Alfred Harwood

When bits of op­ti­miza­tion im­ply bits of mod­el­ing: the Touchette-Lloyd theorem

• Alfred Harwood 27 Nov 2025 12:07 UTC

  3 points

  0

  in reply to: johnswentworth’s comment on: Re­sam­pling Con­serves Re­dun­dancy (Ap­prox­i­mately)

  Hi, Jeremy and I have a couple of updates on this thread. I have put them in a shortform here.

Alfred Har­wood’s Shortform

• Alfred Harwood 27 Nov 2025 12:03 UTC

  26 points

  0

  on: Alfred Har­wood’s Shortform

  Here are some updates on the work that @Jeremy Gillen and I have been doing on Natural Latents, following on from Jeremy’s comment on Resampling Conserves Redundancy (Approximately). We are still trying to prove that a stochastic natural latent implies the existence of a deterministic latent which is almost as good.

  First, if a latent is a deterministic function of X and Y, (ie. $\Lambda =f(X,Y)$), the deterministic redundancy conditions will be satisfied up to errors of $H\left(Y|X\right)$ and H(X|Y). (We don’t actually use this in anything that follows, but think its quite nice.)

  Proof:Proof

  First, write out $H(X,Y,\Lambda )$ using the entropy chain rule:

  $H(X,Y,\Lambda )=H\left(X\right)+H\left(Y|X\right)+H(\Lambda |X,Y)=H\left(X\right)+H\left(Y|X\right)$

  Since $\Lambda$ is a deterministic function of X and Y, we can set $H(\Lambda |X,Y)=0$.

  We can expand $H(X,Y,\Lambda )$ differently using the entropy chain rule.

  $H(X,Y,\Lambda )=H\left(X\right)+H\left(\Lambda |X\right)+H(Y|X,\Lambda )$

  The two expressions for $H(X,Y,\Lambda )$ must be equal so:

  $H\left(X\right)+H\left(Y|X\right)=H\left(X\right)+H\left(\Lambda |X\right)+H(Y|X,\Lambda )$ .

  Re-arranging gives:

  $H\left(\Lambda |X\right)=H\left(Y|X\right)-H(Y|X,\Lambda )\le H\left(Y|X\right)$

  We can repeat this proof with X and Y swapped to get the bound $H\left(\Lambda |Y\right)\le H\left(X|Y\right)$

  Furthermore, we can always consider a latent which just copies one of the variables (for example $\Lambda =Y$). This kind of latent will always perfectly satisfy one of the deterministic redundancy conditions (if $\Lambda =Y$, then $H\left(\Lambda |Y\right)=0$ ). The other deterministic redundancy condition will be satisfied with error $H\left(\Lambda |X\right)=H\left(Y|X\right)$. The mediation condition will also have zero error since $I(X;Y|\Lambda )=H\left(Y|\Lambda \right)-H(Y|X,\Lambda )=0$ (since $H\left(Y|\Lambda \right)=0$).

  We can choose $\Lambda$ to copy either $X$ or $Y$, meaning that using this method we can always construct a deterministic natural latent with $\text{error}=\text{min}\left[H\right(Y|X),H(X|Y\left)\right].$

  Another deterministic latent that can always be constructed is the constant latent $\Lambda =const.$ This will always perfectly satisfy the two deterministic redundancy conditions, since $H\left(\Lambda |X\right)\le H\left(\Lambda \right)=0$. The mediation error for the constant latent is just the mutual information between $X$ and $Y$.

  Between these two types of latent, a deterministic natural latent can always be constructed with error bounded by $\text{min}\left[I\right(X;Y),H(X|Y),H(Y|X\left)\right]$. Loosely: if $X$ and $Y$ are highly correlated you can use the copy latent and if $X$ and $Y$ are close to being independent, you can use the constant latent.

  To test how tight this bound is, Jeremy has numerically found the optimal deterministic and stochastic natural latents for a family of (X,Y) distributions. The distribution is parametrised by $\alpha$ which captures how correlated $X$ and $Y$ are ($\alpha =0$ implies perfect correlation and $\alpha =1$ implies no correlation). A nice graph of these results is below:

  

  The ‘score’ of the latents is the sum of the errors on the mediation and redundancy conditions. The fact that the best deterministic score closely hugs the line $min\left[I\right(X,Y),H(Y|X\left)\right]$ suggests that the bound we found above is pretty tight. When we look at the latents found by the numerical optimizer, we find that it is using the copy latent at low $H\left(Y|X\right)$ and the constant latent when$I(X;Y)<H\left(Y|X\right)$.

