Bar Graphs: Practical Examples & Insights

When is graph G with n vertices is isomorphic to complement $\stackrel{¯<!--\; ¯\; -->}{G}$?
Question : When is graph G with n vertices is isomorphic to complement $\stackrel{¯<!--\; ¯\; -->}{G}$?
So, total no of edges that a n vertex graph can have is $(\genfrac{}{}{0ex}{}{n}{2})$. And if G is isomorphic to $\stackrel{¯<!--\; ¯\; -->}{G}$ then no of edges in G is equal to no of edges in $\stackrel{¯<!--\; ¯\; -->}{G}$. So no of edges in $(\genfrac{}{}{0ex}{}{n}{2})$ . But for this to be an integer, n = 4k or n = 4k+1 for some integer k.
My question is : Is this the only requirement? That is, is n=4k or 4k+1 the sufficient condition to construct such a graph G? Does it have anything to do with the fact that G and $\stackrel{¯<!--\; ¯\; -->}{G}$ can both be bipartite if both of them don't have any triangle in it? The question asks to think about what is the condition of G and $\stackrel{¯<!--\; ¯\; -->}{G}$ simultaneously bipartite.

Prove that G or $\stackrel{¯<!--\; ¯\; -->}{G}$ must be nonplanar if G has 11 vertices.
The exact problem is the following: "Let G be a graph with 11 vertices. Prove that G or $\stackrel{¯<!--\; ¯\; -->}{G}$ must be nonplanar.
I went to my professor to discuss the proof, but she said it didn't quite work, that something was "off" about it. The following is the proof I presented:
Let G be a graph with 11 vertices. It is clear that $|<!--\; |\; -->E(\stackrel{¯<!--\; ¯\; -->}{G})|<!--\; |\; -->$
Assume, for the sake of contradiction, that both G and $\stackrel{¯<!--\; ¯\; -->}{G}$ are nonplanar.
Then
$|<!--\; |\; -->E(G)|<!--\; |\; -->+|<!--\; |\; -->E(\stackrel{¯<!--\; ¯\; -->}{G})|<!--\; |\; -->\le <!--\; \le \; -->(3|<!--\; |\; -->V(G)|<!--\; |\; -->-<!--\; -\; -->6)+(3|<!--\; |\; -->V(\stackrel{¯<!--\; ¯\; -->}{G})|<!--\; |\; -->-<!--\; -\; -->6)=54.\Rightarrow <!--\; \Rightarrow \; -->\Leftarrow <!--\; \Leftarrow \; -->$
Therefore, either G or $\stackrel{¯<!--\; ¯\; -->}{G}$ is nonplanar.
So, where exactly does this proof fail? Any tips on the aesthetics of the proof would also be greatly appreciated.

Eigenvalues of complement of regular graphs
So I have encountered the following fact:
Let G be a d-regular graph with adjacency eigenvalues ${\mathrm{\lambda <!--\; \lambda \; -->}}_1\ge <!--\; \ge \; -->...\ge <!--\; \ge \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_n$. Then its complement graph $\stackrel{¯<!--\; ¯\; -->}{G}$ has eigenvalues $n-<!--\; -\; -->1-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_1,-<!--\; -\; -->({\mathrm{\lambda <!--\; \lambda \; -->}}_2+1),...,-<!--\; -\; -->({\mathrm{\lambda <!--\; \lambda \; -->}}_n+1)$
The proof uses the fact that $A(\stackrel{¯<!--\; ¯\; -->}{G})=J-<!--\; -\; -->A(G)-<!--\; -\; -->I$, where J is the matrix with all ones. So if vi is an eigenvector of ${\mathrm{\lambda <!--\; \lambda \; -->}}_i$, then $A(\stackrel{¯<!--\; ¯\; -->}{G})v_i=J{v}_i-<!--\; -\; -->A(G)v_i-<!--\; -\; -->v_i$. If i=1, then since G is regular, $v_1$ must be the vector of all ones, hence $J{v}_1=n{v}_1$
But for i>1, then does the equation $A(\stackrel{¯<!--\; ¯\; -->}{G})v_i=J{v}_i-<!--\; -\; -->A(G)v_i-<!--\; -\; -->v_i=J{v}_i-<!--\; -\; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_i{v}_i-<!--\; -\; -->v_i=-<!--\; -\; -->({\mathrm{\lambda <!--\; \lambda \; -->}}_i+1)v_i$ imply that $J{v}_i=0$? If it does, then I can't see why it is true.

