seed

• seed 16 Nov 2019 19:48 UTC

  4 points

  on: Topolog­i­cal Fixed Point Exercises

  I am sorry because I cannot figure out how to hide big formulas in a spoiler. Also the spoiler feature is somewhat broken so it adds weird tabs around formulas.

  #1:

  Let’s count the number of blue edge ends. Each blue point inside the segment is the end of two blue edges, and the leftmost blue vertex is the end of one. Therefore, their total number is odd. On the other hand, each bichromatic edge produces one blue edge end, and each unichromatic edge produces an odd number—zero or two—of blue edge ends. Therefore, an odd number of edges are bichromatic.

  #2:

  Suppose $x=inf\{y\in [0,1\left]|f\right(y)\ge 0\}$ . If $f\left(x\right)>0$ then $x\ne 0$and, since f is continuous, f stays positive in some neighborhood of x, and then x is not the infimum. Therefore, f(x) = 0.

  #3:

  Consider the function $g\left(x\right)=f\left(x\right)-x$ . Since $g\left(0\right)\ge 0$ and $g\left(1\right)\le 0$ by exercise 2, there should be a point where g(x) = 0.

  #8.

  Consider the family of functions: $f_t\left(x\right)=min(max(x-t+0.5,0),1)$

  For t < 0.5, the only fixed point is of $f_t$ is 1; for t > 0.5, the only fixed point is 0.

  #9.

  Lemma:

  Suppose a k-dimensional simplex is subdivided into smaller k-dimensional simplices and all vertices are colored into k+1 colors so that there are no vertices of color i on the i-th edge of the big simplex. Then there are an odd number of subdivision simplices whose vertices are colored in k+1 different colors.

  Proof:

  Induction by k. Base k=1 proved in exercise 1.

  Induction step: supposed the lemma is proved for k-1, let’s prove it for k.

  Let us count the number of tuples (X, Y) where X is a k-1-dimensional simplex colored in colors 0, 1, …, k-1,

  Y is a k-dimensional subdivision simplex, and X is on the boundary of Y. Each properly colored simplex X inside the big simplex produces two tuples, and each simplex on the boundary produces one tuple. X can only be on the k-th edge of the big simplex, and by the inductional assumption, there are an odd number of simplices X there. So, the total number of tuples is odd. On the other hand, each k-dimensional simplex Y can be a part of either:

  0 tuples;

  1 tuple if all his vertices are different;

  2 tuples if has vertices of colors 0,1,...,k-1 but not all his vertices are different.

  Therefore, a number of k-simplices Y with all different vertices must be odd.

  #4

  Follows from 9

  #5

  Suppose that center is not in the image of the triangle. Let us call a set of points bichromatic if it doesn’t have points of all three colors. We color each point in the triangle in the same color as its image. Then every point in the image has an open bichromatic neighborhood. Since the map is continuous, the preimage of this neighborhood is also open. So, around every point in the triangle there can be drawn an open bichromatic ball of radius r. These balls are an open cover of the triangle, let us choose a finite subcover out of them. Suppose $ϵ$s the minimum radius in this subcover. Split the triangle into subtriangles so that the diameter of each triangle is smaller than $ϵ/2$ By Sperner’s lemma, there is a trichromatic triangle, but since its diameter is smaller than $ϵ/2$ it lies completely inside one of the bichromatic balls. Contradiction.

  #10

  First, I am going to prove that a function from a unit ball $D^n$o itself has a fixed point, then that any compact convex subset of $R^n$s homeomorphic to a ball.

  Suppose that $f:D^n\to D^n$as no fixed point, n>1 (case n=1 was proved in exercise 3). Then I can build a retraction from $D^n$nto its boundary $S^{n-1}$

  send x to the first intersection of the ray (f(x), x) with $S^{n-1}$ Let us prove that such a rectraction cannot exist. Suppose that such a rectraction $g$ exists. Denote $i:S^{n-1}\to D^n$the inclusion map. Then $g\circ i=id$nd the induced homology group homorphism $g^{\ast }\circ i^{\ast }$ust also be identity: $H^{n-1}\left(S^{n-1}\right){\to }^{i\ast }{H}^{n-1}\left(D^n\right){\to }^{g\ast }{H}^{n-1}\left(S^{n-1}\right)$

  But this is impossible because $H^{n-1}\left(S^{n-1}\right)=Z$ and $H^{n-1}\left(D^n\right)=0$

