What are you trying to imply?
I don’t see any policy in either of the threads you linked, just flame in one of them and nothingness in the other.
You're expressing disbelief at... what exactly? That we have as many as four women? That we have so few? And because of this we need the feature... why?
Actually, let's dwell on that last point. What problem does revealing one's gender solve here? I'm really struggling to come up with a good reason.
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On a more serious note, how has our policy changed since these two threads? Neither thread got very far before the conversation and sentiment turned to, "Oh my goodness, who the hell cares?" I mean, I was curious until I read those threads and then I realized, hey, why on Earth do I care?
I will defer to the wishes of our female members since they would be most affected by the policy, but if it were my choice, I would advise not allowing/requiring users to declare their gender. Let's just be all-inclusive and keep the preferred shape of our sex organs off the table, mmmkay?
(Wishful thinking on my part.)
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ALAKTORN wrote:
GoddessMaria15 wrote:
ALAKTORN wrote:
Oh my God, seriously? This post is eye-opening. We definitely need the feature.
And this right here is why arguments and problems start. You always go out of your way to start them.
I’m not sure what you mean. I was simply expressing disbelief at the new information that I was given. I didn’t know, so I was surprised, and I would’ve known if we had the feature, so it would be good to have.
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Scepheo wrote:
First of all: thank you for the explanation of aleph numbers. Although you started a lot more basic than needed, the last bit was precisely what I needed.
Yeah, I realized as I was typing it that you probably knew a good deal of that already. Even so, I thought it might be nice to make sure anyone could understand it.
Scepheo wrote:
To me this sounds a lot like: "a huge number of the numbers in ℚ aren't in ℤ, so ℤ must be smaller than ℚ". Although intuitive, incorrect (your disjoint set argument goes too, as a nice bonus). I'll stick with my guesstimation of aleph-2.
You're right. It was fallacious of me.
Tub wrote:
Bobo the King wrote:
This hints to me that the cardinality of the set of all simple, closed curves is actually strictly less than 2|ℝ|.
Bobo the King wrote:
I think it's unlikely that the cardinality is less than aleph-2
|2^R| = aleph-2.You can't really assume both at the same time ;)
Oops. You caught me. I am a little careless with my math sometimes, especially analysis, which I rarely use...
p4wn3r wrote:
Bobo the king wrote:
What is the cardinality of the set of all simple, closed curves in two dimensions, excluding all rotations and reflections?
Since a curve is a continuous map, then the cardinality is essentially the same as that of the set of all real continuous functions, which is 2^aleph-0, the same cardinality of the reals.
To prove this, I won't construct an explicit bijection, because this gets messy pretty quickly. It's sufficient to show that you can construct an injection from A to B and an injection from B to A. The Schroeder-Bernstein theorem guarantees that a bijection exists given these conditions.
It's instructive to recall the proof that the set of continuous functions C has the same cardinality of R. For the injection from R to C, it's clear that for every real number a, there's the constant function f(x)=a, which is continuous. For the injection from C to R, we notice that we can sample any continuous function f(x) at rational values of x, and the values of f at rational numbers completely define the function at the real line (if ou have trouble understanding that, just notice that for every irrational number, you can find a sequence of rationals (xn) that converges to it, and the continuity condition will guarantee that all f(xn) sequences will converge to a value, which should be f evaluated at the irrational point). That gives us an injection between C and sequences of reals, which are in bijection with the set of reals, that finishes the proof.
The case is analogous to closed curves. A curve is just the image of a continuous function. The extra condition that it must be simple and closed only implies that the cardinality is at most 2^aleph-0. However, we see that for every real number a, there's a circumference with radius a, so the cardinality is at least 2^aleph-0. So, again, the set of all closed curves has the same cardinality of R.
Bobo the king wrote:
If you can quickly prove that its cardinality is greater than aleph-2, then it also serves as an elegant proof of the popular question, Can One Hear the Shape of a Drum? Do you see why?
The answer is in the article you linked to. If you solve Dirichlet's problem with the closed curve as boundary, you get a sequence of real eigenvalues. So, if the set of boundaries is larger than the set of sequences of reals, two boundaries will necessarily correspond to the same sequence of eigenvalues, giving a negative answer to the drum problem.