Prove that $\stackrel{¯<!--\; ¯\; -->}{G}$ contains a triangle
Prove that If G doesn't contain a triangle then $\stackrel{¯<!--\; ¯\; -->}{G}$ contains one
G is a graph of order at least 6, and $d_G(v)\ge <!--\; \ge \; -->3$ where v is a vertex of G
Any hints ?

Find all solutions of $z^2+{\stackrel{¯<!--\; ¯\; -->}{z}}^2=0$
How do I solve this equation? Also plotting all these solutions on the graph needed.
$z^2+{\stackrel{¯<!--\; ¯\; -->}{z}}^2=0$
I have got something like
$(x+iy{)}^2=\sqrt{-<!--\; -\; -->(x-<!--\; -\; -->iy{)}^2}$
$x+iy=y+ix$

If G is a bipartite Euler and Hamiltonian graph, prove that complement of G, $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulers.
I would like to know if my proof of the statement in the title is correct.
So, I started like this:
As G is a bipartite graph, it has two sets X and Y. Using the condition G is Hamiltonian, then |X|=|Y|.
As G is also Eulerian, stands $d_G(v)$ is even $\mathrm{\forall <!--\; \forall \; -->}v\in <!--\; \in \; -->V(G)$, where V(G) is set of vertices of the graph G.
Now, let's look at some vertex $v\in <!--\; \in \; -->X(G)$. As stated above, it's degree is even. Let's look at two possible cases:
If |X|=|Y| is even too, then in $\stackrel{¯<!--\; ¯\; -->}{G}$, v will be connected with all the remaining vertices from X. That vertex v will also be connected with remaining vertices from Y, with which it wasn't connected in the graph G. That is, $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)=|X|-<!--\; -\; -->1+m$, where m represents number of remaining vertices from Y. As |X| is even, then |X|−1 is odd, and m is also even, because $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is even and |Y| is even, so the remaining number of vertices, m is even too. Sum of an even and an odd number is odd, so $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is odd. That means $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulerian;
If $|X|=|Y|$ is odd, in $\stackrel{¯<!--\; ¯\; -->}{G}$, v will be connected with all the remaining vertices from X and all the remaining vertices from Y, and let's call the number of Y remaining vertices m. As $|Y|$ is odd and $d_G(v)$ is even, then m is odd. Degree of v in $\stackrel{¯<!--\; ¯\; -->}{G}$ is once again $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)=|X|-<!--\; -\; -->1+m$, but this time $|X|-<!--\; -\; -->1$is even, and m is odd. $d_{\stackrel{¯<!--\; ¯\; -->}{G}}(v)$ is odd and that means $\stackrel{¯<!--\; ¯\; -->}{G}$ is not Eulerian.
Thank You!

Automorphism group of a graph = Automorphism group of that graph's adjacency matrix?
Is automorphism group (or set) of a graph G equal to the automorphism group (or set) of adjacency matrix of G?
Example: $G_1,G_2$ are separate graphs where $G_1^{\mathrm{\pi <!--\; \pi \; -->}}=G_2$ and $G={\stackrel{¯<!--\; ¯\; -->}{G}}_1\cup <!--\; \cup \; -->{\stackrel{¯<!--\; ¯\; -->}{G}}_2$. Also, $\mathrm{\pi <!--\; \pi \; -->}$ swaps only 2 vertices of $G_1$, say, it swaps $k^{th}$ vertex with $(k+1{)}^{th}$ vertex.
So, the adjacency matrix of $G_1\ne <!--\; \ne \; -->$ the adjacency matrix of $G_2$.
We will have this dissimilarity in G also.
G is made of 2 non equal matrices of ${\stackrel{¯<!--\; ¯\; -->}{G}}_1,{\stackrel{¯<!--\; ¯\; -->}{G}}_2$. There are only two dissimilar columns/ rows. They are $k^{th}$, $(k+n{)}^{th}$ and $(k+1{)}^{th},(k+1+n{)}^{th}$ vertex. Remaining all pairs of rows/columns (i,i+n) are same where $i\ne <!--\; \ne \; -->k,k+1$
Now, consider a permutation σ of G that swaps $k^{th}$ vertex of Gwith $(k+1+n{)}^{th}$ vertex of G. Note that, $k^{th}$ vertex is in $G_1$ and $(k+1+n{)}^{th}$ vertex is in $G_2$.
here, the adjacency matrix of $G^{\mathrm{\sigma <!--\; \sigma \; -->}}\ne <!--\; \ne \; -->$ the adjacency matrix of G.
So, Automorphism group of a graph $\ne <!--\; \ne \; -->$Automorphism group of that graph's adjacency matrix.
Is it correct?
This post is the source of my argument.
If A is a matrix and π is an automorphism of A, then $A^{\mathrm{\pi <!--\; \pi \; -->}}=A$.

