Is anyone familiar with the road coloring problem?
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The road colouring problem.\n\nYou're driving to a friend's place, and find yourself at a crossroads unfamiliar to you. So, you call up your friend from your mobile phone and ask for directions. She gives you a set of left/right/keep-straight directions. You thank her, hang up and follow the directions until you get to her place. As you drive up her driveway, you realize that you hadn't even told her where you were when you called her. So, how did she manage to give you the right directions? This is an instance of the road-colouring problem. A road-colouring is a colouring of the edges of a directed graph so that at each vertex, the outgoing edges get distinct colours.  Such a colouring is called synchronizing if there is a sequence of colours (directions) that when followed takes you from any vertex to a pre-defined destination vertex. The road-colouring problem is to identify graphs that are synchronizable, i.e., graphs that admit a synchronizing road-colouring.\n\nIt is easy to prove, using a pigeon-hole principle argument, that given a strongly connected digraph and a\nproper multi-coloring of its edges, there exists a sequence of length at most n^3 that achieves its synchronizing\nnumber.\n\nFor more information see: http://www.dmtcs.org/pdfpapers/dmAE0155.pdf
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haha i did it on paint sorta messy here and there but hope it helps you =)\n\nhttp://img108.imageshack.us/my.php?image=math5tw.png
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http://members.tznet.com/donjuan/images/Map%201st%20crusade%20&%20routes%20of%20first%203.jpg
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Think of the primary colors, red, yellow, and blue, and assign each of them to the x, y, and z axis.\n\nIf you look at the planes formed by two of the three, you get a number of arbitrary points with a coordinate made up of the two components...this relates to the hues of colors.\n\nFor instance if x= red and y = yellow, then the point 3,3 would be an orange, but the point 8,1 would be more of a lighter red.\n\nFurthermore, in 3D, a point not lying on a plane, with x,y,z coordinates would be a bit of each color, producing it's own hue.
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Go to www.mapquest.com and it will figure it all out for you, even plan a route by highway.
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Green Flag's route finder is excellent\n\nhttp://www.greenflag.co.uk/routeplanning/
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As fewofmany mentioned, it's just three numbers smashed together, so you could conceivably write a function that compares them:\n\nColorX = 00FF00\nColorY = 00EE00\n\nThen, this means that ColorX has a Green component that is 255, and ColorY has a Green component that is 238. The two colors are identical in their Red and Blue components.\n\nIt's conceivable that you could then write a simple comparison check... Take the two colors' Red values & subtract them, and you'll end up with a measure of how similar their Red components are.  Do the same with Green, and you'll get a similarity measure for Green.  Finally, do the same for Blue, and you'll get a measure of the similarity of the Blue components.\n\nAdd up those "similarity" values, and you'll get some number.  In this case, the number is (0 - 0) + (255 - 238) + (0 - 0) = 17.  It seems that all you'd need to do is prevent that number from falling below a certain threshhold.\n\nI hope that makes sense...
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Here is a great article with sample problems.\n\nHope this helps!
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Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
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take it from here \n\nhttp://www.freedownloadscenter.com/Web_Authoring/Java_Programming_Tools/Larry_s_ Paint.html
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Graph X=-4 with dotted lines\n\nthen color everything below the line.
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Fermat's Last Theorem has been solved ashas the Four Color Theorem. The Poincare conjecture has recently been solved. Trisecting a general angle with unmarked straightedge and compass has been shown to be unsolvable (although it is possible with marked straightedge and compass).\n\nHigh on the list of open problems:\n\nRiemann Hypothesis.\nP=NP?\nTwin prime conjecture.\nAre there any odd perfect numbers?\nProof of existence of solutions to Navier-Stokes equations.
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That's the traffic light on the information superhighway... when it's red, all the information has to stop, it can only go on green.  \n\nThis method is commonly used to prevent data packet collisions and thus minimize loss of data and resent packets.  Nothing clogs up a system more than having to resend a bunch of packets... it can be a real mess.  \n\nEvery once in a while, you will get some rouge program (probably from CANADA) that does not adhere to the red / green convention, and just runs amok.  If this is the case, you have to call the information superhighway patrol (ISHP) and get them to write that guy a ticket.  \n\nCheers!
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150\n\nIf we first ignore that each color be used once, there are 3^5 ways to paint the train.\nThen, subtract 2^5 ways that the train could be painted with only red or blue, subtract 2^5 with only blue or yellow and subtract 2^5 ways with only red and yellow then add 3 back because we subtracted all blue , all red and all yellow twice each.\n=243 - 3*2^5 + 3 \n=243 - 96 + 3\n=150
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Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
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Unless it is a very new road you can use www.mapquest.com\n\nIt will even give you directions.
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Try runnig ethereal.  It will show you the actual traffic and you can look at stuff by port.  There mays be clues in the content.