/fonts/uni" FontPath "/usr/X11R6/lib/X11/fonts/ucs/misc" FontPath "/usr/X11R6/lib/X11/fonts/ucs/75dpi" FontPath "/usr/X11R6/lib/X11/fonts/ucs/100dpi" FontPath "/usr/X11R6/lib/X11/fonts/xtest" EndSection Section "ServerFlags" AllowMouseOpenFail EndSection Section "Module" EndSection # This section is no longer supported # See a template below # Section "XInput" # EndSection Section "Keyboard" Protocol "Standard" XkbRules "xfree86" XkbModel "microsoft" XkbLayout "us" EndSection Section "Pointer" Protocol "PS/2" Device "/dev/psaux" SampleRate 60 BaudRate 1200 Buttons 5 EndSection Section "Monitor" Identifier "Primary-Monitor" VendorName "Unknown" ModelName "Unknown" HorizSync 29-64 VertRefresh 47-90 Modeline "1400x1050" 59.93 1400 1416 1704 1816 1050 1050 1055 1097 Modeline "1280x960" 59.90 1280 1296 1552 1664 960 960 965 1003 Modeline "1600x1000" 59.90 1600 1616 1968 2080 1000 1000 1004 1044 Modeline "1024x864" 59.89 1024 1040 1216 1328 864 864 870 902 Modeline "800x600" 58.55 800 816 928 1040 600 600 608 626 Modeline "1152x864" 59.99 1152 1168 1384 1496 864 864 870 902 Modeline "1280x1024" 59.90 1280 1296 1552 1664 1024 1024 1029 1070 Modeline "640x480" 37.44 640 656 720 832 480 480 486 501 Modeline "1024x768" 59.89 1024 1040 1216 1328 768 768 774 802 Modeline "1600x1200" 59.90 1600 1616 1968 2080 1200 1200 1204 1253 EndSection Section "Device" Identifier "Primary-Card" VendorName "---AUTO DETECTED---" BoardName "---AUTO DETECTED---" EndSection Section "Screen" Driver "fbdev" Device "Primary-Card" Monitor "Primary-Monitor" DefaultColorDepth 16 SubSection "Display" Depth 32 Modes "default" EndSubSection SubSection "Display" Depth 24 Modes "default" EndSubSection SubSection "Display" Depth 16 Modes "default" Virtual 1024 768 EndSubSection SubSection "Display" Depth 8 Modes "default" EndSubSection EndSection Section "Screen" Driver "fbdev" Device "Primary-Card" Monitor "Primary-Monitor" DefaultColorDepth 16 SubSection "Display" Depth 32 Modes "default" EndSubSection SubSection "Display" Depth 24 Modes "default" EndSubSection SubSection "Display" Depth 16 Modes "default" Virtual 1024 768 EndSubSection SubSection "Display" Depth 8 Modes "default" EndSubSection EndSection


text:
# Can you append digits to a variable in c++ or python

• C/++/#
OrbitalPower
Say you have a variable x, that is of type int.

x = 1;

I just want to create a while or for loop so that each time it runs it adds another digit to it, such as:

12
123
1234

or 1, 11, 111, 1111, and so on.

OrbitalPower
I know you can do this:

char y;
y = '1';

char test[2];

test[0] = y;
test[1] = y;
test[2] = '\0';

But I wouldn't know how to get it back to int in C++

Code:
int num=1;
for (int i=2; i<len; ++i)
{
num*=10;
num+=i;
}

Obviously though, you'd have to watch out for integer overflow.

You could also go with the string method, and if you don't need to actually perform operations on the number then it would work fine and you wouldn't have to worry about overflow. I'd use std::string instead of a char array, though.

Also, since you might need to convert a string to an integer or visa versa at some point:
Code:
#include <sstream>

int str_to_int(std::string str)
{
int num=0;
std::istringstream is(str);
is>>num;
return num;
}

std::string int_to_str(int num)
{
std::ostringstream os;
os<<num;
return os.str();
}

These can be easily modified for use with other numeric types as well (for all the types std::*stringstream objects can handle anyways) You could even write a generic template function if you felt like it.