However, it's not that simple. As I proved before, the two sets have the same cardinality.
Good answer! Of course I'd hoped for a straightforward proof based on cardinality, but that's a pretty simple solution to the problem. I suppose that one cannot hear the shape of a drum only "coincidentally".
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Scepheo wrote:
I'm at a complete loss what aleph-2 means, and the Wikipedia page doesn't really help, but I'd imagine the set of all two-dimensional curves to be the same size as the set of all infinite collections of real numbers. I have no idea what its aleph number would be, though.
Anyway, a curve is just an infinite collection of real points (right?), so I'd guess the cardinality of the set of all curves would be 2|ℝ|. I don't see the fact that the curves are closed and simple or the exclusion of rotations and reflections factoring into that.
The aleph numbers are a way of expressing infinite cardinalities. I agree that Wikipedia can be very frustrating when trying to learn a new subject in a technical field, especially math or physics.
If you are not familiar with cardinality, it is essentially the number of elements in a set. Say you have some set:
{5, 7, apple, the entire works of Shakespeare, love}
In the above example, there are five elements and so the cardinality is 5. Just count them up and there it is.
Now let's work with a smaller set (and I'll restrict its elements to numbers for the sake of simplicity):
{0, 5, 7}
The cardinality of this set is 3. This set also has many subsets, including {5}, {0, 5}, and {5, 7}. In fact, we can list all the subsets of this set and place them in a new set. We'll call this set the power set of this set, for reasons that will become clear in a minute. The power set is as follows (I will break up its elements line by line so that it's easier to read):
{{},
{0},
{5},
{7},
{5, 7},
{0, 7},
{0, 5},
{0, 5, 7}}
This is an exhaustive list of all subsets of the set, where the last element I listed is the complete set and the first element is the empty set. The power set has eight elements. You can show with some binary representation that the power set of a set with finite cardinality C is necessarily 2C.
From here, I skip over a few paragraphs (unless you'd like me to fill in some details) and I'll just cut to two important theorems:
1) Two sets share a cardinality if and only if there exists a bijection (a one-to-one and onto function) between them.
2) Given some set, its power set necessarily has a greater cardinality than its own cardinality.
These two facts are pretty much trivial if we stick to finite sets. For example, with the sets {0, 5, 7} and {apple, orange, banana}, we can form the bijection (just one example of six):
0 <--> banana
5 <--> apple
7 <--> orange
For the second theorem, all we have to do is note that the cardinality of any power set is 2C, where C is the cardinality of the original set. If C is a non-negative integer, then 2C is clearly larger than C.
But these theorems are nontrivial (yet still hold) for infinite sets. For example, it is known that there are just as many even integers as there are integers and there are just as many primes as there are rational numbers. All we have to do is form a bijection between these various sets and the proof is complete. Because it is often taken to be the "smallest infinity", aleph-0 is defined to be the cardinality of the set of all integers. Just think of aleph-0 as a fancy way of saying "infinity", except that you're strictly referring to cardinalities and it's a particularly "small" infinity.
However, we can define a larger infinity by taking the power set of the set of all integers. By theorem 2, the cardinality of this set is necessarily strictly greater than aleph-0. We define this new cardinality to be aleph-1. (Incidentally, it is undecidable without a further axiom whether there is another cardinality between aleph-0 and aleph-1. Fortunately, it's irrelevant to the problem at hand.) Aleph-1 is also infinity, but it's a strictly larger infinity than aleph-0. It also happens to be the cardinality of the set of any continuum.
Finally, we are ready to construct aleph-2. To define it, we just take the power set again! That is, we need to take the set of all subsets of the set of all subsets of the integers. This is an even larger cardinality and so is given a new name, aleph-2. You can continue taking power sets to construct aleph-3, aleph-4, and so on.
So the question is whether the set of all simple, closed, two-dimensional curves (or equivalently, the set of all simply connected regions in two dimensions) has cardinality less than, equal to, or greater than aleph-2.
As for your guess that the cardinality of the set of all curves being 2|ℝ|, I wouldn't be so sure. A huge number of those "curves" aren't simple. In fact, if you were to pick a "random" subset of all ℝ, you would almost surely have disjoint set. This hints to me that the cardinality of the set of all simple, closed curves is actually strictly less than 2|ℝ|. I could be wrong, though.
