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Contents 1 September 3 1.1 Calculator 1.2 A blanket for a worm

[edit] September 3

[edit] Calculator

Sorry for asking a question which is not technically maths, and which is not really objective.

I'd like to buy a calculator - I'd like it to do as much as possible, but at the least have: graph drawing functions, complex number handling and programmability. It should also be as cheap as possible.

Any suggestions? —Daniel (‽) 08:27, 3 September 2006 (UTC)

The TI 83+ is an amazing calculator. It graphs in 2D, does complex numbers, programs like a dream, and does many statistical functions. It also handles normal trig, arctrig, hyperbolic trig. You can probably find one used on the internet for around $50 (US).

The TI 89 is pricier, but can do much more, much faster. With more RAM and the fact that it is extremely computer compatible, if you want ease, go with that.

I literally do not go anywhere without my TI 83+. M.manary 16:43, 3 September 2006 (UTC)

I'd suggest supplementing your new graphing calculator with something like [1] - it's extremely easy to use, and can graph almost anything you can think of in your mind, even things like x^2 + sin(xy) = 0. --HappyCamper 17:14, 3 September 2006 (UTC)

You can actually get many calculators 'free' and 'try before you buy' in that Linux calc utility actually mimics various different models including some TI ones - not sure exact models since not on normal machine, and they may be old ones. Rentwa 18:39, 4 September 2006 (UTC)

If you really know what you're doing, wikipedia can be made into a good scientific calculator:) A TI83+ is an indispensable tool, but an 89 is faster, and more versatile. Of course like most things that are faster, and more versatile, it's more expensive. Might I suggest before you buy any calculator, you find someone who has it, and borrow it a while to get a feel for it, see which one fits you better before spending money on it. --VectorPotential^{71.247.243.173} 21:02, 4 September 2006 (UTC)

[edit] A blanket for a worm

Suppose we have a flexible worm of unit length of negligible width, and we wanted to find the smallest blanket which could cover the worm in whatever configuration it was in on the plane.

Now, without doing much thinking, the smallest blanket must be less than a circle of radius 1. However, how much better can we do? And is this problem something that has already been solved? --HappyCamper 17:28, 3 September 2006 (UTC)

A circle of radius 1/2 is also sufficent: you just have to place the circle so that its centre is the midpoint of the worm, as no point of the worm can be farther from that point then 1/2. – b_jonas 18:42, 3 September 2006 (UTC)

If by "flexible" and "negligible width" you mean that we're allowed to curl the worm into arbitrarily tight spirals, then any blanket (with a nonempty interior) will do. So there is no smallest blanket. Melchoir 20:00, 3 September 2006 (UTC)

I remember reading a whole chapter about this problem some years ago. I no longer have the book, but it was something like "Game set and Math", by Ian Stewart. Unfortunately, I cannot recall the final conclusion - sorry. Madmath789 20:11, 3 September 2006 (UTC)

Melchoir, I have the impression that HC thinks of a rigid blanket. HC, correct me if that's not what you think of. – b_jonas 20:15, 3 September 2006 (UTC)

Think of it like this: we have to decide on the shape and size of the blanket before the worm decides what shape it is going to lie in :-) Madmath789 20:29, 3 September 2006 (UTC)

OK, but the orientation of the blanket is what we have left to determine once the worm has installed themselves, right? That's the only way I can make a game come out of this. —Bromskloss 21:30, 3 September 2006 (UTC)

So, a disk of radius 1/2 suffices, but it seems it could be trimmed down more, because there's no way the worm is reaching the edge of the disk except at two points. What if you began with a disk of radius 1/2 centered at the origin of the xy-plane, and then removed the part above the line x+y = 1/2? Would the resulting rug still suffice to cover up the worm, in any contortion? -GTBacchus^(talk) 21:39, 3 September 2006 (UTC)

(Edit conflicts!) As I understand it, the worm is able to be in any configuration at any instant, and our blanket should be able to cover it entirely at all times. The blanket has to be able to cover the worm when it's entirely straight, and when it's say, curled up in a spiral. Suppose we define S to be the set of all continuous paths of unit length. Suppose we overlap these paths in a contiguous manner and ask - "What is the smallest convex hull that will cover the set of overlapping paths?" - maybe this is a better way of phrasing the problem?

