Python Software Foundation | GSoC'11 Students

Benedict Stein: Sterkspruit

Benedict Stein (noreply@blogger.com) — Fri, 10 Feb 2012 21:58:11 +0000

Wer meine Facebook Updates verfolgt hat, der weis vermutlich schon was bei mir so in den letzen Wochen los war. Aber gut fangen wir von vorne an. Hilltop expandiert, konkret spreche ich von einem IT Techniker Trainingsseminar in Sterkspruit.

Größere Kartenansicht
Jonathan mein Mitbewohner und Nozuko - die eigentlich Workshopleiterin bei Hilltop sind nun beide seit ca. 1 Woche in Sterkspruit.

Vor zwei Wochen haben wir vor Ort

an der ersten Graduation, also Abschlussfeier teilgenommem, welche im Vergleich zu den andere einfach genial war. Von vorne bis hinten als PR Aktion hat der Abschluss unsere IT Studys und der lokalen IT Einführung viele Kontakte eröffnet.

Die ganze Stadt feierte Quasi mit, die Polizeiverwaltung stellte ihren großen Sahl für die Feier, die traditional Leaders einen Teil als Küche, das Department of Rural Dev ... Mal ganz abgesehen von den Zahlreichen Spenden die u.a. während der Feier

persönlich abgeholt wurden. Diese kamen für die Kosten die z.B. für den Transport der auftretenden Kids aus der Schule usw. entstanden.

Übernachtet haben wir dann anschließend getrennt im Walaza Village bei verschiedenen bekannten des lokalen Organizators Wesley.

Selbstverständlich wurde auch weitere Pläne geschmiedet, neben den bis zu diesem Zeitpunkt bekannten 2 weiteren interessierten Schulen und dem seit dieser Woche laufenden IT Technical training, erreichten uns eine Woche später noch weitere Interessierte SGBs (SchoolGovernmentBoard) und Principals, sowie Traditional Leader die an ABIT (Abendkurse für Erwachsene mit Regierungsförderung) interessiert sind.

Ach ja nicht zu vergessen - unser Taxi zurück ins Village "Walaza" - normal wären max. 13 Personen - wie viel zählt ihr ? - + 5 die noch nicht im Bild sind !

Benedict Stein: Using local

Benedict Stein (noreply@blogger.com) — Wed, 08 Feb 2012 19:05:24 +0000

Butchery

Benedict Stein: Xhosa in Debian

Benedict Stein (noreply@blogger.com) — Tue, 31 Jan 2012 12:55:13 +0000

Today i had to add a missing US locale which brought up the idea of trying Xhosa as well.

Wojciech Wojtyniak: 137

Wojtek (noreply@blogger.com) — Tue, 31 Jan 2012 00:40:44 +0000

It was Richard Feynman, in fact, who suggested that all physicists should put up a sign in their offices or homes to remind them of how much we don't know. The sign would say simply 137. One hundred and thirty-seven is the inverse of something called the fine-structure constant. This number is related to the probability that an electron will emit or absorb a photon. The fine-structure constant also answers to the name alpha, and it can be arrived at by taking the square of the charge of the electron divided by the speed of light times Planck's constant. What all that verbiage means is that this one number, 137, contains the crux of electromagnetism (the electron), relativity (the velocity of light), and quantum theory (Planck's constant). It would be less unsettling if the relationship between all these important concepts turned out to be one or three or maybe a multiple of pi. But 137?

I tell my undergraduate students that if they are ever in trouble in a major city anywhere in the world they should write "137" on a sign and hold it up at a busy street corner. Eventually a physicist will see that they're distressed and come to their assistance. (No one to my knowledge has ever tried this, but it should work.)

Leon Lederman, "The god particle: if the universe is the answer, what is the question?"

Richard Feymann

Well... To be honest I'm not a physicist actually. In fact neither this blog is about physics. I just like the idea of dimensionless constant which changes in time and describes correlation between three basic parts of nature. Moreover, it may find me help one day. (I'm sure that I'll use a sign with "137" on it.) I like it so much that I've even bought a domain 0x89.net once (which is 137 in hexadecimal) and I simply can't restrain myself from using it one more time.

So here we are, 0x89.net.
Welcome!

Benedict Stein: Renaming Network Devices in Linux

Benedict Stein (noreply@blogger.com) — Mon, 30 Jan 2012 12:37:35 +0000

Reinstalling / Cloning the X2GoServer for Lumanyano Primary School i had to rename the network devices to fit the available configuration.

