On 2 September 2013 02:50, Josè Luis Mietta wrote:

I wanna see if two line segments are connected by a path formed by line segments intersected.

You could build a network where each segment is a node and an intersection is a link. Then, you could use NetworkX connected_components to get groups of segments together.

Hi experts! If I do: G =Graph(M) That is: to use the associated intersection graph, where the vertices are the sticks and there is an edge between the two vertices if they intersect. Two sticks are "connected by a 'intersected-stick' path" if they are in the same connected component of this graph. It turns out that the matrix I consider (M) is the adjacency matrix of this graph. Then, I can do: first_stick inG.connected_component_containing_vertex(second_stick) This is 'True' if 'first_stick' is in the sub-graph formed by 'second_stick' and all sticks 'connected' with 'second_stick'. The problem is that G.connected_component_containing_vertex() explore all the sub-graph. How can I do (what is the code) for stop the iteration when the algorithm found 'first-stick'? Waiting for your answers. Thanks a lot!! ________________________________ De: Robert Kern Para: Discussion of Numerical Python Enviado: lunes, 2 de septiembre de 2013 10:40 Asunto: Re: [Numpy-discussion] Stick intersection path algorithm On Sun, Sep 1, 2013 at 11:55 PM, Josè Luis Mietta wrote:

Hi experts!

I wanna study the intersection between line segments (sticks). I wrote a algorithm that generate a matrix, M, with N rows and N columns. The M-element Mij is 1 if stick number 'i' intersect stick number 'j' (obviously M is symmetric).

Given two arbitrary sticks, i need a simple and effective algorithm that determinate if that two sticks are conected by a 'intersected-sticks' path.

You may find it helpful to reword your problem using standard terminology. If I hadn't read your previous question, I would not have been able to understand what you meant by intersected sticks (or, as Chris thought at first, that you needed help determining the intersections). This will also help in searching Google for background and pre-existing software to solve your problem. You have an "undirected graph" (the sticks are "nodes", and the intersections are "edges") and you want to find if two given nodes are "reachable" from each other. You are currently representing your graph as an "adjacency matrix" `M` where `M[i,j]` is 1 iff nodes `i` and `j` have an edge between them and 0 otherwise. The "transitive closure" of your graph `M` is the graph that connects two nodes with an edge iff the two nodes are reachable from each other in the original graph `M`. There are several graph theory packages out there, like NetworkX, that will do this for you. Depending on the kinds of queries you would like to do, as David points out, want to compute the "connected components" of your graph; a connected component is a subgraph of your original graph such that all of the nodes are reachable from each other. You can also look up Python code for computing the transitive closure of a graph; it's not a complicated algorithm. However, this algorithm is usually implemented using a different representation of a graph than an adjacency matrix, so you may need to do a conversion. -- Robert Kern _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

Thanks experts! Thanks Robert Kern! Two more questions about it: 1. networkx.has_path(G, first_stick, second_stick) stop when find second_stick or compute all the sub-graph and then evaluate if first_stick and second_stick are connected? 2. Using networkx or other tool: how can I obtain the 'clusters size' distribution (that is: number of clusters of size 'D', for all cluster-sizes)? Thanks a lot!! José Luis ________________________________ De: Robert Kern Para: Discussion of Numerical Python Enviado: miércoles, 4 de septiembre de 2013 18:49 Asunto: Re: [Numpy-discussion] Stick intersection path algorithm On Wed, Sep 4, 2013 at 3:17 PM, Josè Luis Mietta wrote:

Hi experts!

If I do:

G = Graph(M)

That is: to use the associated intersection graph, where the vertices are the sticks and there is an edge between the two vertices if they intersect. Two sticks are "connected by a 'intersected-stick' path" if they are in the same connected component of this graph. It turns out that the matrix I consider (M) is the adjacency matrix of this graph.

Then, I can do:

first_stick in G.connected_component_containing_vertex(second_stick)

This is 'True' if 'first_stick' is in the sub-graph formed by 'second_stick' and all sticks 'connected' with 'second_stick'.

The problem is that

G.connected_component_containing_vertex()

explore all the sub-graph.

How can I do (what is the code) for stop the iteration when the algorithm found 'first-stick'?

networkx.has_path(G, first_stick, second_stick) http://networkx.github.io/documentation/latest/reference/generated/networkx.... If you are going to be doing many queries, you should compute all of the path lengths, then you can query the final data structure easily. lengths = networkx.all_pairs_shortest_path_length(G) second_stick in lengths[first_stick] -- Robert Kern On Wed, Sep 4, 2013 at 3:17 PM, Josè Luis Mietta wrote: Hi experts!

