Hoagy

• Hoagy 15 Feb 2019 17:08 UTC

  5 points

  in reply to: ESRogs's comment on: Refram­ing Su­per­in­tel­li­gence: Com­pre­hen­sive AI Ser­vices as Gen­eral Intelligence

  Late to the party but I’m pretty con­fi­dent he’s say­ing the op­po­site—that a 1 PFLOP/​s sys­tem is likely to have 10 or more times the com­pu­ta­tional ca­pac­ity of the hu­man brain, which is rather ter­rify­ing.

  He gives the ex­am­ple of Baidu’s Deep Speech 2 which re­quires around 1 GFLOP/​s to run and pro­duces hu­man-com­pa­rable re­sults. This is 10^6 slower than the 1 PFLOP/​s ma­chine. He es­ti­mates that this pro­cess in hu­mans take around 10^-3 of the hu­man brain, thereby giv­ing the es­ti­mate of a 1 PFLOP/​s sys­tem be­ing 10^3 times faster than the brain. His other ex­am­ples give similar re­sults.

• Hoagy 29 Jan 2019 14:40 UTC

  1 point

  in reply to: rohinmshah's comment on: Ca­pa­bil­ity amplification

  An easy way to deal with this difficulty is to re­place ‘at least as happy with policy A as with policy B (in any situ­a­tion that we think might arise in prac­tice)’ with ‘at least as happy with policy A as with policy B (when av­er­aged over the dis­tri­bu­tion of situ­a­tions that we ex­pect to arise)’, though this is clearly much weaker.

  To me it seems that the rea­son this stronger sense of or­der­ing is used is be­cause we ex­pect this am­plifi­ca­tion pro­ce­dure to be of a sort that pro­duces re­sults such that ${\text{A}}^{+}$ is strictly bet­ter than $\text{A}$ but that even if this wasn’t the case, the con­cept of an ob­struc­tion would still be a use­ful one. Per­haps it would be rea­son­able to take the more re­laxed defi­ni­tion but ex­pect that am­plifi­ca­tion would pro­duce re­sults that are strictly bet­ter.

  I also agree with Chris be­low that defin­ing an ob­struc­tion in terms of this ‘bet­ter than’ re­la­tion brings in se­ri­ous difficulty. There are ex­po­nen­tially many poli­cies $\text{B}$ that are no bet­ter than ${\text{A}}^{+}$ and there may well be a sub­set of these can be am­plified be­yond ${\text{A}}^{+}$ but as far as I can tell there’s no clear way to iden­tify these. We thus have an ex­po­nen­tial ob­sta­cle to progress even within a par­ti­tion, ne­ces­si­tat­ing a stronger defi­ni­tion.

• Hoagy 17 Jan 2019 13:31 UTC

  2 points

  on: The Steer­ing Problem

  When you talk about ‘black-box’ ver­sions of Hugh, do you en­vi­sion that H is able to an­swer ques­tions re­lat­ing to the cog­ni­tive pro­cesses that lead to the an­swer given, or about H’s think­ing in gen­eral? This seems to con­tra­dict the spirit of a black box but self re­flec­tion is an im­por­tant part of Hugh’s cog­ni­tive abil­ity.

  Per­haps they are both use­ful pos­si­bil­ities, my in­tu­ition is that this kind of self re­flec­tion is as far from be­ing pos­si­ble for AI as any hu­man abil­ity and so we should ex­pect that we might have sys­tems pow­er­ful enough to take on wide re­spon­si­bil­ity with­out this abil­ity. If it were pos­si­ble, though, the abil­ity to use loops of self re­flec­tion to check whether a cog­ni­tive pro­cess serves a cer­tain goal would be very helpful.

• Hoagy 20 Nov 2018 20:45 UTC

  11 points

  AF

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

  I’ve re­al­ised that you’ve gotta be care­ful with this method be­cause when you find a trichro­matic sub­tri­an­gle of the origi­nal, it won’t nec­es­sar­ily have the prop­erty of only hav­ing points of two colours along the edges, and so may not in fact con­tain a point that maps to the cen­tre.

  This isn’t a prob­lem if we just in­crease the num­ber n by which we di­vide the whole tri­an­gle in­stead of re­cur­sively di­vid­ing sub­tri­an­gles. Un­for­tu­nately now we’re not re­duc­ing the range of co-ords where this fixed point must be, only find­ing a triad of ar­bi­trar­ily close points that map to a tri­an­gle sur­round­ing the cen­tre. You can, for ex­am­ple, take the cen­tre point of the first of these tri­an­gles (with some method of num­ber­ing to make the func­tion definite) for each value of $n=1,2,\mathrm{3..}$ as a se­quence in $R^2$. This must have a con­ver­gent se­quence which should con­verge to a point that maps to the cen­tre but I can’t prove that last stage.