  We also have a graph which breaks down the individual errors for optimal stochastic and deterministic latents:

  

  It shows quite clearly the point where the optimal deterministic latent switches from ‘copy’ to constant around $\alpha =0.1$. Furthermore, for $\alpha >1/3$ we have shown that the constant latent is at a local minimum in the space of possible stochastic latents. We did this by differentiating with respect to the conditional probabilities $P(\Lambda |X,Y)$ and finding the Hessian. This was done using a symbolic program so we don’t have a compact proof to put here. The point $\alpha =1/3$ is also the point where the errors for the best stochastic and deterministic latents become the same. This seems like good news for the general NxN conjecture, since it shows that when two states are roughly independent, there’s zero cost (to the difference between stochastic and deterministic scores) to ‘merging’ those states from the perspective of the latent.

  Eyeballing the first graph, it seems like the deterministic score is never more than twice the best stochastic score for any given distribution. It is interesting to note that the optimal stochastic latent roughly follows the same pattern as the deterministic latent but it notably performs better than the deterministic latent in situations where both $H\left(Y|X\right)$ and $I(X;Y)$ are high. Bear in mind that these graphs are just for one family of distributions (but we have tested others and got similar results).

  At the moment we are thinking about ways to lower bound the stochastic latent error in terms of $I(X;Y)$ and $H\left(Y|X\right)$. For example if we could show that it is impossible for a stochastic latent to have a total error less than $\frac{1}{n}\text{min}\left[I\right(X;Y),H(Y|X),H(X|Y\left)\right]$ then this would prove that the existence of a stochastic latent with error $ϵ$ implies the existence of a deterministic latent with error $nϵ$ or less. We are also thinking about ways to show that certain latent types have errors which are global minima (as opposed to just local minima) in the space of possible latents.

• Alfred Harwood 16 Sep 2025 8:55 UTC

  1 point

  0

  in reply to: Eva Lu’s comment on: The In­ter­nal Model Prin­ci­ple: A Straight­for­ward Ex­pla­na­tion

  That sounds about right. The extra thing that they are claiming is that these assumptions are things that naturally apply in real life, when a controller is doing its job (ie. they are not just contrived/​chosen to get the result). So (Wonham et al claim) the interesting thing is that you can say that these isomorphisms hold in actual systems. Obviously there are a bunch of issues with this. I intentionally avoided too much discussion and criticism in this post and put it in a separate post.

• Alfred Harwood 10 Sep 2025 10:56 UTC

  1 point

  0

  in reply to: Winter Cross’s comment on: The In­ter­nal Model Prin­ci­ple: A Straight­for­ward Ex­pla­na­tion

  Thanks for your comment. I am glad the post helped you!

  Good questions. The short answer is that you are correct and I was sloppy in that section.

  Could the evolution of the joint states form a cycle?

  Yes! In fact, if S and W are finite sets then the evolution must eventually form a cycle. (If a finite set S has cardinality n, you can only apply a function $\alpha :S\to S$ at most n times before you return to a state you have visited before). I meant this to be implicit in the ‘… etc.’ part but I didn’t make it clear. I have added the following sentence to the post which hopefully clarifies things:

  If the sets $S$ and $W$ are finite with cardinality $n$ then the evolution would eventually cycle around so we would have to specify that the evolution will eventually come full circle ie.$\overline{\alpha }\left(w_n\right):=w_1$ and ${\alpha }_E\left(s_n\right):=s_1$.

  Could multiple joint states evolve to the same joint state?

  This is a good question involving a subtlety that I skipped over. The answer is ‘yes sometimes’. But when it does happen its a little weird and worth thinking about. There are a few ways ways in which multiple joint states could evolve to the same joint state.

  1)Two different environment states with the same controller state evolving to the same environment state. eg.$(s_1,w_1)\to (s_2,w_2)$ and $(s_3,w_1)\to (s_2,w_2)$

  In this case, the Detectability condition is violated, since the controller will do the same thing, regardless of whether the environment is in $s_1$ and $s_3$. The Detectability condition would tell us that this means that s_1 and s_3 are (from the point of view of the controller) identical, so we should coarse grain them so that they are both labelled the same. This means that we wouldn’t expect this kind of joint evolution.

