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Questions about properties of and problems on graphs, discrete data structures that have the form of nodes connected by edges, that is networks.

2
votes
Use DFS to find a cycle in the graph (in linear time). Save the maximum-weight edge while performing DFS, Delete it from the outputted cycle. Repeat 11 times. Done.
answered Dec 24 '13 by Hendrik Jan
1
vote
If you start with a dag then the only thing you need to check is whether adding the edge $U\to V$ will not create a cycle. That is, there should not exist a path from $V$ to $U$ in the original graph. …
answered Apr 14 '16 by Hendrik Jan
6
votes
In a bipartite graph there can be no clique of size three. Not much of a problem that is left I guess? (edit) Sorry, that was too fast, I realize thanks to comment by Nicholas below. To make up for i …
answered Oct 29 '12 by Hendrik Jan
2
votes
Homework? I start with hints. Integer values assumed. Reduction. Make two copies of the vertices in the graph (odd and even) and rewire the graph such that each odd length edge of the graph is betwee …
answered May 20 '13 by Hendrik Jan
1
vote
Reconstructing an unrooted undirected weighted tree from its distance matrix is an important problem in bio-informatics, where it is used in the context of creating phylogenetic trees. A commonly use …
answered Nov 16 '17 by Hendrik Jan
1
vote
Let $\tau_1: x_1, x_2, \dots, x_n$ and $\tau_2: y_1, y_2, \dots, y_n$ be two topological orderings of the vertices of graph $G$. Our aim is to show that $\tau_1$ can be transformed into $\tau_2$ by a …
answered Nov 19 '13 by Hendrik Jan
1
vote
Based on the example I infer the following structure. The pancake graph seems to be defined with flipping pancakes in mind. For order $n$ we take all $n!$ permutations of the numbers $1,\dots,n$ as v …
answered Jan 8 '14 by Hendrik Jan
2
votes
Personally I would not prove the second statement ("the shocker") unless I was explicitly forced to do so by my professor. The weight in a particular leaf is added to the weight of each of the nodes …
answered Nov 23 '15 by Hendrik Jan
2
votes
No need to introduce two arrays. In the original algorithm the distance equals the minimum (over all paths) of the sum of edges (in the path). The new variant defines a distance as the minimum (over a …
answered Sep 16 '13 by Hendrik Jan
5
votes
A clue then. It seems you have to multiply with a multinomial coefficient in each step. Here the size of the subtree comes in handy. Explanation: if children have certain topsorts, then these sequen …
answered Jun 17 '13 by Hendrik Jan
5
votes
Nope. In order to find the distance from $s$ to $t$ it is necessary to determine the length of all paths that are at least at the same distance as $t$ is. If $t$ has a "median" distance half of the di …
answered Nov 15 '14 by Hendrik Jan
5
votes
Every binary tree (with the right number of nodes) has exactly one labelling that satisfies a given postorder labelling. So you need to find the number of binary trees. That is the famous Catalan numb …
answered Apr 8 '16 by Hendrik Jan