EPTAlthough then it would be too easy so it's probably wrong.
Yet Another Math Puzzle - Page 2
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Nytefish
United Kingdom4282 Posts
Although then it would be too easy so it's probably wrong. | ||
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drift0ut
United Kingdom691 Posts
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goswser
United States3548 Posts
+ Show Spoiler + I think there will be a finite number equal to ((n - x)^2)(x) where n is the number of lines on the plane and x is the number of parallel lines on the plane. My logic was that the number of lines squared gives you the number of T's in a plane since each line will intersect with each other non-parallel line once. Since lines can be parallel, each parallel line will intersect with each line non-parallel to it once also, except parallel lines cannot intersect with each other. Anyways not sure if I'm right. | ||
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silynxer
Germany439 Posts
There are at least four rational intervalls [a,b],[a',b'] (on the x-axis) and [c,d],[c',d'] on the y-axis so that the projection of a given spider is within [a,b],[c,d] but not within [a',b'],[c',d'] so that no other spider has this feature. | ||
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jtan
Sweden5891 Posts
On May 13 2009 06:14 drift0ut wrote: any disc will contain a point of QxQ so we can count them You still have to show that you cannot put an uncountable number of spiders in any disk in RxR | ||
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qrs
United States3637 Posts
To keep things simple, let's take a hypothetical spider and connect two of its endpoints with an imaginary line to create an imaginary triangles of finite area. Now divide that triangle into two halves (also triangles of finite area). Place a smaller spider into one of these triangles, then divide the other one in half again, and repeat. Clearly, we can always find a spider to fit inside an arbitrary triangle--just make sure each endpoint lies within the triangle. This procedure shows that for any spider, we can generate a (countably) infinite number of smaller spiders inside it. But each of the (infinite number of) smaller spiders can similarly house an infinite number of spiders smaller than it within its hypothetical triangle. But I got stuck trying to prove that an infinity to the power of an infinity must be uncountable. (Starting writing this about 3 hours ago: what I thought would be a trivial bit ended up stumping me. May as well post what I have; it's a start, but it's not a full answer.) edit: PS- I was trying to prove that the number must be uncountable; it looks like some of the others were trying to prove that it must be countable. | ||
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silynxer
Germany439 Posts
I have to take back what I said: because of closed/open line sections there are at maximum 2 spiders in such intervalls, still it remains countable. | ||
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qrs
United States3637 Posts
I still wasn't able to prove that that is uncountable, but it certainly seems harder to count when you divide by infinity at each step instead of dividing by some constant at each step (which would be equivalent to some subset of rational numbers). | ||
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Muirhead
United States557 Posts
Suppose we have a three-legged spider that looks like this: (with A,B, and C the endpoints of the legs and O the common point of all three legs) A O B C Construct a sequence of three-legged spiders as follows. Choose a sequence of points O_1, O_2, O_3,... approaching B from the lower left. Suppose that O_i is distance 1/i from B. Let A_i be 1/i units to the left of A. Let B_i be a small amount below A_i (exact amount will be clear from context). Let C_i be 1/i units to the left of C. Let S_i be the spider with center O_i and legs ending at A_i,B_i, and C_i. The S_i together with the original spider are all disjoint. Suppose you choose a,b,c,d,a',b',c', and d' for the original spider. It seems that infinitely many of the S_i (not just S) fit into [a,b] times [c,d] but not into [a',b'] times [c',d']. If you explain this it would help. Thanks! | ||
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Muirhead
United States557 Posts
drift0ut your answer is a little convoluted considering its spread out over multiple posts, but I think the same problem as with silynxer's solution occurs? | ||
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qrs
United States3637 Posts
On May 13 2009 06:55 Muirhead wrote: qrs unfortunately your solution still generates only countably many spiders. Your set of spiders is in bijection with the set of finite sequences of natural numbers, which is countable. Oh, it is countable? Well that explains why I was having trouble proving otherwise, at least. Thanks for enlightening me; otherwise I would probably have wasted another hour thinking about it. | ||
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silynxer
Germany439 Posts
And by dividing each subset in constant other subsets in each step you can still reach uncountable limits (see Cantor set, or my example only that the limits are sadly not three-legged spiders). I'm not sure I understand your construction Muirhead but I discovered a flaw myself though I'm still not sure the flaw is a flaw. Well I did a lot of rash guessing this time and now it's getting late so if by tomorrow this is not solved in a satisfiable way I'll give it a more serious shot  Good night. | ||
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ninjafetus
United States231 Posts
Actually, I think the fact that it has 3 disjoint legs is exactly what gives you the fact that it has both a rational x AND y coordinate somewhere. To be more specific, consider a spider with one leg each on an irrational x and irrational y coordinate parallel to the axis. Any 3rd leg (which must be a distinct leg) HAS to have a rational coordinate. And that's where you can associate them with Q*Q so, yeah, countable. | ||
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DeathSpank
United States1029 Posts
http://en.wikipedia.org/wiki/Countable_set http://en.wikipedia.org/wiki/Injective_function | ||
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qrs
United States3637 Posts
On May 13 2009 07:47 DeathSpank wrote: I don't know what you guys are talking about, I already solved this mofo! its an injective function and is therefore countable! Gosh! http://en.wikipedia.org/wiki/Countable_set http://en.wikipedia.org/wiki/Injective_function um, if you're quoting wikipedia, you skipped the words "from S to the natural numbers". When did you show that an injective function from the set of spiders to the natural numbers exists? | ||
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Cascade
Australia5405 Posts
the injective function has to be from the set to the set of NATURAL NUMBERS, which you will see if you read more than the first line of your link.  So please give us an injective function from any possible set of spiders to the natural numbers, and you have solved the problem. EDIT: ok, so what qrs said in other words. on topic: it is enough to show it for a limited set of the plane, since a solution on the infinite plane can be mapped to a bounded space by for example keeping r < 1, and mapping r > 1 to 2 - 1/r, which comntinuosly maps the plane to the r < 2 circle. Not sure if that helps though. can't find a solution straight away, but i'm getting curious about it now. Thanks for the interesting problem, and fuck you for stealing my sleep!!! | ||
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Muirhead
United States557 Posts
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ShOoTiNg_SpElLs
Korea (South)690 Posts
On May 13 2009 07:45 ninjafetus wrote: I'm pretty sure the fact that we have 3 legs and 2 dimensions is absolutely necessary. If we had 2 legged spiders in 2d, we could, for example, have a cantor set running along the line y=x and have a 2 legged spider at each point with one leg pointing up and one leg pointing right. (Actually... if we had k<d legged in n dimensions we could always do something similar.) Actually, I think the fact that it has 3 disjoint legs is exactly what gives you the fact that it has both a rational x AND y coordinate somewhere. To be more specific, consider a spider with one leg each on an irrational x and irrational y coordinate parallel to the axis. Any 3rd leg (which must be a distinct leg) HAS to have a rational coordinate. And that's where you can associate them with Q*Q so, yeah, countable. No, T is a valid spider according to the OP, so if the horizontal line lies on x = sqrt 2, the vertical line on y = sqrt 3, and the center at (sqrt2, sqrt3), then every point on the spider has at least one irrational x or y. | ||
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DeathSpank
United States1029 Posts
HERE spider spider loc a spider - - - ->(x,y) another spider---------(x2,y2) ETC>>>>>>>>>>> GOSH GUYS | ||
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flag
United States228 Posts
it doesnt matter that they are spiders, it can count anything which cannot have the same coordinates for multiple things. | ||
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