sil ver

Karma: 356 (LW), 2 (AF)

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sil ver 2 Feb 2019 21:01 UTC
3 points

on: What kind of information would serve as the best evidence for resolving the debate of whether a centrist or leftist Democratic nominee is likelier to take the White House in 2020?
This does not answer the question, but it seems plausible to me that the leftist-centrist axis only has a very small impact on who is likely to win, which would be consistent with PredictIt’s estimates.

sil ver 1 Feb 2019 17:36 UTC
13 points

on: Drexler on AI Risk
6.7 Systems composed of rational agents need not maximize a utility function There is no canonical way to aggregate utilities over agents, and game theory shows that interacting sets of rational agents need not achieve even Pareto optimality.
Is [underlined] true? I know it’s true if you have agents following CDT, but does it still hold if agents follow FDT? (I think if you say ‘rational’ it should not mean ‘CDT’ since CDT is strictly worse than FDT).

sil ver 4 Dec 2018 13:00 UTC
1 point

in reply to: TheWakalix's comment on: Topological Fixed Point Exercises
$(v, w)$ is defined just for one particular graph. It’s the first edge in that graph such that $f (v) < 0 < f (w)$ . (So it could have been called $(v_{n}, w_{n})$ ). Then for the next graph, it’s a different $v$ . Basically, $x_{1}$ looks at where the first graph skips over the zero mark, then picks the last vertex before that point, then $x_{2}$ looks at the next larger graph, and if that graph skips later, it updates to the last vertex before that point in that graph, etc. I think the reason I didn’t add indices to $(v, w)$ was just that there ar ealready the $v$ with two indices, but I see how it can be confusing since having no index makes it sound like it’s the same value all throughout.

sil ver 25 Nov 2018 22:52 UTC
4 points

in reply to: Charlie Steiner's comment on: How rapidly are GPUs improving in price performance?
That makes perfect sense. Thanks.

sil ver 25 Nov 2018 21:47 UTC
7 points

in reply to: Charlie Steiner's comment on: How rapidly are GPUs improving in price performance?
When trying to fit an exponential curve, don’t weight all the points equally.
Um… why?

