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Graph Theory A Path to Riches?. Jonathan Choate Groton School Jchoate@groton.org www.zebragraph.com. The Birth of Graph Theory. Leonhard Euler solved the famous Seven Bridges of Königsberg problem in 1736. Leonhard Euler (1707 – 1783).
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Graph TheoryA Path to Riches? Jonathan Choate Groton School Jchoate@groton.org www.zebragraph.com
The Birth of Graph Theory Leonhard Euler solved the famous Seven Bridges of Königsberg problem in 1736.
Leonhard Euler (1707 – 1783) The Euler Question: Does the graph have an Euler Path or Circuit? Is there a path that goes along each edge once and only once?
Sir William Rowan Hamilton (1805 - 1865) The Hamilton Question:Does there exist a path or circuit that goes through each vertex once and only once?
The Million Dollar Prize The Clay Institute has offered a $1,000,000 prize to anyone who can solve what is known as the P = NP? problem. The P = NP? problem has to do with how hard problems are.
A problem is considered to be “easy” if it can be solved in polynomial time. For example, How many handshakes are possible in a group of n people? There are n(n-1)/2 possible handshakes so this problem is “easy”.
A problem is “hard” if it takes lots and lots of time to solve or becomes impossible to solve in a finite amount of time.
There is a third type of problem that is referred to as an NP problem. This is a problem that may take huge amounts of time to solve and has no known general solution but for which it is easy to check whether a given solution works.
The P=NP? Problem asks whether or not in fact the set of P problems is equal to the set of NP problems. If you can show that a known hard problem has in fact a general solution you will be well on your way to winning the prize.
http://www.google.com/patents?id=nO8tAAAAEBAJ&dq=martin+cooperhttp://www.google.com/patents?id=nO8tAAAAEBAJ&dq=martin+cooper
In order to serve the most customers the average cellsize is roughly 10 square miles. Each cell can service approximately 76 users at once because - Each cell is allotted 832 frequencies or channels - 42 channels are used for control issues- 790 are available for voice and data transmission.- Cell phones are duplex devises and need 2 frequencies per user unlike walkie talkies.
-Each cell is surrounded by 6 other cells so in order to avoid interference issues there has to be seven separate sets of frequencies. -395/7 is roughly 76 so for each cell there are 76 sets of frequencies so each cell can handle 76 users at once. - In a 7 cell cluster, 526 people can be handled.
http://www.doc.ic.ac.uk/~nd/surprise_96/journal/vol1/pr4/article1.html#Cellshttp://www.doc.ic.ac.uk/~nd/surprise_96/journal/vol1/pr4/article1.html#Cells
Big Problem: How do you determine what the signal coverage zones are when in reality you can’t have perfect hexagonal regions?
One solution to this problem is to construct for each tower the region formed by all the points nearest to that tower. These are called Vornoi Regions.
Fortune’s Method • For more than 3 points, finding Vornoi Regions is very hard. In 1986, Steven Fortune came up with an ingenious way of finding them making use of parabolas.
Given a line D, the directrix and a point F, the Focus, not on D, the set of points equidistant from F and D form a parabola.
Here’s how to implement Fortune’s Algorithm using Geometer’s Sketchpad Step 1. Open a file and select the graph option. Plot the points which represent the tower locations Step 2. Construct a vertical line with a movable point D. Through D construct a horizontal line. This will serve as a movable Directrix. Step 3. Create the function which will plot the parabola with the upper most tower point and plot it.
How Does Your Cell Phone Keep Track of Where You or Your Friends Are?
http://www.u-blox.com/images/stories/Resources/gps_compendiumgps-x-02007.pdfhttp://www.u-blox.com/images/stories/Resources/gps_compendiumgps-x-02007.pdf
The College Mathematics Journal Vol. 42, No. 2, March 2011 The Symmedian Point:...
To construct the Symmedian Point S for triangle ABC • Construct the centroid G 2. Reflect G about the angle bisector of angle BAC creating point G1. Create ray AG1 and hide the angle bisector. 3. Repeat for vertices B and C, creating rays BG2 and CG3. 4. Rays AG1, BG2 and CG3 are concurrent at the Symmedian point S