What is the length of the bar needed to represent 75 kilometers( in centimeters)?
In a bar graph, 1 centimeter represents 30 kilometers. What is the length of the bar needed to represent 75 kilometers( in centimeters)?

Free programs to draw graphs?
allow me to draw graphs, lines, curves such as parabolas, set intervals with sliding bars etc.
I have tried geogebra but it is very limited and is a pain to use. For example, you cannot set a line to have a particular gradient, you can only draw a line and then measure the gradient afterward. And to plot parabolas, you cannot plot equations but rather do it in a microsoft paint style.

Are Linear Complexes and Simple Graphs the same thing?
I refer to the background of the notion of Linear complex by A. A. Zykov in the article entitled "General properties of Linear Complexes"
A Linear complex (or simplex complex) is a finite set L for which the set $(L{)}^2$ of all pairs is split into non- intersection parts: $(L{)}^2=K+\stackrel{¯<!--\; ¯\; -->}{K,}$ with the following conditions:
(I) Two pairs, which differ only in the order of the elements, belongs to one and the same part.
(II) The pairs of equal elements belongs to $\stackrel{¯<!--\; ¯\; -->}{K}$.
The number e(L) of elements of the complex L is called its order.
The elements A and B of L are called adjacent $(A\mathrm{\omega <!--\; \omega \; -->}B)$, if the pair of them belongs to K, and not adjacent $(A\stackrel{¯<!--\; ¯\; -->}{\mathrm{\omega <!--\; \omega \; -->}}B)$, if it belongs to $\stackrel{¯<!--\; ¯\; -->}{K}.$. The condition I expresses the symmetric property of the sign ω, the condition II its anti- reflexiveness.
The complexes $L_1$and $L_2$are equal ($L_1$=$L_2$if they consist of the same elements and if any two elements which are adjacent in $L_1$ are adjacent in $L_2$, and conversely.
The complementary complex to L is that complex $\stackrel{¯<!--\; ¯\; -->}{L}$which consists of the same elements but in which any two distinct elements which are adjacent in L are not adjacent in $\stackrel{¯<!--\; ¯\; -->}{L}$, and conversely.
Now, I wonder as what are the points of difference between a Linear complex and a simple graph? Or are they both same?

Strange math notation, vertical bar with parentheses.
So I was reading "Resistance Distance" (D. J. Klein, M. Randić) (Journal of Mathematical Chemistry 12(1993)81-95) when I came up with strange notation.
From the paper:
The graph adjacency matrix is defined as:
$A_{xy}=(x|A|y)=\{\begin{array}{ll}1/r_{xy}& x\sim <!--\; \sim \; -->y\\ 0& \text{otherwise}\end{array}$
The graph degree matrix of a graph is defined as:
${\mathrm{\Delta <!--\; \Delta \; -->}}_{xy}=(x|\mathrm{\Delta <!--\; \Delta \; -->}|y)=\mathrm{\delta <!--\; \delta \; -->}(x,y)\underset{z}{\overset{\sim <!--\; \sim \; -->x}{\sum <!--\; \sum \; -->}}1/r_{xy}$
$|\mathrm{\varphi <!--\; \varphi \; -->})\equiv <!--\; \equiv \; -->\underset{x}{\sum <!--\; \sum \; -->}|x)$
My main question is what does the combination of the vertical bar parentheses mean. Thanks in advance.

Number of undirected graphs with n vertices and k edges (inclusive of simple, non-simple, isomorphic, and disconnected graphs)
Given the constraints (or non-constraints rather), is there a closed solution on a set of labeled vertices?
One way I looked at this problem is by trying to implement a constrained stars-and-bars technique on the diagonal and diagonal-exclusive half of the $n\times <!--\; \times \; -->n$ adjacency matrix, but I'm not getting any good leads.
It seems there are multiple definitions of a self-loop in undirected graphs. In the context of this question, I define a self-loop to represent a degree of 2 in an undirected graph.