sil ver 24 Nov 2018 23:23 UTC
9 points
AF
on: Topological Fixed Point Exercises
Ex 5 (fixed version)
Let $D$ denote the triangle. For each $n \in N$ , construct a 2-d simplex $S_{n}$ with $(n + 1)^{2}$ nodes in $D$ , where the color of a point corresponds to the place in the disk that $f$ carries that point to, then choose $x_{n}$ to be a point within a trichromatic triangle in the graph. Then $(x_{n})_{n \in N}$ is a bounded sequence having a limit point $x^{*}$ . Let $t$ be the center of the disc; suppose that $f (x^{*}) \neq t$ . Then there is at least one region of the disc that $f (x^{*})$ doesn’t touch. Let $ϵ$ be the distance to the furthest side, that is, let $ϵ = max {d (f (x^{*}), G), d (f (x^{*}), R), d (f (x^{*}), B)}$ . Since the sides are closed regions, we have $ϵ > 0$ . Using continuity of $f$ , choose $δ \in R_{+}$ small enough such that $B_{d} (x^{*}, δ) \subset B_{d} (f (x^{*}), ϵ)$ . Then choose $n \in N$ large enough so that (1) all triangles in $S_{n}$ have diameter less than $\frac{δ}{2}$ and (2) $d (x_{n}, x^{*}) < \frac{δ}{2}$ . Then, given any other point $y$ in the triangle around $x_{n}$ in $S_{n}$ , we have that $d (y, x^{*}) \leq d (y, x_{n}) + d (x_{n}, x^{*}) < \frac{δ}{2} + \frac{δ}{2} = δ$ , so that $d (f (y), f (x^{*})) < ϵ$ . This proves that the triangle in $S_{n}$ does not map points to all three sides of the disc, contradicting the fact that it is trichromatic.
Ex 6
(This is way easier to demonstrate in a picture in a way that leaves no doubt that it works than it is to write down, but I tried to do it anyway considering that to be part of the difficulty.)
(Assume the triangle $T = ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ A B O$ is equilateral.) Imbed $T$ into $R^{2}$ such that $O = (0, 0)$ , $A = (- \frac{1}{2}, \frac{1}{2} \sqrt{3})$ , $B = (\frac{1}{2}, \frac{1}{2} \sqrt{3})$ . Let $f : T \mapsto T$ be continuous. Then $h : T \mapsto R^{2}$ given by $h : x \mapsto f (x) - x$ is also continuous. If $h (x) = O = (0, 0)$ then $(0, 0) = h (x) = f (x) - x ⟹ f (x) = x$ . Let $R$ be the circle with radius 2 around $O$ ; then $h (T) \subset R$ because $| | h (x) | | = | | f (x) - x | | \leq | | f (x) | | + | | x | | \leq 2$ (it is in fact contained in the circle with radius 1, but the size of the circle is inconsequential). We will use exercise 5 to show that $h$ maps a point to the center, which is $O$ , from which the desired result follows. For this, we shall show that it has the needed properties, with the modification that points on any side may map precisely into the center. It’s obvious that weakening the requirement in this way preserves the result.
Rotate the disk so that the red shape is on top. In polar coordinates, the green area now contains all points with angles between $60^{\circ}$ and $180^{\circ}$ , the blue area contains those between $180^{\circ}$ and $300^{\circ}$ , and the red area those between $0^{\circ}$ and $60^{\circ}$ and those between $300^{\circ}$ and $360^{\circ}$ . We will show that $h (¯ ¯¯¯¯¯¯ ¯ A B)$ does not intersect the red area, except at the origin. First, note that we have
$h (¯ ¯¯¯¯¯¯ ¯ A B) = {f (x) - x | x \in ¯ ¯¯¯¯¯¯ ¯ A B} \subset {y - x | x \in ¯ ¯¯¯¯¯¯ ¯ A B, y \in T} = {T - x | x \in ¯ ¯¯¯¯¯¯ ¯ A B} = T - ¯ ¯¯¯¯¯¯ ¯ A B$
Since both $T$ and $¯ ¯¯¯¯¯¯ ¯ A B$ are convex combinations of finitely many points, it suffices to check all combinations that result by taking a corner from each. This means we need to check the points
$A - A$ and $B - A$ and $C - A$ and $A - B$ and $B - B$ and $C - B$ .
Which are easily computed to be
$(0, 0)$ and $(- 1, 0)$ and $(\frac{1}{2}, - \frac{1}{2} \sqrt{3})$ and $(1, 0)$ and $(0, 0)$ and $(- \frac{1}{2}, - \frac{1}{2} \sqrt{3})$
Two of those are precisely the origin, the other four have angles $270^{\circ}$ and $150^{\circ}$ and $90^{\circ}$ and $210^{\circ}$ . Indeed, they are all between $60^{\circ}$ and $300^{\circ}$ .
Now one needs to do the same for the sets $T - ¯ ¯¯¯¯¯¯ ¯ O A$ and $T - ¯ ¯¯¯¯¯¯ ¯ B O$ , but it goes through analogously.

