It is my understanding (and correct me if I'm wrong) that the current model of the universe is that it has finite volume, but has no edge. You can't just traverse to the "edge" of the universe and go outside of it (or be stopped by some strange horizon, where the universe stops existing). On the other hand, at the same time, the geometry of the universe is postulated to be "flat". Whatever that means.

I can't speak to what our actual universe is shaped like, but if you want an example of a finite-volume universe with no boundary and flat curvature, you can think of games like Asteroids, where you warp to the other side of the screen when you hit the edge.
If you still think asteroids has a boundary, you could also consider Defender, where the camera follows the player and there is no left/right boundary at all, not even a "warping" one.
This sort of model easily generalizes to three dimensions.
(Again, I'm not saying this is an actual viable model for our universe - I seem to recall that theories of our universe have curvature and finiteness linked together, such that the only viable finite universes have positive curvature (like the positively-curved surface of a sphere is finite but boundless), and the only infinite universes have flat or hyperbolic curvature. In other words, I'm not sure anyone is postulating a finite, flat universe, even if asteroids shows that such a thing is mathematically consistent.)

At 9:32, could you take damage from the falling spikes to avoid backtracking? (Obviously you'd need to avoid damage later - maybe the black bird in the top left is avoidable?)
12:21 lolwut

Looks good, great effort! Also, sorry to hear about your little one, and glad to hear he's better now.
One thing I'm not sure about: at 5:10 and 6:44 in your youtube encode, you jump before the gravity pill expires, so that you're lower down when gravity reverts to normal and you have less distance to fall. But you're not at the apex (nadir?) of the jump when gravity reverts, so it looks like you could be even lower down if you timed your jump more carefully. Would that save time?
At 12:42 it looks like you could save time by jumping a little earlier, but again it might be an illusion.
I enjoyed when you killed the wall enemy with the power pill at 13:45. At first it just looked like entertainment value, since it might not be that well known that those enemies are killable at all, but then on the return journey it looks like it's an actual timesaver. More generally, I enjoyed the enemies you killed with the "splat" on the wall where it looks like you missed at first, only for the enemy to die anyway.
16:03: most annoying air generator in the game! I died so many times to this one!
This was a favourite game of my childhood, and it's enjoyable to see it nimbly beaten. It's also a bit sad in that although I have fond memories of it, going back and revisiting it shows me that it just isn't as good as I remember, especially compared to contemporaries on SNES/Genesis. Nostalgia's a hell of a thing, isn't it? :)

That was super interesting! I didn't manage to solve it, but my suspicions were on the right case (5 neighbours, blue colours opposite, ie p4wn3r's Figure 2b case). I worked out that the two lines could cross, which the video diagram doesn't make clear, but I didn't spot that recolouring the blue-green region could break the red-green line, thus disallowing the blue-yellow recolouring.
Does the proof say anything useful at all that can be salvaged? For example, is the 4-neighbour argument sound? If it is, can we use it to say that in any minimal non-4-colourable graph, every vertex has at least 5 neighbours?
(Of course, I know that this statement is vacuously true, because no such graph exists; but that requires a proof of the 4-colour theorem. I'm interested in what we can say from the arguments in this failed proof.)

I have had some similar questions about kinetic energy before. I think at high school I asked my teacher this question (or something similar):
Consider a mass of 2, such that kinetic energy is simply velocity squared. Accelerating a mass from 0 to 1 takes 1 unit of energy, and accelerating from 1 to 2 takes 3 units of energy. But an acceleration from 1 to 2 is just an acceleration from 0 to 1 in a different reference frame. When you change reference frame, where does the extra kinetic energy come from?
Years later, I think I have an answer (but I'd be happy to hear other's thoughts). It's based on two principles:
1. momentum is conserved
2. the exact distribution of kinetic energy depends on the reference frame, but the amount of work done in an interaction is frame-independent.
Conservation of momentum is important because you can't accelerate an object in a vacuum; to accelerate an object, the momentum has to come from somewhere, and that will have an effect on the energy requirements of the whole interaction.
So, to reframe the original question, we have to add in the extra parts of the system. To keep it simple, we'll have two equal masses that are accelerating away from each other. Then in the two reference frames, we have:
* mass A accelerates from 0 to -1, mass B accelerates from 0 to 1
* mass A decelerates from 1 to 0, mass B accelerates from 1 to 2
In the first reference frame, we start with 0 kinetic energy, and end with 1+1=2 kinetic energy.
In the second frame, we start with 1+1=2 kinetic energy, and end with 0+4=4 kinetic energy.
The total kinetic energy is different, but the change in kinetic energy (aka the work done) is 2 units of energy, independent of reference frame.
The original question undercounted the energy requirements in the first frame, and overcounted the energy requirements in the second frame, because it ignored the extra energy needed for (or provided by) mass A.
So: changing reference frame doesn't magically make energy appear out of nowhere. It might change the distribution of kinetic energy, it might change the total amount of energy, but it does not change the energy required to accelerate an object by a given amount.