If I do: G =Graph(M)

That is: to use the associated intersection graph, where the vertices are the sticks and there is an edge between the two

vertices if they intersect. Two sticks are "connected by a 'intersected-stick' path" if they are in the same connected component of this graph.

It turns out that the matrix I consider (M) is the adjacency matrix of this graph.

Then, I can do: first_stick inG.connected_component_containing_vertex(second_stick) This is 'True' if 'first_stick' is in the sub-graph formed by 'second_stick' and all sticks 'connected' with 'second_stick'.

The problem is that

G.connected_component_containing_vertex()

explore all the sub-graph.

How can I do (what is the code) for stop the iteration when the algorithm found 'first-stick'?

Waiting for your answers.

Thanks a lot!!

________________________________ De: Robert Kern

Para: Discussion of Numerical Python Enviado: lunes, 2 de septiembre de 2013 10:40

Asunto: Re: [Numpy-discussion] Stick intersection path algorithm

On Sun, Sep 1, 2013 at 11:55 PM, Josè Luis Mietta wrote:

...

Hi experts!

I wanna study the intersection between line segments (sticks). I wrote a algorithm that generate a matrix, M, with N rows and N columns. The M-element Mij is 1 if stick number 'i' intersect stick number 'j' (obviously M is symmetric).

Given two arbitrary sticks, i need a simple and effective algorithm that determinate if that two sticks are conected by a 'intersected-sticks' path.

You may find it helpful to reword your problem using standard terminology. If I hadn't read your previous question, I would not have been able to understand what you meant by intersected sticks (or, as Chris thought at first, that you needed help determining the intersections). This will also help in searching Google for background and pre-existing software to solve your problem.

You have an "undirected graph" (the sticks are "nodes", and the intersections are "edges") and you want to find if two given nodes are "reachable" from each other. You are currently representing your graph as an "adjacency matrix" `M` where `M[i,j]` is 1 iff nodes `i` and `j` have an edge between them and 0 otherwise. The "transitive closure" of your graph `M` is the graph that connects two nodes with an edge iff the two nodes are reachable from each other in the original graph `M`. There are several graph theory packages out there, like NetworkX, that will do this for you. Depending on the kinds of queries you would like to do, as David points out, want to compute the "connected components" of your graph; a connected component is a subgraph of your original graph such that all of the nodes are reachable from each other.

You can also look up Python code for computing the transitive closure of a graph; it's not a complicated algorithm. However, this algorithm is usually implemented using a different representation of a graph than an adjacency matrix, so you may need to do a conversion.

-- Robert Kern

_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

_______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

-- Robert Kern _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

Hi experts! The array ([ 0, 39,  7,  3,  1,  1,  0,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0, 0,  0,  1]) means that in the sistem (graph) are : 4 cluster of size 1, one cluster of size 3, one cluster of size 7 and one cluste of size 39? What does means 'zero' (13 times) in the array? Thans a lot! José Luis ________________________________ De: Daπid Para: Discussion of Numerical Python Enviado: jueves, 5 de septiembre de 2013 11:06 Asunto: Re: [Numpy-discussion] Stick intersection path algorithm On 5 September 2013 13:14, Josè Luis Mietta wrote: 2. Using networkx or other tool: how can I obtain the 'clusters size' distribution (that is: number of clusters of size 'D', for all cluster-sizes)? This is best asked in their mailing list. A possible way is: np.bincount([len(i) for i in nx.connected_components(G)]) For example: np.bincount([len(i) for i in nx.connected_components(nx.erdos_renyi_graph(100, 0.01))]) array([ 0, 39,  7,  3,  1,  1,  0,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0, 0,  0,  1]) _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

Thanks so much!!   Best regards, José Luis ________________________________ De: Daπid Para: Discussion of Numerical Python Enviado: jueves, 5 de septiembre de 2013 12:56 Asunto: Re: [Numpy-discussion] Stick intersection path algorithm On 5 September 2013 17:03, Josè Luis Mietta wrote: The array ([ 0, 39,  7,  3,  1,  1,  0,  0,  0,  0,  1,  0,  0,  0,  0,  0,  0, 0,  0,  1]) means that in the sistem (graph) are : 4 cluster of size 1, one cluster of size 3, one cluster of size 7 and one cluste of size 39? No, it means there are 39 of size 1, 7 of size 2 and so on. David. _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion