# How can I add && in an my if else statement

How can I add a && inside my if else statement.60 && <=.

# What is the term for this type of matrix?

$\begin{pmatrix} c & c & c & c & \cdots & c & c \\ c & a & b & b & \cdots & b & b \\ c & b & a & b & \cdots & b & b \\ c & b & b & a & & b & b \\ \vdots & \vdots & \vdots & & \ddots & & \vdots \\ c & b & b & b & & a & b \\ c & b & b & b & \cdots & b & a \\ \end{pmatrix}$

The matrix contains just 3 different items $a, b, c$:

The first row is $c$.Background: $a, b, c$ can be chosen such that the matrix is orthogonal, but has a constant first row.

# A generalization of Chebyshev polynomials

09216 \\
- & - & - & - & - & 0.09216 \\
- & - & - & - & - & - & 0.

# A question about $O(3,1)$

Recall that $O(3,1)$ is the collection of matrices $A\in M_4(\mathbb R)$ such that
$$A\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}A^T=\begin{pmatrix}1 &&&\\&1&&\\&&1&\\&&&-1\end{pmatrix}.Can we find a Q\in O(3,1) such that$$Q\psi Q^{-1}=\begin{pmatrix} 0&a&&\\-a&0&&\\&&0&b\\&&b&0\end{pmatrix}$$for some a,b\in \mathbb R? # Can homotopy pullbacks of spaces be checked on fibers? simplicial sets) is a homotopy pullback iff the induced map on each homotopy fiber is a weak equivalence.More precisely, consider the following diagram:$$\begin{array}{c}&&&&& A& \longrightarrow & B\\
&&&&&\downarrow && \downarrow \\
&&&&& C &\longrightarrow & D \\
&&&&\nearrow & &\nearrow\\
&&&1 & \longrightarrow & 1 \end{array}$$Suppose that, after having taken homotopy pullbacks on both sides, the resulting front face is a homotopy pullback (i. # Placing numbers 1,2,\ldots,n^3 in a cube so that numbers of any two adjacent unit subcube are coprime Basically, one can start with something like this (if n was 11 and i_0 was 3): \begin{array}{ccccccccccc} 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\\ \vdots & & & & & & & & & & \vdots\\ 1 & 30 & 1 & 2 & 15 & 2 & 3 & 10 & 1 & 6 & 5\\ 30 & 1 & 2 & 15 & 2 & 1 & 10 & 3 & 2 & 5 & 6\end{array} and then fix it so there aren't quite so many 30s, etc.Peter Mueller's answer gives solutions for (4,3,3), (5,3,3) and (4,4,4). # Verifying the correctness of a Sudoku solution We are thus looking for a solution of the following grid:$$\begin{array}{|ccc|ccc|ccc|ccc|ccc|}
\hline
1&&&2&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
&&&1&&&2&&&&&&&&&\\
&&&&&&&&&&&&&&\cdots&\\
&&&&&&&&\ddots&&&&&&&\\
\hline
2&&&&&&&&&1&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
\strut&&&&&&&&&&&&&&&\\
\strut&&&&\vdots&&&&&&&&&&&\\
\strut&&&&&&&&&&&&&&&\\
\hline
\end{array}$$where the upper part is a q'\times q' subgrid, q'=q/2.S is not included in the complement of a Sudoku permutation of one of the following sets: a) \{r_1,r_2\} b) \{r_1,c_1,b_1\} c) \{b_1,b_2,b_4,b_5\} d) \{r_1,r_4,b_1,b_4\} e) \{r_1,c_1,b_2,b_4,b_5\} f) \{b_2,b_3,b_4,b_6,b_7,b_8\} g) \{r_1,r_4,b_1,b_5,b_7,b_8\} h) \{r_1,c_1,b_3,b_5,b_6,b_7,b_8\} Maximal incomplete sets http://math. # Is the operadic butterfly symmetric? The operadic butterfly is a diagram in the category of operads in vector spaces.$$\begin{array}{ccccc}
& Dend & & & & Dias & \newline
\nearrow & &\searrow & &\nearrow & &\searrow\newline
Zinb & & &Ass & & & \quad Leib \newline
\searrow & &\nearrow & &\searrow & &\nearrow\newline
& Comm & & & & Lie & \newline
\end{array}$$Here is a paper by Loday in which some discussion can be found. # why do these characters belong to the first group in this JS regex match? :Dokumentation&&&KKS-Nummer&&&Beschreibung&&&Seite)(&&&([^(&&&)]+)&&&([^(&&&)]+)&&&(\d+))+ The test string: %5B"Deckblatt: Anlagendokumentation&&&Produktdaten&&&KKS-Nummer&&&Hersteller&&&Typ&&&Artikelnummer&&&MA-KF1&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF11&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF12&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF13&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF14&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF15&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF16&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF17&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF18&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF19&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF20&