Adele Lopez

• Adele Lopez 13 May 2019 4:19 UTC

  2 points

  in reply to: Said Achmiz's comment on: Co­her­ent de­ci­sions im­ply con­sis­tent utilities

  Pre­sum­ably, any agent which we man­age to build will be com­putable. So to the ex­tent our agent is us­ing util­ity func­tions, they will be con­tin­u­ous.

  If an agent is only ca­pa­ble of com­putable ob­ser­va­tions, but has a dis­con­tin­u­ous util­ity func­tion, then if the uni­verse is in a state where the util­ity func­tion is dis­con­tin­u­ous, the agent will need to spend an in­finite amount of time (or as long as the uni­verse state re­mains at such a point) de­ter­min­ing the util­ity of the cur­rent state. I think it might be pos­si­ble to use this to cre­ate a more con­crete ex­ploit.

• Adele Lopez 13 May 2019 1:35 UTC

  4 points

  in reply to: Samuel Hapák's comment on: Co­her­ent de­ci­sions im­ply con­sis­tent utilities

  I think you’re right that this is part of where the in­tu­ition comes from. But it’s still ir­ra­tional in a con­text where you ac­tu­ally know the prob­a­bil­ities ac­cu­rately enough.

• Adele Lopez 13 May 2019 1:27 UTC

  3 points

  in reply to: Said Achmiz's comment on: Co­her­ent de­ci­sions im­ply con­sis­tent utilities

  For con­ti­nu­ity, it’s rea­son­able to as­sume this be­cause all com­putable func­tions are con­tin­u­ous. See the­o­rem 4.4 of https://​​eccc.weiz­mann.ac.il/​​re­sources/​​pdf/​​ica.pdf

  Edit: I re­al­ized that the con­ti­nu­ity as­sump­tion is differ­ent (though re­lated) from as­sum­ing the util­ity func­tion is con­tin­u­ous. My guess is that com­putabil­ity is still a good jus­tifi­ca­tion for this, but I’d have to check that that ac­tu­ally fol­lows.

• Adele Lopez 15 Dec 2018 18:29 UTC

  6 points

  on: What is ab­strac­tion?

  Ab­strac­tion can be un­der­stood as a re­la­tion­ship be­tween mod­els. You can have a model of what it means to “brush your teeth”, and you can have mod­els of what it means to “pre­pare tooth­brush”, etc… The mod­els of the sub­tasks can be com­posed into a model of the whole se­quence, and the ab­strac­tion re­la­tion­ship tells you how this model is a re­al­iza­tion of the ab­stract “brush your teeth” model. Similarly, we can have a model of what an “an­i­mal” is, and mod­els for “dogs”, “hu­mans”, “cats”. The ab­strac­tion re­la­tion­ship tells you how each of these mod­els are re­al­iza­tions of the “an­i­mal” model. Re­mov­ing prop­er­ties is an easy way to cre­ate mod­els with this re­la­tion­ship, but it’s not the only way. For ex­am­ple, you can also re­place con­tin­u­ous time with dis­crete time.

  You can chain this re­la­tion­ship to cre­ate a hi­er­ar­chy, and for hu­mans, the most con­crete mod­els we typ­i­cally use are the ones that are “mod­el­ing” our raw sen­sory ex­pe­riences. I think that ad­e­quately ex­plains why peo­ple use it this way, but that this isn’t what ab­strac­tion is.

• Adele Lopez 24 Nov 2018 7:01 UTC

  3 points

  AF

  in reply to: lbThingrb's comment on: Topolog­i­cal Fixed Point Exercises

  Awe­some! I was hop­ing that there would be a way to do this!

• Adele Lopez 24 Nov 2018 6:25 UTC

  4 points

  AF

  in reply to: JacobKopczynski's comment on: Di­ag­o­nal­iza­tion Fixed Point Exercises

  Cur­ry­ing doesn’t pre­serve sur­jec­tivity. As a sim­ple ex­am­ple, you can eas­ily find a sur­jec­tive func­tion $2\times 2\to 2$, but there are no sur­jec­tive func­tions $2\to 2^2$.

• Adele Lopez 22 Nov 2018 4:17 UTC

  14 points

  AF

  on: Iter­a­tion Fixed Point Exercises

  Ex 6:

  If at any point $f^n\left(b\right)=f^{n-1}\left(x\right)$, then we’re done. So as­sume that we get a strict in­crease each time up to $n=|P|$. Since there are only $|P|$ el­e­ments in the en­tire poset, and $f$ is mono­tone, $f^{n+1}\left(x\right)$ has to equal $f^n\left(x\right)$.

  Ex 7:

  For a limit or­di­nal $\alpha$, define $f^{\alpha }\left(x\right)$ as the least up­per bound of $f^n\left(x\right)$ for all $n<\alpha$. If $\alpha >|L|$, then the set $f^n\left(x\right)$ for $n<\alpha$ is a set of size $\alpha$ that maps into a set of size $L$ by tak­ing the value of the el­e­ment. Since there are no in­jec­tions be­tween these sets, there must be two or­di­nals $n<m$ such that$f^n\left(x\right)=f^m\left(x\right)$. Since $f$ is mono­tone, that im­plies that for ev­ery or­di­nal $l>n$, $f^l\left(x\right)=f^n\left(x\right)$and thus is a fixed point. Since $n<\alpha$ this proves the ex­er­cise.

  Ex 8:

  Start­ing from $x$, we can cre­ate a fixed point via iter­a­tion by tak­ing $\alpha >|L|$, and iter­at­ing $\alpha$ times as demon­strated in Ex 7. Call this fixed point $f_x$. Sup­pose there was a fixed point $k$ such that $x\le k$ and $k\le f_x$. Then at some point $f^n\left(x\right)\le f^n\left(k\right)=k$, but $f^{n+1}\left(x\right)\ge f^{n+1}\left(k\right)=k$, which breaks the mono­ton­ic­ity of $f$ un­less $k=f_x$. So $f_x$ gen­er­ated this way is always the small­est fixed point greater than $x$.

