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Hmmm... Perhaps I didnt state it right.
The pipes cannot be redirected from the houses. They will need to be pulled from the source itself :-)) otherwise, cool google plot! :) |
This makes it a whole new riddle... hehe.. Hard one actually.
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The problem you describe is to draw a bipartite graph of 3 nodes connected in all ways to 3 nodes, all embedded in the plane. The graph is called K3,3. A famous theorem of Kuratowsky says that all graphs can be embedded in the plane, EXCEPT those containing a subgraph that is topologically equivalent to K3,3 or K5 (the complete graph on 5 vertices, i.e., the graph with 5 nodes and 10 edges). So your problem is a minimal example of a graph that cannot be embedded in the plane. The proofs that K5 and K3,3 are non-planar are really quite easy, and only depend on Euler's Theorem that F-E+V=2 for a planar graph. For K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and E is 10, so F = 7. In this case each face has at least 3 edges, so E >= (F*3)/2 = 10.5, contradiction. The difficult part of Kuratowsky is the proof in the other direction! A quick, informal proof by contradiction without assuming Euler's Theorem: Using a map in which the houses are 1, 2, and 3 and the utilities are A, B, and C, there must be continuous lines that connect the buildings and divide the area into three sections bounded by the loops A-1-B-2-A, A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around* whichever loop is the outer edge of the network.) C must be in one of these three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of the loop that rings its area and hence is inaccessible to C. The usual quibble is to solve the puzzle by running one of the pipes underneath one of houses on its way to another house; the puzzle's instructions forbid crossing other *pipes*, but not crossing other *houses*. |
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here's a riddle for you not even Google can help you answer:
why france? the first one to come up with an answer will win a free ticket to belgium. |
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Give my ticket to someone else...thanks. |
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Welcome to Greenland |
Ok, last riddle from me this eve... :Graucho
I know a drifter.... He smokes used cigarettes! When he collects 7 cigaretstumps, he can make them together as one, and smoke a whole one. One day, he collects 49 stumps. How many whole cigarettes can he smoke? <br><br> |
Let me make a small riddle of my own here. :)
Joe and bob are enthuastic dart players. They decide to make a bet. They wanna measure which of the 2 is a better darts player. But since they live in different places and cannot meet regularly they decide to throw the darts in a local pub. The pubkeeper will keep records. The competition would last 2 months. When the first month is over Joe and Bob go to the pubkeeper and ask which of them has a better average on the throws for the first month. Joe is the winner with a better average. They continue into the second month. When the month is over Joe and Bob meet at the bar again and ask which of the 2 has a better average, and is a winner for the second month. Joe again has a better average. So he's declared as the winner. As Joe and Bob are leaving the pubkeeper stops them and says that Joe didn't win after all. When the 2 months are counted together Bob has a better average, and should be declared winner. Is this possible, or is the pubkeeper just drunk? If it is, how? :glugglug edit: misleading typo. |
Hey! Not a decient shot?
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7 x 49
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So, it is possible. How? |
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Any one else? |
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ok, got one for everyone...
What's black and white, and has trouble getting through revolving doors..? :) |
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My riddle too hard?
Can't find the answer in google? No surprise there.. I made the riddle up. It still is possible. |
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And it's 8 cigarettes. |
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There's 3 people involved. Pubkeeper Joe Bob Nobody else. |
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To make it easier..
the answer lies within the word 'average' |
scary thread.
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In that case here's a scenario that would work (keep in mind I have no idea how darts is scored).
In the first Month, Joe played 10 games and scored 50 points in each of them for an average of 50 points in the first month. Bob played 1 game and scored 49 points for an average of 49. Joe wins 50 to 49. In the second month, Joe played 1 game and scored 100 points for an average of 100 points. Bob played 100 games and scored 99 points for an average of 99 points. Joe wins 100 to 99. Overall however, Joe played 11 games and scored 600 points for an average of 54.5 points. Bob played 101 games and scored 9949 points for an average of 98.5 points. Bob wins 98.5 to 54.5. |
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That's the correct answer. now, wasn't that fun?! :) |
I love word puzzles and other mind games.
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What's the difference between a truckload of bowling balls, and a truckload of live babies? or How do you make a dead baby float? Glad you like the photos. That's pretty much why I do what I do... |
HERE'S ONE
There are two rooms... In the first room there is 3 light switches with a door that leads to the second room. In the second room there is 3 light bulbs. In the first room each light switch controls one light bulb in the second room. The light switches and bulbs are random. Being able to go into either room ONLY once how can you figure out which light switch controls which lightbulb? |
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