sil ver 23 Nov 2018 21:22 UTC
13 points
AF
on: Topological Fixed Point Exercises
I’m late, but I’m quite proud of this proof for #4:
Call the large triangle a graph and the $n^{2}$ triangles simply triangles. First, note that for any size, there is a graph where the top node is colored red, the remaining nodes on the right diagonal are colored green, and all nodes not on the right diagonal are colored blue. This graph meets the conditions, and has exactly one trichromatic triangle, namely the one at the top (no other triangle contains a red node). It is trivial to see that this graph can be changed into an arbitrary graph by re-coloring finitely many nodes. This will affect up to six triangles; we say that a triangle has changed iff it was trichromatic before the recoloring but not after, or vice versa, and we shall show that re-coloring any node leads to an even number of triangles being changed. This proves that the number of trichromatic triangles never stops being odd.
Label the three colors $x, y$ , and $z$ . Let $v$ be the node being recolored, wlog from $x$ to $y$ . Suppose first that $v$ has six neighbors. It is easy to see that a triangle between $v$ and two neighbors has changed if and only if one of the neighbors has color $z$ and the other has color $x$ or $y$ . Thus, we must show that the number of such triangles is even. If all neighbors have color $z$ , or if none of them do, then no triangles have changed. If exactly one node has color $z$ , then the two adjacent triangles have changed. Otherwise, let $j$ and $k$ denote two different neighbors of color $z$ . There are two paths using only neighbors of $v$ between $j$ and $k$ . Both start and end at a node of color $z$ . By the 1-D Sperner Lemma (assuming the more general result), it follows that both paths have an even number of edges between nodes of color $z$ and ${x, y}$ ; these correspond to the triangles that have changed.
If $v$ is a node on one of the graph’s boundaries changing color from $x$ to $y$ , it has exactly 4 neighbors and three adjacent triangles. The two neighbors that are also on the boundary cannot have color $z$ , so either none, one, or both of the ones that aren’t do. If it’s none, no triangle has changed; if it’s one, the two neighboring triangles have changed; and if it’s both, then the two triangles with two nodes on the graph’s boundary have changed.

sil ver 22 Nov 2018 20:12 UTC
10 points
AF
on: Topological Fixed Point Exercises
Ex 1
Let $V = {v_{0}, . . ., v_{n}}$ and $e_{k} := (v_{k - 1}, v_{k})$ . Given an edge $e$ , let $ϕ_{e}$ denote the map that maps the color of the left to that of the right node. Given a $k \in {1, . . ., n}$ , let $ϕ_{k} = ϕ_{e_{k}} \circ . . . \circ ϕ_{e_{1}}$ . Let $b$ denote the color blue and $g$ the color green. Let $c (k)$ be 1 if edge $e_{k}$ is bichromatic, and 0 otherwise. Then we need to show that $ϕ_{n} (b) = g ⟹ (\sum_{k = 1}^{n} c (k)) is odd$ . We’ll show $(\sum_{k = 1}^{n} c (k)) is even ⟺ ϕ_{n} (b) = b$ , which is a striclty stronger statement than the contrapositive.
