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This is not possible. As is commonly a good approach, let us suppose that the least number of primary points anyone gets is $m$ and the most is $M$. We have that someone who earned $m$ primary points can earn no more than $mM$ total points, which occurs if they beat only players with $M$ primary points. Conversely, someone who earned $M$ primary points can ...

3

Everything looks good except for when $p$ is odd, the part where you say it has at most $MN$ edges if the parts have sizes $M$ and $N$ is very good. I would proceed as follows: Clearly $M+N=p$ so we can assume $M\geq N$ and $M=\frac{p+k}{2}$ and $N=\frac{p-k}{2}$, for some positive integer $k$. The options for $k$ are $0,2,\dots,p$ if $p$ is even and $1,3,... 3 As already noted in comments:$\alpha$and$\beta$are explicitly specified to be "one-to-one correspondences", which is a way of saying "bijections". Since there is a bijection between$V_G$and$V_H$, they have the same number of elements. Similarly, since there is a bijection between$E_G$and$E_H$, they have the same number of elements. 2 Yes. If it has no directed cycles, you can sort it topologically. However, because this is a tournament (i.e. there is an edge between every pair of vertices), then this has to be a total (strict) linear order, which is isomorphic to the one you asked about. I hope this helps$\ddot\smile$2 HINT: If you remove the edge$\{1,2\}$from such a tree, you get a pair of trees, the subtrees rooted at$1$and at$2$; call these$T_1$and$T_2$. You can split the remaining three vertices,$3,4$, and$5$, arbitrarily between$T_1$and$T_2$. For$k=0,1,2,3$, how many ways are there to assign$k$of these three vertices to$T_1$(and the rest to$T_2$)? ... 1 The proof is by a form of induction, namely maximal counterexample. Let$S$denote the set of all graphs on$n$vertices, that violate the theorem. If$S$is empty, we are happy as the theorem is true. If$S$is nonempty, it must be finite, as there are only finitely many graphs on$n$vertices. Hence, if we consider the ordering on$S$based on number ... 1 Recall that $$\chi(t)=\det(tI-L)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{k=1}^n\left(t[k=\sigma(k)]-L_{k,\sigma(k)}\right)\;,\tag{1}$$ where$[k=\sigma(k)]$is an Iverson bracket, equal to$1$if$k=\sigma(k)$and to$0$otherwise. The terms in the summation on the righthand side of$(1)$that contribute to the$t$term in the polynomial$\chi(...

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I want to disagree with the answer by Newb slightly, but mainly just to point out some issues with formalism. Also if you answered $O(|V+E|)$ on my test you would get zero points and I'm sure there are other teachers like that. Since $V$ and $E$ are sets (exactly what kind depends on your particular definition of graph but the most prevalent definition in ...

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I think you're describing a graded poset and you can use the term "graded graph". Eric's condition is correct and the homomorphism is usually denoted $\rho: V \rightarrow \mathbb{Z}$ where $(x,y) \in E \implies \rho(y) = \rho(x) + 1$ $\rho$ is called a rank function and is sometimes written $|x|$.

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To give an idea of the difficulty of this problem here is the case of $d=2,$ which means we have a multiset of cycles. As there is just one cycle on $n$ nodes and the smallest cycle is a triangle these have generating function $$C(z) = \frac{z^3}{1-z}.$$ We get for the set of non-isomorphic $2$-regular graphs the generating function (Polya ...

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I'm not sure you need to use removal lemma to prove your claim, you can prove it using simple combinatoric argument: every edge in the graph can participate, at most, in $n-2$ triangles. so, removing any edge will reduce the number of triangles in the graph by at most $n-2$ triangles. If you remove $\epsilon n^2$ edges, it will reduce the number of ...

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Take a vertex with three edges. Assume you do NOT start there. Then there is a first time you will reach this vertex, through one of the possible paths leading to it. It wonâ€™t be the end of your drawing because two other paths are not yet traversed. Thus, you will leave this vertex, through a second adjacent path. As a result, when you traverse the third ...

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