  Now let us prove that any compact convex subset X of $R^n$s homeomorphic to a ball. Let us select a maximum set of affinely independent points in X. They form some k-dimensional simplex, all X lies in the affine space spanned by this simplex, and all the interior of this simplex belongs to X, because X is convex. I’ll take a ball $D^k$f radius $d$ side this simplex and build a homeomorphism between X and $D^k$. Taking the center of the ball as the center of coordinates, define

  $f\left(x\right)=x\ast d/r\left(x\right))$ where $r\left(x\right)=r\left(||x||\right)$s the distance to the farthest point of X in the $x$ direction, if $x\ne 0$, 0 if

  Let us prove that f and its inverse are continuous. Since X is compact, it is bounded, so there is a $R>0$ such that $r\left(x\right)\le R$ It follows that f and its inverse $f^{-1}\left(x\right)=x\ast r\left(x\right)/d$ are continuous in zero: $\forall ϵ>0$ if $|x|<R/d\ast ϵ$ $|f\left(x\right)|<ϵ$ $\forall ϵ>0$ if $|x|<d/R\ast ϵ$$|f^{-1}\left(x\right)|<ϵ.$

  Now let us prove that functions are continuous in all other points. It is sufficient to prove that r(x) is the continuous on the unit sphere. (Since composition and product of continuous functions is continuous, division by bounded from below (by d) continuous function r is continuous, ||x|| is a continuous function).

  Since X is convex, the tangent cone from any point of X to $D^k$ lies in X. So if we take a point at the distance $R$ from the center, draw a tangent cone, and go down its boundary, we get the steepest possible rate of change of r(x) with respect to x. Therefore, r is continuous.

  #6, #7: follow from #10.

  #11:

  Suppose f has no fixed point. Distance d(a, B) is a continuous function of a, and a continuous function reaches its minimum on a compact. TxT and the graph of f are nonitersecting compact sets, therefore the Hausdorff distance $\delta$ between them is positive. It is easy to see that Hausdorff metric is indeed a metric, i.e. that a triangle inequality holds for it. So if we take any continuous function g at a distance less than $\delta$ from f, its Hausdorff distance to TxT will be positive, so it can have no fixed points.

  #13:

  Suppose $h:S\to 2^S$ is a Kakutani function. We already know that any compact convex subset of $R^n$s homeomorphic to a simplex. Denote $g:T\to S$he homeomorphism between a simplex T and S.

  Denote $y_1^k,\dots ,y_t^k$ the k-th barithentric subdivision of T. For each $y_i^k,$ choose an element $h_i^k\in h\left(g\right(y_i^k\left)\right)$

  Define $f_k\left(x\right)$ ${\sum }_{i=1}^{n+1}{p}_i^k\left(g^{-1}\right(x\left)\right)h_i^k$ where $p_i^k$ are the baricentric coordinates of point $g^{-1}\left(x\right)$n its subdivision simplex. Function $f_k$s continuous, $f_k\left(g\right(y_i^k\left)\right)=h_i^k$ and, since S is convex, the image of$f_k$ lies in S.

  By the Brouwer fixed point theorem, $f_k$as a fixed point. Since S is compact, from the infinite sequence of fixed points of $f_k$e can choose a convergent subsequence.

  Suppose $x_k\to x^{\ast }$s this subsequence, $g^{-1}\left(x_k\right)$ lies in the simplex $y_1^k,\dots ,y_{n+1}^k$and has baricentric coordinates $p_1^k,\dots ,p_{n+1}^k$. Then $g^{-1}\left(x_k\right)=\sum p_i^k{y}_i^k$ and $f_k\left(x_k\right)=\sum p_i^k{h}_i^k$so

  $g^{-1}(\sum p_i^k{h}_i^k)=\sum p_i^k{y}_i^k$ (1).

  Since simplices go down in diameter, $g\left(y_i^k\right)\to x^{\ast }$as $x_k\to x^{\ast }$ Each $p_i^k\in [0,1]$s a bounded sequence, so we can, sequentially, choose a convergent subsequence out of each of them, so we can assume that $p_i^k\to p_i^{\ast }$ Similarly, we can choose a convergent subsequence out of $h_i^k$ so we assume $h_i^k\to h_i^{\ast }$ The sequence $\left(g\right(y_i^k),h_i^k)$ belongs to the graph of h and converges to the point $(x^{\ast },h_i^{\ast })$ Since the graph is closed, $h_i^{\ast }$ must belong to the image of $x^{\ast }$ Since $\sum p_i^k=1$ for every k, $\sum p_i^{\ast }=1.$ince the image is convex, $\sum p_i^{\ast }{h}_i^{\ast }$lso belongs to the image of $x^{\ast }$ On the other hand, as we remember, since equality (1) held for every k, it also holds in the limit: $g^{-1}(\sum p_i^{\ast }{h}_i^{\ast })=\sum p_i^{\ast }{g}^{-1}\left(x^{\ast }\right)$. Hence, $\sum p_i^{\ast }{h}_i^{\ast }=x^{\ast }$ So, $x^{\ast }$ is the fixed point of h.

