Mupen refuses to load savestates for ROMs that aren't the one currently playing. Additionally, instead of using the ROM's checksum to determine if it's the same ROM, it uses its own checksum.
Is there any way to force Mupen to load a wrong savestate anyways?
So, how is that memory watch coming along? The cheat search still doesn't update values in real time, doesn't allow 64-bit searches, and doesn't allow for searches in the form of floats or doubles.
According to the Wikipedia page about Russell's paradox, not all sets contain themselves. The example of the set of all squares is given (because the set itself is not square).
Not all sets are elements of themselves, but all sets contain themselves. A contains B if all elements of B are also elements of A, which is distinct from B being an element of A.
Ok, time for a Portal 2 reference:
Does the set of all sets contain itself?
All sets contain themselves, so yes. Of course, one can point out that the set of all sets doesn't exist. But then it's still true that it contains itself, as all statements are true of nonexistent objects.
Portal 2 badly misquoted Russel's Paradox there.
A game that uses a true RNG would be impossible to TAS but no harder to speedrun than a game that uses a PRNG. However, I don't know of any games that do this.
So, is this Zero Mission built in the engine of Super Metroid, or some sort of blending of the two? Also, is it a remake of either or does it have entirely new maps?
I don't have time to work on this now, but it looks like a good approach would be to prove angle FCD is a right angle, where F is the other point where AE intersects the smaller circle.
Obviously Stewart's theorem is faster, but...
Derakon wrote:
Shouldn't the center of the new circle have a distance of R + 1 from (-2, 0), R +2 from (1, 0), and 3 - R from (0, 0)? Two of those should give you two legs of a triangle, the remaining leg of which is known already (lying on the X axis); from there you ought to be able to do some trig to get the triangle's angles.
Easier to solve for the x and y coordinates of the center.
(R+1)2 = (x+2)2 + y2
(R+2)2 = (x-1)2 + y2
(3-R)2 = x2 + y2
Subtract the third equation from the first two.
(R+1)2 - (3-R)2 = x2 + 4x + 4 + y2 - x2 - y2
(R+2)2 - (3-R)2 = x2 - 2x + 1 + y2 - x2 - y2
Simplify
8R - 8 = 4x +4
10R - 5 = -2x + 1
Add double the second equation to the first
28R - 18 = 6
Solve and back-substitute
R = 6/7
x = -9/7
y = 12/7
My above analysis leads to the following. If I is the area of the inscribed polygon, and c is the cosine of half the central angle of the polygon, then
I' = I/c
c' = sqrt((1+c)/2)
Starting with a triangle has I = 3*sqrt(3)/4 and c = 1/2. Starting with a square has I = 2 and c = 1/sqrt(2). At any stage of the iteration, I/c < pi < I/c^2
Starting with a triangle, this gives...
2.598, 3.000, 3.106, 3.133, 3.139, 3.141
5.196, 3.464, 3.215, 3.160, 3.146, 3.142
As you can see, we're getting about 1 digit every 2 repetitions. Note that the last pair is approximately (223/71,22/7) and corresponds to a polygon of 96 sides, as before.
Just wanted to point out there's an easier way to do this one. Actually, I remember this problem coming up in a math contest I did in high school.
1) Show that, if p is a prime larger than 3, then p^2 - 1 is divisible by 24.
p^2 - 1 = (p+1)(p-1)
Since p is odd, both p+1 and p-1 must be even, and as they are consecutive even numbers, one of them must be divisible by 4. So p^2 - 1 is divisible by 8.
Further, since p-1, p, and p+1 are three consecutive integers, 3 must divide one of them. Since p is prime and greater than 3, it must be p+1 or p-1, and so 3 divides p^2 - 1.
p^2 - 1 is divisible by 3 and 8, and thus is divisible by 24.
Warp wrote:
This is a quite easy one, but I was thinking about it and couldn't come up with the most obvious answer. (It has been too long since my last math class with derivatives and such... I can't remember almost anything about it because of lack of need and usage.)
Demonstrate that the area of the unit circle is 3.14159265...
(Obviously you can't use pi itself, nor any function defined in terms of pi, because that would be a circular proof.)
Inscribe and circumscribe a hexagon in the circle, then keep doubling the number of sides. You can recalculate the area each time using the half-angle formulae. It doesn't converge particularly fast, and the algebra gets really nasty, but it will converge in the end.
This is how Archimedes came up with 223/71 < pi < 22/7. He used a 96-sided polygon for this, though. A hexagon gives the much worse approximation of 1.5*sqrt(3) < pi < 2*sqrt(3), which in rational terms is about 18/7 < pi < 7/2.
If you jump the same frame that you land, samus keeps ALL of her dash speed. much like "skipping" stones, except you only stop when you hit a wall.
However, can you change direction and keep that speed? Dash Jumps go extremely far already, and there aren't a ton of places we need to move very far in a straight line. Not that this won't be useful...it's just that with the ability to change directions it'd be a lot more useful.
For a specific example, tan((pi/2)*x) maps [0,1) to [0,Inf). Since tan is an invertible function on this domain, that means [0,1) and [0,Inf) have equal cardinality.
Here's a fun question. Prove that there is no infinite set smaller than the integers.
I take it that means you're on Sand Ocean. I did a lot of work with that level when I was speedrunning the game, so I'd really like to see what you've come up with. I may even be able to give you a few pointers.
^I'm inclined to agree, seeing as the sequence matches the partial sums of 1/k^2 exactly. I can't manage to prove the sequence transforms into that, but the agreement is perfect for the first 20 terms.