  2)Two different controller states with the same environment state lead evolve to the same joint state. eg $(s_1,w_1)\to (s_2,w_2)$ and $(s_1,w_3)\to (s_2,w_2)$

  In this case, Detectability is satisfied. As far as I can tell, this kind of evolution does not violate any of the conditions for the IMP so it is valid. However, notice what this would imply. There are two controller states (w_1 and w_3) which both do the same thing to the system (cause it to evolve to s_2). After either of these states, the controller then evolves to w_2 and from then on behaves identically forever. It seems to me that for our purposes w_1 and w_3 are ‘the same’ controller state so I would be inclined to coarse grain them and label them as the same, removing this kind of evolution. However, since there are no assumptions which explicitly require this kind of coarse graining over the controller states, this kind of evolution is technically allowed within the IMP.

  3)Two joint states with different environment and controller states evolve to the same joint state. eg.$(s_1,w_1)\to (s_2,w_2)$ and $(s_3,w_3)\to (s_2,w_2)$.

  Again, I think that this is allowed within the IMP. But notice that after the evolution both trajectories will behave the same. This means that if the environment and controller state sets are finite then at most one of these joint states will be involved in any kind of repeating cycle. The other will be ‘transient’ ie. it will occur once and never again.

  What links here?

  • The In­ter­nal Model Prin­ci­ple: A Straight­for­ward Ex­pla­na­tion by Alfred Harwood (12 Apr 2025 10:58 UTC; 23 points)

Ap­ply for the 2025 Dove­tail fellowship

• Alfred Harwood 16 Aug 2025 8:21 UTC

  3 points

  3

  in reply to: Linch’s comment on: Linch’s Shortform

  I think that reading the whole soliloquy makes my reading clearer and your reading less plausible. I can maybe see that if you:

  1. Ignored the whole context of the rest of the play

  2. Ignored the lines in the soliloquy which specifically mention the family names

  3. Ignored the fact that there is a convention in English to refer to people by their first names and leave their family name implicit

  Then maybe you would come to the conclusion that Juliet has an objection to specifically Romeo’s first name and not the fact that his name more generally links him to his family. But if you don’t ignore those things, it seems clear to me that Juliet is lamenting the fact that the man she loves has a name (‘Romeo Montague’) which links him to a family who she is not allowed to love.

• Alfred Harwood 15 Aug 2025 12:11 UTC

  14 points

  12

  in reply to: Linch’s comment on: Linch’s Shortform

  I strongly agree with the general sentiment of ‘don’t be afraid to say something you think is true, even if you are worried it might seem stupid’. Having said that:

  I don’t agree with your analysis of the line. She’s not upset that he’s named Romeo. She is asking “Why does Romeo (the man I have fallen in love with) have to be the same person as Romeo (the son of Lord Montague with whom my family has a feud)?”. The next line is ‘Deny thy father and refuse thy name’ which I think makes this interpretation pretty clear (ie. if only you told me you were not Romeo, the son of Lord Montague, then things would be ok). The line seems like a perfectly fine (albeit poetic and archaic) way to express this.

  This works with your modern translation (“Romeo, why you gotta be Romeo?”). Imagine an actor delivering that line and emphasising the ‘you’ (‘Romeo, why do you have to be Romeo?’) and I think it makes sense. Given the context and delivery, it feels clear that it should be interpreted as ‘Romeo (man I’ve just met) why do you have to be Romeo (Montague)?’. It seems unfair to declare that the line taken out of context doesn’t make sense just because she doesn’t explicitly mention that her issue is with his family name. Especially when the very next line (and indeed the whole rest of the play) clarifies that the issue is with his family.

  Sure, the line is poetic and archaic and relies on context, which makes it less clear. But these things are to be expected reading Shakespeare!

• Alfred Harwood 11 Aug 2025 21:40 UTC

  9 points

  0

  in reply to: Elizabeth’s comment on: Eliz­a­beth’s Shortform

  It is also fairly common for directors/​writers to use a book as a inspiration but not care about the specific details because they want to express their own artistic vision. Hitchcock refused to adapt books that he considered ‘masterpieces’, since he saw no point in trying to improve them. When he adapted books (such as Daphne du Maurier’s The Birds) he used the source material as loose inspiration and made the films his own.