I think b_jonas is on the right track. The words "convex hull" probably make a difference for the question. If it is absent, then I suppose, it is quite possible that the area of the blanket is rigorously zero, but I'd like a proof of this if there is one. I vaguely remember something that is called the "Kakela problem" (or at least something that sounds like this) - but I haven't been able to find this on Google. Maybe this helps identify the problem too? --HappyCamper 21:43, 3 September 2006 (UTC)

(I'm coming to this really late, but for the benefit of future eyes: the Kakeya needle problem is not the same thing as Moser's worm problem or Lebesgue's universal cover for sets of unit diameter. Lunch 04:51, 26 June 2007 (UTC))

Your rephrasing seems correct, HappyCamper, except I'm a little unclear what you mean by "Supose we overlap these paths in a contiguous manner". -GTBacchus^(talk) 21:51, 3 September 2006 (UTC)

Yes, that does not make much sense. I'm trying say that there should be some sort of connectivity. There should be precisely one pile of overlapping paths. --HappyCamper 22:01, 3 September 2006 (UTC)

Ok, but we're free to move each path around isometrically before adding it to the pile, right? Otherwise the disk of radius 1/2 would certainly be the smallest. So, we get to pile up the paths in some efficient way, and then draw a convex hull around that pile, right? -GTBacchus^(talk) 22:09, 3 September 2006 (UTC)

According to one of my teachers circa 1979, the smallest cover then known was a sector of a circle, a bit less than a semicircle. I don't see anything relevant in Mathworld. —Tamfang 22:22, 3 September 2006 (UTC)

Pin the center of the worm, and consider positions like the hands of a clock. We must be able to cover 180° and 90° and 30° and 0° and anything between. Assume we are allowed to rotate the cover. What does that tell us? --KSmrq^T 23:36, 3 September 2006 (UTC)

If the worm only bent at its center, and were restricted to straight lines for its two halves, then a semicircle would do. However, the worm could coil about in a way that we would need 360° of coverage close to the center, and only a small arc would need to be covered when you get close to 1/2 unit from the center. -GTBacchus^(talk) 23:44, 3 September 2006 (UTC)

Some results and references are given here, unfortunately it's in German, but this text should also be in the corresponding issue of Scientific American. In particular, Stewart states that a semi-circle is sufficient and that the best-known solution in 1997 has area 0.27523, reference: Rick Norwood, George Poole and Michael Laidacker: The Worm Problem of Leo Moser. Discrete and Computational Geometry 7/2 (1992) 153–162 (PDF).--gwaihir 00:46, 4 September 2006 (UTC)

Awesome! That last PDF file is quite useful. Would this problem be worthy of an article on Wikipedia? Sort of a neat problem, and I can sort of see this become a featured article one day if written up properly. The shape proposed is rather unintuitive I think. --HappyCamper 01:53, 4 September 2006 (UTC)

Sounds like the Sofa Problem to me. Rentwa 11:05, 4 September 2006 (UTC)

While it has a similar feel, it is an entirely different problem. Gerver's sofa is not convex.[2] Here the best known solution is convex (and my intuition tells me that the best solution, once we find it, will prove to be convex). For the Moving Sofa problem we try to maximize the area. Here we try to minimize the area. --Lambiam^Talk 16:34, 4 September 2006 (UTC)

Not sure how convexness affects problem, but I'll give you the benefit of the doubt. How is the proof done? Is it at all elegant, or is it an horrific kludge? Any ideas? Good joke btw. Rentwa 18:47, 4 September 2006 (UTC)

(and Lambiam was silent) What? Too horrific to mention? :) Rentwa 08:41, 5 September 2006 (UTC)

Seeing as that these worms may as well be called "snakes", and that this takes place on the plane, we may call this the Snakes on a Plane problem. --Lambiam^Talk 16:20, 4 September 2006 (UTC)

Lol. – b_jonas 17:17, 4 September 2006 (UTC)

lol - Snakes on a Plane... --HappyCamper 23:15, 6 September 2006 (UTC)

-Well done. I will be working on this, though I may not post my results. M.manary 21:56, 4 September 2006 (UTC)

I think the right statement is this: Find a set, B, such that for any connected path of unit length, W, there is either (a) a congruent copy of the path W that is a subset of B or (b) a congruent copy of the reflection of W that is a subset of B. This allows that the blanket may cover if it is translated or rotated and we can flip the blanket over. If we don't wish to allow flipping the blanket, then we disallow the case (b).

Note however, that we have freedom in the classification of W. Is it smooth? differentiable? continuous? connected? May it self-intersect? Does it have bounded curvature? I think there are a couple of easy things to prove in some of these cases...

Connected is relatively weak. Almost any fractal path is allowed, so the blanket will have to have positive measure somewhere (to cover the zoo of potential fractals of dimension increaseingly near 2, but of decreasing convex hull).

If W is smooth and not self-intersecting, then at the origin, only a patch homeomorphic to the line need be included. ... and if the curvature is bounded, then we have an upper bound on the curvature of that patch. -- Fuzzyeric 03:34, 6 September 2006 (UTC)

Are you sure in the "any fractal path" thingy? The worm must still have unit length which a fractal path usually doesn't. – b_jonas 07:56, 7 September 2006 (UTC)

-I agree that fractals are key to the problem, but say a terminated fractal path instead.

By the way, can we keep this forum going somewhere else so that it doesn't get eaten a couple days from now? M.manary 00:02, 8 September 2006 (UTC)