Tub wrote:
Bobo the King wrote:
Since you hinted that it must be greater than aleph-2 I'll accept that as proof by authority and state...
Actually, I'm not sure what the answer is. I only hinted that it is likely to be greater than or equal to aleph-2, but this is just a guess. It's based on two observations:
1) There are a lot of simple, closed curves.
2) There are some pairs of drumheads that produce the same timbre.
I think it's unlikely that the cardinality is less than aleph-2, since this would imply that even though you cannot hear the shape of a drum, there are many more timbres than there are drums. This would be highly "coincidental".
I haven't had a chance to look too closely at the rest of your post. Looks like an interesting start, though.
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All the previous talk of cardinalities got me thinking about something I once asked my real analysis professor many years ago: What is the cardinality of the set of all simple, closed curves in two dimensions, excluding all rotations and reflections? If you cannot prove it easily, can you provide a lower bound on its cardinality? If you can quickly prove that its cardinality is greater than aleph-2, then it also serves as an elegant proof of the popular question, Can One Hear the Shape of a Drum? Do you see why?
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Not to derail Warp's question, but I have a question of my own that concerns practical topology.
My girlfriend recently bought a wonderful looking top (the garment, not the kind you spin). The problem is there is a twist in the back. Let me do my best to describe it. There are five "holes" in this top: two for her arms, one for her neck, one for her torso, and then a fifth hole on the back. On a "normal" top, the fifth hole would not make the top unusual in any way and if you were to fill in the holes with fabric, you'd have something homeomorphic to a sphere's surface. The problem is that there is a full twist on the strip of fabric between the hole for the neck and the hole in the back. That particular strip of fabric is twisted 360 degrees before it is rejoined to the opposite side. (I'll draw a picture if necessary, but it will be time-consuming and I warn you that I'm not much of an artist.)
Although it could be a fashion statement and she could do something like put a bow over the twist, my girlfriend and I both agreed that it looks kind of silly and we'd love to see how it looks without the twist. We both thought that it might be invertible such that the twist can be undone but after just a few minutes of playing with it, we quickly concluded that it just doesn't seem possible to get rid of the twist. My girlfriend's plan is to cut the top where the twist is and sew it back together.
But the topology question itself is enticing to me! I know next to nothing about topology, so I'll defer the following questions to the experts here:
• What is the top's topological class? (Is this the right term?)
• What is the topological class of the shape we're trying to turn the top into?
• If neither of the above is easy to define, can it still be elegantly shown that the twist cannot be undone?
• If it twist can be undone, how would we do so?
Like I said, I know next to nothing about topology, so the simpler the explanation, the better. A brief overview of the techniques used to characterize and/or solve the problem would be greatly appreciated. The only thing I thought to use was the Euler characteristic, but if I'm not mistaken, the Euler characteristic of the twisted and untwisted top are exactly the same (I calculated -3). Does this perhaps imply that the twist could be removed if the top were embedded in four dimensions? What techniques can and should I use?
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Warp may (or may not-- I don't really know) be touching upon a subject known as Ehrenfest's paradox. I was interested in it back in high school, but since I never became comfortable with general relativity, I don't have any insights to offer. I can't even properly interpret Wikipedia's resolutions of the paradox.
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BioSpark wrote:
Bobo the King wrote:
Great run! Of course the ending was awesome (I guess it counts as beating the game), but my absolute favorite part came at 26:33 in the encode. You freeze an otherwise inconsequential enemy then use it to build up a wavedash. It's not much, but it's pretty clever and funny.
that trick was actually in the 100% tas. and it's called a shinespark, this isn't melee. :P
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Great run! Of course the ending was awesome (I guess it counts as beating the game), but my absolute favorite part came at 26:33 in the encode. You freeze an otherwise inconsequential enemy then use it to build up a wavedash. It's not much, but it's pretty clever and funny.
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I have mixed feelings. In this movie's favor is... well, that it's a Sunset Riders TAS. It was clearly well-executed in most essential ways and I was fairly entertained. I liked that it was a pacifist run as it required skillful dodges and creative solutions to some problems (mainly in the last level).
