I moved all code under github here: https://github.com/stefanutti
- I organized the folders a little betters (different github repos)
- I’m experimenting docker to deliver & deploy the Java and Python software
- I’ll also try to integrate a deep learning module, based on Tensorflow, to help me find the “missing piece”
For the decomposition of a graph representing a map, I’m trying to use different algorithms to select the edge to remove.
The question is:
- When you have multiple valid choices, which is the best edge to select … if any?
Some basic rule are:
- Avoid new F5 to be created
- Get rid of F5 whenever is possible, selecting the edge of a face with 2, 3 or 4 edges
- Consider to remove an edge from an F5 face only if no faces with 2, 3, 4 edges exist
Here are all possibilities. Last 2 groups (removing edges from F5 and F6) don’t have to be taken into account. I just wanted to see how it continued after the cases with faces made of 2, 3, 4 and 5 edges:
This is what I want to try:
- Select an F2, F3 or F4 preferably when it is near an F5 and remove the edge that joins the two faces
- Analyse also the balance of the Euler’s identity when removing edges, to decide if it is better to choose other couple: F2 near the highest or other possibilities
- Let Deepmind DQN (or Tensorflow) learn the best approach for selecting faces and remove the edges
- The selection of the face and the edge based on the characteristics of the neighbor faces
- Avoid the F5 worst case (in the rebuilding phase), when it is necessary to apply the random Kempe’s color switches to solve the impasse
The rule: avoid F5 or at least do not create them.
- Remove F2 when near F5 –> Good (will get rid of the F5 and generate an F3)
- Remove F3 when near F5 –> Good (will get rid of the F5 and generate an F4)
- Remove F4 when near F5 –> NOT Good (will end up with another F5)
- Remove F5 when near F5 –> Good (will get rid of the F5 and generate an F6)
Other to avoid:
- Remove F2 when near F7 –> NOT Good (will end up with another F5)
- Remove F3 when near F6 –> NOT Good (will end up with another F5)
- It is also important to check what appens to f3 and f4
The algorithm I use to color graphs works pretty well … BUT:
- Sometimes (very rarely) it gets into an infinite loop where also random Kempe color switches (around the entire graph) do NOT work. The good is, if I reprocess the same graph from the beginning (since in a part of the code I choose edges randomly), the algorithm
usually works fine
So, what now?
I want to try other strategies while removing edges from the original graph, to avoid the above condition. I want to eliminate the need of random swithes.
The sequence I used so far is:
- Select an F2 and if an F2 does not exist, select an F3 an so on: F4 and then F5. Once selected the face, I remove a random edge from it, only paying attention not to chose the edge if the resulting graph is 1-connected
I want to try:
- Change the order of selecting the faces
- 2, 3, 4, 5 –> This is the default in the current version of the program (22/Nov/2016)
- 3, 2, 4, 5
- 4, 2, 3, 5
- I want to try all possible combinations to see if … may be … if I’m really lucky, some does not require random swithes
- Once selected the face, I want to try other strategies to select the edge to remove
- random edge –> This is the default in the current version of the program (22/Nov/2016)
- Among all the edges of the face, select the one that is shared with the face that has less edges respect to the other F
- … that has more …
A planar cubic graphs to visualize what I wrote in this post:
I transformed the really bad case into a rectangular map to play with the Java program and Kempe random switches.
Here is the graph with the two edges to connect:
The two edges marked with the X, have to be joined by a new edge that form a new F5 face.
To try the Java program, rebuild this graph and play with the switches, it is possible to use this string:
- 1b+, 2b+, 3b+, 4b+, 5b+, 15b+, 14b-, 13b-, 5e-, 6b-, 12b-, 4e-, 8b-, 13e-, 6e-, 9b-, 12e-, 11b-, 3e-, 14e-, 10b-, 7b-, 2e-, 8e-, 9e-, 11e-, 7e-, 10e-, 15e+, 1e+
Next is the graph that shows how the graph has been converted:
A new edge 12-7 that connects the edges 6-10 and 32-15.
With this case seems that infinite random switches throughout the entire graph do not solve the impasse, which is really really really bad.
Experimenting with the Python program, I have found that if you have an F5 impasse, a single switch may not be enough to solve the impasse. It means: counterexample found.
This is a bad news, since I believed that a single switch, on an appropriate edge, could have been the key to solve the theorem.
Since a limited number of switches still work, I am going to study if two switches, on appropriate edges, may work.