I don't think so...
Roots
Flower
Of course, you're correct. I was thinking of a Romanesco Cauliflower. I have no idea why I confused the two
Let's discuss Mathematics
- Thread starter ParadigmShifter
- Start date
Of course, you're correct. I was thinking of a Romanesco Cauliflower. I have no idea why I confused the two
IdiotsOpposite
Boom, headshot.
That's good. Thanks for the info.
ThinkTank
RL Addict
Correct.
Also correct, this in fact occurred to me this morning in the bus to work ...
ThinkTank
RL Addict
Hmmm. Maybe we have an uncomputable sequence then? This does not follow of course, but it would explain why it is not in the encyclopedia. I'm not a professional mathematician, but I do know that tiling problems of known complexity or (un-)decidability are often used in reduction arguments establishing the complexity or (un-)decidability of other problems.
I was sort of guessing that the symmetry restrictions were so strong as to make the problem at least decidable. That guess is still on, but ...
ThinkTank
RL Addict
We probably need to dive into this http://en.wikipedia.org/wiki/Polyomino ...
What do you mean by "uncomputable"?
For any given n, we can give an upper bound on the time it will take to know if n is in our set or not. In less precise terms, this is a "finite" problem. It is decidable. It might still be a "hard" problem.
As for the strong restrictions making the problem decidable, this is a fallacy.
Adding restrictions can make the problem harder.
For example, if we remove the symmetry conditions, then the problem is trivial. EVERY n has the property that there exist a tile with n squares than can tile the plane (take the 1 by n rectangle for example). Adding the symmetry condition makes the problem harder, because it interacts in non-trivial way with the tiling property.
Doesn't seem to be much there that is applicable, except that there are rather few polyomino's with full dihedral symmetry group, which is obvious by simply playing with them.
(note also the easily explainable numerical fact that these only exist when n is 0 or 1 mod 4).
Possibly the references from "Tiling the plane with copies of a single polyomino" might have something of interest.
By the way, was there an interesting motivation to this problem?
ThinkTank
RL Addict
There's a number of (equivalent) ways ways to make that precise, but let's go for this one: the enumeration function of the sequence (f(n) = the n-th number in the sequence) is recursive (i.e. μ-recursive). But I'll settle for any of the equivalent characterizations of computable function.
I'm not so sure. This is exactly the problem for it to be computable/decidable. For that there must be a uniform procedure that works in all cases. I do not think you can give that upper bound without such a procedure. But I'm open to suggestions. Proofs even better.
I think one approach would be to isolate some form of periodic property any such tiling must have; and then predict that if you have not found such periodicity in some number of steps (that number depending on the size and form of the tile), you will not find it anymore.
I'm not so sure here either. Adding restrictions can also make the problem easier. Certainly the assumption that all tiles are squares is stronger than our current assumption - and it also makes the problem trivial.
ThinkTank
RL Addict
Yes, I was especially thinking of that but maybe some of the concepts in other sections may also be useful.
Well, the two 13 tiles in my first post came out of discussions on civ 3 city placement. I somehow wondered if it was possible to characterize the class of similar shapes that also tile the plane - but remove the civ related restictions.
I'm not so sure. This is exactly the problem for it to be computable/decidable. For that there must be a uniform procedure that works in all cases. I do not think you can give that upper bound without such a procedure. But I'm open to suggestions. Proofs even better.
My suggestion would be to try to prove that if there is a tiling, then there is a lattice one, with a rather small fundamental area (in terms of number of tiles in it).
I actually believe this to be true, and I believe it would imply decidability.
All I said was that "Adding restrictions CAN make the problem harder."
It can also make the problem easier, as you say. What are you "not so sure" about?
ParadigmShifter
Random Nonsense Generator
Tonight, I was wondering if there is a simple relationship between circles and coins.
You can surround a circle of radius 1 with 6 coins of radius 1, all touching.
What is the radius of the centre coin for 7, 8, etc. surrounding coins all touching? Is there an analytical solution?
EDIT: Must be easy to calculate from the vertices of regular n-gons.
You can surround a circle of radius 1 with 6 coins of radius 1, all touching.
What is the radius of the centre coin for 7, 8, etc. surrounding coins all touching? Is there an analytical solution?
EDIT: Must be easy to calculate from the vertices of regular n-gons.
Quackers
The Frog
does anybody have a sure-fire proof method of re-arranging equations? like some rules...?
*anticipates abuse from PS*
*anticipates abuse from PS*
PS: Draw the big circle with radius R and center A, and small circles touching each others and the big one with radii r. Take two small circles that touch each others, and call their centers B and C. Then draw a triangle ABC.
The vertices AB and AC are R+r and BC is 2r, the angle at A is 2pi/n. Split the triangle at A, you'll get a right triangle with hypotenuse R+r, angle pi/n and the opposite side r. From that you get
tan(pi/n)=r/(R+r),
and furthermore
(R+r)tan(pi/n)=r
and
R=r [1-tan(pi/n)]/tan(pi/n).
So any R bigger than that will do.
I bet you didn't even try
Quackers:, can you be little more specific, and also tell what you've been told at school?
EDIT: sin of course, not tan
The vertices AB and AC are R+r and BC is 2r, the angle at A is 2pi/n. Split the triangle at A, you'll get a right triangle with hypotenuse R+r, angle pi/n and the opposite side r. From that you get
tan(pi/n)=r/(R+r),
and furthermore
(R+r)tan(pi/n)=r
and
R=r [1-tan(pi/n)]/tan(pi/n).
So any R bigger than that will do.
I bet you didn't even try
Quackers:, can you be little more specific, and also tell what you've been told at school?
EDIT: sin of course, not tan
Do you mean for solving them? Like [wiki]Gaussian elimination[/wiki]?
ParadigmShifter
Random Nonsense Generator
I think quackers means just algebra, rearranging the equation so it says x=<stuff> rather than y=<stuff> or similar.
An example would be handy to show the general principle of it.
Wolfram alpha will do it though
e.g.
http://www.wolframalpha.com/input/?i=solve+y=12x^2+2x+for+x
An example would be handy to show the general principle of it.
Wolfram alpha will do it though
e.g.
http://www.wolframalpha.com/input/?i=solve+y=12x^2+2x+for+x
Mise
isle of lucy
"Whatever you do to one side you must do to the other"?
IdiotsOpposite
Boom, headshot.
Partial fraction decomposition is the devil!
Quackers
The Frog
yeah thats what I mean.
will have a look at the link
some easy ones I can work out. but stuff like
express x in terms of y when
y=x-3/x+2
IdiotsOpposite
Boom, headshot.
Multiply both sides by (x+2) to get a linear equation in x. Then solve for x.
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