combinatoire – A question about "the universality of the graphs commuting"?

assume $ G $ is a group and $ S subassembly G $ is his finite subset. Let's define the commuting graph of $ G $ in respect of $ S $ as $ Comm (G, S) $ – a graph, where the vertices are the elements of $ S $, and they are connected with edges if they commute.

Let's define the universality of displacements of $ G $ as $ UGT (G) $ – the maximum $ k in mathbb {N} $ such as sny graph with k $ vertices is a commuting graph of $ G $ for some people $ n $.

What is the minimum possible order of $ G $, such as $ UGT (G) = n $?

The constructs of the questions "Group Commutation Graphs" and "What is the minimum possible size of a graph $ n $ -universal?" We provide such a command group $ 2 ^ {(1 + o (1) 2 ^ { frac {n-1} {2}}} $.

However, a much smaller example (of order $ (2n (n-1))! $) is possible. C & # 39; $ S_ {2n (n-1)} $.

Indeed, let's suppose $ A = {a_ {ij} } _ {1 leq i, j leq n} cup {b_ {ij} } _ {1 leq i, j leq n} $. For a chart $ Gamma (V, E) $, or $ V = {1, …, n } $, we can take $ S_ Gamma subset Sym (A) $, defined by the following formula S $ {ji}) | i in V } $.

However, I do not know if there is still room for improvement …

graphs – improve at Monte Carlo

Can I improve a Monte Carlo search for the problem described?

enter the description of the image here

I therefore have a graph / network consisting of segments a1, a2, …, b1, b2, …, and c1, c2, …

For all the underlying segments, there is some weighting, for example: a1 = 2, b3 = 0, c3 = 4

In addition, I have a matrix of distances from each segment to all other segments of the network, e.g.

|from/to | a1  | a2  | a3  | b1  | b2  | b3  | c1  | c2  | c3  |
|--------|-----|-----|-----|-----|-----|-----|-----|-----|-----|
|   a1   |  0  | 4   | 6   | 84  | 82  | 80  | 150 | 148 | 146 |
|   a2   | ... | 0   | ... | ... | ... | ... | ... | ... | ... |
|   a3   | ... | ... | 0   | ... | ... | ... | ... | ... | ... |
|   b1   | ... | ... | ... | 0   | ... | ... | ... | ... | ... |
|   b2   | ... | ... | ... | ... | 0   | ... | ... | ... | ... |
|   b3   | ... | ... | ... | ... | ... | 0   | ... | ... | ... |
|   c1   | ... | ... | ... | ... | ... | ... | 0   | ... | ... |
|   c2   | ... | ... | ... | ... | ... | ... | ... | 0   | ... |
|   c3   | ... | ... | ... | ... | ... | ... | ... | ... | 0   |
| ...    | ... | ... | ... | ... | ... | ... | ... | ... | ... |

I want to place not agents (in the picture n = 3) so as to cover as many segments as possible by weighting, over a distance of 50. And to be able to optimize any combination of parameters, e.g. all not and any distance.

Until here I have tried:

  • a greedy approach: place an agent where most segments are covered (local optimum), then place the next agent to cover most segments, etc. not.
  • an approach monte carlo: selection not random segments and evaluate, repeat several times and choose the best solution.

In reality, this network and the number of agents not can be much bigger and more complex.

I wonder what other approaches might work, better than a monte carlo?

graphs – Separate overlapping groups

Suppose I have multiple data points in 2d (x, y) labeled A, B, C, or D. I find a minimum bounding box for points labeled A and call it cluster A. I can do the same thing for B, C, and D. After that, there is a good chance that group A will overlap group B, C, or D. My goal is to find an algorithm to ignore points so that

  1. There will be more overlap between the clusters
  2. Discard as few points as possible
  3. The "reduced" area of ‚Äč‚Äčeach cluster should be as similar as possible. This means that the old minimum delimitation zone minus the new minimum delimitation zone must be similar between the different groups.

We can assume that the initial group makes sense, which means that labeling is not random. The clusters overlap at about 20%. To potentially simplify the problem, we can assume that the surface of each cluster is convex. Although not 100% of my data sets are convex from each group, I would be happy if I could solve this problem first. If there is no optimal solution for this, I am also interested in greedy solutions.

I also have trouble understanding what kind of keyword should I use on Google to find similar problems to the solution. I would appreciate all pointers. Thank you very much.

graphs – How does the algorithm slow All-pairs-shortest-path?