• seed 22 Dec 2019 11:58 UTC

  1 point

  in reply to: Ben Pace’s comment on: Topolog­i­cal Fixed Point Exercises

  Thank you!

• seed 29 Dec 2019 11:11 UTC

  1 point

  in reply to: EvelynM’s comment on: Book Re­view: Naïve Set The­ory (MIRI course list)

  By the way, the Shen’s book takes a different route to the Zorn’s lemma: first he introduces well-ordered sets, then uses tranfinite recursion to prove Zermelo’s theorem (that any set can be well-ordered), then he uses Zermelo’s theorem and tranfinite recursion to prove Zorn’s lemma. Thus the proof of Zorn’s lemma is reduced from two pages to a few lines. I personally found it easier to follow and remember.

• seed 29 Dec 2019 16:01 UTC

  3 points

  on: Di­ag­o­nal­iza­tion Fixed Point Exercises

  Ex 1.

  Suppose there is a surjection f : S → P(S). Consider the set $X=\{x\in S|x\notin f(x\left)\right\}$ . Since f is a surjection, X = f(y) for some y in S. Does y lie in X? If $y\in X$ , then $y\in f\left(y\right)$ , so by definition of X, $y\notin X$ . If $y\notin X$ , then $y\notin f\left(y\right)$, so y must belong to X. Contradiction.

  Ex 2.

  Since there is a function $f:T\to T$ without fixed points, T must have at least two elements. Hence, there is a surjection $s:T\to \{0,1\}$ , which induces a surjection $T^S\to 2^S$ (a function $g:S\to T$ goes to $s\circ g$ ). So, if there were a surjection $S\to T^S$ , there would also be a surjection $S\to 2^S$ , which cannot be by previous exercise.

  Ex 4.

  Suppose $f:S\to \text{Comp}(S,\{T,F\left\}\right)$ is a computable surjective function. Consider the function $g:S\to \{T,S\}$defined by $g\left(s\right)=\neg f\left(s\right)\left(s\right)$. The function g is computable, therefore there should exist an $s^{\ast }\in S$: $f\left(s^{\ast }\right)=g$ .

  Then $f\left(s^{\ast }\right)\left(s^{\ast }\right)=g\left(s^{\ast }\right)=\neg f\left(s^{\ast }\right)\left(s^{\ast }\right)$ . Contradiction.

  Ex 5.

  Suppose halt(x,y) is a computable function. Consider the function $f:S\to \{T,F\}$: $f\left(s\right)=\neg s\left(s\right)\text{if halt(s,s)}$; T if $\neg \text{halt}(s,s)$

  Suppose $s^{\prime }$ is a Turing code of f. Since f halts everywhere, halt(s’, s’) = T. But then $s^{\prime }\left(s^{\prime }\right)=f\left(s^{\prime }\right)=\neg s^{\prime }\left(s^{\prime }\right)$ . Contradiction.

  Ex 6.

  Suppose that $f:X\to \text{Cont}(X,S)$ is a continuous surjection. Consider the function $g\in \text{Cont}(X,S)$ $g\left(x\right)=-f(x,x)$ (here—f(x, x) is a point diametrically opposed to f(x, x)). f is surjective, hence g = f(y), but then g(y) = f(y,y) = - f(y,y). Contradiction.

  Ex 7. A quine in python3:

  code = “”“code = {}{}{}{}{}{}{}
  print(code.format(‘”’,‘”’,‘”’,code,‘”’,‘”’,‘”’))”””
  print(code.format(‘”’,‘”’,‘”’,code,‘”’,‘”’,‘”’))

  Ex 8. In python:

  import inspect
  def f(string):
  return string[::-1]
  def applytoself(f):
  source = inspect.getsource(f)
  return f(source)
  applytoself(f)

  ‘\n]1-::[gnirts nruter \n:)gnirts(f fed’

  Ex 9.

  The formula for $\psi$ is $\forall x\left(\right(x=┌\psi ┐)\to \varphi (x\left)\right)$

  Ex 11.