For $n = 1$ , the LHS is equivalent to $c (ϕ_{1}) = 1$ , and indeed $ϕ (e_{1}) (b)$ equals $g$ if $e_{1}$ is bichromatic, and $b$ otherwise. Now let $n > 1$ and let it be true for $n - 1$ . Suppose $\sum_{k = 1}^{n} c (k) is even$ . Then, if $c (e_{n}) = 1$ , that means $(\sum_{k = 1}^{n - 1} c (k)) is odd$ , so that $ϕ_{n} (b) = ϕ_{e_{n}} (ϕ_{n - 1} (b)) = ϕ_{e_{n}} (g) = b$ , and if $c (e_{n}) = 0$ , then $(\sum_{k = 1}^{n - 1} c (k)) is even$ , so that $ϕ_{n} (b) = ϕ_{e_{n}} (ϕ_{n - 1} (b)) = ϕ_{e_{n}} (b) = b$ . Now suppose $\sum_{k = 1}^{n} c (k) is odd$ . If $c (e_{n}) = 1$ , then $(\sum_{k = 1}^{n - 1} c (k)) is even$ , so that $ϕ_{n} (b) = ϕ_{e_{n}} (ϕ_{n - 1} (b)) = ϕ_{e_{n}} (b) = g$ , and if $c (e_{n}) = 0$ , then $(\sum_{k = 1}^{n - 1} c (k)) is odd$ , so that $ϕ_{n} (b) = ϕ_{e_{n}} (ϕ_{n - 1} (b)) = ϕ_{e_{n}} (g) = g$ . This proves the lemma by induction.
Ex 2
Ordinarily I’d proof by contradiction, using sequences, that $inf {x \in [0, 1] | f (x) > 0}$ can neither be greater nor smaller than 0. I didn’t manage a short way to do it using the two lemmas, but here’s a long way.
Set $x_{0} = 0$ . Given $n \in N_{+}$ , let $G_{n}$ be a path graph of $n + 1$ vertices ${v_{n, 0}, . . ., v_{n, n}}$ , where $v_{n, k} = \frac{k}{n}$ . If for any $n$ and $k$ we have $f (v_{n, k}) = 0$ , then we’re done, so assume we don’t. Define $c (v, v^{'})$ to be 1 if $f (v)$ and $f (v^{'})$ have s different sign, and 0 otherwise. Sperner’s Lemma says that the number of edges $e$ with $c (e) = 1$ are odd; in particular, there’s at least one. Let the first one be denoted $(v, w)$ , then set $x_{n} = max {v, x_{n - 1}}$ .
Now consider the sequence $(x_{n})_{n \in N}$ . It’s bounded because $x_{k} \in [0, 1]$ . Using the Bolzano-Weierstrass theorem, let $(x_{n}^{'})_{n \in N} \mapsto x^{*}$ be a convergent subsequence. Since $f (x_{k}) > 0$ for all $k \in N_{+}$ , we have $f (x^{*}) \geq 0$ . On the other hand, if $f (x^{*}) > 0$ , then, using continuity of $f$ , we find a number $x^{'} > x^{*}$ such that $f (x^{*}, x^{'}) > 0$ . Choose $n$ and $k$ such that $v_{n, k} \in (x^{*}, x^{'})$ , then $c (v_{n, j}) = 0$ for all $j < k$ , so that $x_{n} \geq v_{n, k} > x^{*}$ and then $x_{ℓ} \geq x_{n}$ for all $ℓ \geq k$ , so that $x^{*} \geq x_{n} \geq v_{n, k} > x^{*}$ , a contradiction.
Ex 3
Given such a function $f$ , let $g : [0, 1] \mapsto [- 1, 1]$ be defined by $g (x) = f (x) - x$ . We have $g (0) = f (0) - 0 \geq 0 \geq f (1) - 1 = g (1)$ . If either inequality isn’t strict, we’re done. Otherwise, $g (0) > 0 > g (1)$ . Taking for granted that the intermediate value theorem generalizes to this case, find a root $x^{*}$ of $g$ , then $f (x^{*}) = f (x^{*}) - x^{*} + x^{*} = g (x^{*}) + x^{*} = x^{*}$ .