  Say we have fixed points $x_i$. Then let $x$ be the least up­per bound of $x_i$, and gen­er­ate a fixed point from $f_x$. So $f_x$ will be greater than each el­e­ment of $x_i$ since $f$ is mono­tone, and is the small­est such fixed point as shown in the above para­graph. So the poset of fixed points is semi-com­plete with up­per bounds.

  Now take our fixed points $x_i$ again. Now let $x$ be the great­est lower bound of $x_i$, and gen­er­ate a fixed point $f_x$. Since $x\le x_i$ and $f$ is mono­tonic, $f^{\alpha }\left(x\right)\le f^{\alpha }\left(x_i\right)=x_i$, and so $f_x$ is a lower bound of $x_i$. It has to be the great­est such bound be­cause $x$ it­self is already the great­est such bound in our poset, and $f$ is mono­tonic.

  Thus the lat­tice of fixed points has all least up­per bounds and all great­est lower bounds, and is thus com­plete!

• Adele Lopez 22 Nov 2018 2:21 UTC

  13 points

  AF

  on: Di­ag­o­nal­iza­tion Fixed Point Exercises

  Ex 8: (in Python, us­ing a re­ver­sal func­tion)

  def f(s):
  re­turn s[::-1]
  
  dlmt = ’”“”‘
  code = “””def f(s):
  re­turn s[::-1]
  
  dlmt = ’{}’
  code = {}{}{}
  code = code.for­mat(dlmt, dlmt, code, dlmt)
  print(f(code))”″”
  code = code.for­mat(dlmt, dlmt, code, dlmt)
  print(f(code))

• Adele Lopez 21 Nov 2018 18:42 UTC

  13 points

  AF

  in reply to: Davidmanheim's comment on: Topolog­i­cal Fixed Point Exercises

  You can use 1d-Sperner to deal with all the cases effec­tively.

• Adele Lopez 21 Nov 2018 18:11 UTC

  16 points

  AF

  on: Topolog­i­cal Fixed Point Exercises

  Found a nice proof for Sperner’s lemma (#9):

  First some defi­ni­tions. Call a d-sim­plex with ver­tices col­ored in (d+1) differ­ent col­ored chro­matic. Call the par­ity of the num­ber of chro­matic sim­plices the chro­matic par­ity.

  It’s eas­ier to prove the fol­low­ing gen­er­al­iza­tion: Take a com­plex of d-sim­plices that form a d-sphere: then any (d+1)-col­or­ing of the ver­tices will have even chro­matic par­ity.

  Proof by in­duc­tion on d:

  Base d=-1: vac­u­ously true.

  As­sume true for d-1: Say you have an ar­bi­trary com­plex of d-sim­plices form­ing a d-sphere, with an ar­bi­trary d+1-col­or­ing. Choose a ver­tex. W.L.O.G. we will call the color of the cho­sen ver­tex blue.

  Take the com­plex of sim­plices that con­tain this ver­tex. Since a sphere has no bound­ary or branches, this com­plex will be a d-ball. Delete the cho­sen ver­tex, and keep only the op­po­site (d-1)-sim­plices that are left, which will form a (d-1)-sphere, call it the shell.

  We need to choose a sec­ond color, say red. We’ll call a (d-1)-sim­plex with ver­tices of all d+1 col­ors ex­cept red an R-chro­matic face, and similarly with blue.

  Now, re­place all the red ver­tices in the shell with blue ver­tices, so that the shell is now R-chro­matic. By in­duc­tion it has an even num­ber of R-chro­matic faces. Con­sider what hap­pens when we re­con­vert a ver­tex on the shell back to red: since the ver­tex was pre­vi­ously blue, any R-chro­matic faces will get turned into B-chro­matic faces. Let r be the num­ber of R-chro­matic faces on the shell, and b be the num­ber of B-chro­matic faces. The par­ity of r-b will re­main even as we con­tinue this pro­cess.

  Let’s go back to the ver­tex in the cen­ter of the shell. All cur­rently chro­matic sim­plices with this ver­tex have op­po­site faces which are B-chro­matic, since this ver­tex is blue. We’ll flip the ver­tex to red, which de­stroys chro­matic sim­plices with op­po­site B-chro­matic faces and cre­ates chro­matic sim­plices with op­po­site R-chro­matic faces. Since r-b is even, the chro­matic par­ity is pre­served!

  Since we’ve shown that ar­bi­trary re­col­or­ing of ver­tices pre­serves the chro­matic par­ity, it’s clear that the chro­matic par­ity will be even for any col­or­ing.
  

  Corol­lary: Sperner’s lemma

  Start with a d-sim­plex which has been di­vided into d-sim­plices, and where each face of the large sim­plex has one color which ver­tices on it are for­bid­den from tak­ing. Take a point of each color, and match it with a face of the sim­plex that that color is al­lowed on. Then con­nect that ver­tex to each point on that face. This will cre­ate a bunch of non-chro­matic sim­plices. Fi­nally, cre­ate a sim­plex of all of the new points. This will cre­ate one chro­matic sim­plex.

  This will form a d-sphere, and thus will have an even chro­matic par­ity. That means the origi­nal sim­plex must have had odd chro­matic par­ity.

• Adele Lopez 27 May 2018 6:06 UTC

  2 points

  on: Edi­tor Mini-Guide

  Is there a way to make drafts available to spe­cific peo­ple, for re­view?