- Road coloring problem
In

graph theory the**road coloring**, known until recently as thetheorem **road coloring**, deals with synchronizedconjecture instruction s. The issue involves whether by using such instructions, one can reach or locate an object or destination from any other point within anetwork (which might be a representation of city streets or amaze ). [*cite news | last =Seigel-Itzkovich | first =Judy | title =Russian immigrant solves math puzzle | pages = | publisher =The Jerusalem Post | date =*] In the real world, this phenomenon would be as if you called a friend to ask for directions to his house, and he gave you a set of directions that worked no matter where you started from. This theorem also has implications in2008-02-08 | url =http://www.jpost.com/servlet/Satellite?cid=1202246348334&pagename=JPost%2FJPArticle%2FShowFull | accessdate = 2008-03-21symbolic dynamics .The theorem was first conjectured in 1970 by Benjamin Weiss and Roy Adler.R.L. Adler, B. Weiss. Similarity of automorphisms of the torus. Memoires of the American Mathematical Society, Vol 98. 1970.] It was proved by

Avraham Trahtman in September 2007. [*cite web | last =Trahtman | first =Avraham | title =The road coloring problem | url=http://front.math.ucdavis.edu/0709.0099 | accessdate = 2008-03-21*]**Example and intuition**The image to the right shows a

directed graph on eight vertices in which each vertex has out-degree 2. (Each vertex in this case also has in-degree 2, but that is not necessary for a synchronizing coloring to exist.) The edges of this graph have been colored red and blue to create a synchronizing coloring.For example, consider the vertex marked in yellow. No matter where in the graph you start, if you traverse all nine edges in the walk "blue-red-red—blue-red-red—blue-red-red", you will end up at the yellow vertex. Similarly, if you traverse all nine edges in the walk "blue-blue-red—blue-blue-red—blue-blue-red", you will always end up at the vertex marked in green, no matter where you started.

The road coloring theorem states that for a certain category of directed graphs, it is always possible to create such a coloring.

**Mathematical description**Let "G" be a finite

directed graph where all the vertices have the sameout-degree "k". Let "A" be the alphabet containing the letters 1, ..., "k". A "synchronizing coloring" (also known as a "collapsible coloring") in "G" is a labeling of the edges in "G" with letters from "A" such that (1) each vertex has exactly one outgoing edge with a given label and (2) for every vertex "v" in the graph, there exists a word "w" over "A" such that all paths in "G" corresponding to "w" terminate at "v".The terminology "synchronizing coloring" is due to the relation between this notion and that of a

synchronizing word infinite automata theory.For such a coloring to exist at all, it is necessary that "G" be both strongly connected and aperiodic. [

*http://www.emis.de/journals/DMTCS/pdfpapers/dmAE0155.pdf*] The road coloring problem states that these two conditions are also "sufficient" for such a coloring to exist. Therefore, the road coloring problem can be stated briefly as::"Every finite strongly-connected aperiodic directed graph of uniform out-degree has a synchronizing coloring."

**Previous partial results**Previous partial or special-case results include the following:

*If "G" is a finite strongly-connected aperiodic directed graph with no

multiple edges , and "G" contains asimple cycle of prime length which is a proper subset of "G", then "G" has a synchronizing coloring. (O'Brien 1981)*If "G" is a finite strongly-connected aperiodic directed graph (multiple edges allowed) and every vertex has the same in-degree and out-degree "k", then "G" has a synchronizing coloring. (Kari 2003)

**References***

Jarkko Kari , "Synchronizing finite automata on Eulerian digraphs", "Theoretical Computer Science " 295 (2003), 223–232.

*G. L. O'Brien , "The road-coloring problem", "Israel Journal of Mathematics ", Vol. 39, 1981.**ee also***

Four color theorem

*Graph coloring **Footnotes**

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