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Road Coloring Problem Solved
ArieKremen writes "Israeli Avraham Trakhtman, a Russian immigrant mathematician who had been employed as a night watchman, has solved the Road Coloring problem. First posed in 1970 by Benjamin Weiss and Roy Adler, the problem posits that given a finite number of roads, one should be able to draw a map, coded in various colors, that leads to a certain destination regardless of the point of origin. The 63-year-old Trakhtman jotted down the solution in pencil in 8 pages. The problem has real-world implementation in message and traffic routing."
The guy is actually a mathematician who had to work as a security guard right after he immigrated to Israel, which is common for most immigrants. This guy had lots of formal training solving equations like this.
How soon can we see this implemented in my Garman GPS unit or Google Maps?
Life is not for the lazy.
No, today's XKCD is about the traveling salesman problem. They are very different.
Unfortunately for him, a cheap, safe, mass-market flying car was announced an hour later.
Slashdot Burying Stories About Slashdot Media Owned
Since the one in the write-up is borken. Here it is.
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He's in Israel, not the US, dude. Save your diatribe for another time.
from Wikipedia: 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. http://en.wikipedia.org/wiki/Road_Coloring_Conjecture/
not fast, but reliable. if you don't care what time you get in to cleveland, this guarantees you'll never *end* up in pittsuburgh.
wait, ohio to cleveland?
Apart from the referenced paper being some months old, the author has an extended paper with an efficient algorithm. See A Subquadratic Algorithm for Road Coloring.
1. Here's wtf the problem even is, for those of us who aren't all up in the "mathematical curiosities" business. Basically the question is, for a specific kind of graph (where you can go from point A to a finite number of points B, C, or D, etc) can you label the possible paths from each point so that, starting from anywhere, you can follow an invariable rule that will get you to a specific destination point. (Check the link, the picture makes much more sense).
2. Apparently his proof was published last September. It's "news" because it's just now hitting the semi-mainstream press. You people fail at nerddom.
Freedom isn't free; its price is the well-being of others.
I am not a mathematician either, but I think being able to provide a pattern that allows you to reach any given point in the graph would allow for faster switching at the nodes (once you reach certain speeds switching becomes the bottleneck). The problem isn't concerned with efficiency but reachability, anyway.
"Some people think they need to be complicated. I think they need to be nice and simple."
The man is a True Genius. Insight to all of out out there, that is the proper way of thinking.
Do not look at laser with remaining good eye.
Actually, if he were to put it on a SD card, it'd take up less than 1 square inch...
Where's my PEDANTIC:-1 option?
- Despite popular opinion, I am not perfect.
In Soviet Russia they say that because Americans were so poor mathematicians, they had to invent the computer...
The Wise adapts himself to the world. The Fool adapts the world to himself. Therefore, all progress depends on the Fool.
If you RTFA (yes, I must be new here) he worked as a night watchman when he first moved to Isreal. He's been working in mathematics for over a decade.
Originally from Yekaterinburg, Russia, Trahtman was an accomplished mathematician when he came to Israel in 1992, at age 48. But like many immigrants in the wave that followed the breakup of the Soviet Union, he struggled to find work in the Jewish state and was forced into stints working maintenance and security before landing a teaching position at Bar Ilan in 1995.
You might as well say the 2002 Nobel prize in economics went to a lieutenant in the army. It's just a minor detail that he was a lieutenant 50 years before winning the prize.
According to Wikipedia, he is a mathematician at Bar-Ilan University in Israel. Here is his homepage hosted by the university. Maybe he was a night watchman, but it looks to me like he's a professor now...
Forty-two.
Justice is the sheep getting arrested while an impartial judge declares the vote void.
Yeah... this would only apply to the XKCD comic if the road coloring problem was NP-complete.
The question behind road coloring is this: Given a directed network with out-degree 2 (from any place you can get to exactly two other places), we want to color the edges leading out of each node red or blue so the following "universal directions" condition exists: For any final destination A, there is a set of directions (e.g. take red three times, then blue twice, then red again) that gets you to A no matter where you started from. It may not be the shortest path, but you'll get there. There is one obvious obstruction and one slightly less obvious obstruction: If your network is disconnected so you can't get to A from your starting point no matter what path you follow, you're clearly stuck. On the other hand, there's also a "periodicity" obstruction if all of the cycles of the graph are multiples of the same number. For example, suppose that you were trying to give universal directions for a square, where the roads in question connected every vertex to its two neighbors. If I want to go from my starting point to an adjacent vertex, I have to take an odd number of steps. If I want to go from my starting point to the opposite vertex, I have to take an even number of steps. This means I can't even know how many steps I have to take (let along which steps) unless I knew where I started. It was conjectured, and Trahtman showed, that these two are the only possible obstructions. In particular, he even gives an algorithm for figuring out how to label the roads quickly. I guess the applications I'd see in this are for algorithms and the design of autonomous systems. The idea here is that, if the robot gets stuck somewhere in a multi-step procedure, you may want it to restart from the beginning. However, this can be difficult if the robot does not know where it is in the procedure, or even where it is physically. Trahtman's algorithm could allow you to exchange some computation at the beginning of the procedure (which can be done before the robot goes out, for example), for this "reset" functionality. I don't know whether this is feasible or not though.
