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This is only true because it is a complete binary tree. There are twice as many nodes in each succeeding level (starting from the root at level $k=0$). Thus, at level $k$, there are $2^k$ nodes and the index of the first node is $2^0 + 2^1 + \dots + 2^{k-1} + 1 = 2^k$. At level $k$, the children of the first node (index $N = 2^k$) will be labeled as soon ...


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When talking about indexing, it provides a structured way of accessing the array elements. It's not so much a mathematical property of a binary tree (at least, that's not how I'd look at it), as more of a way of modeling the data for efficient lookup and storage in an array. An array based solution allows for a compact representation of the tree. You ...


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You can think of each connected component as a graph by itself and talk about its associated spanning trees. What it is saying is that you build a spanning forest by choosing a spanning tree from each connected component, therefore the number of trees in a spanning forest is the same as the number of connected components.


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If every edge has weight $1$, then every spanning tree of $G$ has the same weight: $\lvert V\rvert-1$. So, all that you actually have to do here is find any $O(\lvert V\rvert+\lvert E\rvert)$-time algorithm for finding ANY spanning tree for $G$. For instance: breadth-first search can be used to do this, with worst-case performance $O(\lvert V\rvert+\lvert ...


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We will show that if T is not a tree, then there is an integer $k>0$ such as $p^{k}(y)=y$ for a $y \in T$. This doesn't agree with your definition. You want to show that there is no periodic point. By definition, a tree is acyclic. So it suffices to show that $p$ will end up traversing a cycle. That's how you should go about contrapositive. ...


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Count all the nodes up to and including the second (highest numbered) child of node $n$; the number of such nodes is the number attached to this second child. But each node so counted is either the root node, or one of the two children of a parent node whose number is in the range $1,\ldots,n$. Conversely for any such parent both children are among the ...



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