David Coudert's profile - activity

Another exponential time solution is to compute all shortest simple paths: def find_k_paths(G, start, end, length):

Closing parenthesis are not well placed. This is working p = MixedIntegerLinearProgram(maximization=False, solver="GLPK

Improving a method that is called many many times makes a difference. However, it would be better to reduce the number o

To count the number of connected components in G-S, the following method avoids a copy of the graph and is faster. I obt

To count the number of connected components in G-S, the following method avoids a copy of the graph and is faster. I obt

To count the number of connected components in G-S, the following method avoids a copy of the graph and is faster. I obt

You can use the lowerbounds from https://alco.centre-mersenne.org/item/10.5802/alco.197.pdf $t(G) \geq \frac{\mu_n\mu_2}

You can use the lowerbounds from https://alco.centre-mersenne.org/item/10.5802/alco.197.pdf $t(G) \geq \frac{\mu_n\mu_2}

Thanks for reporting this issue. With https://github.com/sagemath/sage/pull/37260, we extend the method to work on both

With https://github.com/sagemath/sage/pull/37260, we extend the method to work on both Graph and DiGraph.

I proposed a more efficient method for both versions in #37028. It implements a integer linear programming formulation u

I proposed a more efficient method for both versions in #37028

See also #53887

See also #53887

See also #53887

You can at least use graphs.nauty_gentreeg("5 -D4") to enumerate trees of order 5 with degree at most 4 and then filter

You can at least use graphs.nauty_gentreeg("5 -D4") to enumerate trees of order 5 with degree at most 4 and then filter

You can use graphs.trees(n) to enumerate undirected trees of order n. sage: for g in graphs.trees(1): ....: print(

You can use graphs.trees(n) to enumerate undirected trees of order n. sage: for g in graphs.trees(1): ....: print(

You can use graphs.trees(n) to enumerate undirected trees of order n. sage: for g in graphs.trees(1): ....: print(

This seems a side effect of the default type for real values. sage: a = 0.2; a; type(a) 0.200000000000000 <class 's

This is a NP-hard problem, so the running time can be very long (see https://en.wikipedia.org/wiki/Longest_path_problem

The usefulness of the property and augment parameters is not clear to me. I'm curious to know if some users find these p

The documentation of augment also says that: If for any graph G satisfying the property, every subgraph, obtained from G

The documentation of augment also says that: If for any graph G satisfying the property, every subgraph, obtained from G

The documentation of augment also says that: If for any graph G satisfying the property, every subgraph, obtained from G

According the documentation of graphs, when parameter augment='vertices' all graphs up to n=vertices are generated. sag

According the documentation of graphs, when parameter augment='vertices' all graphs up to n=vertices are generated. #py

The documentation of augment also says that: If for any graph G satisfying the property, every subgraph, obtained from G

The documentation of augment also says that: If for any graph G satisfying the property, every subgraph, obtained from G

According the documentation of graphs, when parameter augment='vertices' all graphs up to n=vertices are generated. sag

I added this idea in the wishlist: issue 35857.

I added this idea in the wishlist 35857.

It's not obvious to modify the current code of connected_subgraph_iterator to return graphs with exactly k vertices. Mor

This is now https://github.com/sagemath/sage/pull/35824

Check method faces sage: G = graphs.Grid2dGraph(3, 3) sage: G.faces() [[((0, 0), (0, 1)), ((0, 1), (0, 2)), ((0, 2)

The first step is to seach for keyword bicyclic in this forum. You will find 46577/maximum-algebraic-connectivity-from-a

The first step is to seach bicyclic in this forum. You will find https://ask.sagemath.org/question/46577/maximum-algebra

Your problem is not well defined. From the definition of a dominating set, a vertex in D is at distance at most two from

Your problem is not well defined. From the definition of a dominating set, a vertex in D is at distance at most two from

Your problem is not well defined. From the definition of a dominating set, a vertex in D is at distance at most two from

Read again constraint p.add_constraint(bv[v,c] + sum(be[e,c] for e in G.edges_incident(v, labels=False)) <= 1). It en

This experimental code has been removed (https://github.com/sagemath/sage/issues/24279) but other standard methods have

The call to p.index(x) returns the first occurence of x in p (so 1) and it takes linear time in the size of the list. s

So for your initial example, you can write sage: [x*1.2^-i for i, x in enumerate(p)] [-100.000000000000, 8.33333333333

The call to p.index(x) returns the first occurence of x in p (so 1) and it takes linear time in the size of the list. s

This is now Issue 35461.

You can use method minimal_dominating_sets to list the minimal distance-2 dominating sets on a modified graph H. The mod