• Hoagy 20 Nov 2018 15:30 UTC

  20 points

  AF

  on: Topolog­i­cal Fixed Point Exercises

  Clean­est solu­tion I can find for #8:

  $f_t\left(x\right)=\frac{1}{1+e^{10(t-x)}}$

  Also, if we have a proof for #6 there’s a pleas­ant method for #7 that should work in any di­men­sion:

  We take our closed con­vex set $S$ that has the bounded func­tion $h:S\to S$ . We take a tri­an­gle $T$ that cov­ers $S$ so that any point in $S$ is also in $T$ .

  Now we define a new func­tion $h^{\prime }:T\to T$ such that $h^{\prime }\left(x\right)=h\left(c_s\right(x\left)\right)$ where $c_s\left(x\right)$ is the func­tion that maps $x$ to the near­est point in $S$.

  By #6 we know that $h^{\prime }$ has a fixed point, since $c_s$ is con­tin­u­ous. We know that the fixed point of $h^{\prime }$ can­not lie out­side $S$ be­cause the range of $h^{\prime }$ is $S$. This means $h^{\prime }$ has a fixed point within $S$ and since for $x\in S$, $h\left(x\right)=h^{\prime }\left(x\right)$, $h$ has a fixed point.

• Hoagy 20 Nov 2018 15:21 UTC

  11 points

  AF

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

  Yeah agreed, in fact I don’t think you even need to con­tinu­ally bi­sect, you can just in­crease n in­definitely. Iter­at­ing be­comes more dan­ger­ous as you move to higher di­men­sions be­cause an n di­men­sional sim­plex with n+1 colours that has been coloured ac­cord­ing to analo­gous rules doesn’t nec­es­sar­ily con­tain the point that maps to zero.

  On the sec­ond point, yes I’d been as­sum­ing that a bounded func­tion had a bounded gra­di­ent, which cer­tainly isn’t true for say sin(x^2), the fi­nal step needs more work, I like the way you did it in the proof be­low.

• Hoagy 18 Nov 2018 18:49 UTC

  12 points

  AF

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

  Here’s a messy way that at least doesn’t need too much ex­haus­tive search:

  First let’s sep­a­rate all of the red nodes into groups so that within each group you can get to any other node in that group only pass­ing through red nodes, but not to red nodes in any other group.

  Now, we trace out the paths that sur­round these groups—they im­me­di­ately look like the paths from Ques­tion 1 so this feels like a good start. More pre­cisely, we draw out the paths such that each ver­tex forms one side of a tri­an­gle that has a blue node at its op­po­site cor­ner. Note that you can have mul­ti­ple paths stem­ming from the same group if the group touches the side of the larger tri­an­gle, or if it has in­ter­nal holes.

  Now we have this set of paths we can split them into three kinds. The first is loops, which arise when you have a group which never touches the edge of the larger tri­an­gle, or in­side ‘holes’ in large groups. Th­ese can be seen as a path start­ing and finish­ing at the same node. They there­fore have an even num­ber of b-g ver­tices. The sec­ond kind is those that be­gin at the edge of the large tri­an­gle and end at the same edge. Th­ese paths be­gin and end on the same colour and there­fore also have an even num­ber of b-g ver­tices. Fi­nally and most im­por­tantly there is a kind of path that goes from one edge to the other -in the case of the reds, the left edge to the right edge. This will hap­pen once with the group that in­cludes the top red node, and if any other group spans the larger tri­an­gle then it will gen­er­ate two more of these paths. Sperner’s lemma tells us that these will have an odd num­ber of b-g ver­tices and we know that there will be an odd num­ber of such paths, so this fi­nal type gen­er­ates an odd num­ber of to­tal b-g ver­tices.

  By the way that we have defined these paths, the to­tal num­ber of r-g-b tri­an­gles is equal to the num­ber of g-b ver­tices on the paths in the set gen­er­ated above. This num­ber is the sum of an odd num­ber from the span­ning paths and a se­ries of even num­bers from the other paths, giv­ing an odd over­all num­ber of r-g-b ver­tices, prov­ing num­ber 4 (as long as I haven’t made an er­ror in cat­e­go­riz­ing the paths).

  I hope this makes sense, let me know if it doesn’t or has er­rors :)

• Hoagy 18 Nov 2018 18:20 UTC

  17 points

  AF

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

  I was able to get at least (I think) close to prov­ing 2 us­ing Sperner’s Lemma as fol­lows:

  You can map the con­tin­u­ous func­tion f(x) to a path of the kind found in Ques­tion 1 of length n+1
  by eval­u­at­ing f(x) at x=0, x=1 and n-1 equally spaced di­vi­sions be­tween these two points and set­ting a node as blue if f(x) < 0 else as green.