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF21&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF22&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF23&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF24&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF25&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF26&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF27&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF28&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF29&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF30&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF31&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF32&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF33&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF34&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF35&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF36&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF37&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF38&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF39&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF40&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&MA-KF41&&&Beckhoff&&&EK1100&&&BECK%2EEK1100&&&Dokumentation&&&KKS-Nummer&&&Beschreibung&&&Seite&&&all&&&Vorwort&&&6&&&all&&&Produktübersicht&&&7&&&all&&&Grundlagen&&&8&&&all&&&Montage und Verdrahtung&&&9&&&all&&&Inbetriebnahme%2FAnwendungshinweise&&&10&&&all&&&Fehlerbehandlung und Diagnose&&&11&&&all&&&Anhang 1&&&12&&&all&&&Anhang 2&&&13&&&all&&&Anhang 3&&&14&&&all&&&Anhang 4&&&15&&&all&&&Anhang 5&&&16&&&all&&&Anhang 6&&&17&&&all&&&Anhang 7&&&18&&&all&&&Anhang 8&&&19&&&all&&&Anhang 9&&&20&&&all&&&Anhang 10&&&21&&&all&&&Anhang 11&&&22&&&all&&&Anhang 12&&&23&&&all&&&Anhang 13&&&24&&&all&&&Anhang 14&&&25&&&all&&&Anhang 15&&&26&&&all&&&Anhang 16&&&27&&&all&&&Anhang 17&&&28&&&all&&&Anhang 18&&&29&&&all&&&Anhang 19&&&30&&&all&&&Anhang 20&&&31&&&all&&&Anhang 21&&&32&&&all&&&Anhang 22&&&33&&&all&&&Anhang 23&&&34&&&all&&&Anhang 24&&&35&&&all&&&Anhang 25&&&36&&&all&&&Anhang 26&&&37&&&all&&&Anhang 27&&&38&&&all&&&Anhang 28&&&39&&&all&&&Anhang 29&&&40&&&all&&&Anhang 30&&&41&&&all&&&Anhang 31&&&42&&&all&&&Anhang 32&&&43&&&all&&&Anhang 33&&&44&&&all&&&Anhang 34&&&45&&&all&&&Anhang 35&&&46&&&all&&&Anhang 36&&&47&&&all&&&Anhang 37&&&48&&&all&&&Anhang 38&&&49&&&all&&&Anhang 39&&&50&&&all&&&Anhang 40&&&51&&&all&&&Anhang 41&&&52&&&all&&&Anhang 42&&&53"%5D The regex I wrote should get a first group, which appears after /Artikelnummer/ and before /Dokumentation&&&/ (etc), as well as a second group, which is what I'm having trouble with: It should consist of repetitions of this pattern: (&&&([^(&&&)]+)&&&([^(&&&)]+)&&&(\d+)+ By my reckoning, that should capture the entire substring: &&&all&&&Vorwort&&&6&&&all&&&Produktübersicht&&&7&&&all&&&Grundlagen&&&8&&&all&&&Montage und Verdrahtung&&&9&&&all&&&Inbetriebnahme%2FAnwendungshinweise&&&10&&&all&&&Fehlerbehandlung und Diagnose&&&11&&&all&&&Anhang 1&&&12&&&all&&&Anhang 2&&&13&&&all&&&Anhang 3&&&14&&&all&&&Anhang 4&&&15&&&all&&&Anhang 5&&&16&&&all&&&Anhang 6&&&17&&&all&&&Anhang 7&&&18&&&all&&&Anhang 8&&&19&&&all&&&Anhang 9&&&20&&&all&&&Anhang 10&&&21&&&all&&&Anhang 11&&&22&&&all&&&Anhang 12&&&23&&&all&&&Anhang 13&&&24&&&all&&&Anhang 14&&&25&&&all&&&Anhang 15&&&26&&&all&&&Anhang 16&&&27&&&all&&&Anhang 17&&&28&&&all&&&Anhang 18&&&29&&&all&&&Anhang 19&&&30&&&all&&&Anhang 20&&&31&&&all&&&Anhang 21&&&32&&&all&&&Anhang 22&&&33&&&all&&&Anhang 23&&&34&&&all&&&Anhang 24&&&35&&&all&&&Anhang 25&&&36&&&all&&&Anhang 26&&&37&&&all&&&Anhang 27&&&38&&&all&&&Anhang 28&&&39&&&all&&&Anhang 29&&&40&&&all&&&Anhang 30&&&41&&&all&&&Anhang 31&&&42&&&all&&&Anhang 32&&&43&&&all&&&Anhang 33&&&44&&&all&&&Anhang 34&&&45&&&all&&&Anhang 35&&&46&&&all&&&Anhang 36&&&47&&&all&&&Anhang 37&&&48&&&all&&&Anhang 38&&&49&&&all&&&Anhang 39&&&50&&&all&&&Anhang 40&&&51&&&all&&&Anhang 41&&&52&&&all&&&Anhang 42&&&53 But, for some reason, the only string in group 2 is: &&&Anhang 42&&&53 Why is this happening?You get &&&all&&&Anhang 42&&&53 in Group 2 because the (pattern)+ is a repeated capturing group that stores only the value captured at the last iteration. # Simplicial “universal extensions”, the hammock localization, and Ext$$

