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**Forum Games / Re: Drunk Mafia 8: Quarantine Edition**

« **on:**April 17, 2020, 12:09:16 am »

I can maybe join later on if things go for a while.

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I can maybe join later on if things go for a while.

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This sounds very up my alley. What platforms is it on?

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I just had a classic example of failing to do basic math: I went for Masterpiece-Feodum in a game with Bandit Fort. I was thinking enough Feodum points would outweigh the Bandit Fort penalty. Then I actually did the math and realized I would need at least 7 Feodums for that to happen... which was no longer possible because my opponent was pursuing the same boneheaded strategy I was!

Fortunately in this situation you can always just trash the feodums and pivot to a new strategy.

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Because the sails are different?

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I stopped quite a few screens before the door. I was in one of the first screens with the blowfish, I was supposed to dash into it, boost myself across some spikes, then dash back to that fish so it would blow me into another and so on. I kept not being able to bounce hard enough from dashing into the fish to clear the spikes without using the dash I needed to go back to the fish afterwards. After feeling rather frustrated with some of the screens in Core (and beating them through sheer persistence rather than skill), I thought it was a good spot to end.

This is the exact one I’m stuck on as well. The great thing about this game is that each screen is two challenges: the puzzle of figuring out what to do to beat it, and then executing the platforming moves. In this one though it seems pretty obvious what to do, but very hard to pull off.

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Snark aside, I don’t really see this. We’re talking about a two card combo, and it was asserted that hunting party is no better than menagerie with patron. How do menagerie and patron constitute an engine that will reliably draw your deck?

Well, I'll try hard to keep this thread in mind the next time I play Dominion with a 2-card kingdom and the cards happen to be Hunting Party and Patron. After all, there is no way I could possibly figure out on the spot that there's anything going on there.And why is the +buy “easier to fit in?”

Because you're playing an engine, not a Hunting Party stack.

All right, I did say "snark aside," but cool.

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This isn't really a combo; there isn't much of anything that makes hunting party differ from eg menagerie or any other reveal card here, and hunting party makes it difficult to actually find all your patrons

combo threads are generally for things that completely change the game when present: hermit / market square, lurker / hunting grounds, etc

(hunting party is much better than menagerie here because you reveal the patrons that you don't draw as well as the ones in your hand; point taken that the word "combo" is controversial)

Yeah, this is the entire point. Maybe try it out before declaring it "not a combo?" In a Hunting Party strategy you're revealing your *whole deck* *many times* *every turn* (edgecasesedgecases). There is no comparison with Menagerie.

In a typical Menagerie strategy (also known as an "engine"), you can easily be revealing your *whole deck* *many times* *every turn* and the +buy is a lot easier to fit in.

Snark aside, I don’t really see this. We’re talking about a two card combo, and it was asserted that hunting party is no better than menagerie with patron. How do menagerie and patron constitute an engine that will reliably draw your deck? And why is the +buy “easier to fit in?”

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This isn't really a combo; there isn't much of anything that makes hunting party differ from eg menagerie or any other reveal card here, and hunting party makes it difficult to actually find all your patrons

combo threads are generally for things that completely change the game when present: hermit / market square, lurker / hunting grounds, etc

(hunting party is much better than menagerie here because you reveal the patrons that you don't draw as well as the ones in your hand; point taken that the word "combo" is controversial)

Yeah, this is the entire point. Maybe try it out before declaring it "not a combo?" In a Hunting Party strategy you're revealing your *whole deck* *many times* *every turn* (edgecasesedgecases). There is no comparison with Menagerie.

It could really use some +Buy, but it's uh, really good.

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bah, /out, sorry

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How would you describe math to a layperson, in such a way as to try to break through their preconceived notions of what it is?

This doesn't precisely answer the question, but this classic essay by Thurston is a great read

https://arxiv.org/pdf/math/9404236.pdf

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People were able to compute plenty of derivatives perfectly well far before any of the modern rigorous notions of limits, which was primarily due to Cauchy. They did so mostly with "handwaving."

