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A little game to relax your mind.


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Not exactly a science project or homework, but we can try and solve it scientifically.


Where do we start from?


1- Upper corners

2- Lower corners

3- Upper junctions

4- Lower junction

5- Mid row left/right junction

6- Middle row middle junction


Use all the possibilities. If you find none, then this is impossible.

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This could be solved if you took a sledge hammer and made another door to the outside from one of rooms with an odd number of doors :naughty:


I imagined putting a chair in each room and one outside, labeled the chairs A to thru F. Each time you enter a room or the outside sit in the chair. I made a simplified schematic with points A-F representing the chairs and connected them with lines representing each possible path thru a door and checked off the doors one by one. Then I made a table counting the number of lines at each point. A 4, B 5, C 4, D 5, E 5,F 9. Same thing. You can have no more than 2 points with an odd number of paths, start and finish. Every other point has to be an even number of connecting paths, pairs entering & leaving. Either that or form a closed loop and all the rooms have to have an even number of doors. Cool puzzle.


Can you use a hyperdimensional pencil?

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You will be forever trying to solve this one. It has been around at least 100 years. No one has solved it yet. Tens of Thousands and perhaps millions have tried this.


My Dad showed me this one when I was a kid. He learned it when he was a kid. I used to have fun with a few people who thought they were smart. This puzzle would stump them every time.:naughty:

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This problem is analogous to the problem of Koenigsberg bridges, solved by Euler.

If you let each field and the area outside be points/vertices and doors be represented by lines/edges connecting these points, then you get a so called graph. Euler proved that in order for the graph to have the property described above, i.e. in order to be able to go through each door once, each point/vertex must have an even number of edges meeting in this vertex. Now, in our case we've got only 2 such vertices out of 6 ... and hence we can be sure it won't work.

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