Bridge+of+Königsberg

=Is there a walk around the city that crosses each bridge exactly once?=

Question: Can the Bridge of Königsberg problem be solved? Explain your reasons with the aid of illustration(s). Due date: 10 September 2009 Due time: 11.55pm Group members: Your usual group Submit through the class wikispace

Answer:

Konigsberg is a city which was the capital of East Prussia but now is known as Kaliningrad in Russia. The city is built around the River Pregel where it joins another river. An island named Kniephof is in the middle of where the two rivers join. There are seven bridges that join the different parts of the city on both sides of the rivers and the island. All the bridges is needed that used to cross across all the cities. So, any city could not be reached if not using the bridge.



People tried to find a way to walk all seven bridges without crossing a bridge twice, but no one could find a way to do it. In 1735, Euler presented the solution to the problem before the Russian Academy. He explained why crossing all seven bridges without crossing a bridge twice was impossible.

To answer the question first of all, we need to convert it into graph,

media type="youtube" key="iVb0dD_OQoo" height="344" width="425" So, it is proved that there is no way we can go through all the bridges only once from one city to another city. Next, understand the question is actually asking for the Euler circuit. Euler circuit definition by Euler : The graph has a simple circuit that contain every edges. Theorem : A multigraph has Euler circuit if and only if each vertex has even degree edges.

Euler reasoned that anyone standing on land would have to have a way to get on and off,which mean thats the number of enter and out from the route is the same, Thus each land would need an even number of bridges. But in Konigsberg, each land mass had an odd number of bridges. For example, the number of bridge that can used to reach R1 is 3,which is 2,3,4. This was why all seven bridges could not be crossed without crossing one more than once.

Anyway, if add another Arc or edges into the graphic, either connecting R1 and IS2, or R2 and IS2, there'll be a solution. maybe the solution can be: R2 --> 7 --> IS2 --> 8 --> R1 --> 4 -->IS2 --> 1 --> IS1 --> 3 --> R1 --> 2 --> IS1 --> 5 --> R2 --> 6

Chieng Lung Hei Carolyne Alphonsus Ng Choon See Ng Boon Ding