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3

Yes because from a single preorder sequence you can create multiple trees. This is of course true in general case, BSTs have a unique inorder sequence. for example preorder: acdef a / \ c d / \ e f has lot's of inorder.

2

First, why the claim is true. Take any vertex. If it's matches to the same vertex in both perfect matching, then it had degree zero on the symmetric difference. Otherwise it has degree two. Second, after removing all isolated vertices in the symmetric difference, all vertices have degree two. Take any such vertex and follow its two edges. What you get is a ...

1

Normalization should have no impact on the performance of a decision tree. It is generally useful, when you are solving a system of equations, least squares, etc, where you can have serious issues due to rounding errors. In decision tree, you are just comparing stuff and branching down the tree, so normalization would not help.

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Hint: How many edges does a spanning tree on $2014$ vertices have? How many edges are in $C_{2014}$? I don't know what a bipartite graph has to do with this.

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I actually think there are 2 false statements in there. I hope the following hints will help you. EDIT : since OP has figured it out, I'll post the answers. a) If $G$ is a complete graph, can you find an ordering of the vertices of $G$ that can not correspond to a DFS visiting order ? Can you find one that can't be done by a BFS ? EDIT : a) is TRUE. ...

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You know that for a tree it holds $$\sum_{v\in V} d(v) = 2|E| = 2(|V|-1)$$ Now you know that there are $|V|-(33+25+15)$ vertices of degree $4$ and therefore: $$33\cdot 1 + 25 \cdot 2 + 15\cdot 3 + [|V|-73]\cdot 4 = 2|V| - 2$$ Now just find the value of $|V|$.

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This follows from $d(\theta^{-1}\sigma\theta(v),v) = d(\sigma(\theta(v)),\theta(v))$.

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Use Proof by contradiction. Assume that at least three pairwise common vertices are distinct, then show that they form a cycle.

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