On the other hand, I notice that there are quite a few missed shots, most notably on the second boss. Was this perhaps for luck manipulation? It looks very sloppy. Also, why was the dual-wield powerup not used until the final boss? (If you are forced to have it for that fight, I may have failed to notice or misremembered.) Finally, I think it was a stylistic error to score zero points on the bonus levels. I think that the additional entertainment would easily offset the fact that the goal is compromised. I suppose you might change the branch name to "minimum score" to emphasize that the bonus rounds had to be thrown.
I could see this going either way.
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FractalFusion wrote:
I seem to remember the problem differently.
Was it the height of the remaining sphere (after drilling the hole top-down) that was fixed, rather than the radius/diameter of the hole?
Edit: I think that's what Flip means by "cylindrical hole, 6cm deep". A cylindrical hole is drilled top-down through the center of a sphere all the way through, such that the length of the cylindrical hole inside the remaining sphere is 6cm. In other words, the height of the remaining sphere is 6cm.
I think you are correct. The width of the drill is such that the hole it bores has a length of 6 cm, but it goes all the way through the sphere. Of course, the sphere must be at least 6 cm in diameter.
There is a very similar problem in two dimensions that I'm more familiar with.
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Flip wrote:
A cylindrical hole, 6cm deep is drilled through the middle of a solid sphere or radius r. Find the remaining volume in the sphere.
You may find this page useful. What's interesting about the solution?
I think I remember this problem. I'll leave the answer for others to determine.
Are you sure the hole is 6 cm deep and not 6 cm in radius (or diameter) and it's bored all the way through? My recollection is that that's the "usual" problem statement.
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thommy3 wrote:
2) Middle:
mid(a,b,c) = max (min(a,b) , min(b,c) , min (b,c) ) = min (max (), max(), max())
3) Second largest is:
second(a,b,c,d) = max ( mid(a,b,c), mid(a,b,d), mid(a,c,d), mid(b,c,d))
edit: OK, mister already got it.
Your solution to the third problem matches my method. You actually don't need all four mid functions in there, since the two middle numbers are surely represented by three of the four functions.
Also, as a very minor technicality, min and max each take two arguments. You'd need to use them iteratively, defining something like
max(a,b,c) = max(a, max(b,c)).
For the second problem, I used
mid(a,b,c) = a+b+c - max(a,b,c) - min(a,b,c).
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RGamma wrote:
Bobo the King wrote:
This problem has been on my mind a little lately. I haven't given it a whole lot of thought, but to the extent that I have, I'm stumped.
1) Construct a function of two real numbers that, using only addition/subtraction, multiplication/division, and absolute values that takes the value of the larger of the two numbers. (So construct the max function using only absolute values. This should be very easy.)
2) Construct a function of three numbers with the same requirements as above (using only a/s, m/d, and absolute values) that takes the value of the middle number. (This is only a bit harder.)
3) Construct a function of four numbers with the same requirements as above that takes the value of the second-largest number. (I don't know off the top of my head how one might do it.)
...
1)
f_1(a,b) := (a + b + |a - b|) / 2 = max(a,b)
Proof:
Let a>= b:
f_1(a.,b) = (a + b + |a - b|) / 2 = (a + b + a - b) / 2 = a = max(a,b)
Let a < b:
f_1(a,b) = (a + b + |a - b|) / 2 = (a + b - a + b) / 2 = b = max(a,b)
Likewise: f_2(a,b) := ( a + b - |a - b| ) / 2 = min(a,b)
For the others: Maybe somehow construct the function from min()/max() functions?
This matches what I have for the first problem. I believe the answer is essentially unique, unlike the other problems. You are also on the right track with your idea to use the min() and max() functions to produce the second and third functions.
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This problem has been on my mind a little lately. I haven't given it a whole lot of thought, but to the extent that I have, I'm stumped.
1) Construct a function of two real numbers that, using only addition/subtraction, multiplication/division, and absolute values that takes the value of the larger of the two numbers. (So construct the max function using only absolute values. This should be very easy.)
2) Construct a function of three numbers with the same requirements as above (using only a/s, m/d, and absolute values) that takes the value of the middle number. (This is only a bit harder.)
