When you have a Laptop with Windows 8.1 preinstalled, then you will find yourself having a hard time installing a clean copy of Windows 8 on said Laptop. That, however, might be desirable for various reasons and so I am telling you how it's done. In my case, I am doing it with the firm intention to encrypt the system partition with TrueCrypt Setup 7.1a, which requires me to have an MBR rather than a GPT. There are probably ways to change this in-place, but there's really no point because I want a clean install of Windows anyway.
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Member variables in python are horrible. They are not visible in the layout of the class which is instantiated, but instead the
__init__ function of a class creates certain member variables for the instance. I have never liked this about python, to be honest. For a recent project, I devised the following solution. Assume you would want this behaviour:
>>> class test(Base):
... # Variables
... number = 4
... string = "hodor"
... # Functions
... def stringmult(self):
... return self.number * self.string
...
>>> test().stringmult()
'hodorhodorhodorhodor'
>>> test(number=2).stringmult()
'hodorhodor'
>>> test(string="Na",number=8).stringmult() + " - Batman!"
'NaNaNaNaNaNaNaNa - Batman!'
>>>
>>> test(end="Batman!")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ctypes.ArgumentError
>>>
Base can be constructed with keyword arguments who must match exactly the correct class variables which you specify. This one does it:
from ctypes import ArgumentError
class Base(object):
def __init__(self, **kwargs):
# "given" is the list of keyword arguments passed to the constructor
# of this object, "needed" is the list of class variables which belong to
# the base class of the object which is being created, which do not end
# with two underscores and which are not a function. Trust me, we do not
# want to meddle with those.
given = list(kwargs.keys())
needed = [attr for attr in dir(self.__class__) if attr[-2:] != '__' \
and type(self.__class__.__dict__[attr])!=type(lambda:0) ]
# Check if keyword arguments have been provided which are not among the
# required arguments and throw an exception if so. Remove this check for
# a less restrictive base class. I wouldn't recommend it.
if not set(given) <= set(needed):
raise ArgumentError()
# First, initialize the attribute dictionary of the object being created
# with a list of default values, indicated by the values of the class
# variables. Then, update the attribute dictionary again with the values
# provided to this constructor.
self.__dict__.update({k: self.__class__.__dict__[k] for k in needed})
self.__dict__.update(kwargs)
It causes me unspeakable agony to see that my post about why sudoku is boring is one of the most frequented posts in this blog, mostly because most of my readers clearly disagree with the title. I recently received an email titled "why sudoku is not all that boring" by an old friend, and he taunted me that the sudoku
S = [
0, 0, 0, 0, 6, 0, 0, 8, 0,
0, 2, 0, 0, 0, 0, 0, 0, 0,
0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 7, 0, 0, 0, 0, 1, 0, 2,
5, 0, 0, 0, 3, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 4, 0, 0,
0, 0, 4, 2, 0, 1, 0, 0, 0,
3, 0, 0, 7, 0, 0, 6, 0, 0,
0, 0, 0, 0, 0, 0, 0, 5, 0 ]
real 51m3.656s
user 50m32.260s
sys 0m2.084s
Opera was updated. They messed up. Let's not talk about this any more. Let's talk about how to make Firefox your new Opera 12. To get the elephant out of the room right away: Yes, you guessed right, I did not switch to Chrome because I am scared of Google harvesting all my data and browsing habits. So that leaves only Firefox, and I think that they did good with the recent releases. We probably can not get back everything we miss from Opera 12 times, but some of it is possible, and even better in Firefox. First of all, Firefox has bookmarks and you can import them from an HTML export of your old bookmarks. For everything else, continue reading. If you are still missing a feature, feel free to post it with or without solution - in the latter case, I will try and solve it myself.
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Let $G$ be an affine, connected, reductive group and $X$ a $G$-module. Choose a maximal torus $T\subseteq G$, a Borel $B\subseteq G$ containing $T$ and let $U$ be the unipotent radical of $B$. Denote by the character group of $T$. Let $\Lambda\subseteq\mathbb{X}$ be the set of dominant weights of $G$ with respect to these choices. We can decompose the graded coordinate ring $\mathbb{C}[X]=\bigoplus_{\lambda\in\Lambda} V_{(\lambda)}$ into its isotypic components $V_{(\lambda)}$ of weight $\lambda$. Let
\[ \Lambda_X=\{ \lambda\in\Lambda \mid V_{(\lambda)}\ne\{0\}\} \]
be the set of weights that occur in $\mathbb{C}[X]$.
Let $V_{(\lambda)}\cong V_\lambda^{\oplus n_\lambda}$, where $V_\lambda$ is the irreducible module of highest weight $\lambda$. Each $V_\lambda$ has a highest weight vector which is unique up to scaling - let $f_{\lambda 1}, \ldots, f_{\lambda n_\lambda}\in V_{(\lambda)}$ be linearly independent highest weight vectors.
Claim. *For each $1\le k\le n_\lambda$, if the function $f:=f_{\lambda k}\in\mathbb{C}[X]$ is reducible, then there exist weights $\lambda_1,\ldots,\lambda_r\in\Lambda_X$ such that $\lambda$ is an $\mathbb N$-linear combination of the $\lambda_i$.*
Indeed, about a year ago this statement was completely unclear to me. However, it's actually not that hard to see and I felt like sharing my proof. (more…)
For the group $G=\mathrm{GL}_n(\mathbb{K})$ of invertible matrices over any field, there is the well-known determinant character $\det:G\to\mathbb{K}^\times$ which, well, maps any matrix to its determinant. It has the property that
\[ [G,G]=\mathrm{SL}_n(\mathbb{K})=\{ g\in G \mid \det(g)=1 \}, \]
where $[G,G]$ denotes the derived subgroup of $G$. In more geometric terms, $[G,G]$ is the vanishing of the regular function $\det-1$ on $G$. Because I find it rather cute, I will show that you can find a similar function for all linear, reductive groups.
