Given a rectangular grid of dimensions $5 \times 9$, find if it contains any Hamiltonian cycle in it or not. Can the result be generalized for dimensions $m \times n.$  Example    I have tried making grids on paper and found that Hamiltonian circuits are easy to build if any(or both) of the dimensions are even, however I couldn't prove and generalize my observation.  Any help is much appreciated.
120855
How to prove "Prove $P_m X P_n$ (graph cartesian product) is Hamiltonian if and only if at least one of $m,n$ is even"  Graph cartesian product is Grid graph, if I figure out $P_2 X P_2$ it will be $C_4$ (it is Hamiltonian). How to state this prove ?
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Show that $ G = (V, E)$ has no Hamiltonian cycle, where the vertices are $ V = \{ a, b, c, d, e, f, g \} $ and the edges are $E = \{ ab, ac, ad, bc, cd, de, dg, df, ef, fg \}$.  From my working out, the vertices $ a, b, c, d, e, f$ are odd degrees of 3 and 1. Moreover $g $ being a even vertices of 2.   There were three points that were made in my textbook to show that a graph does not contain a Hamilton circuit:   A graph with a vertex of degree one cannot have a Hamilton circuit. Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit. A Hamilton circuit cannot contain a smaller circuit within it.   According to the definition graph G does not have a Hamiltonian cycle because of the first definition.   However, I am confused about 2 &amp; 3 definitions and I am not sure if this graph involves them or not.
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Let $G$ be a connected planar graph with a planar embedding where every face boundary is a cycle of even length. Prove that $G$ is bipartite.  Any hints/tips will be greatly appreciated.
46747
Suppose we write each number from $1$ to $n^2$ in an $n\times n$ grid exactly once. Prove there are two adjacent cells such that the absolute value of their difference is not less than $n$.   I think it should be a pigeonhole problem, but the work I did so far was basic observations like if we assume all differences are less than $n$, then neither of $1,2,\dots,2n-2$ could be in the row or the column $n^2$ is placed.  Any hints would be appreciated.
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Assume $G$ is a , i.e. a (finite) directed graph such that between any two vertices, $a$ and $b$, there is at least one edge in one of the two directions, $a\rightarrow b$ or $b\rightarrow a$.   Show that there is at least one , i.e. a path, following the direction of the edges and visiting all vertices exactly once. (This part is a classical exercise and it is here to prepare for the second part of the problem). Consider all possible losers, i.e. the set of all vertices that can be end-points of Hamiltonian paths of the tournament, see Part (1). Show that the subgraph of all losers has a , i.e. a cycle that follows the direction of the edges and passes through every vertex exactly once and returns to the initial vertex.   For example     these are all tournaments and the $\color{red}{\text{red}}$ dots are the losers. Notice how there is always a cycle joining them.
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For what positive real numbers $x,y$ can an $x\times y$ rectangle be partitioned into a finite number of squares?  When $\dfrac{x}{y}$ is a rational number, it is not hard to see that we can partition it.  Also, we can scale the width to $1$, so, say we're dealing with a $1\times c$ rectangle. For what values of $c$ can we do the partition?
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I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.  Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem.
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Here's a quickie:  Let $r\ge2$. Does every connected $r$-regular bipartite graph contain a Hamiltonian cycle?  I've been playing around with this for almost an hour, but I can't prove it.
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I'm trying to find out if it is possible to construct a connected Hamiltonian and a connected non-Hamiltonian graph using the same degree sequence. For disconnected graphs it would be easier, I could choose a degree sequence of $(2,2,2,2,2,2)$ and have $C_6$, which has a Hamiltonian path, and two disjoint 3-cycles, which are non-Hamiltonian.  How to do it with two connected graphs?
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I found the following question in a paper I was trying to solve:  The following figure shows a $3^2 \times 3^2$ grid divided into $3^2$ subgrids of size $3 \times 3$. This grid has $81$ cells, $9$ in each subgrid.    Now consider an $n^2 \times n^2$ grid divided into $n^2$ subgrids of size $n \times n$. Find the number of ways in which you can select $n^2$ cells from this grid such that there is exactly one cell coming from each subgrid, one from each row and one from each column.  My try:  Since we have $n^2$ rows, $n^2$ columns and $n^2$ subgrids in total, we have to choose one and only one cell from each of them. Let's choose them one at a time. We can choose the first cell in $n^4$ many ways. Then, we'll have to avoid that subgrid, that column and that row that we've chosen the first one from when choosing the second cell. So, we have $n^4-n^2-2n(n-1)$ choices. We can continue this to get the total number of possible ways. But, I think there's a hole. Say, we've chosen the first cell from the subgrid of the up-left corner and the second from the subgrid just right to it so that it doesn't violate any rules. Then, when finding the number of ways we can choose the third cell, we would have substracted some of the cells twice. I think you get it. Please, if anyone can help me solving this problem, it'd be greatly appreciated.
