Two disclaimers: IANAL, and I don't know about copyright laws in the netherlands.
But as far as I know, takedown notices only apply to server hosters that have no knowledge about the illegal content on the servers, i.e. the guys with a few hundred server racks. They're not accountable for the infringement until they know about it.
If you're the webmaster of a site explicitely designed to distribute music ripped from commercial, copyrighted games, you might have lost that right. Copyright holders will consider you accountable for the infringement, and that's where the fun starts.
In germany, you'd get an "Abmahnung" ("written warning"), which contains:
a) details about the copyright infringement you did and an order to desist within a certain deadline.
b) a bill for their lawyer (fees are regulated by law, but not cheap)
c) an "Unterlassungserklärung", a contract they need you to sign. It basically says "I acknowledge that I infringed on their copyright and will never do it a gain, or I'll pay 19827483€ in contract penalties".
If you don't pay the bill, they'll take you to court over the bill.
If you don't sign the contract, they'll take you to court to get a court order telling you to do what the contract said.
If you don't remove the content within the deadline, court it is.
Going to court when not being innocent is always expensive. If you lose, you'll have to pay their lawyers (quite a bit more than the original sum), your own lawyers and the court. Even if you get a partial win, it's still expensive.
You can go to court, trying to argue that you had no knowledge of your user's deeds and thus aren't accountable. But how would you argue that?
The easiest way out is to desist, sign the contract and pay the bill. But signing the contract may result in them trying to force license fees out of you. You've signed this thing saying that you're guilty, right?
You can pay your own lawyer to design a different contract to get around that issue, but unless the other party agrees with that contract, court it is.
And contract or court order, both are obviously timebombs. One of your users uploads the same music again? You're in trouble.
There's also no limit to the amount of "Abmahnungen" (and thus lawyer's bills) you can receive, and the deadlines pretty much prohibit you from taking longer vacations. Generally, anything more than one letter per lawyer/client-combination might be considered abusive, but again, you'd have to go to court to prove that.
What I'm saying is: don't bet your financial existence on the assumption that "it'll never happen". Find a local copyright lawyer, discuss the legal backgrounds and worst-case-scenarios with him. He may charge 200€, but that's a good investment. And if it isn't viable in your country of residence, then don't do it, but leave it to someone in a country with different copyright laws.
the first cut is across a 8x3 area, the second across 2x5.
The angles of the cuts are:
atan(3 / 8) = 20.56°
atan(5 / 2) = 68.20°, which is ~21.80° after turning.
So no matter how many 90° or 180° transformations you do, edges from the different cuts won't fit together.
If you look closely at the image, you can see that the outlines are getting thicker in the middle, conveniently hiding the gap in the rectangle.
The school-explanation I was given was this:
Imagine that the magnet (and any magnetic material) contains millions of very tiny mini-magnets, each having their own north- and south pole. In a permanent magnet, all these mini-magnets are aligned in the same direction, all the tiny fields add up to a stronger magnetic field that's measurable outside the magnet.
If you snap a magnet in half, you get two parts, each consisting of half the amount of mini-magnets, but they're still aligned, adding up to two new magnetic fields. Each of the parts is a full magnet, just with half the strength. You don't get a part with a north pole and a part with a south pole, but two whole magnets with two poles each.
In unmagnetized (but magnetic) material, like iron, these mini-magnets are ordered randomly. Their fields go in random directions and cancel each other out.
Now what happens when you put iron near a magnet? The magnetic field enters the iron, aligning its mini-magnets in the direction of the magnetic field. Now the iron is a magnet of its own, and it's attracted the same way two magnets would.
Remove the magnet, and the iron's mini-magnets reposition themselves randomly, losing the magnetic properties of the iron. (though I'm sure you've heard of ways to keep the magnetism inside the iron for a while longer, right?)
So hold any magnetic material near a magnet, it'll align with the field and will always attract.
Of course I've been lying.
There actually are some very rare materials that align themselves against the magnetic field and get repelled.
Why would they do that? At this point, let me mumble something about quantum mechanics and refer you to the more advanced literature. :p
I don't know the physical representation of those "mini-magnets", nor the proper term, nor the actual physical and quantum physical effects that are working on them. I tried reading up on it once and got a headache, so I'll stick to the simplified model.
that individually made little sense, but if you could figure one of them out, it would provide a clue to help decode another
Not that I've gotten the idea from another recent thread in this forum, but you could use a variation of the logic puzzle. Wrap each present in a different color, one for each person. Assign each person a few hidden attributes ("secret number", "soul animal" etc).