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Ich habe letzte Woche versucht, für einige Studenten das Master Theorem zu beweisen. Das ist geringfügig daneben gegangen, deswegen habe ich eine korrigierte Version des Beweises erstellt. Ich frage mich nun aber schon, wie allgemein man das eigentlich machen sollte. Grundsätzlich könnte man sich für eine Laufzeitfunktion $T:\mathbb N\to\mathbb N$ ja Rekurrenzgleichungen der Form $T = \alpha \circ T \circ \beta$ ansehen, wobei in meinem Fall $\beta(n)=n/b$ und $\alpha(n)=a\cdot n + f(n)$ ist. Ich habe leider keine Zeit, mir dazu mehr Gedanken zu machen.
Well. In case you have not stumbled across the corresponding StackOverflow post, and if you have always wondered why vim does not work properly in cygwin, just
[rattle@ALICE:~]$ cat .vimrc
set nocompatible
set backspace=indent,eol,start
I asked a question on mathoverflow, and while researching I stumbled upon a theorem which I did not know before. I consider it quite powerful and the proof is short and sweet, I copied it from this expository paper by Bryden Cais:
Theorem. Let $X$ be a normal, separated, Noetherian scheme and $U\subseteq X$ a nonempty affine open subset. Then, $X\setminus U$ is pure of codimension one.
Proof. We need to show that for each generic point $y\in Y=X\setminus U$, the local ring $\mathcal{O}_{X,y}$ has dimension one. Let $y$ be the generic point of a component of $Y$. Denote by $\mathfrak{m}$ the maximal ideal of the local ring $A:=\mathcal{O}_{X,y}$. Let $S:=\mathrm{Spec}(A)$. The inclusion morphism $f:S\to X$ is affine because $X$ is separated (Lemma 1). Thus, $V:=f^{-1}(U)=S\setminus\{ \mathfrak{m} \}$ is an affine, open subscheme of $S$. The morphism on global sections $A=\mathcal{O}_S(S)\to\mathcal{O}_{V}(V)$, corresponding to the inclusion $V\to S$, is therefore injective. It can not be surjective because $V\ne S$. Pick some $a\in\mathcal{O}_{V}(V)\setminus A$. If we now had $\dim(A)>1$, then the prime ideals of height one $\mathfrak{p}\subseteq A$ would satisfy $\mathfrak{p}\ne\mathfrak{m}$, i.e. they would correspond to points contained in $V$. Consequently, we would have $a\in \mathcal{O}_{V,\mathfrak{p}} = A_{\mathfrak{p}}$. Because $A$ is a normal Noetherian domain, it is the intersection of all localizations at prime ideals of height one - this is Corollary 11.4 in Eisenbud's book. This yields the contradition that $a\in A$.
Lemma 1. If $f:X\to Y$ is a morphism of schemes with $Y$ separated and $X$ affine, then $f$ is an affine morphism.
Proof. This is exercise 3.3.6 in Qing Liu's book. Let $V\subseteq Y$ be an open affine subset. Proposition 3.9(f) in the same book tells you that there is a closed immersion $f^{-1}(V)\cong X\times_Y V\to X\times V$, so $f^{-1}(V)$ is isomorphic to a closed subscheme of an affine scheme, hence affine.
I have been commuting between cities quite a lot for quite a while now, and I have given the issue of travel practicability quite a lot of thought. I was urged to share these thoughts, and so I obey. Let's start with the motivation: The worst thing that can happen to me on a trip is to be missing that one essential item, say a charging cable. This happened to me a lot because I usually do not have my charging cable with me at all times. It's at home, where I charge my phone/laptop. The solution for this kind of problem is redundancy.
I present the concept of the onebag: It is the one bag which always contains everything you really need. That essential assortment of items should be in your onebag at all times. Mind you, though: Your onebag contains the items you'll need on every trip, it does not contain every item you will need on a trip. Its purpose is to never leave you pondering if you forgot something important. My onebag is broken down into three levels:
* Level 1: I have a USB stick attached to my key ring which contains all my most precious data. Together with my wallet and my phone, this is level 1. This stuff is with me at all times.
* Level 2: I use an organizer casket for all my electronic devices and cables that I usually need for work, together with my laptop this is level 2. I take these items to work every day, but I do not take them with me when I go out to buy groceries.
* Level 3: A red backpack. It contains duplicate copies of all my charging cables in this cable organizer, hygiene products ((I have been thinking about putting a towel in there, but I never really need one. Hotels and friends give you towels, and one of those is usually where I stay.)), one change of underwear and binbags. Together with levels 1 and 2, this is my onebag.
This is just an example of course. The point is to minimize thinking about "what to pack". If I need to go to another city for 3 days, I grab level 2 and stuff it into the bag, take 3 shirts and 3 changes of underwear and I am essentially done. If you did it right, about 98% of your travel preparations should be as simple as packing the clothes you need for the trip into your onebag. For some trips, you might need an additional bag for more clothes or you might need to pack some items which are very specific to the trip you are about to take: But your essential gear is already in the bag.
I decided to switch to the new design made by Roman for us, even though he protests that it is not finished. After a long sequence of emails, I managed to pester him into letting me fix the remaining technical wordpress issues with several dirty hacks myself, and there we are. This layout is responsive and will work properly on smartphones. Just two months ago, I could not have cared less about such things, but now I have a smartphone myself.