The key issue here is more subtle, and it's more about being circular than being handwavy. The approximation e^h = 1+h when h is small isonlyvalid for the number e. It's equivalent to the fact that the tangent line to the graph of e^x at x=0 is y=1+x, which can't be established without knowing that the derivative of e^x is e^x in the first place.

I don't think that's a different issue from what I said. Being handwavy is what allows the argument to get away with whatever wrong thing it's actually doing, like being circular. If it spelled out the argument explicitly, we would see what exactly it relies on.

Taking shortcuts is never a mistake in itself, it's always a black box where real mistakes may or may not be hiding.

My point was just that the answer the question:

The physicist's proof for the derivative of e^{x}:

(e^{x + h}- e^{x})/h = e^{x}(e^{h}- 1)/h. Because h is small, e^{h}= 1 + h, so e^{x}(e^{h}- 1)/h = e^{x}(1 + h - 1)/h = e^{x}(h/h) = e^{x}.

What prevents that from working with any number other than e?

is something we can specifically identify in the argument. It's exactly contained in the statement "Because h is small, e

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The substitution step could be done rigorously if you use the infinite series definition of the exponential. That would give a different result if you have a number other than “e”.

I think without the infinite series definition of the exponential function (or the definition as the inverse of the logarithm), the farthest you can go with this is that the derivative of exp is proportional to exp.

EDIT: what cuzz said

Right, I claimed that you can't use that substitution without already knowing the derivative of e^x is itself, but in fact you can if you

In fact, it's worth thinking about what your original definition of the exponential function even is, as this can make a difference as to whether any given fact about it is a theorem or a tautology. For example, it's also perfectly reasonable to define the exponential function as the unique function satisfying f'(x) = f(x) and f(0) = 1, once you prove that such a function exists and is unique.

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People were able to compute plenty of derivatives perfectly well far before any of the modern rigorous notions of limits, which was primarily due to Cauchy. They did so mostly with "handwaving."

The key issue here is more subtle, and it's more about being circular than being handwavy. The approximation e^h = 1+h when h is small is*only* valid for the number e. It's equivalent to the fact that the tangent line to the graph of e^x at x=0 is y=1+x, which can't be established without knowing that the derivative of e^x is e^x in the first place.

The key issue here is more subtle, and it's more about being circular than being handwavy. The approximation e^h = 1+h when h is small is

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This recipe is the best frozen cocktail I’ve ever had. Not a trashy syrupy beach drink, not too sweet, perfect for summer. You can batch the mix in the freezer and then just blend with ice on demand. It’s amazing. https://www.seriouseats.com/2016/07/frozen-gin-and-tonic-blender-summer-drink-cocktail.html

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I'll take Things-I'll-Never-Have for $800, Alex.

Same, and I utterly don't care. It bothers anyone looking over my shoulder so much more than it bothers me.

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I don't like staring at source code to learn something, so I made a texed version of your explanation.

You even explained how to do exponentiation of two complex numbers, which I've wondered about for a while. You'd think they'd teach that in Analysis lectures, but nope.

They do in complex analysis courses! It's an incredibly beautiful subject that I highly recommend taking a course in if you get the chance.

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Thanks! This is a quality explanation.

Your ln(0)s have to be ln(1)s, right?

yep, fixed

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I'm somewhat confused about complex numbers right now. Or rather about how WolframAlpha deals with them, which is presumably the correct way.

If I ask for solutions to the equation x^5 = 1, it gives me all the complex numbers with angles n * 72°, n = 0,1,2,3,4. (Left.) If I ask for solutions to the equation x^{5/2} = 1, it gives me the ones with n = 0, 2, 3. (Right)

So what exactly is the property that the points with angles 72° and 288° don't have?

It’s because the function which raises a complex number to the fifth power is a real function, whereas the function which raises a complex number to the 5/2 power is not, it’s a doubly valued multifunction. In all 5 of these cases, one of the two values of the multifunction is 1, but only in the three given solutions is 1 the *principal* value of the multifunction.

