When I type in any plaintext to an Enigma machine () the substitution does not seem to follow any pattern. In other words, there is no cycle in the use of enciphering alphabets (defined as the number of switches done when turning a plaintext letter to a ciphertext letter - which can vary from 0 to 26).   But since the key is the alignment of the rotors, there must be a finite key length even though it probably is longer than any plaintext ever would be. I imagine if you keep writing long enough, the rotors will eventually align in a way they already were. If I'm right - how long is the key?  And if the key is much longer than any plaintext, wouldn't that mean the Enimga is a one-time pad? From what I can read, the OTP is defined as an algorithm with a key longer than the plaintext. Is this wrong?
I'm trying to calculate the amount of settings the enigma machine has. I have found several sites regarding this topic, but it seems like there are two answers to my question. The first answer is 158,962,555,217,826,360,000 and the second is 107,458,687,327,250,619,360,000. Where the difference is that in the second answer, it has been taken into account that the notched rings on the enigma could be set as well, giving 26 * 26 = 676 more settings than the first answer.   But which answer is the right one, and why? How many possible different Enigma machine settings are there and how can one determine this number?
Using package nomencl QM without complex numbers Quotient of the ring of integers of a number field by a prime ideal Is it okay to have a multi-language deck? I want to buy a few booster pack from a local shop that only gets them in French. Is it generally acceptable to show up to a tournament with cards in different languages? Android "M" was officially announced on at the Google I/O conference on May 28, 2015 and revealed to be 6.0 Marshmallow on August 17, 2015.  You can review the high-level changes at the  and the announcement on the official .  Of course, not all devices will get this update right away and others won't get it at all. Each manufacturer and carrier often choose to add their custom modifications which tend to delay the official release.  When will each device get Android 6.0, either officially or unofficially, if ever?  See also: Six simple tips to get Stack Overflow reputation fast Darth Vader threw Emperor Palpatine to a pit. Well, , but unlike other times, Palpatine wasn't injured or unconscious when he was thrown.  Why didn't he simply prevent his fall using his lightsaber (I have seen such action in Star Wars: The Clone Wars TV series lots of time) and make his way out? Prove that the sum to $n$ terms of a sequence:  $$\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+\cdots $$ is  $$ \frac{n(n+1)}{2(2n+1)} $$ Notify us when the question has been edited after posting an answer Consider this situation. I post a legitimate and correct answer to a question and receive a few upvotes. Then the OP realizes that's not what they wanted to ask, and significantly rewrites the question, rendering my answer irrelevant and useless. Ideally the OP would comment on every answer saying "check my edit" but some people don't know to do that.  This leaves two possibly bad situations:   I never know, the answer stays, and the votes don't reflect the usefulness of the answer anymore. All of a sudden I get an influx of downvotes because my answer isn't helpful anymore. If I'm checking and notice my rep has dropped, I come back after the fact, edit the answer, and bring it back to the original quality. This is great, but 9 times out of 10, the downvoters never come back and notice the answer has been edited. It might get more upvotes, but it still doesn't reflect the true value of the answer.   If we can get notified when the OP gets edited, we can nip it in the bud and go back to edit our answers to match - thus maintaining the quality of the answers on SO. How do they decide Blu-ray/DVD release date? What is the difference between DX format and FX format lenses, and which to choose for what purpose? What is the difference between DX format and FX format lenses, and which to choose for what purpose?  