  Suppose $\varphi$ is the formula $Bew\left(x\right)\to \psi$ . By the diagonal lemma, there exists a formula A such that $⊢A\leftrightarrow \left(Bew\right(A)\to \psi )$ .

  Therefore, $⊢Bew(A\to (Bew\left(A\right)\to \psi \left)\right)$

  By property c, $⊢Bew\left(A\right)\to Bew\left(Bew\right(A)\to \psi )$

  Again by property c, $⊢Bew\left(Bew\right(A)\to \psi )\to \left(Bew\right(Bew\left(A\right))\to Bew(\psi \left)\right)$

  Combining previous two implications, $⊢Bew\left(A\right)\to \left(Bew\right(Bew\left(A\right))\to Bew(\psi \left)\right)$

  Since $⊢Bew\left(A\right)\to Bew\left(Bew\right(A\left)\right)$ , we have

  $⊢Bew\left(A\right)\to Bew\left(\psi \right)$

  Combining this with $⊢Bew\left(\psi \right)\to \psi$ , we get

  $⊢Bew\left(A\right)\to \psi$

  From this we get $⊢A$ , therefore, $⊢Bew\left(A\right)$ and $⊢\psi$ . QED.

• seed 16 Jan 2020 9:11 UTC

  3 points

  on: From Ra­tion­al­ity to Power in 3 Steps

  So, it’s been two years… How much power did you get?

• seed 24 Mar 2020 14:57 UTC

  1 point

  on: Against Dog Ownership

  Okay, this essay convinced me that dogs can have depression. I also think that dogs probably have real feelings and don’t just act the part like this creepy robot child, although I wonder how can one actually test this.

  I am not at all convinced, though, that a dog can have preferences, long-term goals, or “a meaningful life”. I don’t think I’ve ever seen a dog work on a long-term goal. And if dogs really preferred to take their chances alone in the alien world like the author suggests, a lot more would run away.

  A dog’s mind is different. Just because I wouldn’t enjoy being a pet, doesn’t mean a dog doesn’t. The author acknowledges this, but still says that “it’s reasonable to say that dogs have some sort of conception of meaning that rises above moment-by-moment pleasures, and that the unfulfillment of this meaning has a negative effect on the happiness of dogs.” Well, why do you believe this?

• seed 9 Apr 2020 8:30 UTC

  1 point

  in reply to: riceissa’s comment on: Open & Wel­come Thread—March 2020

  Lewak et al (1985) studied 81 couples and found no correlation between IQ and marital satisfation. https://​​www.sciencedirect.com/​​science/​​article/​​abs/​​pii/​​0191886985901400

  And, well, everything in life is potentially value-corrupting, or value-improving, depending on whether you judge from your past self’s or present self’s point of view. I think the more experience you have in dating, the better judgement you can make about what makes you happy. And if a girl seduced you with great sex, that’s a predictor of a good relationship, don’t see anything illegitimate about that. There are known failure modes in relationships: you don’t want to end up with an abuser, an alcoholic or a drug addict. If you’re not in one of these, it’s probably fine. From personal experience, I married a man with a lower IQ, and I’m happy.

• seed 14 Apr 2020 14:14 UTC

  1 point

  on: Pre­ma­ture death

  Life expectancy tables may overestimate on your death day, but they underestimate some people’s lives on some other days, so it’s not like they always overestimate. It seems like you’ve explained it all pretty well, I don’t see any paradox left.

• seed 20 Apr 2020 16:21 UTC

  1 point

  in reply to: habryka’s comment on: Please Press “Record”

  This is so sad.

• seed 25 Apr 2020 13:24 UTC

  3 points

  on: Iter­a­tion Fixed Point Exercises

  #1

  $d\left(f^n\right(x),f^{n+1}(x\left)\right)\le q^{n}d(x,f(x\left)\right)$ - can show by induction.

  $\forall m>n,d\left(f^n\right(x),f^m(x\left)\right)\le q^n+q^{n+1}+\dots +q^{m-1}\le \frac{q^n}{1-q}{\to }_{n\to \infty }0$

  Therefore, $f^n\left(x\right)$ is a Cauchy sequence, and since (X, d) is complete, it must have a limit in X. Suppose $y={lim}_{n\to \infty }{f}^n\left(x\right)$ . Then $d(y,f^{n+1}(x\left)\right)\le qd(y,f^n(x\left)\right)$ , therefore $d(y,f^n(x\left)\right)\le q^{n}d(x,y)$

  #2

  Suppose $y={lim}_{n\to \infty }{f}^n\left(x\right)$ . Let’s show that y is a fixed point. Indeed, for any n, $d\left(f^n\right(x),f(y\left)\right)\le qd\left(f^{n+1}\right(x),y)$ , and if we take the limit in both sides, we get $d(y,f(y\left)\right)\le qd(y,y)=0$ .