  François Truffaut: Your own works include a great many adaptations, but mostly they are popular or light entertainment novels, which are so freely refashioned in your own manner that they ultimately become a Hitchcock creation. Many of your admirers would like to see you undertake the screen version of such a major classic as Dostoyevsky’s Crime and Punishment, for instance.

  Alfred Hitchcock: Well, I shall never do that, precisely because Crime and Punishment is somebody else’s achievement. There’s been a lot of talk about the way in which Hollywood directors distort literary masterpieces. I’ll have no part of that! What I do is to read a story only once, and if I like the basic idea, I just forget all about the book and start to create cinema. Today I would be unable to tell you the story of Daphne du Maurier’s The Birds. I read it only once, and very quickly at that. An author takes three or four years to write a fine novel; it’s his whole life. Then other people take it over completely. Craftsmen and technicians fiddle around with it and eventually someone winds up as a candidate for an Oscar, while the author is entirely forgotten. I simply can’t see that.

  FT: I take it then that you’ll never do a screen version of Crime and Punishment.

  AH: Even if I did, it probably wouldn’t be any good.

  FT: Why not?

  AH: Well, in Dostoyevsky’s novel there are many, many words and all of them have a function.

  FT: That’s right. Theoretically, a masterpiece is something that has already found its perfection of form, its definitive form.

  AH: Exactly, and to really convey that in cinematic terms, substituting the language of the camera for the written word, one would have to make a six- to ten-hour film. Otherwise, it won’t be any good.

  (From Hitchcock/​Truffaut, quoted here).

  Alfonso Cuaron also liked the idea of Children of Men (the book) but disliked almost all the specific details, so he used his film as a chance to make all of the changes he wanted to see.

• Alfred Harwood 4 Jul 2025 8:52 UTC

  18 points

  3

  in reply to: Kabir Kumar’s comment on: Kabir Ku­mar’s Shortform

  In the post ‘Can economics change your mind?’ he has a list of examples where he has changed his mind due to evidence:

  1. Before 1982-1984, and the Swiss experience, I thought fixed money growth rules were a good idea. One problem (not the only problem) is that the implied interest rate volatility is too high, or exchange rate volatility in the Swiss case.

  2. Before witnessing China vs. Eastern Europe, I thought more rapid privatizations were almost always better. The correct answer depends on circumstance, and we are due to learn yet more about this as China attempts to reform its SOEs over the next five to ten years. I don’t consider this settled in the other direction either.

  3. The elasticity of investment with respect to real interest rates turns out to be fairly low in most situations and across most typical parameter values.

  4. In the 1990s, I thought information technology would be a definitely liberating, democratizing, and pro-liberty force. It seemed that more competition for resources, across borders, would improve economic policy around the entire world. Now this is far from clear.

  5. Given the greater ease of converting labor income into capital income, I no longer am so convinced that a zero rate of taxation on capital income is best.

  6. The social marginal value of health care is often quite low, much lower than I used to realize. By the way, hardly anyone takes this on consistently to guide their policy views, no matter how evidence-driven they may claim to be.

  7. Mormonism, and other relatively strict religions, can have big anti-poverty effects. I wouldn’t say I ever believed the contrary, but for a long time I simply didn’t give the question much attention. I now think that Mormonism has a better anti-poverty agenda than does the Progressive Left.

  8. There are positive excess returns to some momentum investment strategies.

  I don’t know enough about economics to tell how much these meet your criteria for ‘I was wrong’ rather than ‘revised estimates’ or something else (he doesn’t use the exact phrase ‘I was wrong’) but it seems in the spirit of what you are looking for.

• Alfred Harwood 21 Jun 2025 23:28 UTC

  1 point

  0

  in reply to: Linda Linsefors’s comment on: A Straight­for­ward Ex­pla­na­tion of the Good Reg­u­la­tor Theorem

  I know you asked for other people (presumably not me) to confirm this but I can point you to the statement of the theorem, as written by Conant and Ashby in the original paper :

  Theorem: The simplest optimal regulator R of a reguland S produces events R which are related to the events S by a mapping $h:S\to R$
  

  Restated somewhat less rigorously, the theorem says that the best regulator of a system is one which is a model of that system in the sense that the regulator’s actions are merely the system’s actions as seen through a mapping h.