Start with a few of the usual hints like "The person with the secret number 8 gets the yellow present". Include a few personal traits like "The person with the largest secret number has long hair", possibly including some of last year's notable events ("The owl is the soul animal of the person who gave birth").
Give just enough hints to deduce one assignment of person to present. Inside the present is an additional hint (or additional hints) to solve the next, etc.
Of course you don't just give out the hints, but you make them do little puzzles to determine the hint. From math puzzles to figure out a secret number, to someone's very own quest to find their soul animal, go nuts.
It's not the most interesting overarching theme, but it'll give you a lot of freedom with the individual puzzles, and every now and then someone gets to open a present and share their treats.
But what is actually inherent to TASing? I say, only the fact of editing the replay data.
Indeed. The fact that the input is not generated by holding a controller and playing the game, but by artificially constructing the best possible input one can imagine using any tools available.
Which is why I dislike the idea of a replay-controller for the AGDQ presentation: it limits the input generation to holding a controller and playing the game.
Sure, none of the tools I mentioned are required to make a good TAS. There are other tools I didn't list. But holding a controller is probably the worst tool we have, it's absolutely insufficient in terms of precision and efficiency when aiming for perfection, and I disagree that there should be much focus on this way of TASing.
AnS wrote:
Of course, this process would take hundreds of years to produce a publish-worthy TAS
Remember that the timeslot during AGDQ is somewhat limited, so it's important to bring the point across as quickly as possible. That leaves more time to show off our best TASes :)
feos wrote:
BTW, if such tool gets widely available, the SDA's mark "verified, no cheating" would finally lose its sense :P
<ot>That mark is no more than a marketing gag. There have been cheated runs officially sanctioned with that mark. Many runs are encoded by the players themselves, so you have the situation of runs being submitted for verification that already contain the mark.</ot>
A controller that can replay the button presses you previously entered? Sounds cool. But I doubt it's the right tool to explain to others what a TAS is. Unless that controller can do savestates, slowdown, memory watch and LUA scripting. ;)
Thanks for the detailed descriptions, dwangoAC. I'm purely a software guy and can't claim to understand half of it, but it's interesting to read.
Allowing 180 degree turns makes the solution fairly simply: If there is an Eulerian path, you simply turn around at the end and traverse it in the other direction.
Except there is no Eulerian path in the undirected graph, where you traverse every street only once. There are four nodes (street corners) with an odd number of edges, so there cannot be a path.
There is an eulerian path (even a circle) in the directed graph, where you traverse every street twice, once in each direction. Bobo has already mentioned the proof, and it's easy to construct a circle as well. I can't get below two 180° turns, though, and I suspect it's impossible without.
/edit: found a circle with only one 180° turn. If you happen to start at that position, you'll have a 180°-free route.
I realized that there are 8 street corners, but actually only 4 nodes that matter, since the upper two and lower two corners only give you the choice to carry on, if you want to avoid the 180. Thus it's much easier to play around and brute force a bit.
If you look closely at the two vertical roads AC and BD and consider when to travel them and what to do in between, I doubt you'll find a true circle, but I still lack rigorous proof.
and my next step is to figure out how best to interface with a Raspberry Pi
I was under the impression that the previous bots were simple things that couldn't do anything but send a few electrical signals, and they had to be reprogrammed from a PC anytime you wanted to changed movies or something.
But the Raspberry Pi can run a complete OS with a GUI. So.. would it be possible to make the bot completely self-sufficient, by installing an emulator on the Raspberry, attaching keyboard and monitor, and working from there?
I think that'd be cool for a live demo. Show the Raspberry's screen on stream, play around with the emulator.. this is slowdown, we can do frame advance, here's a savestate.. just record a little movie while explaining the basic tools we use. Then send it to the real console. Live. And when the viewers understand what's going on, go for the real movies.
Possible? Feasible?
500 GB should be sufficient for all currently published avis. I'm sceptical about all "unlimited bandwidth" offers, though. Once you actually start using that much bandwidth, there's a good chance for your contract to be terminated, or for your bandwidth to be throttled.
Then again, nothing is stopping you from setting up a shell, installing a bt client and something similar to btfriend.pl, and start seeding.
If you're already a client, make sure to keep the accounts separate though. Just in case they decide to terminate the contract; you wouldn't want that to extend to your other server(s). And with regards to copyright laws, I think getting an US server is the safest thing at the moment, due to their "fair use" clauses.
Guys, relax. The poll has never mattered and will never matter.
For the easy choices with 100 yes and 1 no vote (there'll never be 0, right?) and the other easy choices with 20 no votes, the judges would've figured it out on their own without any votes.
On the controversial submissions, what matters are the arguments posted in the thread, not the amount of votes.