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I'm thinking of making a puzzle game where the objective is to fill a grid with shaped puzzle pieces (for example, the classic Tetris shapes).  How can I go about generating a set of pieces that can be guaranteed to be used to fill the grid, leaving no gaps? How could I adapt this algorithm to scale the difficulty of the resulting puzzle?
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let $\alpha$ be a cycle of length $s$, say $\alpha = (a_1, a_2, \ldots, a_s)$  Prove $\alpha^2$ is a cycle if and only if $s$ is odd.  Let me start off by saying I am in my 5th week of Group Theory.  I often have trouble getting these problems started.  This is my first proof based course.    I believe $\alpha^2 = (a_2, a_3, \ldots, a_1)$  Any tips on where to go from here would be great.  Also...if there are any tips for starting proofs like these in general, I could really use them! My teacher teaches as if a proofing class was a pre-req, which it was not.
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I'm trying to show, for the Hamiltonian $H = \vec{P}^2/2m + V(\vec{X})$, that $[\vec{L}^2,H]=0$ if $V(\vec{X}) = V(|\vec{X}|)$, and I pretty much almost have it, there's just one thing I'm getting messed up on.  So what I did was: $$ [\vec{L}^2,H] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(\vec{X})] = [L_x^2 + L_y^2 + L_z^2, \frac{\vec{P}^2}{2m} + V(|\vec{X}|) ] $$ $$ = [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] + [L_y^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] + [L_z^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] $$ Looking at the $L_x^2$ component: $$ \rightarrow [L_x^2,\frac{\vec{P}^2}{2m} + V(|\vec{X}|)] = \frac{1}{2m}[L_x^2,P^2] + [L_x^2,V(|\vec{X}|)] $$  $$ = \frac{1}{2m}[L_x^2,P_x^2+P_y^2+P_z^2] + [L_x^2,V(|\vec{X}|)] $$  $$ = \frac{1}{2m} \bigg( [L_x^2,P_x^2]+[L_x^2,P_y^2]+[L_x^2,P_z^2] \bigg) + [L_x^2,V(|\vec{X}|)] $$  Now for the momentum parts, that's a lot of tedious work that I won't type up, but at the end of the day, I get that $$ [L_x^2,P_x^2]=[L_x^2,P_y^2]=[L_x^2,P_z^2]=0 $$ and similarly for $[L_y^2,P^2]=0$ and $[L_z^2,P^2]=0$. My problem is this, for this to work out, I need $$ [L_x^2,V(|\vec{X}|)] = 0 $$ and similarly $$ [L_y^2,V(|\vec{X}|)]=[L_z^2,V(|\vec{X}|)]=0 $$ but I don't really understand why that would be the case? Any insight would be appreciated.
100645
Let $a = (123)(456)$ be in $S_{10}$. Find a permutation $b$ in $A_{10}$ such that the disjoint cycle form of does not contain any $3$ cycles and $a$ is in $\langle b \rangle$.  Attempt: I'm assuming that $b$ must also assign each integer to the following one in each of it's cycles (or part of a cycle) containing $1$ to $6$. But I'm not sure how to rebracket $a$ so that $b$ doesn't have a cycle of the form (###) while some other element in it's generated group sends $3$ back to $1$ and $6$ back to $4$. Is this right?
85735
I've seen two questions on solving the Rubik's cube but none of the answers have given a complete solution using mainly mathematical techniques. Furthermore, I've not seen a good explanation of general techniques for solving permutation puzzles in general, including those of the Rubik's cube family, without involving any memorized move sequences. So I've decided to write up my method below as an example of how group theory can be applied to solving such puzzles.
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Source is in , but it has no proof. Also, I am not completely sure what "unbalanced" means here. Non-equal number of even and uneven degrees?  What I am interested in, are there some universal properties of (connected) graphs that prevent hamiltonicity? Something to do with number of vertices versus edges, or average degrees or such? Something that is easy to demonstrate?  I will (hopefully) be using them as counterexamples of hamiltonian graphs in my thesis.
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Show that if $G$ is a graph with $n$ vertices, then no more than $n/2$ edges can be colored in the same color, in any edge coloring of $G$.  I know there are  satisfy the statement, e.g a circuit of $n$ vertices with an even $n$ .   However, I want the proof of above statement. As we known, cases can only falsify a statement instead of verify it.