The key element of Floyd-Warshall's operation (All-pairs-shortest-path) is very different from the key Bellman-Fords element.

Floyd-Warshall solves the following problem using dynamic programming. Calculate the shortest path from each pair of vertices using only the first k summit as intermediate jumps. At first, it's the initial weight of the existing edges and $ infty $ elsewhere (this is the basic case).

Suppose you have calculated this matrix for the first time $ k summit and try to calculate it for the first $ k + 1 $ Mountain peak. We should take into account the paths that use first $ k summit and passes by summit $ k + 1 $. Note by $ dp_k (u, v) $ the shortest way to $ u $ at $ v $ using first $ k top, then it follows that the optimal path of $ u $ at $ v $ using $ k + 1 $ vertex does not use this new vertex, nor even use it, so we get the minimum of both values:

$$ dp_ {k + 1} (u, v) = min (dp_k (u, v), dp (u, k) + dp (k, v)) $$

In practice, the same table is reused at each stage. The algorithm only requires $ O (n ^ 2) $ spatial complexity. The temporal complexity is $ O (n ^ 3) $.

Two apparently identical but different graphs!

I would like to add the following two seemingly identical graphs, but it seems that they are different. I can not understand what is the difference between them and how to make them really identical

Graph[List[X, x14, x17, Y] , List[UndirectedEdge[X, x14], UndirectedEdge[x14, x17], UndirectedEdge[x14, x17], UndirectedEdge[x17, x14], UndirectedEdge[x17, Y]]]

and

Graph[List[X,x14,x17,Y] , List[UndirectedEdge[X,x14],UndirectedEdge[x17,x14],UndirectedEdge[x17,Y],UndirectedEdge[x14,x17],UndirectedEdge[x14,x17]]]

I can see that the order of UndirectedEdges is different in the graphics. But is this a problem for the Undirected graph?

graphs – Another question about vertex coverage?

This is the equivalent of the coverage problem. You can consider each vertex on the right side as a set of its neighbors on the left. In your example $ e, f, g $ correspond to the sets $ {d }, {d }, {a, b, c, d }. Now, a minimum vertex coverage in your problem equals a defined minimum coverage.

graphs – Unique property of a complete binary tree

I read a statement that was not clear to me and I hoped to get clarification.

He said that considering a complete binary tree with $ n> $ 2 leaves, there is an internal node such as one-third to two-thirds of all $ n $ the leaves in the tree are his descendants.

From my understanding, I know that every internal node in a complete binary tree has 2 children. That said, the whole tree has $ 2n – $ 1 nodes, which means $ n – $ 1 of them are internal nodes (no leaves).

I can offer examples from cases where this is always the case, but I do not know how to formally reason. Any help would be greatly appreciated.

Waiting value of complete bipartite graphs in a random graph G (n, p)

I'm trying to calculate the expected number of complete bipartite graphs $ K_ {r, s} $ in a random graph $ G (n, p) $.

I've calculated that the probability that a given subset is a complete bigraph is $$ P (X = 1) = p ^ {rs} (1-p) ^ { frac {r (r-1) + s (s-1)} {2}} $$

and the expected number of bigraphs will simply be the number of ways to arrange $ r $ summits in $ n $ and $ s $ summits in $ n-r $ give
$$
mathbf {E} (X) = binom {n} {r} binom {n-r} {s}
p ^ {rs} (1-p) ^ { frac {r (r-1) + s (s-1)} {2}
$$

However, when I run a code to check this, the number of subgraphs observed is different, for a low value of $ n $ he predicts too few bigraphs and for the big ones $ n $ he predicts too much, but the shape and the top of the graph correspond to the theory. The code works correctly for the case where $ r = $ 1 that is to say that the star graphs and the number of observed graphs correspond to the theory.

So I wonder where, in my derivation, I forgot something that will only affect the number of bigraphs for when $ r $ or $ s $ is not one.

graphs and networks – Bug in NearestNeighborGraph

Update

Not corrected in 12.0.

NearestNeighborGraph generates a poorly formatted output on a rasterized input. Reported to support Wolfram CASE: 4093797

$Version
(* "11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)" *)

NearestNeighborGraph(
 Rasterize(Style(#, 20), "Image") & /@ Alphabet(), 
 2,
 VertexLabels -> "Name"
)

enter the description of the image here

The labels are too big.

Is there a job around?

How to combine two graphs

Anyone can help me combine these graphics without taking into account the graph built from the imported file and have tried many options but I still can not work.enter the description of the image here