It's a theorem which states that if a graph is strongly connected and aperiodic, then there exists a road coloring. It was conjectured by Weiss and proven by Trahtman 30 years later, which is what the article is about. A graph that is periodic may still have a road coloring, but that's not the scope of the theorem.
Most roads are strongly connected but periodic (not aperiodic), in the sense that you can drive around a block in 4 segments, and drive around 2 blocks in 6. The period in this case is 2. You can't guarantee a computer network to be aperiodic either. These graphs may still have a road coloring, but the theorem doesn't apply. Therefore, this theorem has little application in practice.
I haven't read the paper, but there are generally two ways to prove the theorem: (1) show a coloring algorithm that makes only the assumptions of strong connectivity and aperiodicity, or (2) show the contraposition, that if there is no road coloring, then it implies either the graph is not strongly connected or there is period. Only (1) is useful in practice because you have a method to generate road coloring and instructions to reach all vertices, but it's harder to verify that (1) is correct. (2) is not very useful because although you have a proof, but you don't end up with an algorithm that generates road coloring.
In practice, algorithms get things done. Proofs are only certificates.
I once had a signature.
Proof at last, that that guy who kept saying, "You can't get there from here" is a bloody liar.
Mir tut es leid, Menschen daß Einfältigfehlersuchenbaumfolgendenaffen sind.
Based upon his ground-breaking research on the road coloring problem, night watchman Trakhtman has been promoted to the Department of Transportation pavement striping crew.
Have gnu, will travel.
More Twoson than Cupertino
The road colouring problem is not 'an equation'.
'deterministic automaton' basically means: computer program which doesn't have any form of randomness involved. If you're in this particular state in the program and you do this particular thing then always the same thing happens.
The road colouring problem amounts to: For any program, I claim that there is one single sequence of actions so that if you do this sequence of actions, from whatever start point, when you finish you will be in a specific state (`target', say) of the program.
If you take any real normal program, it's `obvious' what to do: for example if you're looking at 'edit' and you want a typed copy of Shakespeare in Times 12pt, you hit backspace as many times as edit can have characters (lots, but it's a finite number) then you go through the menus by keypress and set each style to the correct thing, then you start typing in the works of Shakespeare which eventually gives you the target state. And it doesn't matter if the monitor was off, you know that this method definitely works, whatever state edit was in when you got there, whether it was in the blank just loaded state, or whether it was editing font size in War and Peace, whatever.
It is not so easy to prove that in fact for any program you can find such a sequence.
In fact, it isn't even true. A really simple example is the program which every time you press a key changes the screen between red and blue. Now you're supposed to give a sequence of actions - keystrokes - which are guaranteed to end up with the screen blue (target state). But really what the keystrokes are is irrelevant. This program doesn't care if you press 'a' or 'ESC', it just changes screen colour every stroke. So really your sequence of actions comes down to 'press keys however many times'. How many times? Well, if the screen starts red (and remember the monitor is off, so you don't know if that's true) then you'd need to press keys an odd number of times to get it to end up blue. So you have to do that. But then what happens if the screen started blue? It ends up red. This sequence of actions doesn't always reach the target state - and there cannot be any sequence of actions which does.
So you have to impose a condition to make the result true. In the example I gave, the program has two states and it loops between them. It's a program with exactly one loop, of length two, and the greatest common divisor (GCD) of these loops is of course 2. Basically, having the ability to loop between states like this (or in a loop of length 3, or 4, or whatever) is the only 'obvious' barrier to having a sequence of actions which goes to a target state. So you impose a condition: the GCD of all the loops is 1 (this doesn't mean there are no loops, but the guy proves it means you can find a way out of all the loops). If you don't have that, then you can find automatons (programs) which work like the example and you cannot possibly have the sequence of actions you need. If you do have it, there is no 'obvious' barrier to the sequence of actions existing. And what this paper shows is that there isn't any hidden catch: once you get rid of the 'obvious' barrier then you can find a sequence of actions which definitely puts the program you're looking at into the target state.