# sil ver

Karma: 356 (LW), 2 (AF)

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6.7 S

**ystems 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).

is defined just for one particular graph. It’s the first edge in that graph such that . (So it could have been called ). Then for the next graph, it’s a different . Basically, looks at where the first graph skips over the zero mark, then picks the last vertex before that point, then 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 was just that there ar ealready the 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.

That makes perfect sense. Thanks.

When trying to fit an exponential curve, don’t weight all the points equally.

Um… why?

Ex 5 (fixed version)

Let denote the triangle. For each , construct a 2-d simplex with nodes in , where the color of a point corresponds to the place in the disk that carries that point to, then choose to be a point within a trichromatic triangle in the graph. Then is a bounded sequence having a limit point . Let be the center of the disc; suppose that . Then there is at least one region of the disc that doesn’t touch. Let be the distance to the furthest side, that is, let . Since the sides are closed regions, we have . Using continuity of , choose small enough such that . Then choose large enough so that (1) all triangles in have diameter less than and (2) . Then, given any other point in the triangle around in , we have that , so that . This proves that the triangle in 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 is equilateral.) Imbed into such that , , . Let be continuous. Then given by is also continuous. If then . Let be the circle with radius 2 around ; then because (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 maps a point to the center, which is , 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 and , the blue area contains those between and , and the red area those between and and those between and . We will show that does not intersect the red area, except at the origin. First, note that we have

Since both and 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

and and and and and .

Which are easily computed to be

and and and and and

Two of those are precisely the origin, the other four have angles and and and . Indeed, they are all between and .

Now one needs to do the same for the sets and , but it goes through analogously.

I’m late, but I’m quite proud of this proof for #4:

Call the large triangle a graph and the 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 , and . Let be the node being recolored, wlog from to . Suppose first that has six neighbors. It is easy to see that a triangle between and two neighbors has changed if and only if one of the neighbors has color and the other has color or . Thus, we must show that the number of such triangles is even. If all neighbors have color , or if none of them do, then no triangles have changed. If exactly one node has color , then the two adjacent triangles have changed. Otherwise, let and denote two different neighbors of color . There are two paths using only neighbors of between and . Both start and end at a node of color . 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 and ; these correspond to the triangles that have changed.

If is a node on one of the graph’s boundaries changing color from to , it has exactly 4 neighbors and three adjacent triangles. The two neighbors that are also on the boundary cannot have color , 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.

Ex 1

Let and . Given an edge , let denote the map that maps the color of the left to that of the right node. Given a , let . Let denote the color blue and the color green. Let be 1 if edge is bichromatic, and 0 otherwise. Then we need to show that . We’ll show , which is a striclty stronger statement than the contrapositive.

For , the LHS is equivalent to , and indeed equals if is bichromatic, and otherwise. Now let and let it be true for . Suppose . Then, if , that means , so that , and if , then , so that . Now suppose . If , then , so that , and if , then , so that . This proves the lemma by induction.

Ex 2

Ordinarily I’d proof by contradiction, using sequences, that 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 . Given , let be a path graph of vertices , where . If for any and we have , then we’re done, so assume we don’t. Define to be 1 if and have s different sign, and 0 otherwise. Sperner’s Lemma says that the number of edges with are odd; in particular, there’s at least one. Let the first one be denoted , then set .

Now consider the sequence . It’s bounded because . Using the Bolzano-Weierstrass theorem, let be a convergent subsequence. Since for all , we have . On the other hand, if , then, using continuity of , we find a number such that . Choose and such that , then for all , so that and then for all , so that , a contradiction.

Ex 3

Given such a function , let be defined by . We have . If either inequality isn’t strict, we’re done. Otherwise, . Taking for granted that the intermediate value theorem generalizes to this case, find a root of , then .

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.

Ex 4

Given a computable function , define a function by the rule . Then is computable, however, because for , we have that and .

Ex 5:

We show the contrapositive: given a function halt, we construct a surjective function from to as follows: enumerate all turing machines, such that each corresponds to a string. Given a , if does not decode to a turing machine, set . If it does, let denote that turning machine. Let with input first run halt; if halt returns , put out , otherwise will halt on input ; run on and put out the result.

Given a computable function , there is a string such that implements (if the turing thesis is true). Then , so that is surjective.

Ex 6:

Let be a parametrization of the circle given by . Given and , write to denote the point , where . First, note that, regardless of the topology on , it holds true that if is continuous, then so is for any , because given a basis element of the circle, we have which is open because is continuous.

Let be a continuous function from to . Then is continuous, and so is the diagonal function . Fix any , then given by is also continuous, but given any , one has and thus . It follows that 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);}}}

Ex 1

Exercise 1: Let and let . Suppose that , then let be an element such that . Then by definition, and . So , a contradiction. Hence , so that is not surjective.

Ex 2

Exercise 2: Since is nonempty, it contains at least one element . Let be a function without a fixed point, then , so that and are two different elements in (this is the only thing we shall use the function for).

Let for nonempty. Suppose by contradiction that is surjective. Define a map by the rule . Given any subset , let be given by Since is surjective, we find a such that . Then . This proves that is surjective, which contradicts the result from (a).

Ex 3

Exercise 3: By (2) we know that , and so and where . That means for any . and .

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.

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?

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.

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.

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.

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?

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.

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?

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.

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.