When you say double valued multifuction, you mean a function p : \C -> \powerset(\C) defined by p(c) = {[first square root of c^5], [second square root of c^5]}?

So what's the definition / consensus on what is the principal value?

Yeah that's one way to think about it (though the more proper and technical setting for this involves Riemann surfaces).

To define principal values, we need to talk about arguments, logarithms, and power functions.

The

Then we note that we can write a complex number z in exponential notation, as z = r*e^(i*a), where r=|z| is the modulus and a is one of the arguments of z. Here we can choose any of the arguments of z, since the complex exponential function is periodic with period 2*pi*i (for the same reason that adding 2*pi to an angle gives the same angle). This means in particular that the complex exponential function is no longer one-to-one as the real exponential function is, and hence the inverse function (the logarithm) is not a real function.

But we decide that we want to talk about logarithms anyway, so we define it as a multifunction--the log of a nonzero complex number z is now the

This now lets us define power functions. Given z and another complex number c, we can define z^c by z^c = e^(c*logz) = e^(c*ln|z| + i*c*arg z). Since log is multi-valued, the power function typically is as well, but we can single one of these out by using the principal branch of the log function, e^(c*Log z) = e^(c*ln|z| + i*c*A). It happens to be the case that the power function is infinitely-valued unless c is rational, in which case it has as many values as the denominator of c written in lowest terms. When c is an integer, the denominator is 1, so power functions with integer exponents are in fact all single-valued.

So for example, let's apply this to some of your numbers. The number z_1 at angle 72 degrees has principal argument of 2pi/5 radians, and a modulus of 1, so the principal value of z_1^(5/2) is

z_1^(5/2) = e^(5/2*ln(1) + i*(5/2)*(2pi/5)) = e^(i*pi)=-1.

The number z_3 at angle 3*72 = 216 degrees has argument of 6pi/5 radians, but its

z_3^(5/2) = e^(5/2*ln(1) + i*(5/2)*(-4pi/5)) = e^(i*(-2pi))=1.

But remember that this whole principal business is rather arbitrary. Using the argument 12pi/5 for z_1 gives

z_1^(5/2) = e^(5/2*ln(1) + i*(5/2)*(12pi/5)) = e^(i*6pi)=1,

and using the argument 6pi/5 for z_3 gives

z_3^(5/2) = e^(5/2*ln(1) + i*(5/2)*(6pi/5)) = e^(i*3pi)=-1.

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I'm somewhat confused about complex numbers right now. Or rather about how WolframAlpha deals with them, which is presumably the correct way.

If I ask for solutions to the equation x^5 = 1, it gives me all the complex numbers with angles n * 72°, n = 0,1,2,3,4. (Left.) If I ask for solutions to the equation x^{5/2} = 1, it gives me the ones with n = 0, 2, 3. (Right)

So what exactly is the property that the points with angles 72° and 288° don't have?

It’s because the function which raises a complex number to the fifth power is a real function, whereas the function which raises a complex number to the 5/2 power is not, it’s a doubly valued multifunction. In all 5 of these cases, one of the two values of the multifunction is 1, but only in the three given solutions is 1 the *principal* value of the multifunction.

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I am researching how to apply for a US visa right now... The example visa application that the US government provides is both cute and somewhat embarrassing:

There's a typo. Should be Spouse/Cousin.

Spouse/Aunt

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2,3,5,7,11,13,17,19,23,29,31,33,36,40,44,47,51,57,59,62,65,69,73,83,86,89,100,101,104,107,110,116,117,113,115,117,114,119,124,124,125,126,128,132

Can you guess where I stopped?

Man this thing sucks at computing powers of 2.

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My wife is pregnant with our first child! Due in December. Got to tell our families on mother's day, which was fun

Congrats! It is a wild ride which will reprogram your life and brain in ways you can’t possibly conceive! But it’s good times.