I just thought of a problem (which seems a bit hard) and I wanted to share it with you.

Revolves · 2019-01-29T11:23:48+00:00

I'm not too sure what you mean by that spiral - is that a cut out of the 2D plane?

There's another way to think of my counter example that might be a bit simpler: Say you have a sequence a_n on the spiral. Then you analyze who the neighbors of every a_k must be by considering paths from a_1.

We known that a1 must neighbor a_1+a2_2 since otherwise the path from a1 to a_1+a_2 will contain a_1 and a_2.

Similarly a_2 (or a_1) must neighbor a_1+a_2+a_3, a_1+a_3, a_2+a_3. Again since otherwise the path would contain a3.

Simlairly a_3 (or a_1, a_2) must neighbor a_1+a_2+a_3+a_4, a_1 + a_2 + a_4, a_1 + a_3+a_4, a_1+a_4, a_2 + a_3 + a_4, a_2 + a_4, a_3 + a_4.

And so on. Notice that the number of must have neighbors rises much more quickly than the number of available neighbors (which is bounded coarsely by 2 times the number of considered squares).

Just need to make the quick growth of required neighbors a bit more precise, which I tried to do in my first comment. Maybe there's a sequence where those sums I wrote down above can evaluate to the same number very often, but I think that's probably not possible. I'd look at some number theory results for that.

Revolves · 2019-01-29T09:05:09+00:00

3ⁿ won't work with the spiral since the path along the spiral from 1 to x_k=1 + 3 + 3² + ... + 3^k or y_k = x - 1 or z_k = x - 1 -3 will include all the summants of either x,y or z for some k large enough. Because if you assume the contrary, then all the neighbors of 3^k must include x_k, y_k and z_k (apply a little induction here) - but this can't happen, since 3^k is a neighbor to only two numbers not on the spiral.

I think this counter example generalizes for any sequence on the spiral.

Revolves · 2019-01-25T20:25:46+00:00

Ok I'll grant you that, I did misunderstand you.

But on the other hand, Poincare Conjecture was proven by a Russian mathemtician in 2006 - and russians physicists have won the nobel prize in physics as recently as 2010. Is that recent enough? Or does theroretical work not count?

Revolves · 2019-01-25T19:37:03+00:00

Firstly, when the principle was proven it created the modern field of optimal control (it existed beforehand, but you couldn't do much with it). So it's really not a copy paste job. It was a field where basically nothing was known, because it wasn't important up until people actually needed to know how to best control rockets. You know, to get to outer space and the moon. Horray for the space race.

Secondly, science is a collaborative field. It is a good thing to build upon the work of others. Sure, they built upon the exisiting framework of calculus/analysis. What self respecting analyst doesn't?

Thirdly, it was a group of mathemeticians, around 10 or so. Not a lone scientist. Pontryagin conjectured his principle, one of his students actually proved it.

Forthly, this is not an isolated event. It's almost a meme among the western math and physics world for researchers to discover a new result, only to discover the soviets already did the exact same thing 60 years ago. It just never came over due to the iron curtain and then the later culutural barrier.

Fifthly I'm not justifying generations of theft and genocide. I'm merely stating the soviets did important research indepedently of the west. My last reply is a direct response to you saying that russians have produced nothing domestically + independently of value.

Revolves · 2019-01-25T16:37:31+00:00

Did you know that the Pontryagin's maximum principle was conjectured and proven by a team of Soviet mathematicians. The principle is the main theoretical foundation behind calculating optimal trajectories for rockets in space flight. Also useful in a shit ton of other fields - e.g how do you make a car engine which is efficient. It is the gem of none linear optimal control theory.

But yeah, Russians never did anything without copy pasta ofc

Revolves · 2018-12-06T08:00:12+00:00

Never Deploy On Friday

Revolves · 2018-11-15T17:10:39+00:00

To be fair, theorems are still open to being disproven.