A morphism of $n$-extensions of $Y$ by $X$ is defined to be a hammock

$$\begin{matrix} &&A_1&\to&A_2&\to&A_3&\to&\ldots&\to &A_{n-2}&\to &A_{n-1}&\to& A_{n}&&\\ &\nearrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\searrow&\\ X&&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&&Y\\ &\searrow&\downarrow&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow&&\downarrow&\nearrow&\\ &&B_1&\to&B_2&\to&B_3&\to&\ldots&\to &B_{n-2}&\to &B_{n-1}&\to& B_{n}&&\end{matrix}$$

This detetermines a category $n\operatorname{-ext}(Y,X)$.Questions:

Why can we get $\pi_1$ of the function complex by looking at maps into $N[1]$?

# Graphs and hypercubes

\end{array} $$C_3:$$ \begin{array}{ccccccc}
X_{000} & \rightarrow & \rightarrow & \rightarrow & X_{010} && \\
\downarrow & \searrow & & & \downarrow & \searrow & \\
\downarrow & & X_{100} & \rightarrow & \rightarrow &\rightarrow & X_{110} \\
\downarrow & & \downarrow & & \downarrow & & \downarrow \\
X_{001} & \rightarrow & \downarrow & \rightarrow & X_{011} && \downarrow \\
& \searrow & \downarrow & & & \searrow & \downarrow \\
& & X_{101} & \rightarrow & \rightarrow & \rightarrow & X_{111}
\end{array} $$And so on, inductively over n \in \mathbb{N}.) Example 1: I = \{ X \overset{f}{\underset{g}\rightrightarrows} Y \} can be arranged on C_2:$$ \begin{array}{ccc}
X & \overset{f}\rightarrow & Y \\
{\scriptstyle g}\downarrow && \downarrow \\
Y & \rightarrow & 0.

# formation and interpretation of a dual LP

The coffee company makes $p_{i,j}$ dollars for each kilogram it produces at farm $i$ and ships to restaurant $j$.$\begin{array}{l*{12}{r}lr} \mbox{Maximize} & p_{1,1}x_{1,1} & + & p_{1,2}x_{1,2} & + & p_{1,3}x_{1,3} & + & p_{2,1}x_{2,1} & + & p_{2,2}x_{2,2} & + & p_{2,3}x_{2,3} \\ \mbox{Subject to} & x_{1,1} & + & x_{1,2} & + & x_{1,3} & ~ & ~ & ~ & ~ & ~ & ~ &\leq & s_1 \\ ~ & ~ & ~ & ~ & ~ & ~ & ~ & x_{2,1} & + & x_{2,2} & + & x_{2,3} & \leq & s_2 \\ ~ & x_{1,1} & ~ & ~ & ~ & ~ & + & x_{2,1} & ~ & ~ & ~ & ~ & \geq & d_1 \\ ~ & ~ & ~ & x_{1,2} & ~ & ~ & ~ & ~ & + & x_{2,2} & ~ & ~ & \geq & d_2 \\ ~ & ~ & ~ & ~ & ~ & x_{1,3} & ~ & ~ & ~ & ~ & + & x_{2,3} & \geq & d_3 \\ &&&&&&&&&&&x_{i,j} & \geq & 0 \end{array}$

Here we can think of each $x_{i,j}$ as the amount (in kg) that is shipped from farm $i$ to restaurant $j$.

