# sil ver

Karma: 356 (LW), 2 (AF)
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• 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.

• 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).

• 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.

• 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.