  Let’s show uniqueness: suppose x and y are fixed points, then $d(x,y)=d\left(f\right(x),f(y\left)\right)\le qd(x,y)$ , therefore d(x,y) = 0.

  #3

  $X=[1,+\infty )$ , f(x) = x + 1/​x.

  #4

  Suppose $f\left(x\right)=ϵ||x|{|}^2+h\left(x\right)$ , where h is some convex function and $ϵ<1/2$. Take $x,y\in R^n$. Since h is convex on segment [x,y], its directional derivative is nondecreasing. Its directional derivative is a projection of gradient of g on the [x,y] line. Therefore, we have $⟨\nabla h\left(y\right),y-x⟩\ge ⟨\nabla h\left(x\right),y-x⟩$ , or $⟨\nabla h\left(y\right)-\nabla h\left(x\right),y-x⟩\ge 0$ . Hence,

  $||x-\nabla f\left(x\right)-y+\nabla f\left(y\right)||=||x-2ϵx-\nabla h\left(x\right)-y-2ϵy+\nabla h\left(y\right)||\le (1-2ϵ)||x-y||$

  Therefore, g is a contraction mapping, and from problem 1 it follows that the gradient descent converges exponentially quickly.

  #5

  Suppose A is an NxN positive matrix, and e is its minimal entry. (Then e < 1/​N). Then we can write A = eJ + (1 - Ne)Q, where J is a matrix whose entries are all 1, and Q is a matrix whose entries are all nonnegative and the sum of each column is 1 (because the sum of each column is 1 in A and Ne in J). Suppose x and y are probability distributions, i.e. N-dimensional vectors with nonnegative entries whose sum is 1. Then $||Ax-Ay|{|}_1=||eJ(x-y)+(1-Ne)Q(x-y)|{|}_1=(1-Ne)||Q(x-y)||$

  Denote $x^{+}=max(x-y,0)$ , $x^{-}=min(x-y,0)$ (pointwise max/​min). Then $x-y=x^{+}-x^{-}$ ,$||Q(x-y)|{|}_1=||Q{x}^{+}-Q{x}^{-}|{|}_1\le ||Q{x}^{+}|{|}_1+||Q{x}^{-}|{|}_1=||x^{+}|{|}_1+||x^{-}|{|}_1=||x-y|{|}_1$ ,

  so $||Ax-Ay|{|}_1\le (1-Ne)||x-y|{|}_1$ . The space of all probability distributions with metric induced by—$||.|{|}_1$norm is a compact subset of $R^n$, so it is a complete metrics space, therefore, the sequence $A^n\left(x\right)$ converges to a unique fixed point.

  #6

  Let us assume $x\le f\left(x\right)$ (the proof for $x\ge f\left(x\right)$ is the same). Then, from monotonicity of f, $x\le f\left(x\right)\le f\left(f\right(x\left)\right)\le \dots$ is an ascending chain. This sequence cannot have more that |P| distinct elements, so an element of this sequence is going to repeat: $f^m\left(x\right)=f^n\left(x\right),m<n,m<|P|$ . Then all the inequalities in $f^m\left(x\right)\le f^{m+1}\left(x\right)\le \dots \le f^n\left(x\right)$ must be equalities, so $f\left(f^m\right(x\left)\right)=f^m\left(x\right)$ , $f^m\left(x\right)$ is a fixed point.

• seed 10 May 2020 9:41 UTC

  −5 points

  in reply to: TurquoisePrincess’s comment on: What are good defense mechanisms against dan­ger­ous bul­let bit­ing?

  I don’t need any defense mechanisms against these ones, because I can just see the fallacy in the arguments.

  In one description, Blaise Pascal is accosted by a mugger who has forgotten his weapon. However, the mugger proposes a deal: the philosopher gives him his wallet, and in exchange the mugger will return twice the amount of money tomorrow. Pascal declines, pointing out that it is unlikely the deal will be honoured. The mugger then continues naming higher rewards, pointing out that even if it is just one chance in 1000 that he will be honourable, it would make sense for Pascal to make a deal for a 2000 times return. Pascal responds that the probability for that high return is even lower than one in 1000. The mugger argues back that for any low probability of being able to pay back a large amount of money (or pure utility) there exists a finite amount that makes it rational to take the bet – and given human fallibility and philosophical scepticism a rational person must admit there is at least some non-zero chance that such a deal would be possible. In one example, the mugger succeeds by promising Pascal 1,000 quadrillion happy days of life. Convinced by the argument, Pascal gives the mugger the wallet.