What are the pros and cons of both? If $|f(z)|\le |g(z)|$, can we say $f$ is constant? Let $f$ and $g$ be entire functions and $g(z)\neq 0$ for all $z\in \mathbb{C}$. If $|f(z)|\le |g(z)|$, can we say $f$ is constant?  Liouville theorem says that an entire bounded function is constant, but g is not given bounded here. So I think it should be false. What else can we say about $f$? I was surprised to discover my dictionary had this entry for dilemma:     a situation in which a difficult choice has to be made between two or more alternatives, esp. equally undesirable ones   The notion of dilemma meaning two or more flies against what I was taught about the word. The very idea of a  is specifically based on the number two.  Has my dictionary merely updated its definition to encapsulate the many people who use dilemma for more than two equal choices? Or was someone in my youth being unnecessarily pedantic? find a sequence converging to zero but not the element of lp space for every 1<_p<infinity The title says it all, just wondered if accessing the iOS will count towards your consecutive visited days. Every Function in a Finite Field is a Polynomial Function Repeated Iteration of a 2x2 matrix Suppose I am given a $2$x$2$ matrix $A=$ \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix}  And an initial vector $x_n$ = \begin{pmatrix} x_0 \\ y_0\end{pmatrix}.  Under repeated iteration $x_{n+1} = Ax_n$, we find that $x_{n+1}$ converges to a line in the direction of the $eigenvector$  of $A$ of the form $X$= $[u,mu]$ where $m$ is the slope of the line.  I know that this process somehow also involves the largest $eigenvalue, \lambda,$ of $A$, though I am unsure how it connects.  My question is: How does the largest eigenvalue connect to this repeated iteration, how do I find the slope of the line of convergence $m$, and what is meant by the "direction of the eigenvector"? Determining the Orbits/Orbit Space of $O(3)$ on Real 3 by 3 Traceless Symmetric Matrices Let $O(n)$ be the set of $3 \times 3$ orthogonal matrices, and $S$ be the set of traceless real symmetric $3 \times 3$ matrices.  Now  $O(n)$ will act on $S$ by conjugation. Note that by the spectral theorem $ \forall A \in S$ in the orbit of $A$ there is a diagonal matrix $D=diag(\lambda_1, \lambda_2,\lambda_3)$ with $\lambda_1\le\lambda_2 \le\lambda_3 $.  Determine the all the orbits of the action &amp; the orbit space of the action.  What I've done:  Since all elements of $S$ are diagonlizable that the orbit of any element should be the orbit of its' diagonalized form because $\forall A \in S,   \exists P \in O(3)$ st  $PAP^t=D $ with $D$ a diagonal matrix and moreover for any other  $Q\in O(3)$ we get $QDQ^t=QPAP^tQ^t=(QP)A(QP)^t=RAR^t$.   And so would that imply the orbit space is just the set of traceless diagonal matrices as $tr(QDQ^t)=tr(DQ^tQ)=tr(Q)$ (cyclic property of trace) and $Q\in S$.  Any help is appreciated. The following code gives me an error:  \documentclass[12pt]{article} \usepackage[margin=2cm]{geometry} \usepackage [english]{babel} \usepackage{amsmath, amssymb} \usepackage{enumitem, array} \begin{document} {\begin{minipage}[t]{\textwidth} \begin{enumerate}[label=(\roman*)] {\setlength\itemindent{10pt}\item  \begin{equation*} \begin{array}{&gt;{\centering\arraybackslash}m{1.5cm}&gt;{\centering\arraybackslash}m{1.5cm}&gt;{\centering\arraybackslash}m{1.5cm}&gt;{\centering\arraybackslash}m{1.5cm}&gt;{\centering\arraybackslash}m{1.5cm}}  \hline x &amp; l_{[x]} &amp; d_{[x]} &amp; l_{x+1} &amp; \mathring{e}_{[x]} \\ \hline \end{array} \end{equation*}} \end{enumerate} \end{minipage}} \end{document}   Note: for various reasons, I have to keep minipage there. For some reason, this code gives me an error when it compiles. It gives the following output:     The errors either say there is a } missing or a $ missing. My eyes don't see such a thing! Time dependent Schrödinger equation with time independent potential and separation of variables suppose we have a potential that's independent of time $V(x,t) = V(x)$ so in Schrödinger equation we get: $$i\hbar \frac{\partial \Psi (x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi (x,t)}{\partial x^2}+V(x)\Psi(x,t)$$ since the LHS involves a variation of $\Psi$ with $x$ and the LHS involves a variation of $\Psi$ with $t$ can we say that a basis for the solutions will be with the form $\Psi(x,t)=\psi(x)T(t)$?