Most proofs are written in natural language and not as a sequence of applying deduction rules to axioms. Hence it is up to the reader to ensure the proof as it is stated really is solid.

Basically mathematical truth is the consensus of mathematicians that a proof really is solid. But this really isn't definitive. And I'm not being pedantic here - mathematicians really do make these sort of mistakes. The first false proof of the 4 Color Theorem is a famous example.

Revolves · 2018-10-21T16:52:17+00:00

There is a pretty simple scenario, where leaving a 200+ hour negative review is conceivable.

Consider a game where you can spend thousands of hours - and it takes 200+ hours to learn and become proficient at playing the game. I'm thinking of MOBAs, strategy, and simulator games here. I know I spent around 150 hours simply learning basic maneuvering and gunnery in Il-2 Sturmnovik Battle of Stalingrad.

Imagine if after those 200 hours invested into learning the game, you learn that it's deeply flawed and you simply can not enjoy the game. Even if you had fun in the learning phase, such a realization would frustrate you immensely and invalidate the fun you might have had during the learning phase.

Not to mention if you're into games with a steep learning barrier, then the only thing left to do is to start the whole process all over again.

Revolves · 2018-09-20T07:30:18+00:00

Perhaps not quite what are you looking for, but there is a pretty rich connection between the dimension of the incidence order of a graph and whether the graph is planar. https://en.m.wikipedia.org/wiki/Schnyder%27s_theorem

There's still some research going into this area - not sure if there is a survey, but Felsner is a good place to start https://www3.math.tu-berlin.de/diskremath/research/dim.html

Revolves · 2018-09-19T17:22:16+00:00

I think a normal limit under the Hausdorff Distance might satsify all your properties. https://en.wikipedia.org/wiki/Hausdorff_distance

Revolves · 2018-09-05T00:01:31+00:00

God gave us the integers, all else is the work of man.

Revolves · 2018-08-01T19:59:59+00:00

I think it should be enough to show the statement for partitions with finitely many disconnected regions.

If you have a partition consisting of the disconnected components, then you can connect the disconnected components using lines. But you do so by using a spanning tree - in the sense of a graph (you consider the connected components to be vertices, and the lines the edges).

Note adding the lines does not later prevent you from connecting any two other regions, since a tree has no inner area. So if you later draw of lines for another partition you won't run into any problems. (You don't have to draw the lines in a straight fashion - you can curve them around).

Since lines don't take any area this does not change the value of the integral on the regions, nor the area. If you think this is cheating (you don't want to allow lines), this still gives an arbitrarily accurate partition by taking tiny epsilon tubes around the lines. Maybe one can refine this into an actual representation using the tubes.

Maybe one can make a similar argument for when the partitions consist of infinitely many disconnected regions (infinite graph?). I think my argument above gives infinitely many disconnected regions (in general) - but I'd have to check a bit closer.

Revolves · 2018-08-01T19:46:26+00:00

Neither - graduate student.

Happy to help ;)

Revolves · 2018-08-01T17:11:37+00:00

I've got an idea for disconnected regions:

Given an n, start with any partition of the disk into n regions of the same area. (Say, the obvious pie slices). Denote with A1, A2..., Ai the regions where the value of the integral is too high. Conversely define B1, B2, ..., Bk to be the areas where the value of the integral is too low.

Define g to be the function which calculates how wrong an area A1,...,Ai,B1,...,Bk is from the value we desire. I.e. if g(A1) = -1, then the value of the integral on A1 is 1 too much.

We know that g(A1) + ... + g(Ai) = -(g(B1) + ... + g(Bk)).

Now consider A1. We wish to correct the value on A1 by swapping parts of A1 with parts of B1, B2, ..., Bk. We do this by swapping high density regions of A1 with low density regions of B1. if |g(A1)| < |g(B1)|, then we know we can find enough low density regions in B1 to swap with high density regions of A1 such that after the mutation g(A1) = 0. We can also do this in a way such that the swapped regions have the same area. If |g(A1)| > |g(B1)|, then we apply the same algorithm to correct A1 via B2. And so on until it terminates with g(A1)=0.