sil ver 21 Nov 2018 18:03 UTC
3 points

on: Preschool: Much Less Than You Wanted To Know
Mostly agreed. But I think the obvious counter point is that you’re arguing for a slightly different standard. Like, if the question is ‘does pre-school basically make sense’ then you’re right, it doesn’t, and the black box approach is weird. But if the question is ‘should you send your children to pre-school’ then the black box approach seems solid. Even if you could come up with something better in five minutes, you can’t implement it, so the standard for it being worthwhile might be really low.

sil ver 18 Nov 2018 23:11 UTC
13 points
AF
on: Diagonalization Fixed Point Exercises
Ex 4
Given a computable function $ϕ : S \mapsto Comp (S, {T, F})$ , define a function $f : S \mapsto {T, F}$ by the rule $f (s) = {\begin{matrix} T & ϕ (s) (s) = F F & ϕ (s) (s) = T \end{matrix}$ . Then $f$ is computable, however, $f \notin ϕ (S)$ because for $s \in S$ , we have that $ϕ (s) (s) = F ⟹ f (s) = T ⟹ ϕ (s) \neq f$ and $ϕ (s) (s) = T ⟹ f (s) = F ⟹ ϕ (s) \neq f$ .
Ex 5:
We show the contrapositive: given a function halt, we construct a surjective function $f$ from $S$ to $Comp (S, {T, F})$ as follows: enumerate all turing machines, such that each corresponds to a string. Given a $t \in S$ , if $t$ does not decode to a turing machine, set $f (t) \equiv F$ . If it does, let $[t]$ denote that turning machine. Let $f (t)$ with input $s \in S$ first run halt $(t, s)$ ; if halt returns $F$ , put out $F$ , otherwise $[t]$ will halt on input $s$ ; run $[t]$ on $s$ and put out the result.
Given a computable function $g : S \mapsto {T, F}$ , there is a string $t^{*} \in S$ such that $[t^{*}]$ implements $g$ (if the turing thesis is true). Then $g = f (t^{*})$ , so that $f$ is surjective.
Ex 6:
Let $h : R \mapsto S$ be a parametrization of the circle given by $h (r) \mapsto (cos (r), sin (r))$ . Given $p \in S$ and $r \in R$ , write $p + r$ to denote the point $h (p^{'} + r)$ , where $p^{'} \in h^{- 1} ({p})$ . First, note that, regardless of the topology on $X$ , it holds true that if $f : X \mapsto S$ is continuous, then so is $f_{r} : x \mapsto f (x) + r$ for any $r \in R$ , because given a basis element $(h (a), h (b))$ of the circle, we have $f_{r}^{- 1} (B) = f^{- 1} (h (a) - r, h (b) - r)$ which is open because $f$ is continuous.
Let $ϕ$ be a continuous function from $X$ to $Cont (X, S)$ . Then $ϕ^{'} : X \times X \mapsto S$ is continuous, and so is the diagonal function $h : x \mapsto ϕ (x, x)$ . Fix any $r \in (0, 2 π)$ , then $f : X \mapsto S$ given by $f : x \mapsto (ϕ^{'} (x, x) + r))$ is also continuous, but given any $x \in X$ , one has $ϕ (x) (x) = ϕ^{'} (x, x) \neq ϕ^{'} (x, x) + r = f (x)$ and thus $ϕ (x) \neq f$ . It follows that $ϕ$ is not surjective.
Ex 7:
I did it in java. There’s probably easier ways to do this, especially in other languages, but it still works. It was incredibly fun to do. My basic idea was to have a loop iterate 2 times, the first time printing the program, the second time printing the printing command. Escaping the ” characters is the biggest problem, the main idea here was to have a string q that equals ” in the first iteration and ” + q + ” in the second. That second string (as part of the code in an expression where a string is printed) will print itself in the console output. Code:
package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?“”+(char)34:“”;String q=“”+(char)34;q=i==1?q+“+q+“+q:q;String e=i==1?o+“+e);}}}“:”System.out.print(o+“;System.out.print(o+“package maths;public class Quine{public static void main(String[]args){for(int i=0;i<2;i++){String o=i==1?“+q+“”+q+“+(char)34:“+q+“”+q+“;String q=“+q+“”+q+“+(char)34;q=i==1?q+“+q+“+q+“+q+“+q:q;String e=i==1?o+“+q+“+e);}}}“+q+“:”+q+“System.out.print(o+“+q+“;”+e);}}}

sil ver 18 Nov 2018 20:57 UTC
11 points
AF
on: Diagonalization Fixed Point Exercises
Ex 1
Exercise 1: Let $f : S \mapsto P (S)$ and let $A = {x \in S | x \notin f (x)}$ . Suppose that $A \in f (S)$ , then let $x \in S$ be an element such that $f (x) = A$ . Then $x \notin f (x) = A ⟹ x \in A$ by definition, and $x \in A = f (x) ⟹ x \notin A$ . So $x \notin A ⟺ x \in A$ , a contradiction. Hence $A \notin f (S)$ , so that $f$ is not surjective.
Ex 2
Exercise 2: Since $T$ is nonempty, it contains at least one element $t_{0}$ . Let $h : T \mapsto T$ be a function without a fixed point, then $t_{0} \neq h (t_{0}) =: t_{1}$ , so that $t_{0}$ and $t_{1}$ are two different elements in $T$ (this is the only thing we shall use the function $h$ for).
Let $Ψ : S \mapsto T^{S}$ for $S$ nonempty. Suppose by contradiction that $Ψ$ is surjective. Define a map $f : S \mapsto P (S)$ by the rule $f (x) = {s \in S | Ψ (x) (s) = t_{0}}$ . Given any subset $U \subset S$ , let $g : S \mapsto T$ be given by $g : x \mapsto {\begin{matrix} t_{0} & x \in U t_{1} & x \notin U \end{matrix}$ Since $Ψ$ is surjective, we find a $s^{*} \in S$ such that $Ψ (s^{*}) = g$ . Then $f (s^{*}) = U$ . This proves that $f$ is surjective, which contradicts the result from (a).
Ex 3
Exercise 3: By (2) we know that $T = {x}$ , and so $f = {(x, x)}$ and $g = {(s, h) | s \in S}$ where $h = {(s, x) | s \in S}$ . That means $x = g (s) (s^{'}) = f^{n} (g (s) (s^{'}))$ for any $s, s^{'} \in S$ . and $n \in N$ .