Strilanc
Say you have a variable x, that is of type int.

x = 1;

I just want to create a while or for loop so that each time it runs it adds another digit to it, such as:

12
123
1234

or 1, 11, 111, 1111, and so on.

To append at the low end, multiply by 10 and add [digit].
To append at the high end, add [digit] * 10 ^ floor(log_10(n)).

You can compute the log relatively quickly using shifts and a constant for 1/log_2(10).

OrbitalPower
Yes, thank you both JaWiB and Strilanc. I guess to do the 1111+ I'd so something like:

Code:
	int x;
x = 1;

int num=1;
for (int i=2; i<8; i++)
{
num = num * 10;
num = num + 1;

}

I guess I was so busy thinking about trying to convert between types I didn't think of the simple arithmetic solution.

I also like the append at the high end method.

Dale
Mentor
2021 Award
I don't know how big you want your numbers, but if you want them really big you will need to download an arbitrary precision arithmetic library. There are a bunch available for C++, but I don't know about Python.

In Python, the above would become:
Code:
maxappends = 8   # actually its max appends -2
num = 1
for i in range(2,maxappends):
num = num*10 + 1
You get the added benefit of arbitrary length integers: maxappends could be 1001 or more.

To add digits to the LHS (only makes sense if the digits are different, and if you are not prepending a zero):
Code:
maxappends = 20; num = 0
for i in range(maxappends):
num = num + ((i % 9) + 1) * 10**(i)
print i,num

OrbitalPower
In Python, the above would become:
Code:
maxappends = 8   # actually its max appends -2
num = 1
for i in range(2,maxappends):
num = num*10 + 1
You get the added benefit of arbitrary length integers: maxappends could be 1001 or more.

This is built in natively? What I was doing was constructing a program that return the remainder in a number composed entirely of ones with 50 or so digits. The idea was to determine the remainder by using the divisor only a few times into the dividend, as it was also composed entirely of ones. Obviously, to do that in C++ I'd perhaps need one of those libraries needed above. So, basically I just tested my method with smaller numbers.

*Also fixed an error I made, should be:

char test[3];

test[0] = y;
test[1] = y;
test[2] = '\0';

and not char test[2].


text:
# Divided square

• January 18th 2011, 10:41 PM
qwertyuiop
Divided square
Hey guys i found a problem in the MENSA book, i've tried to use basic algebra to this this problem but i think i'm on the wrong track.
Can anyone suggest any techniques to do this problem?

The answers are at the back of the book although i do want to know how to do it.

THE IMAGE IS ATTACHED.

This field measures 177m x 176m. It has been split up into 11 squares that exactly equal the total area. The squares are only roughly drawn to scale. All new squares are in whole yards. Can you calculate the size of each square ?

- much appreciated. (Wink)
• January 19th 2011, 02:43 AM
Ackbeet
It's a question of setting up the right system of equations. At least, that's how I solved it. Let each variable name be the length of a side of the corresponding square. Then the equations I have are the following:

$A+B=177$

$A+D=176$

$C+F=B$

$E+G=D$

$I+K=F$

$J+H=G$

$B+F+K=176$

$K+J+G+D=177$

$H+I=J$

$D+E+F=177$

$A+C+F=177$

$D=J+H+E$

$B+C+E+G=176$

At this point, admitting laziness, I turned the solution over to Mathematica, which spit out the result:

Spoiler:

A=99, B=78, C=21, D=77, E=43, F=57, G=34, H=9, I=16, J=25, K=41.

In principle, though, you would probably employ Gaussian elimination with back substitution to solve the system. You have a decently sparse system there, so it might not be all that messy, actually.