3) Construct a function of four numbers with the same requirements as above that takes the value of the second-largest number. (I don't know off the top of my head how one might do it.)
So for example, without loss of generality, for any a<b<c<d, we seek functions f, g, and h (corresponding to cases 1, 2, and 3 above) that only involve addition, subtraction, multiplication, division, and the absolute value, such that:
f(a,b) = f(b,a) = b
g(a,b,c) = g(b,c,a) = g(c,a,b) = g(c,b,a) = g(b,a,c) = g(a,c,b) = b
h(a,b,c,d) = {23 other permutations of the arguments} = c
Can this be done?
Edit: I found one way to do it. It's a little tricky and I'm sure there are many equivalent ways. I also suspect the process can be generalized to yet more variables.
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It just occurred to me that one reason this run might be so appealing is that it cannot use zip glitches, which are overwhelming other Sonic runs. It plays like a very well executed "traditional" speedrun.
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Do you have access to a Lua function that inputs a bitmap (or other image format) and outputs an array with indices x, y, and RGB? If so, what is the array's format? Are x and y read from left to right and top to bottom respectively?
(Just to be clear, I'm still not yet committed to writing this script, but having access to a function like that would make my life much easier.)
Edit: Also, you should crop the map and move it to one of the corners when it comes time to do the real thing.
Edit 2: A quick search reveals that all three of addresses 7E0052, 7E005E, and 7E008C are correlated with one of the camera's coordinates. (I believe that their previous addresses are involved as well because they seem to be the least significant byte.) Would it be possible to freeze one or all of these to a position a fixed distance from the karter, effectively expanding the field of view? Why do everything in Lua when we can take advantage of the SNES's direct ability to use Mode 7?
Edit 3: Try cheating in values for 7E0052 or 7E005E (or best of all, both simultaneously). The position of the camera isn't fixed like I'd hoped, but the results are hilarious. Fixing the value for 7E008C doesn't seem to have an effect but if it is indeed the less significant byte in a word, then it might not be noticeable. I tried fixing 7E008B and 7E008D to no noticeable effect.
Anyway, the fact that altering 7E0052 and 7E005E both affect the camera gives me hope that this can be done entirely within the SNES. Once we find the position addresses, all you would need is a small Lua script to extrapolate back the position of the camera.
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Wikipedia says that this is Mode 7's transformation principle:
That's very simple to implement in Lua. The hard part (for me, at least) would be fetching the pixels from the source image.
Now, you used a track from Mario Kart as your "example" image, which is a little bit harder to replicate, since it would require some perspective effects, which Wikipedia states goes a bit beyond what Mode 7 alone can do. I still think it wouldn't be so bad, but I would need to sit down and give it some careful thought.
I'm intrigued by the idea, but I'm very busy at the moment and probably will remain so through the month (though I tend to get a lot of work done on side projects when I'm putting off bigger work).
Also, I will never forgive you for causing Phil Hartman's death.Edit: Since I'm guessing you want this for one of the Atlas encodings, I'm a little confused as to how you plan on using it. Are you sure you don't wish to undo the Mode 7 transformations? After all, the analogous video for Mario Kart would impose the top-down view on the karters' positions. If that's your goal, I would instead recommend building the encode from the ground up: First, get all karters' x, y, and z positions and orientations every frame and have them output to some text file. Then have a script draw the karters on the map you've provided (I assume a program like Adobe Aftereffects can do this). Finish off the encode by syncing the video with the run's audio.
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Warp wrote:
thatguy wrote:
Not really Warp, because to a mathematician π is so much more than the ratio of a circle's circumference to its diameter. In this case, the sum is linked to Taylor series of trigonometric functions, whose solutions involve π.
Given that sines and cosines are closely related to circles...
No, not really. What's going on is more like this...
• Circles are geometric objects.
• Sine functions are geometrically related to circles.
• The Taylor series of a sine function (which lacks an intuitive geometric interpretation) follows a nice pattern.
• A piece of that pattern is followed by the sum of the reciprocals of square numbers.
Just as thatguy said, the connection is in the pattern of the series. You can see from my outline above that the connection to geometry is quite tenuous.
I mean, if you think you can view it in terms of circles, read the proof and try to come up with a geometric way of thinking of it. I just don't think it's reasonably possible.