  When a mugger promises me to return twice my money tomorrow, I can see that it is almost certainly a hoax. There is maybe a one in a million chance he’s saying the truth. The expected value of the wager is negative. If he promises a 2000x return, that’s even less likely to be true. I estimate it as one in two billion. The expected value is still the same, and still negative. And so on, the more lavish reward the mugger promises, the less likely I am to trust him, so the expected value can always stay negative.

  Roko’s basilisk

  Why don’t I throw themselves into a research institute dedicated to building the basilisk? Because there is no such institute, and if someone seriously tried to build one, they’d just end up in prison or a mental asylum for extortion. Unless they are keeping their work secret, but then it’s just an unnecessarily convoluted way of building an AI that kills everyone. So there is no reason why I would want to do that.

• seed 10 May 2020 11:27 UTC

  1 point

  in reply to: eapache’s comment on: Win­ning vs Truth – In­fo­haz­ard Trade-Offs

  It’s a relief to know you aren’t advocating self-deception, and you may want to choose your phrasing in the
  post not to give that impression. “Epistemic rationality” means knowing the truth for yourself. Been honest with others is a different virtue.

  That said, I think telling the truth almost always does more good than harm, and my policy is to only lie to defend myself or others from violence. In this particular case, I don’t see how the CDC post is going to hurt the average person, since the readers are not average people, but LW community.

• seed 11 May 2020 11:58 UTC

  3 points

  on: Ma­hatma Arm­strong: CEVed to death.

  My problem with CEV is the arbitrariness of what it means to “know more”. My brain cannot hold all the knowledge about the universe, so the AI has to somehow choose what information to impart and in what order, and this would significantly influence the outcome. E.g. maybe hearing 100 heartwarming stories would make me care more about others, while hearing 100 stories about people being bastards to each other would make me care less, hearing all evidence supporting some political theory would sway me towards it, et cetera.

• seed 13 May 2020 12:59 UTC

  2 points

  in reply to: handoflixue’s comment on: Teach­able Ra­tion­al­ity Skills

  So, the girl challenge is to get a girl to accompany you to one of your hobbies.

• seed 15 May 2020 19:08 UTC

  1 point

  on: Crav­ing, suffer­ing, and pre­dic­tive pro­cess­ing (three char­ac­ter­is­tics se­ries)

  So, do you not need painkillers now thanks to meditation? How did it impact your motivation, do you get more things done?

• seed 25 May 2020 12:12 UTC

  1 point

  in reply to: Blueberry’s comment on: Ureshiku Naritai

  I don’t mind people talking to me on public transportation, as long as they immediately believe me when I say I’m not interested, and leave me alone.

• seed 30 May 2020 18:28 UTC

  1 point

  in reply to: ChristianKl’s comment on: Some­thing’s Wrong

  Most critiques from radicals that I read don’t contain an analyses of the root courses of the problem they are criticizing.

  Then calling them “radicals” is a misuse of the word, I think.

• seed 1 Jun 2020 13:31 UTC

  2 points

  on: How ra­tio­nal is the path of least re­sis­tance?

  It depends on whether you want to end up where the flow is going, or somewhere else… I think your question is too vague to give any useful answer, though. As ch. 27 of HPMOR teaches us, it’s best to know context before giving out sage life advice.

• seed 9 Jun 2020 18:59 UTC

  4 points

  on: GAN Discrim­i­na­tors Don’t Gen­er­al­ize?

  Getting a validation accuracy of 50% in a binary classification task isn’t “surprisingly well”. It means your model is as good as random guessing: if you flipped a coin, you would get the right answer half the time, too. Getting 0% validation accuracy would mean that you are always guessing wrong, and would get 100% accurate results by reversing your model’s prediction. So, yes, just like the article says, the discriminator does not generalize.

  What links here?

  • GAN Discrim­i­na­tors Don’t Gen­er­al­ize? by tryactions (8 Jun 2020 20:36 UTC; 18 points)

• seed 9 Jun 2020 19:59 UTC

  1 point

  in reply to: tryactions’s comment on: GAN Discrim­i­na­tors Don’t Gen­er­al­ize?

  Oh, I see, sorry.