An upper bound for a graph Ramsey number
I am trying to prove the following result, given as an exercise in my book:
$r(K_m+\stackrel{¯<!--\; ¯\; -->}{K_n},K_p+\stackrel{¯<!--\; ¯\; -->}{K_q})\le <!--\; \le \; -->(\genfrac{}{}{0ex}{}{m+p-<!--\; -\; -->1}{m})n+(\genfrac{}{}{0ex}{}{m+p-<!--\; -\; -->1}{p})q$
Here r(G,H) denotes the Ramsey number for the graphs G and H, i.e. the smallest positive integer t, such that any graph F of order t either contains G or $\stackrel{¯<!--\; ¯\; -->}{F}$ (the complement of F) contains H. The join of graphs G+H is defined as the graph obtained by first drawing $G\cup <!--\; \cup \; -->H$ and then filling out all possible edges between the vertices of G and H.

How can I describe the following numbers?
0.2
0.4
0.6
0.8
Can I call them "even tenths"?
For example:
"If the maximum value in the data set is 1, then the values displayed in the bar graph are the ______." I am pertaining to the numbers above.

Have I incorrectly set up the integral for this center of gravity/centroid problem?
Consider the bounded region R between the graphs of $g(x)=3x$ and g(x)=3x. Sketch R, which appropriate labels. Then set up all the integrals required do not evaluate.
My solution:
Definitions:
$A={\displaystyle {\int <!--\; \int \; -->}_a^b(f(x)-<!--\; -\; -->g(x))dx}$
$M_x=y-<!--\; -\; -->\text{bar}=\frac{1}{A}{\displaystyle {\int <!--\; \int \; -->}_a^b\frac{1}{2}(f(x{)}^2-<!--\; -\; -->g(x{)}^2)dx}$
$M_y=x-<!--\; -\; -->\text{bar}=\frac{1}{A}{\displaystyle {\int <!--\; \int \; -->}_a^{b}x(f(x)-<!--\; -\; -->g(x))dx}$
….
$A={\displaystyle {\int <!--\; \int \; -->}_0^3(x^2-<!--\; -\; -->3x)dx=\frac{9}{2}}$
$M_x=y-<!--\; -\; -->\text{bar}=\frac{2}{9}{\displaystyle {\int <!--\; \int \; -->}_0^3\frac{1}{2}((x^2{)}^2-<!--\; -\; -->(3x{)}^2)dx}$
$M_y=x-<!--\; -\; -->\text{bar}=\frac{2}{9}{\displaystyle {\int <!--\; \int \; -->}_0^{3}x(x^2-<!--\; -\; -->3x)dx}$

One graph a subgraph of another?
Consider a graph G on n vertices with minimum degree $\mathrm{\delta <!--\; \delta \; -->}$ and with its largest independent set $a>\mathrm{\delta <!--\; \delta \; -->}$. Consider the graph ${\stackrel{¯<!--\; ¯\; -->}{K}}_a\otimes <!--\; \otimes \; -->K_{n-<!--\; -\; -->a-<!--\; -\; -->1}$ (in other words, take a set of a points and add every edge relation between that set and $K_{n-<!--\; -\; -->a-<!--\; -\; -->1}$. Intuitively, this graph is a $K_{n-<!--\; -\; -->1}$ with a missing $K_a$).
How does one prove that that no matter how I take a vertex $v$ and connect it to $\mathrm{\delta <!--\; \delta \; -->}$ vertices in the ${\stackrel{¯<!--\; ¯\; -->}{K}}_a$, I will always get a copy of G.
I think this is trivially true for bipartite graphs, but I dont know how to prove it in general. Any help is nice!

Shift graph towards mean with time?
Say I have a bar graph with one dependent variable and one independent variable. Given a time t, I essentially want to modify the given graph so that as $t-<!--\; -\; -->>\mathrm{\infty <!--\; \infty \; -->}$, all the bars become equal to the mean.
In other words, I essentially want to move the bars of the graph, so that the bars above the mean lower so that they are closer to it and bars below it rise so that they are also closer to it.
I would like to the transformation to happen in negative exponential time (In other words, the bars move a lot initially, and they only converge to the mean when after an infinite amount of time)
At any time, the mean value of the graph must be equal to the original mean.
Any ideas how I can do this? I've never really worked with statistics before.