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Warp wrote:
Bobo the King wrote:
As a very simple example, we know from physics that the kinetic energy is equal to 1/2mv2. Does that mean that lurking behind every moving object is a square with a side whose length varies proportional to the velocity? Or perhaps even more to the point, I know that when I drop an object from a height h, its speed will be v = sqrt(2gh) just before it hits the ground. Does that mean that "somewhere" there is a square I can draw with great physical significance that has a side of length sqrt(2gh)?
I think you are succumbing to a fallacy of equivocation. You are confusing the two meanings of the word "square" (one of them being "a geometric shape with certain characteristics" and the other being "raised to the power of 2". The same word may be used for both, but that doesn't mean they are synonyms.)
Granted, there's a relationship between the geometric shape and the power-of-2, namely that the area of the former is calculated using the latter, and that's the etymology of the latter. However, that doesn't make them synonyms.
There may be a connection between the area of a square and a formula like mv2/2, but not a direct one. Rather, they both have a common underlying geometric reason why they both use a power of 2.
A clearer example would be the intensity of light being proportional to the inverse of the square of the distance from the light source. Are there literal geometric squares involved? No. However, both share same underlying geometric principles (namely, that light intensity is measured by area, and areas involve powers of two.)
The point I was trying to make was not that squaring a number and the geometric object of a square are the same thing, but rather to highlight that there are "familiar" (based in geometry or more familiar concepts) mathematical tools, functions, and constants that "coincidentally" show up in other places. For example, why does the square-root of π so frequently show up in the gamma function of half-integers? Are there circles involved here? As far as I can tell, the answer is simply "no". Your original question (if I understand it) was why the particular ratio of the circumference of a circle to its diameter shows up so conspicuously in the sum of the reciprocals of the squares of the natural numbers. Well, you're invoking geometry! It sounds like you want to know "where the circles come in". Maybe you can create some squares with sides 1, 1/2, 1/3, 1/4, etc., then cut them all up into pieces of a very special shape, rearrange them, and by a very careful argument show that they create a circle whose radius is exactly sqrt(π/6). It sounds like that's exactly the sort of argument you're looking for.
Well, I know of no such argument. Establishing that sort of connection between the conspicuous constant (π) in the theorem and the geometric object of a circle is nigh-- if not outright-- impossible. It just is because the proof says so.
Look, let me put it this way. Suppose all you know is that numbers define quantities. You are only aware that you can use them to count things: "One apple, two apples, three apples..." and so on. You know nothing of rational fractions, negative numbers, or other such things. You are, however, taught the decimal expansion of π. So you notice that the tenth digit of pi is 3 and you lay awake at night wondering, "Where did that 3 come from?" Perhaps, somewhere deep in the derivation of π, there is something that says π's tenth digit represents three of something. What is that something? Does the universe have ten apples stowed away somewhere? Maybe you're clever enough to have a sense of scale and say that π is "3 big things, plus 1 thing one-tenth as big, plus 4 things one-tenth as big as that, plus 1 thing one-tenth as big as that..." But where do these things really come in?
They don't. You can't compare apples to circles.
That's very simple to implement in Lua. The hard part (for me, at least) would be fetching the pixels from the source image.
Now, you used a track from Mario Kart as your "example" image, which is a little bit harder to replicate, since it would require some perspective effects, which Wikipedia states goes a bit beyond what Mode 7 alone can do. I still think it wouldn't be so bad, but I would need to sit down and give it some careful thought.
I'm intrigued by the idea, but I'm very busy at the moment and probably will remain so through the month (though I tend to get a lot of work done on side projects when I'm putting off bigger work).
Also, I will never forgive you for causing Phil Hartman's death.
Edit: Since I'm guessing you want this for one of the Atlas encodings, I'm a little confused as to how you plan on using it. Are you sure you don't wish to undo the Mode 7 transformations? After all, the analogous video for Mario Kart would impose the top-down view on the karters' positions. If that's your goal, I would instead recommend building the encode from the ground up: First, get all karters' x, y, and z positions and orientations every frame and have them output to some text file. Then have a script draw the karters on the map you've provided (I assume a program like Adobe Aftereffects can do this). Finish off the encode by syncing the video with the run's audio.