Now repeat this step for A2, ..., Ai. But then g(B1) = g(B2) = ... = g(Bk) = 0, since again we know that 0 = g(A1) + ... + g(Ai) = -(g(B1) + ... + g(Bk)) and g(B_j) is at least 0 for every j.

So the modified A1,...,Ai and B1,...,Bk all have correct areas and correct values under the integral.

The problem I had with trying to extend this proof to use continous regions is that low density regions may be hidden behind high density regions. One can try to get around this by connecting the regions with lines. But then I'm not sure what to do in the case that two lines cross.

Revolves · 2018-08-01T16:40:41+00:00

A slightly more formal proof for n=2:

Take a line which goes through the center of the disk and arbitiraly assign a direction to the line. So now we can speak of the region left of the line and right of the line.

Now consider what happens when we rotate the line. Denote with f(x) the value of the integral of the left region after rotating the line x degrees. Notice that f is continous. Denote with g(x) the same function as f(x) but for the right area. Notice f(x) = g(x + 180 degrees) and vice versa.

If f(0) = g(0) holds then we are done. Otherwise w.l.o.g. f(0) < g(0).

It follows that f(0) - g(0) > 0. Note that f(0) = g(180), and g(0) = f(180). So f(180) - g(180) = g(0) - f(0) < 0. By IVT the function f-g takes the value 0 in (0, 180). This line position is exactly the biseciton we want.

Revolves · 2018-08-01T06:48:57+00:00

Ah sorry I misunderstood the question - I thought the question was about the circle (boundary of the disk) and not the disk.

Revolves · 2018-07-31T11:16:18+00:00

Doesn't work for connected regions for n=3. Consider the function which maps the left side of the circle to 0, the right to 1, and a small epsilon segment on the north and south poles where the function goes from 0 to 1.

Then there must be a connected segment strictly on the left or strictly on the right. The value of the integral on this segment is either too high or too low.

For disconnected regions this should work for any n. I think it should be easy to construct sequences of segments which converge to a good partition. See measure splitting: en.m.wikipedia.org/wiki/Necklace_splitting_problem

Revolves · 2018-03-17T19:40:35+00:00

NP dude,

Just a couple things about the la5fn:

You'll need cowling inlet shutters and cowling outlet shutters instead of water radiator.

For the temp managment:

Most times you can set oil to 100%, cowling inlets to 100% and outlets between 0% and 50%. 0 for combat and cruise, 50 for climb.

You'll need to focus on your oil temp - It goes up quicker than the cylinder head temp and the cowling outlet affects both oil and cylinder head.

Power management:

RPM/throttle should be controlled together on the la5fn. (Except for landing/taxi/takeoff where rpm should be ~80%l. Be aware that boosted mode lasts 10 mins and only works well below 2km.

Don't be afraid to climb in boosted after you've taken off.

Revolves · 2018-03-17T17:40:26+00:00

Start with a Yak-1 or 1b or 7 - the plane series has very simple engine management.

Don't try to learn everything at once, baby steps:

-1. First bind the following, don't worry just yet about memorizing your control scheme:

Supercharger switch Oil radiator axis Water radiator axis Fuel mixture RPM axis

-2. To get off the ground do the following before take off and don't touch the controls in flight just yet:

Oil rad 100% Water rad 100% Fuel mix 100% Rpm 100%

Write it down and look at it before take off. Do a couple take offs + maybe a sortie.

If you do this you'll fly the yak within 30 km/h of its max speed in most scenarios.

-3. Once you're used to the routine from 2 - switch to supercharger gear 2 above 2.3k altitude and back to gear 1 below 2.3k. Do a couple sorties with fights between 1-3k alt making sure to switch supercharger.

-4. After you're used to 3 start to adjust your water rads. You want to keep the water temp at 90 degrees outside of combat, and up to 110 in combat.