So please stop worrying about the votes. Just pick one. If you feel like your opinion cannot be sufficiently expressed by the three choices contained within, write a post. The judges will figure it out.
Someone who's interested on the ongoing discussion regarding Input time vs avi time comment about this?
Haven't read the discussion, but in this case optimizing for input time would have an ugly consequence: the most optimized run would only build as much of the city as they can get away with, then stop input. Consequently it'd take hours to reach 600k.
So in the interest of keeping the run watchable, the only reasonable goal is the time it takes to reach 600k. And I do wonder: isn't there anything that can be done to speed up that goal by extending input? Clicking away text boxes? Reducing lag? Maybe manipulate luck to get the city to grow faster?
Kudos for finding your own layout and improving the money glitch, but I don't think this is as optimized as it could be.
The thing I disliked about the old video was that it didn't even try to educate the viewer. At no point does the video mention that these tricks are done using emulators, or how they are done.
I like the idea of grouping games based on the techniques used (precision, luck manipulation, bug abuse, doing things for entertainment, ..), but I'd suggest prefixing each group with a short explanation.
Warp wrote:
The depth of the tree is five, the leaf nodes is at level five.
So there's 3^5 = 243 characters?
henke37 wrote:
Now I wonder what the chance is for blindly guessing A all of the battles and B just the final winner.
B is very easy to answer: Each character has a 1/243 chance of being the final winner. If you blindly pick one, your chance is 1/243.
A is a bit more difficult. There are two interpretations:
A1) for each battle, you blindly pick a winner from the set of characters that could participate. You may guess for Mr X to lose his first poll, but win the finals. Which is stupid, but certainly "blind".
For each poll, your chances are 1/[number of characters that could participate]. The chance of winning all of them is
(1/3)^81 * (1/9)^27 * (1/27)^9 * (1/81)^3 * (1/243)
A2) You're a bit smarter than that. Whenever you pick a winner, you bet on him to win all his previous polls, too.
Let's work backwards:
(1/243) chance for the winner to be correct. If you got him correct, you've also guessed his previous four polls correctly.
(1/81)^2 chance for the other two semi-finals to be correct, which includes their previous three polls.
(1/27)^6 chance for the remaining tier 3 polls to be correct. At this point you'll likely want to draw the tree and cross out any paths you've already guessed. There's 9 polls in tier 3, and you already know that three of them are correct, so 6 remain. For the remaining two tiers, it'll get a bit more finicky. In fact, I'll leave that as an exercise to the reader :p
Once you've understood the structure, you can try to derive a general formula that can tell you your chances of guessing a 4-per-match, 7-tier poll. If you feel like it.
I think it might be. Or am I wrong when I say that (0.4999..., 0.9999...) (which maps to 0.4999...) and (0.5, 0.0) (which maps to 0.5) map to the same point?
Well, many numbers have two decimal representations. You'll have to agree which one to use in your function, otherwise you'll have up to four different function values for each set of (x, y) coordinates, and your function stops being a function.
f(0.5, 0.5) = 0.55
f(0.5, 0.5) = 0.4499999999
f(0.5, 0.5) = 0.5409090909
f(0.5, 0.5) = 0.4590909090
pick one and stick to it. I suggest the first, if you value your sanity in the upcoming proofs. ;)
Warp wrote:
What I mean is that it's counter-intuitive[..]
Maybe your intuition just sucks ;)
After learning about countability, whenever you encounter a new infinite set, your newly trained intuition should kick in and ask: "So how would I number these elements without missing any?". If you can't come up with a numbering off the top of your head, it's probably uncountable.
Until you retrain your intuition to do so (maybe Pavlov can help), your intuition will keep getting wrong results. Don't rely on the part of your brain made for counting apples to understand advanced math.
Warp wrote:
Looking for the Z-order curve?
Granted, that's interleaving binary digits instead of decimals (thus resulting in a much more useful curve), but the curve is spacefilling in any base.
Proving that your function is indeed a bijection is easy.
Warp wrote:
This means that there is no one-to-one mapping between the set of natural numbers and the set of all possible combinations of natural numbers... which is very counter-intuitive.
Is it? All you have is a mapping of infinite elements from A to a single element form B. That's nowhere near a bijection, and is useless in constructing a bijection. Just because there's some kind of a relationship doesn't mean there's a bijection.
Warp wrote:
One would think that you could number each combination uniquely.
Why would you? Can you come up with an algorithm that would do so?
Once TASing was raised to the level it is now, I'm rarely shocked by a run as a whole[..]