# Finding the Smallest Convex Hull of an Adjacency Matrix

How do I know if the convex hull is the smallest possible without using brute force?Here is an example:

We have the 9X9 adjacency matrix,

\begin{Vmatrix}
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
\end{Vmatrix}

which with dots is:
\begin{Vmatrix}
& & & & \cdot & & & & \\
& & & & & \cdot & & & \\
\cdot & & & & & & & & \\
& & & & & \cdot & & & \\
& & & & & \cdot & & & \\
& & & & & & & & &\\
\cdot & & & & & & & & \\
\cdot & & & & & & & & \\
& & & \cdot & & & & & \\
\end{Vmatrix}

then rearranged is:
\begin{Vmatrix}
& \cdot & & & & & & & \\
& & \cdot & & & & & & \\
\cdot & & & & & & & & \\
& & \cdot & & & & & & \\
& & \cdot & & & & & & \\
\cdot & & & & & & & & \\
\cdot & & & & & & & & \\
& & & \cdot & & & & & \\
& & & & & & & & \\
\end{Vmatrix}

with the convex hull of the latter clearly having the smaller area.

# How to compute probability distribution functions?

Then your probability space looks like this

\begin{align}
1&&2&&3\\
M&&M&&M\\
M&&M&&H\\
M&&H&&M\\
M&&H&&H\\
H&&M&&M\\
H&&M&&H\\
H&&H&&M\\
H&&H&&H
\end{align}

You can regroup these according to the number of hits.Let $A_i$ denote a successful shot of player $i$, $i=1,2,3$.

# Is this matrix totally unimodular?

Check whether the following matrix is totally unimodular.$$A=\pmatrix{0&0&0&1&0&0&0\\ 1&-1&0&0&1&0&1\\ 0&1&-1&-1&0&0&0\\ 0&0&0&1&-1&0&-1\\ 0&-1&0&0&0&0&0\\ 0&0&1&0&0&1&0\\ -1&0&0&0&0&-1&0}$$

Well, when you use the Laplace expansion for $A$, you can leave out the first row without changing the result.

# How do you calculate the expected value of geometric distribution without diffrentiation?

Is there any way I can calculate the expected value of geometric distribution without diffrentiation?)
$$\begin{array}{cccccccccccccccccccccccc} & 0 & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 \\ \hline & & & p^1 & + & 2p^2 & + & 3p^3 & + & 4p^4 & + & 5p^5 & + & 6p^6 & + & \cdots & {} \\[12pt] = & & & p^1 & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\ & & & & + & p^2 & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\ & & & & & & + & p^3 & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\ & & & & & & & & + & p^4 & + & p^5 & + & p^6 & + & \cdots \\ & & & & & & & & & & + & p^5 & + & p^6 & + & \cdots \\ & & & & & & & & & & & & + & p^6 & + & \cdots \\ & & & & & & & & & & & & & & + & \cdots \\ & & & & & & & & & & & & & & \vdots \end{array}$$
First sum each (horizontal) row.

# Generate a matrix where entries are non-zero and determinant is non-zero

$$A=\begin{bmatrix}2 & 1 & 1& \dots& 1 \\1& 2 & 1 &\dots& 1\\\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\dots&2\end{bmatrix}$$
i.You can use the matrix

A =
\begin{pmatrix}
t & t & t & \cdots & t & t \\
1 & 2 & 1 & \cdots & 1 & 1 \\
1 & 1 & 2 & & 1 & 1 \\
\vdots & \vdots & & \ddots & & \vdots\\
1 & 1 & 1 & \cdots & 2 & 1 \\
1 & 1 & 1 & \cdots & 1 & 2
\end{pmatrix}
\,,

which has determinant $t$.

# Suppose we have an $n \times n$ matrix such that $M^k = I_n$ (see details)

Suppose we have a matrix $M$ of the following form:

$$\pmatrix{ a&1&\cdots&&0&0\\ &a&1&\cdots&0&0\\ &&\ddots &\ddots &\vdots&\vdots\\ &&&a&1&0\\ &&&&a&1\\ &&&&&a }$$

That is, a constant multiple of the identity matirx with ones on the super diagonal.Denote $M$ as $M_a$, so we can talk of $M_0$, which is a nilpotent matrix: $M_0^n=0$.

# Can I develop a LQR if the system is NOT controllable but stable?

77& 300& 0& 0&\\ 0& 0& 1& 0& 0& 0& \\ 8.09& 0& \\ 0& 0& 0& 0& 0& 1& \\ 0& 0& 179249.

\begin{pmatrix}
2na & -a & -a & -a & -a & -a& -a\\
-a& a+b & 0 & 0 & -b & 0 & 0\\
-a& 0 & a+b & 0 & 0 & -b &0 \\
-a& 0 & 0 & a+b & 0 & 0&-b \\
-a& -b & 0 & 0 & a+b & 0 & 0\\
-a& 0&-b & 0 & 0 & a+b &0 \\
-a& 0& 0&-b & 0 & 0 & a+b
\end{pmatrix}

I represent the matrix when $n=3$.I think the characteristic polynomial of this matrix for any natural number $n$ is calculated efficiently.