Yup, I got the same answer. I make a google sheet, translating the map into spreadsheet cells through judicious use of merging. That got me to a place where I could explore options by highlighting cells in particular colours.
The easy places to start the analysis are:
If we start with 24 = red and 21 = green as you did, then:
green needs another odd number, and its only options are the 9 or the right-hand 3. If green takes the 9, most choices are now forced, and you end up with 2 possible colourings.
If green takes the 3, the next obvious place to split the search is: who takes the remaining 3 and 9? They must be taken together as they are the only odd numbers. The only choices are red or a new colour (without loss of generality, blue).
If red takes the 9 and 3, it's a bit tricky to see but there is exactly one way to finish.
If blue takes the 9 and 3, there are no more obvious ways to prune the search space. I decided to focus on the 6s. After a lot of thought, there are only three viable options: a fourth colour (yellow) takes all 6s, blue takes 2 and green takes 2, or blue takes 2 and red takes 2. These end up with 3, 2, and 1 colourings respectively.
Adding up all the colourings, you get 9 total colourings.

Noob question but: when does the bell get rung in that plan? I confess I don't fully understand the mechanics of the ritual, but don't you have to spend an action ringing the bell before reading the book? (or is that what you mean by "action 1" and "action 2"?)
EDIT: never mind, found the answer earlier in the thread

I came here just to point out the trivial but surprising (to me, at least) corollary:
sin(18°) = (φ-1)/2
where φ is the golden ratio. Alternatively, we can define φ in terms of sin(18°):
φ = 2sin(18°) + 1

It depends what you mean by "smooth", but you probably want C^{n} continuity for n=1 or n=2. (See wiki on smoothness for more on this.)
I'd probably use a B-spline which, as the article states, is a curve which "has minimal support with respect to a given degree, smoothness, and domain partition". In this sense it is the optimal solution to your problem (to the extent that your problem is well-defined at all).

I don't know what "induction hypothesis" means here.

It's a proof by induction: it works by assuming that z_{n}^{*} = z^{*}_{n} and using this assumption to show that z_{n+1}^{*} = z^{*}_{n+1}. The step labelled "using induction hypothesis" is the part where I rely on the assumption that z_{n}^{*} = z^{*}_{n}.

EDIT: The key reason why this argument doesn't work on the other axis is that the corresponding operation (negation or "unary minus") doesn't commute with multiplication/squaring: (-x)(-y) ≠ -(xy) in general, while x^{*}y^{*} = (xy)^{*}.

I'm not sure I understand that.

What I'm saying is: if you try to do my proof by induction, but using negation instead of complex conjugation, the step labelled "[conjugation commutes with squaring]" won't work any more, because negation doesn't commute with squaring: (-x)^{2} ≠ -x^{2}.
Admittedly saying this proof doesn't work isn't the same as saying that there definitely isn't horizontal symmetry. I'm pretty sure this is only one or two steps away from proving that though.

As an additional question: The z_{n+1}=z_{n}^{3}+c fractal is (or, more precisely, appears to be) symmetric on both the imaginary and the real axes. Why is that the case?

If you understood what I wrote above, you'll see this is directly related to the following series of identities:
(-x)^{2} = x^{2}
(-x)^{3} = -x^{3}
(-x)^{4} = x^{4}
(-x)^{5} = -x^{5}
...
For any odd natural number n, negation commutes with raising to the nth power, and so you can use the above proof by induction to show that there is horizontal symmetry. For every even number, negation does not commute with raising to the nth power, so the proof fails.

Let z^{*}_{n} denote the nth iteration of the sequence using c^{*} instead of c. Then we can see that z^{*}_{n} = z_{n}^{*} (proven below). In words, we can calculate the nth member of the sequence for c^{*} just by computing the nth iteration for c and taking its conjugate. Membership of the Mandelbrot set is based on the magnitude of z_{n} going to infinity. Given that z and z^{*} have the same magnitude, z^{*}_{n} goes to infinity if and only if z_{n} goes to infinity, and so c is a member of the Mandelbrot set if and only if c^{*} is, which gives exactly the symmetry you describe.
Proof sketch: z_{n}^{*} = z^{*}_{n} holds because complex conjugate commutes with addition and multiplication, and the sequence z_{n} is based only on addition and multiplication.
Proof by induction:
Base case:
z_{0}^{*} = 0 = z^{*}_{0}
Induction step:
If z_{n}^{*} = z^{*}_{n}, then:
z_{n+1}^{*}
= (z_{n}^{2} + c)^{*} [by definition]
= z_{n}^{2*} + c^{*} [conjugation distributes over addition]
= z_{n}^{*2} + c^{*} [conjugation commutes with squaring]
= z^{*}_{n}^{2} + c^{*} [using induction hypothesis]
= z^{*}_{n+1} [by definition]
EDIT: The key reason why this argument doesn't work on the other axis is that the corresponding operation (negation or "unary minus") doesn't commute with multiplication/squaring: (-x)(-y) ≠ -(xy) in general, while x^{*}y^{*} = (xy)^{*}.