Existence of solution for matrix equation $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)\stackrel{¯<!--\; ¯\; -->}{x}=\stackrel{¯<!--\; ¯\; -->}{b}$
This is my first question in here and I would be really thankful if someone could help me with understanding the matter.
I am solving a matrix equation $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)\stackrel{¯<!--\; ¯\; -->}{x}=\stackrel{¯<!--\; ¯\; -->}{b}$for a positive vector $\stackrel{¯<!--\; ¯\; -->}{x}$. I don't know anything about the signs of the scalar $\mathrm{\alpha <!--\; \alpha \; -->}$ and vector $\stackrel{¯<!--\; ¯\; -->}{b}$ (but I can always split the solution into several cases). Matrix A is a symmetric matrix with tr(A)=0 (basically it is an adjacent matrix of an undirected graph so it is built only of 0 and 1 with 0 on its diagonal).
My approach would be just to write it down as $\stackrel{¯<!--\; ¯\; -->}{x}=(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A{)}^{-<!--\; -\; -->1}\stackrel{¯<!--\; ¯\; -->}{b}$ whenever $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)$ is nonsingular (as far as I understand that happens for a finite number of α and thus Lebesgue measure of this set of "inappropriate" $\mathrm{\alpha <!--\; \alpha \; -->}$ is 0).
In most of the literature, however, it is required that $1>\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ for the convergence of $I+\mathrm{\alpha <!--\; \alpha \; -->}A+{\mathrm{\alpha <!--\; \alpha \; -->}}^2{A}^2+\dots <!--\; \dots \; -->$ and for the solution $\stackrel{¯<!--\; ¯\; -->}{x}$ to exist (where ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ is a maximum eigenvalue of A) . Moreover, then $det(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)=1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->det(A)=1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->\underset{i}{\prod <!--\; \prod \; -->}{\mathrm{\lambda <!--\; \lambda \; -->}}_i>0$ and hence it is also invertible.
So I have two questions here:
Q1: Why do I need convergence of $I+\mathrm{\alpha <!--\; \alpha \; -->}A+{\mathrm{\alpha <!--\; \alpha \; -->}}^2{A}^2+\dots <!--\; \dots \; -->$ in here? I understand that if it converges then $\underset{k=0}{\overset{+\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\alpha <!--\; \alpha \; -->}}^k{A}^k=(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A{)}^{-<!--\; -\; -->1}$, but why do I need that for the existence of a solution x¯? For instance, if we take a simple example:
$A=\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|\mathrm{\alpha <!--\; \alpha \; -->}=30$
${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)=1$ and hence, convergence condition is violated: 30>1. However, I can still calculate $\stackrel{¯<!--\; ¯\; -->}{x}$. Inverse of $(I-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}A)$ exists and is equal to $-<!--\; -\; -->\frac{1}{899}\left|\begin{array}{cc}1& 30\\ 30& 1\end{array}\right|$. If, for instance, $\stackrel{¯<!--\; ¯\; -->}{b}$ was negative, then it would give me a positive $\stackrel{¯<!--\; ¯\; -->}{x}$. Or is the condition $1>\mathrm{\alpha <!--\; \alpha \; -->}\cdot <!--\; \cdot \; -->{\mathrm{\lambda <!--\; \lambda \; -->}}_{max}(A)$ needed to guarantee that inverse will give a positive solution with both positive α and $\stackrel{¯<!--\; ¯\; -->}{b}$?
Q2: What will happen if $\mathrm{\alpha <!--\; \alpha \; -->}<0$ (and potentially $\stackrel{¯<!--\; ¯\; -->}{b}<0$)? What are the conditions for existence of solution $\stackrel{¯<!--\; ¯\; -->}{x}$? What are the conditions for it to be positive?

G is a simple undirected, regular, and planar graph with 9 vertices. Prove or disprove that $\stackrel{¯<!--\; ¯\; -->}{G}$ must have a Hamiltonian cycle.
Original question: G is a simple undirected planar graph with 9 vertices. Suppose that every vertex in G has the same degree (regular). Prove or disprove that the complementary graph $\stackrel{¯<!--\; ¯\; -->}{G}$ must have a Hamiltonian cycle.
This is part of a past paper for a discrete math course, but the answers were not provided. I have tried to do it for a while now, but I don't even have any idea how to start this proof. Could someone please point me in the right direction, or even better, show me how to do this?

Prove that if a graph has six vertices, then at least one of G or $\stackrel{¯<!--\; ¯\; -->}{G}$ has a subgraph isomorphic to $K_3$
I think this proof is related to proving to Theorem on friends and strangers which can be proved with the pigeonhole principle. But I am at a loss as to what are the holes and pigeons in this case.
I have tried many different graphs and they all work out to have a subgraph that is isomorphic to the $K_3$.