I did manage to prove it converges to pi^2/6 using my above relation and the fact that H(n) = ln(n) + y + d(n), where d(n) < .5/n for all n. Let S(f(k), n) represent the sum of f(k) from k = 1 to n
H(n) - H(n - k) = ln(n) - ln(n-k) + d(n) - d(n-k)
S((.5/n)/k, n) = .5*H(n)/n -> 0 as n -> Inf, so S(d(n)/k,n) -> 0 as n -> Inf.
S((.5/(n-k)/k,n) = .5/n*S(1/k + 1/(n-k), n) = H(n)/n -> 0 as n->Inf.
So, clearly, S(d(n)/k, n) -> 0 and s(d(n-k)/k, n) -> 0 as n->Inf.
Meanwhile, -ln(1-k/n) = sum(i=1 to Inf) (k/n)i/i. So we have...
S(-ln(1-k/n)/k, n) = S((sum(i=1 to Inf)(k/n)i/i)/k, n) = sum(i=1 to Inf) S(ki-1,n)/(i*ni)
Now as n->Inf, S(ki-1,n) -> ni/i. So, for n->Inf, S(-ln(1-k/n)/k, n) -> sum(i=1 to Inf) (ni/i)/(i*ni) = sum(i=1 to Inf)(1/i2) = pi^2/6.
That was incredibly inelegant, but I believe it's valid.
Through some manipulation of sequences, that expression is equal to...
sum(k=1 to n) (H(n) - H(n-k))/k
Unfortunately, I have no real way to sum this series, and it doesn't obviously diverge. Still, it looks like it'll be easier to work with than the one you posted.
^your proof for part 2 doesn't really work. He didn't even specify that the function was continuous, and it could very well be nowhere differentiable.
Since the image of that interval under f is a set of real numbers that is bounded above, there is a least upper bound. Call it M. Now choose any real number e. Since M is the least upper bound, there must exist some number xe such that f(xe) > M - e. Let de = a - xe, which means xe = a - de. For all x > xe, we have f(x) > M - e. We also have f(x) < M from before. Thus, |f(x) - M| < e for all x > xe in (-Inf, a). This is equivalent to saying that |f(x) - M| < e for all x in (a-de, a).
So, for any e, I can find a de such that |f(x) - M| < e for all x in (a-de, a). By definition, this means LIM f(x)(x->a-) = M, and so the limit exists,
Yeah, a feature that is programmed into the game and can be accessed during normal gameplay, even without tool assistance. That sure sounds like cheating to me. It's certainly no different than using an external device to forcibly change RAM addresses.
They are called "cheat codes" for a reason.
Why is the konami code banned? Why are gamegenie codes banned? The answers to these questions are, in fact, pretty much the same. If you start allowing the former, it's only a small step to allowing the latter.
You really don't see the difference? Fine then, let me spell it out for you.
Konami code isn't used because it doesn't make for an interesting run. Action Replay codes are banned because they are not a part of the game. Using a cheat code to unlock hard mode would be perfectly acceptable, but using an Action Replay code to unlock it will never be. Even starting from a save file for a New Game + is essentially an in-game cheat code, and yet it still isn't anywhere close to an Action Replay code.
...You know, now I'm not entirely convinced I would reject the run if it used the code at the start. It'd be like a New Game+ for Super Metroid, and that might actually be pretty fun to watch.
The step from that to allowing gamegenie cheat codes is disturbingly small.
I still endorse splitting the site into two: The purist side with strict "no hacking, no cheating" rules, and the free side with a "do whatever you want as long as it's funny" mentality (iow. a machinima site).
Yeah, a feature that is programmed into the game and can be accessed during normal gameplay, even without tool assistance. That sure sounds like cheating to me. It's certainly no different than using an external device to forcibly change RAM addresses.
And good luck deciding on what the "purist" side actually means. Is save file corruption "impure"? What about New Game+ runs that start from SRAM? What about playarounds?
It's pretty clear that if this debug code was usable from the very start of the game, the run would not be accepted. Does it make any difference that the code becomes usable only mid-way through the run?
(Imagine that the code was usable right from the start, but the author delayed using it until later, in the same way as here. The result would be identical to this, but it would most probably be rejected because the usage of the cheat was completely arbitrary. Why would it make a difference when it becomes possible to use the code?)
I'd say there is a difference. In your parenthetical case, I personally would reject the run solely because it is suboptimal. If he's going to use the code, he'd better use it as soon as it saves time to do so.
But in this case, it's not available from the start. In fact, it's only available in the Golden Torizo's room, a room which is quite frankly about as far as possible from the start. So we start by seeing an unusual route which aims to reach that room as fast as possible. After that, we get a rather interesting hybrid of the 100% and any% runs, where we get to see all the various speed tricks you can do with the items without constantly stopping for missile expansions. As well as a few you don't see in the 100%, like Plasma+Xray vs Draygon.
...You know, now I'm not entirely convinced I would reject the run if it used the code at the start. It'd be like a New Game+ for Super Metroid, and that might actually be pretty fun to watch.
Fast, highly entertaining run. Sufficiently different from the unglitched any%, I would say.
Also, Redesign must be submitted. It's not my favorite mod of the game, but your TAS of it is beautiful.
Darkdata wrote:
Earthbound, Pokemon green and yellow.
Why debate?
Yellow didn't use a debug code. It used save corruption. Very different.
For the GCN controller, 6 sliders and 12 buttons would work perfectly fine. The 6 sliders are of course for AX, AY, CX, CY, L, and R, while the 12 buttons are for A, B, X, Y, Z, S, L, R, and the D-Pad. This gives us independent control over every possible input.
A little circular display showing the current angle/displacement of the two sticks would be nice, but it isn't really necessary.
Further in the "unnecessary but nice to have" features would be scripted macros. Turbo input is the most basic, but the ability to have the plugin automatically go through predefined input sequences is really quite nice.