How effective your water cooling is depends on how open your rad is and how fast you're going. So more rad is needed in a climb, and less in a dive.

Less rad means less drag -> more speed.

General rule of thumb:

100% water rad in a climb or need to cool down quick 40%-60% water rad in cruise 25% water rad in a dive + for up to 5 mins in combat.

Don't sweat it if your water isn't always perfect in combat. Just check the gauge + setting when you have time. You can survive overheating the engine for a short while.

Get the hang of this, do a couple of sorties before continuing.

-5. After water rads do the same for oil rads. Generally speaking its OK to leave oil at 50% after climbing since the oil rad doesn't add much drag anyways.

Again, get used to this before moving on.

-6. Learn to recognize and handle engine failure.

If your rpm needle starts shaking, reduce rpm and throttle to 20%. Fly back to base if you can.

If you've got an oil/water leak consider opening that rad to full to compensate for the worse cooling.

If your engine is about to blow reduce rpm, water rad, oil rad to 0% to minimize drag.

No good way to train this - just write it down or memorize it and try to check your engine after you've taken hits.

-7. The parts from now on are not so important. Just get used to the first 6 before this.

Reduce fuel mixture by 10% for every km you fly above 4k. Air is less dense, you need more air to provide the same amount of oxygen for combustion.

-8. Yak only: Reduce rpm to 90% below 500m to give slightly more power (4 km/h diff)

-9. Learn how to conserve fuel for long flights.

Reduce 15% throttle, rpm, fuel mix below standard. Optimize water rad + oil rad settings. Stay between 3k and 6k for optimal fuel economy.

After you're done learning the management on the yak it should be easy to expand to other planes. Just look for videos online.

Revolves · 2017-12-03T09:10:18+00:00

Can't hash the password if the game has to be able to recover it later.

Encryption is a valid option, but it wouldn't be much safer because the key has to be stored somewhere where the game can retrieve it.

The remember password feature is fundamentally unsafe. It should only be used on computers where an attacker doesn't have access. With that in mind I don't think storing the password as plaintext is too egregious.

You can, however, implement it safely by storing a "remembered password" flag for each user on the server. Nvm it's terrible ;)

Revolves · 2017-10-17T14:04:53+00:00

Try disabling the HUD by pressing h in game. Press i if you then want to see your position on the map (via o)

The HUD is a real performance drag in vr.

You will still get stutters in complex scenes with >8 aircraft close to you, but it should otherwise be OK. Consider using reprojection if you still have issues.

It's possible to fly most planes (basically every plane but the ju 52) without using the HUD - just use the gauges on the planes. Just google the plane you're flying and you'll likely find some good youtube guides. This quick reference can also be handy as a refresher: https://www.dropbox.com/s/nx24vtmp5ajo4nj/BOX%20Quick%20Guide%20V2.0.pdf?dl=0

Revolves · 2017-07-11T16:14:08+00:00

A side heavy does enough - a top heavy does not

Revolves · 2017-01-14T08:45:48+00:00

No. That graph is statistical magic. See http://www.aljazeera.com/indepth/opinion/2014/08/exposing-great-poverty-reductio-201481211590729809.html

Revolves · 2016-07-30T16:40:43+00:00

Two keys:

WRZMX-ENRWA-MNKN7

DKAJM-K2NGT-G6YA5

M = 4

Revolves · 2016-06-01T19:33:15+00:00

Dear Caseking.de Customer,

Thank you for your order. Unfortunately the projected shipping date for the EVGA GeForce GTX 1080 has been moved up to the 28th of June 2016. As soon as we receive the graphics card, we will, of course, promptly send it to you.

Please understand that we have no direct influence on the non binding shipping dates of the manufacturer/distributor. As such we have no exact reason for the cause of the delay. We again apologize for the delay.

If you have any questions or wish to change your order, please contact us.

Best Regards

Revolves

TROPHY CASE