EventHorizon

The solution for the second problem follows. Good job, both seem right. Your answer to number 1 was the first one I came up with, then I was able to reduce the number of discontinuities to 4. Same kind of thing for number 2.

First a ground rule, no imaginary numbers (as it makes the questions too easy). So x can be any real number, and f(x) will stay a real number. 1. Can you find a function f such that f(f(x)) = -x? 2. Can you find a function f such that f(f(x)) = 1/x?

I'm not positive this is completely correct. I analyzed the score for a single frame by itself and multiplied by 10. There may be correlation between frames that throw things off, but here's my work/guess...

Yup, that's usual way the rationals are shown to be countable. Here's two interesting things regarding this. If you think of it as simply tuples (ordered pairs of numbers) instead of a numerator and denominator, you can replace the column labelling with the tuple mapped to by the number, and get a way to count 3-tuples (ordered triples). Continuing in this way, you can show all N-tuples are countable for any finite N. So, now that we know all N-tuples are countable, we can turn the tuple into a function (like division for the initial tuples), and still the cardinality of those numbers created will be Aleph0. (e.g., for a 5-tuple (a,b,c,d,e) you could map it to the function (a/b)+(c/d)^-e. The countable set created includes all rationals and many irrationals, but still is not "larger" than natural numbers.) This essentially shows that Aleph0 * Aleph0 = Aleph0. Something pointed out by Octopuppy when he mentioned the square and cube etc with edges being 1 on a number line all have the same density. Yeah, that didn't prove it. Notice that between any two unequal irrational numbers are infinitely many rationals as well. To see this we'll use the repeating decimal definition of rational numbers that Bonanova used. First, find the first difference in the decimal expansions of the two irrational numbers. Move a few decimal places to the right, then either round up the lower or round down the higher from there (I might need to be more specific here for a solid proof, but you get the idea). You can now tack on the decimal expansion for any rational number, and the result will be a rational number between the two given irrational numbers.

The counting numbers / natural numbers are trivially countable. The definition of countable is that there exists a 1-to-1 mapping between the set and a subset (including the possibility of the whole set) of the natural numbers. Just some quick trivia, but the rational numbers (all numbers that can be expressed as the division of one whole number by another) is also countable. The set of all numbers between 0 and 1 is not countable. No matter how you try and create a 1-to-1 mapping between them and the natural numbers, you'll miss some (a lot... but who's counting? pun not initially intended). I'd say that the answer is undefined as given, but that the range of possibilities are as you say (infinity, negative infinity, or any real number). If given two formulae, sets, limits, etc that spawned the two infinities, I'd be inclined to say that you could calculate the difference and get the (possibly finite) result. I guess it's because I disagreed with a calculus teacher back in the day that integrating 1/x between -1 and 1 was undefined... it should be 0, right? It's rotationally symmetric! grr! (hmm....now that I think about it....could it be proven with Lesbesgue integration? I know very little about that, but do know it integrates by cutting up the space differently (horizontal cuts vs. vertical cuts)) This is sort of the same problem as shown in the following example... 0 = (0+0+0+...) = ((-1+1) + (-1+1) + (-1+1) + ... ) = (-1 + (1+-1) + (1+-1) + (1+-1) + ...) = (-1+0+0+0+...) = -1 But this also equals ((1+1+1+...) + (-1+-1+-1+...)) = infinity + -infinity = infinity-infinity = ....? Arranging two infinite sums in different ways can give you drastically different answers. So I guess without knowing not only the degree but the actual "value" of infinity, a difference may be impossible to find. You know.....Godel became mentally unstable after thinking about this kinda stuff too long... (I know, I know, correlation != causation, single data point, etc.). It is my intention to sit down and play video games for several hours (for my sanity's sake, of course).

Yes, the set of numbers in the series 1, 2, 3, 4, ... does have infinite cardinality, but the set of numbers counted will always be finite and therefore have a cardinality less than Aliph0. So you can count towards infinity, but you cannot count to infinity. "Towards" implies only the direction, but "to" implies reaching the destination. Notice that Aliph0 / 2 = Aliph0. If you happen to ever reach infinity in a finite amount of time, you should have reached it in half the time. Repeating this logic, you would need to count an infinite amount of numbers in an infinitesimal amount of time, or, effectively, less than the amount of time required to count to 1.

I posted youtube videos of all my interesting record runs on Maze Stopper and Maze Stopper 2. I thought I should add a link here. youtube videos I do have strategies with better scores (for this posted problem). Though I am still not sure about a limit (currently seems like O(log(n)) where n is the length of the edge of the square map).

I just happened to check brainden today. I modified my code to print out all the solutions (actually half of them, the rest are just reversed).

Nice proof CaptainEd! Wish I'd thought of it.

Yeah, you all got it. I didn't think it would last long.

I did a quick search and didn't find this one already posted. I thought it was pretty good...so far as riddles go (I'm more of a puzzle guy). I saw this riddle on a student's assignment (I was recording scores for a class), so I can't claim credit for it. In which can you see farther: during the day, or at night?

Edit: Oops...sorry EH if you had the same solution...I was actually thinking about this when I was tossing and turning in bed last night... that is not the exact one I had, but it's equivalent.

Alright. Can I hold the straight edge fixed at one point to create a circle? If so, then the area is 1. Otherwise, I'm stuck at 3 for the moment.