THIS PUZZLE HAS NO SOLUTION!!!
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THIS PUZZLE HAS NO SOLUTION!!!
I don't have a lot of patience for simple puzzles that probably have an obvious solution.... you try!
PUZZLE!
PUZZLE!
THE DEBATE IS OVER...
JESUS SAVES!!!
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Re: THIS PUZZLE HAS NO SOLUTION!!!
jay_a2j wrote:I don't have a lot of patience for simple puzzles that probably have an obvious solution.... you try!
PUZZLE!
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Re: THIS PUZZLE HAS NO SOLUTION!!!
dammit its blockedjay_a2j wrote:I don't have a lot of patience for simple puzzles that probably have an obvious solution.... you try!
PUZZLE!
i'll try it either when i go to clark or go home
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- Daring Overlord5
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Daring Overlord5 wrote:a 3D cylinder will do the trick
Actually, a cylinder will not do the trick (assuming you mean the surface of a cylinder). A cylinder is topologically identical to a plane with a hole. Removing a point from a plane will not allow this problem to be solved. No solution to this puzzle exists in any 2-d geometry.
Here is a simple proof that no 2-d solution exists.
Let A1, A2, A3 be houses; B1, B2, B3 be utilities.
Connect B1 with A1 and A2.
Connect B2 with A1 and A2, without crossing previous lines.
Thus, A1 - B1 - A2 - B2 - A1 represent a closed path encircling some region of space; call this region R1.
Case 1 - One of A3/B3 is inside R1, the other is outside R1.
Thus, since R1 represents an enclosed region, there is no path which connects A3 and B3 without crossing the boundary.
Case 2 - Either both A3 and B3 are inside R1, or both are outside. WLOG, assume both are inside.
Connect B1 and B2 to A3. Thus, R1 has been bifurcated into two new regions, R2 = A1 - B1 - A3 - B2 - A1; and R3 = A2 - B1 - A3 - B2 - A2.
Then B3 must be inside either R2 or R3; WLOG assume B3 is inside R2.
Note that A2 is outside R2. Thus, there is no path which connects B3 and A2 without crossing a boundary.
Therefore, in all cases, there will be at least one connection which cannot be made.
QED.
This proof works for any 2-d geometry. It's not quite as rigorous as it could be; if I had the inclination and the time, I could build a similar proof working from the definitions of open and closed sets in a 2-d metric space, but frankly I don't care. This is a problem than any high-school geometry student ought to be able to solve by construction; it disappoints me that the answer is not readily apparent to all here.
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