Suppose I have two values of the form A=a*10^{b} and B=c*10^{d}

Given your use of the term "mantissa" in this context, I'm guessing you're talking about a base-10 floating point number? In which case we have the extra constraints that 1 ≤ a,c < 10 and that b and d are integers (I'm assuming A and B are strictly positive for this, for simplicity).
Restating the problem, we're looking for a function f(A,B,t) such that
Such a function has striking discontinuities in it. For example:
but:
These discontinuities occur every time one of the exponent values b or d goes up by 1. Intuitively it feels to me like the simplest way to achieve this behaviour is by splitting A and B into mantissas and exponents, like you originally had, because the functions that split A and B into mantissas and exponents provide precisely the discontinuities that f(A,B,t) has. It's hard to imagine how you would replicate this discontinuous behaviour without the splitting functions.
Incidentally, this is an odd form of interpolation, in that it can easily have a result that is outside the range from A to B. Consider that f(9,10,0.5) = 5√10 ≅ 15.81

Yes, the formula with y in it only needs to hold for y < 1, and then I take a limit (which can give you a proof that the formula holds for y = 1, i.e. x = -1, but by then I'm already done). This means that you do need a brief justification for why taking the limit is valid here, which I do give but don't fully spell out.

Thanks, that makes a lot of sense.
I didn't manage to solve Bobo the King's original problem, but here are some things I discovered along the way while trying to attack it:
Here's the original again:
We can transform the product index to "simplify" the product:
And we can add an extra sum term:
This second step works because the k=0 case always equals 1 (if we accept that an empty product equals 1 vacuously). I like this form because summing to zero looks more elegant, but seems to take you further away from the original binomial series.

The binomial series tells us that, if |x| < 1, then we have

Most likely I'm missing something here, but if you require |x| < 1 to use this series, then later let y → 1, doesn't that then imply that you've used x=-1 and hence |x|=1?
Does this all work out because you're taking limits? (Is this similar to finding f(0) when f(x) = sin(x)/x ?)

I think the answer is π+1.
Essentially, there are two phases to the duck's movement: in the first phase, the duck tries to stay on the radius directly opposite the fox, no matter what the fox's movements are. The duck continues like this until she can't move fast enough to keep the fox directly opposite. At this point, she enters a second phase, where she travels directly to the water's edge, along the shortest straight line path.
The first phase lasts as long as the duck can move faster around the pond (in the φ-direction, in polar coordinates) than the fox can. Call the duck's speed d and the fox's f, measured in units defined such that the radius of the pond is 1. If the duck is r from the centre, then the duck can move around the pond faster than the fox (and thus stay directly opposite him) as long as r<d/f.
Having sustained this movement, responding to whatever movements the fox makes by staying directly opposite him, the duck can now start the second phase in the best possible position: she is at a point d/f along a radius toward the edge, with the fox at the end of the opposite radius. Since she can no longer outmanoeuvre him, she might as well make a run for it. She has to move a distance of (1 - d/f) at a speed d, while he has to move a distance of π at a speed f. At the point where the fox just manages to catch the duck, they will cover these distances in the same time, so
(1 - d/f)/d = π/f
Simplifying:
(1/d - 1/f) = π/ff/d - 1 = π
f/d = π+1
I suspect it might be possible to have an even better strategy in phase 2 than the straight-line dash, but I can't see it myself.

yup, that. See also my posts on castlevania's item drop algorithm and soul drop algorithm - some of the soul drops and item drops have rather low probabilities (eg final sword has a 0.5% chance of dropping, ectoplasm soul has a 0.57% chance of dropping).

lolwut.
A short run, and with enough wtf-factor to keep it entertaining.
I particularly like when you're forced to carefully climb down from midair rather than just falling.
(The scrolling in this game is a little grating and just shows what a bad port this is.)

I guess it's about understanding your audience.
You already accept that you can assume the reader knows things like the Pythagoras Theorem.
I don't think it's a big step to assume the reader can prove, for themselves, some intermediate lemma, even if they don't already know the lemma to be true. Of course, it depends on the lemma and the reader.
If you don't assume any background knowledge in the reader, and try to prove everything from first principles, you end up needing hundreds of pages to prove 1+1=2. The proof is very difficult to follow, because it spends so much time in the detail that you don't get a sense of the bigger picture.
So instead mathematicians tend to use a "proof sketch" - a short summary, where the reader can fill in the details as necessary. The proof sketch is useful because, to the appropriate reader, it gives exactly the right amount of information, and because it operates at the higher level, it's easier to understand. The cost of this is that it shuts out readers who aren't as familiar with the required background. (As someone who was studying for a PhD for some time, let me tell you that I found this just as frustrating as you.)
The right level for a proof is on a sliding scale depending on the reader. For example, I mostly remember the infinite prime proof as "contradiction. assume finite list of primes, construct another prime not in the list.". If I need to, I can fill in the details as required, by redoing the work.
Some well-written papers write both a sketch and a fuller proof (although even the fuller proof will assume some background). Some readers will only need the sketch, while others will slog through the fuller proof.
Amaraticando wrote:

if there're only finitely many primes, the sum of the inverse of all primes would not diverge. But it does diverge. Therefore, there're infinitel many primes.

The proof that the sum of the inverses of all primes diverges (probably?) depends on the proof of the infinity of primes, so you can't use it to prove there are infinitely many primes. That'd be a circular argument.