sil ver 16 Nov 2018 19:32 UTC
5 points

on: Sam Harris and the Is–Ought Gap
Good post. I think Carroll’s PoV is correct and Sam’s is probably correct. Thinking about it, I would have phrased that one very differently, but I think there’d be zero difference on substance.
Edit: Having caught myself referencing this to explain Harris position, I amend my post to say that the way you put is actually exactly right, and the way I would have put it would at best have been a mildly confused version of the same thing.

sil ver 4 Sep 2018 13:04 UTC
2 points

in reply to: ChristianKl's comment on: Open Thread September 2018
I wouldn’t call the answer obvious. I’m not even sure if I could have guessed the majority view on this beforehand. Why do you think it’s obvious? Are there no upsides to changing or are the downsides too significant?

sil ver 24 Aug 2018 19:29 UTC
7 points

on: Turning Up the Heat: Insights from Tao’s ‘Analysis II’
There’s a lot I wanted to say here about topology, but I don’t think my understanding is good enough to break things down—I’ll have to read an actual book on the subject.
I’m working through Munkres’ book on topology at the moment, which is part of Miri’s reserach guide. It’s super awesome; rigorous, comprehensive, elegant, and quite long (with lots of exercises). I’m planing to do a similar post once I’m done, but it’s taking me a while. if you get to it eventually, you’ll probably beat me to it.

sil ver 28 Jul 2018 15:22 UTC
1 point

in reply to: cousin_it's comment on: Anthropic Probabilities: a different approach
1000:1 on tails (with tails → create large universe). It’s a very good question. My answer is late because it made me think about some stuff that confused me at first, an I wanted to make sure that everything I say now is coherent with everything I said in the post.
If god flipped enough logical coins for you to be able to make the approximation that half of them came up heads, you can update on the color of your logical coin based on the fact that your current version is green. This is a thousand times as likely if the green coin came up tails vs heads. You can’t do the same if god only created one universe.
If god created more than one but still only a few universes, let’s say two, then the chance that your coin came up heads is a bit more than a quarter, which comes from the heads-heads case. The heads-tails case is possible but highly unlikely.

sil ver 25 Jul 2018 14:35 UTC
1 point

in reply to: cousin_it's comment on: Anthropic Probabilities: a different approach
Yes.
I’m not used to the concept of a logical coin, but yes, that’s equivalent.
You need the consciousness condition & that god only does this once. Then my theory outputs the SSA answer.

sil ver 24 Jul 2018 21:28 UTC
1 point

in reply to: cousin_it's comment on: Anthropic Probabilities: a different approach
No; like I said, procedures tend to be repeatable. Maybe there is one, but I haven’t come up with one yet. What’s wrong with the presumptuous philosopher problem (about two possible universes) as an example?

sil ver 24 Jul 2018 14:56 UTC
1 point

in reply to: TAG's comment on: Anthropic Probabilities: a different approach
Ok, so I think our exchange can be summarized like this: I am operating on the assumption that numerical non-identity given qualitative identity is not a thing, and you doubt that assumption. We both agree that the assumption is necessary for the argument I made to be convincing.

sil ver 24 Jul 2018 14:23 UTC
1 point

in reply to: TAG's comment on: Anthropic Probabilities: a different approach
I was under the impression that the opposite was the case, that numerical non-identity given qualitative identity is moonshine. I’m not a physician though, so I can’t argue with you on the object level. Do you think that your position would be a majority view on LW?

sil ver 24 Jul 2018 13:27 UTC
1 point

in reply to: TAG's comment on: Anthropic Probabilities: a different approach
You mean The two bodies aren’t the selfsame body (numerical identity)
You mean identity of particles? I was just assuming that there is no such thing. I agree that if there was, that would be a simpler explanation.