Cheers.
• January 20th 2011, 07:27 AM
qwertyuiop
Thank you.
• January 20th 2011, 07:49 AM
Ackbeet
You're very welcome. Have a good one!
• January 20th 2011, 12:12 PM
wonderboy1953
To mention, Beiler covers this type of problem in his book on number theory recreations.


text:
# Properties

 Label 44T4 Order $$44$$ n $$44$$ Cyclic No Abelian No Solvable Yes Primitive No $p$-group No Group: $D_{22}$

## Group action invariants

 Degree $n$ : $44$ Transitive number $t$ : $4$ Group : $D_{22}$ Parity: $1$ Primitive: No Nilpotency class: $-1$ (not nilpotent) Generators: (1,25)(2,26)(3,28)(4,27)(5,24)(6,23)(7,22)(8,21)(9,19)(10,20)(11,17)(12,18)(13,15)(14,16)(29,42)(30,41)(31,43)(32,44)(33,39)(34,40)(35,38)(36,37), (1,6)(2,5)(3,7)(4,8)(9,42)(10,41)(11,44)(12,43)(13,40)(14,39)(15,37)(16,38)(17,35)(18,36)(19,33)(20,34)(21,31)(22,32)(23,29)(24,30)(25,27)(26,28) $|\Aut(F/K)|$: $44$

## Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_2^2$
22:  $D_{11}$

Resolvents shown for degrees $\leq 29$

## Subfields

Degree 2: $C_2$ x 3

Degree 4: $C_2^2$

Degree 11: $D_{11}$

Degree 22: 22T2, $D_{22}$ x 2

## Low degree siblings

22T3 x 2

Siblings are shown with degree $\leq 29$

Data on whether or not a number field with this Galois group has arithmetically equivalent fields has not been computed.

## Conjugacy Classes

 Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22) (23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)(37,38)(39,40)(41,42)(43,44)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $11$ $2$ $( 1, 3)( 2, 4)( 5,42)( 6,41)( 7,43)( 8,44)( 9,40)(10,39)(11,37)(12,38)(13,35) (14,36)(15,33)(16,34)(17,31)(18,32)(19,30)(20,29)(21,28)(22,27)(23,26)(24,25)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $11$ $2$ $( 1, 4)( 2, 3)( 5,41)( 6,42)( 7,44)( 8,43)( 9,39)(10,40)(11,38)(12,37)(13,36) (14,35)(15,34)(16,33)(17,32)(18,31)(19,29)(20,30)(21,27)(22,28)(23,25)(24,26)$ $22, 22$ $2$ $22$ $( 1, 7,12,16,20,23,28,31,35,40,42, 2, 8,11,15,19,24,27,32,36,39,41) ( 3, 6,10,14,18,22,25,30,33,37,44, 4, 5, 9,13,17,21,26,29,34,38,43)$ $11, 11, 11, 11$ $2$ $11$ $( 1, 8,12,15,20,24,28,32,35,39,42)( 2, 7,11,16,19,23,27,31,36,40,41) ( 3, 5,10,13,18,21,25,29,33,38,44)( 4, 6, 9,14,17,22,26,30,34,37,43)$ $22, 22$ $2$ $22$ $( 1,11,20,27,35,41, 8,16,24,31,39, 2,12,19,28,36,42, 7,15,23,32,40) ( 3, 9,18,26,33,43, 5,14,21,30,38, 4,10,17,25,34,44, 6,13,22,29,37)$ $11, 11, 11, 11$ $2$ $11$ $( 1,12,20,28,35,42, 8,15,24,32,39)( 2,11,19,27,36,41, 7,16,23,31,40) ( 3,10,18,25,33,44, 5,13,21,29,38)( 4, 9,17,26,34,43, 6,14,22,30,37)$ $11, 11, 11, 11$ $2$ $11$ $( 1,15,28,39, 8,20,32,42,12,24,35)( 2,16,27,40, 7,19,31,41,11,23,36) ( 3,13,25,38, 5,18,29,44,10,21,33)( 4,14,26,37, 6,17,30,43, 9,22,34)$ $22, 22$ $2$ $22$ $( 1,16,28,40, 8,19,32,41,12,23,35, 2,15,27,39, 7,20,31,42,11,24,36) ( 3,14,25,37, 5,17,29,43,10,22,33, 4,13,26,38, 6,18,30,44, 9,21,34)$ $22, 22$ $2$ $22$ $( 1,19,35, 7,24,40,12,27,42,16,32, 2,20,36, 8,23,39,11,28,41,15,31) ( 3,17,33, 6,21,37,10,26,44,14,29, 4,18,34, 5,22,38, 9,25,43,13,30)$ $11, 11, 11, 11$ $2$ $11$ $( 1,20,35, 8,24,39,12,28,42,15,32)( 2,19,36, 7,23,40,11,27,41,16,31) ( 3,18,33, 5,21,38,10,25,44,13,29)( 4,17,34, 6,22,37, 9,26,43,14,30)$ $22, 22$ $2$ $22$ $( 1,23,42,19,39,16,35,11,32, 7,28, 2,24,41,20,40,15,36,12,31, 8,27) ( 3,22,44,17,38,14,33, 9,29, 6,25, 4,21,43,18,37,13,34,10,30, 5,26)$ $11, 11, 11, 11$ $2$ $11$ $( 1,24,42,20,39,15,35,12,32, 8,28)( 2,23,41,19,40,16,36,11,31, 7,27) ( 3,21,44,18,38,13,33,10,29, 5,25)( 4,22,43,17,37,14,34, 9,30, 6,26)$