Our Preinstalled Network Settings offer the PXE Boot Server on eth0 - which should be the onboard RJ45 Plug. In addition eth1 is configured for static internet usage, eth2 as a DHCP client for internet and WLAN for static Internet.

Inserting an Intel NetBios Network Card this one got eth0 - but the change is simple - see screenshot on my private PC below - just change the name eth0 → eth1 ...

Vlad Niculae: Nash-Williams theorem on the Hamiltonian property of some regular graphs

vene — Sun, 29 Jan 2012 20:39:07 +0000

I have been digging on the internet for the proof of this theorem for the last couple of days without success. The result was published by Sir Crispin Nash-Williams as Valency Sequences which force graphs to have Hamiltonian Circuits. Interim Rep, University of Waterloo Res Rep., Waterloo, Ontario, 1969. However, this old paper is unavailable online but I have a proof in some lecture notes from my class, that I want to share here.

Theorem. Let $G=(V, E)$ be an $n$-regular graph with $|V| = 2n + 1$. Then, $G$ is Hamiltonian.

Proof. We first remark that $n$ must be even, since $$\sum_{x \in V} d(x) = n(2n + 1) = 2|E|$$ We might try to apply Dirac’s theorem, which would give us a Hamiltonian cycle if $ \forall x \in V, d(x) \geq \frac{|V|}{2}$. But in the current case, $\forall x \in V, d(x) = n \frac{2n+1}{2}$.

So we force Dirac by adding an extra vertex $w$ and connecting it to all $ x \in V $. In this new graph $G'$, $d(x) = n + 1 \forall x \in V$ and $d(w) = 2n + 1$. Therefore we have a Hamiltonian cycle that passes through $w$ and in which, $w$ is adjacent to two vertices $x$ and $y \in V$. Therefore this cycle induces a Hamiltonian path in $G$: $$P = [x = v_0, v_1, ..., v_{2n-1}, v_{2n}=y] $$

Suppose that $G$ is not Hamiltonian. It follows that if $ v_0v_i \in E $, then $ v_{i-1}v_{2n} \notin E$ and also that if $ v_0v_i \notin E $, then $ v_{i-1}v_{2n} \in E$.

We have two cases. If $v_0$ is adjacent to $v_1, ..., v_n$ then it follows that $v_{2n}$ is adjacent to $v_n, v_{n+1}, ..., v_{2n-1}$, since it cannot be adjacent to any $v_i, i n$ without creating a Hamiltonian cycle. But in this case, in the graph induced by the first half $G[\{v_0, v_1, ... v_n\}]$, $v_n$ cannot be adjacent to all the others, since in $G$ it has degree $n$ and it already has $2$ outgoing edges. So there is at least one vertex $v_i, i n$ that isn't adjacent to it, which means $v_i$ is adjacent to some $v_j, j > n$, thus forming a Hamiltonian cycle.

In the second case, we have a vertex $v_i, 2 \leq i \leq 2n – 1$ such that $v_0v_i \notin E$ and $v_0v_{i+1} \in E$. This also means that $v_{i-1}v_{2n} \in E$.

We therefore have a cycle of length $2n$ in $G$ that excludes $v_i$. Let’s rename this cycle $C=[y_1, y_2, ..., y_{2n}, y_1]$ and $v_i=y_0$.

$y_0$ cannot be adjacent to two consecutive vertices $y_i$ and $y_{i+1}$ because this will give a Hamiltonian cycle. But we know that $deg(y_0) = n$. It follows that it’s adjacent to all of the even or odd numbered vertices. We assume the latter, without loss of generality. Let $2k$ be some even index. Notice that we have $\{y_0y_{2k-1}, y_0y_{2k+1}\} \subset E$ and we can follow the cycle $C$ from $y_{2k+1}$ all the way back to $y_{2n-1}$ giving us a new cycle $C’ = [y_1, y_2, ..., y_{2n-1}, y_0, y_{2k+1}, ..., y_{2n}, y_1]$ also of length $2n$. So by repeating the same reasoning for every even vertex, by placing it in the middle and building a cycle around it, it follows that every even vertex is adjacent to all the odd vertices. But there are $n+1$ even indices, so it follows that the degree of any odd vertex is at least $n+1$, contradicting the initial conditions of the theorem. $\square$