Exactly, we're all getting jaded. Back then these runs blew my mind. I was wondering how such precise play was even possible, it looked unlike anything I'd ever seen.
Today I'm looking at them, thinking "oh hey, that's sloppy, he's losing frames there".
While the question was about runs that blew our minds, a newcomer list should contain runs that blow their minds, and our nostalgia is no factor in that.
For the list, I still can't think of a run better than SM64 0%.
Are there technical ways to prevent invalid movies from being submitted?
No, but that's already the case. If someone doesn't bother to make sure their movie actually syncs, there's nothing we can do about that.
I'm sorry, I didn't mean "invalid", but "cheated".
Imagine me submitting a movie that
plays the game normally
saves
uses memcard alterations in the middle of the saving process to create a broken savegame, like the chrono trigger run does with resetting
later reloads from that save under the pretence of save-warping
What I've been doing to the SD card is certainly cheating. Altering the SD card in between frames is impossible on a real console (no matter what kind of robot you use to pull and reinsert the cards), and I don't think the state of the SD card is clearly defined if it's pulled while written to.
Assuming that the effects of the save manipulation are subtle enough that they're not immediately apparent to the casual viewer, how would you detect it?
So you want to
a) load a savegame
b) remove the memory card from the console
c) restore the memory card to a previous version (with no save file)
d) reinsert the memory card
?
What happens to the real console when you remove the SD card while a game is running? Does the game or operating system notice? Is there some OS overhead for handling the changes that would need to be emulated?
How is this going to affect recording and replaying? When loading a savestate, the emulator has to ensure that not only the SD card contents are in the correct state, but also that any backups are present and in the correct state as well. Can this be done in a safe and robust way without noticeably slowing down quicksaves/quickloads? And without introducing incompatibilities between emulator versions for savestate- or movie files?
How is this going to affect movie verification? Allowing arbitrary writing of SD contents in a movie file is obviously a no-go, since you could just manipulate your savegames without anyone noticing. But even a simple API that only allows you to "save SD", "restore SD" or "format SD" is going to be problematic if someone calls them during saving or loading. Are there technical ways to prevent invalid movies from being submitted?
Next up is console verification. NESBot's grand grand grandchildren are going to have trouble replaying a movie that does fancy things with a memory card.
All in all, if it's just about skipping a confirmation dialog, I very much doubt that it's worth all the trouble. A hackish solution for a single game won't do, and a robust and accurate general-purpose implementation is a lot of effort, both on the emulator and the site's backend, and it'd need more valid use cases before I can consider it a good idea.
it would be a 3X4 rectangle filled with an orange "L" a green "s" and another orange "L".
Of course. That satisfied the requirements as stated, though for my unstated purpose of "looks like tetris when painted on stuff", it won't really do. ;) Thanks!
Twelvepack wrote:
You can prove that this is at least tied for the optimal solution
There's tie at 2x6
#----%
###%%%
which I also didn't consider due to the reasons mentioned above.
Twelvepack wrote:
Are your requiring that all tetronimos appear in the solution? in that case this answer is no good.
Oh how I'd like to have a rectangle with each tetronimo exactly once! Unfortunately, that isn't possible. You need an even amount of T shapes. And "every tile once and T twice" is at least 8x4.
More of a puzzle than a math question..
As you all know, there are seven tetrominos, the five shown here and the mirrored versions of the two asymmetric ones:
I was wondering: What's the smallest rectangle you can completely fill with tetrominoes so that no two equal tetrominoes touch each other?
(Let's skip the two trivial solutions and require two or more tetrominoes)
Wikipedia shows a solution to a different tiling problem, but by removing the rightmost 4x1 tile it yields a solution to mine with 4x9:
Unfortunately, I couldn't find anything smaller. The corners are tricky. Are you more lucky? Otherwise I'll probably try to write a small program over the weekend.
And I'm aware there's a lot of stuff on tetromino tiling on the net, but I couldn't find anything matching my specific problem.
If you look closely at the two vertical roads AC and BD and consider when to travel them and what to do in between, I doubt you'll find a true circle, but I still lack rigorous proof.
I was wondering: What's the smallest rectangle you can completely fill with tetrominoes so that no two equal tetrominoes touch each other?
(Let's skip the two trivial solutions and require two or more tetrominoes)
Wikipedia shows a solution to a different tiling problem, but by removing the rightmost 4x1 tile it yields a solution to mine with 4x9:
Unfortunately, I couldn't find anything smaller. The corners are tricky. Are you more lucky? Otherwise I'll probably try to write a small program over the weekend.
And I'm aware there's a lot of stuff on tetromino tiling on the net, but I couldn't find anything matching my specific problem.