## Group invariants

 Order: $44=2^{2} \cdot 11$ Cyclic: No Abelian: No Solvable: Yes GAP id: [44, 3]
 Character table:  2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 11 1 1 . . 1 1 1 1 1 1 1 1 1 1 1a 2a 2b 2c 22a 11a 22b 11b 11c 22c 22d 11d 22e 11e 2P 1a 1a 1a 1a 11b 11b 11d 11d 11e 11e 11c 11c 11a 11a 3P 1a 2a 2b 2c 22c 11c 22e 11e 11b 22b 22a 11a 22d 11d 5P 1a 2a 2b 2c 22e 11e 22a 11a 11d 22d 22b 11b 22c 11c 7P 1a 2a 2b 2c 22d 11d 22c 11c 11a 22a 22e 11e 22b 11b 11P 1a 2a 2b 2c 2a 1a 2a 1a 1a 2a 2a 1a 2a 1a 13P 1a 2a 2b 2c 22b 11b 22d 11d 11e 22e 22c 11c 22a 11a 17P 1a 2a 2b 2c 22e 11e 22a 11a 11d 22d 22b 11b 22c 11c 19P 1a 2a 2b 2c 22c 11c 22e 11e 11b 22b 22a 11a 22d 11d X.1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X.2 1 -1 -1 1 -1 1 -1 1 1 -1 -1 1 -1 1 X.3 1 -1 1 -1 -1 1 -1 1 1 -1 -1 1 -1 1 X.4 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 X.5 2 2 . . A A D D E E B B C C X.6 2 2 . . B B E E A A C C D D X.7 2 2 . . C C A A B B D D E E X.8 2 2 . . D D B B C C E E A A X.9 2 2 . . E E C C D D A A B B X.10 2 -2 . . -A A -D D E -E -B B -C C X.11 2 -2 . . -B B -E E A -A -C C -D D X.12 2 -2 . . -C C -A A B -B -D D -E E X.13 2 -2 . . -D D -B B C -C -E E -A A X.14 2 -2 . . -E E -C C D -D -A A -B B A = E(11)^4+E(11)^7 B = E(11)^5+E(11)^6 C = E(11)^2+E(11)^9 D = E(11)^3+E(11)^8 E = E(11)+E(11)^10


text:
01/09/20

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