  By Sperner’s Lemma there is an odd, and there­fore non-zero num­ber of b-g ver­tices. You can then take any b-g pair of nodes as the start­ing points for a new path and re­peat the pro­cess. After k iter­a­tions you have two val­ues of x—only one where f(x) is be­low zero—that are 1/​(n^k) away from each other. We thus can find ar­bi­trar­ily close points that strad­dle zero. By tak­ing the se­quence f(x) of ini­tial nodes x we get a se­quence that, by B-W, has a sub-se­quence which con­verges to zero. By con­ti­nu­ity we have proved the ex­is­tence of an x such that f(x)=0.

  We can be sure that the sub-se­quence does in fact con­verge to zero, rather than any other value be­cause if it con­verges to any num­ber |a|>0, the gra­di­ent of f(x) would have to be ar­bi­trar­ily high to dip back be­low/​above 0 for a value of x ar­bi­trar­ily close by and there­fore would not be a con­tin­u­ous func­tion.

  Com­ments to tighten up/​poke holes in the above ap­pre­ci­ated :)

• Hoagy 28 Aug 2018 0:29 UTC

  2 points

  on: Sub­si­diz­ing Pre­dic­tion Markets

  For long term bets, where the op­por­tu­nity cost of ty­ing money up in these bets be­comes high, I would have thought that the bets should be de­nom­i­nated in US bonds (or other agreed min­i­mal-risk in­ter­est rate as­set) to min­i­mize this cost.

  Even if the bet does not pay out one way or an­other, the money still ac­cu­mu­lates in­ter­est.

  Other than be­ing in­com­pat­i­ble with Augur, are there any the­o­ret­i­cal or prac­ti­cal hur­dles to us­ing this? It would hope­fully re­duce the sub­sidy re­quired to make an at­trac­tive mar­ket with­out in­cur­ring cost in and of it­self.

• Hoagy 7 May 2018 20:29 UTC

  1 point

  in reply to: Said Achmiz's comment on: Every­thing I ever needed to know, I learned from World of War­craft: In­cen­tives and rewards

  Thanks for re­ply­ing :)

  If the join­ing bonus were large enough to give a new mem­ber enough DKP to get the choice items, then older mem­bers would (quite rightly) com­plain. If it were smaller, it wouldn’t work.

  I guess my cen­tral ques­tion is, a new player will have in­finite EP/​GP af­ter they first re­ceive EP. They can there­fore wait un­til their perfect item comes up, and choose that. This to me seems ex­tremely similar to giv­ing an un­cer­tain but po­ten­tially very large join­ing bonus. After los­ing this in­finite ra­tio sta­tus, the situ­a­tion then seems very similar to a free mar­ket one. In par­tic­u­lar I don’t un­der­stand why hav­ing col­lected lots of points (ie abil­ity to claim fu­ture value) would lead to your in­cen­tive drop­ping off, while ac­cu­mu­lat­ing a high ra­tio (which you’d pre­sum­ably need to ‘save’ for a while for re­ally top items) doesn’t have this prob­lem.

• Hoagy 7 May 2018 13:13 UTC

  4 points

  on: Every­thing I ever needed to know, I learned from World of War­craft: In­cen­tives and rewards

  I’m cu­ri­ous but a bit con­fused about some of the benefits of EP/​GP over the straight free-mar­ket model, but if EP/​GP did in­deed take over then I’m sure there’s some­thing I’m miss­ing.

  1: Pre­sum­ably, in both mod­els, in the long run it takes roughly the same av­er­age amount of time (mod­u­lated by your effi­ciency of pro-so­cial ac­tivity) to get an item of qual­ity >x, but it seems that in EP/​GP you get your first al­most im­me­di­ately, while in DKP your timer starts from 0. Was there the is­sue of in­di­vi­d­u­als jump­ing around guilds to try and get that first item?

  2: Is there any sys­tem by which one can defer the re­ceiv­ing of items in EP/​GP so that you don’t end up get­ting some­thing that is of low or nil value to you (es­pe­cially since they can’t be traded)? The main ad­van­tage of the free-mar­ket, at least in sys­tems where in­di­vi­d­u­als have similar abil­ity to earn cur­rency, is usu­ally that items go to those who value them most, so you’d ex­pect DWP to have a big effi­ciency ad­van­tage if you can’t choose whether to ac­cept. On the other hand, if this defer­ral is pos­si­ble, would this de­gen­er­ate into some­thing like a free mar­ket, ex­cept where new en­trants have first dibs over ev­ery­thing?

  The power of at­tract­ing new play­ers is a valuable ad­van­tage I’m sure but it’s the only one that I re­ally see from the 3 given above, and I can’t see how this isn’t pos­si­ble in a similar way by, say, a free mar­ket sys­tem where a new mem­ber gets some kind of join­ing bonus.