  I agree that it has nothing to do with modelling and is not very interesting! But the simple theorem is surrounded by so much mysticism (both in the paper and in discussions about it) that it is often not obvious what the theorem actually says.

• Alfred Harwood 17 Jun 2025 9:39 UTC

  3 points

  0

  in reply to: Richard_Kennaway’s comment on: A Straight­for­ward Ex­pla­na­tion of the Good Reg­u­la­tor Theorem

  The diagram is a causal Bayes net which is a DAG so it can’t contain cycles. Your diagram contains a cycle between R and Z. The diagram I had in mind when writing the post was something like:

  

  which is a thermostat over a single timestep.

  If you wanted to have a feedback loop over multiple timesteps, you could conjoin several of these diagrams:

  

  Each node along the top row is the temperature at successive times. Each node along the bottom row is the controller state at different times.

Seven ways to Im­prove the In­ter­nal Model Principle

• Alfred Harwood 24 Apr 2025 12:42 UTC

  4 points

  0

  in reply to: johnswentworth’s comment on: $500 Bounty Prob­lem: Are (Ap­prox­i­mately) Deter­minis­tic Nat­u­ral La­tents All You Need?

  Thanks for the clarifications, that all makes sense. I will keep thinking about this!

• Alfred Harwood 23 Apr 2025 14:52 UTC

  23 points

  0

  on: $500 Bounty Prob­lem: Are (Ap­prox­i­mately) Deter­minis­tic Nat­u­ral La­tents All You Need?

  This seems like an interesting problem! I’ve been thinking about it a little bit but wanted to make sure I understood before diving in too deep. Can I see if I understand this by going through the biased coin example?

  Suppose I have 2^5 coins and each one is given a unique 5-bit string label covering all binary strings from 00000 to 11111. Call the string on the label $\Lambda$.

  The label given to the coin indicates its ‘true’ bias. The string 00000 indicates that the coin with that label has p(heads)=0. The coin labelled 11111 has p(heads)=1. The ‘true’ p(heads) increases in equal steps going up from 00000 to 00001 to 00010 etc. Suppose I randomly pick a coin from this collection, toss it 200 times and call the number of heads X_1. Then I toss it another 200 times and call the number of heads X_2.

  Now, if I tell you what the label on the coin was (which tells us the true bias of the coin), telling you X_1 would not give you any more information to help you guess X_2 (and vice versa). This is the first Natural Latent condition ($\Lambda$ induces independence between X_1 and X_2). Alternatively, if I didn’t tell you the label, you could estimate it from either X_1 or X_2 equally well. This is the other two diagrams.

  I think that the full label $\Lambda$ will be an approximate stochastic natural latent. But if we consider only the first bit^[1] of the label (which roughly tells us whether the bias is above or below 50% heads) then this bit will be a deterministic natural latent because with reasonably high certainty, you can guess the first bit of $\Lambda$ from X_1 or X_2. This is because the conditional entropy H(first bit of $\Lambda$|X_1) is low. On the other hand H($\Lambda$ | X_1) will be high. If I get only 23 heads out of 200 tosses, I can be reasonably certain that the first bit of $\Lambda$ is a 0 (ie the coin has a less than 50% of coming up heads) but can’t be as certain what the last bit of $\Lambda$ is. Just because $\Lambda$ satisfies the Natural Latent conditions within $ϵ$, this doesn’t imply that $\Lambda$ satisfies $H\left(\Lambda |X_1\right)<ϵ$. We can use X_1 to find a 5-bit estimate of $\Lambda$, but most of the useful information in that estimate is contained in the first bit. The second bit might be somewhat useful, but its less certain than the first. The last bit of the estimate will largely be noise. This means that going from using $\Lambda$ to using ‘first bit of $\Lambda$’ doesn’t decrease the usefulness of the latent very much, since the stuff we are throwing out is largely random. As a result, the ‘first bit of $\Lambda$’ will still satisfy the natural latent conditions almost as well as $\Lambda$. By throwing out the later bits, we threw away the most ‘stochastic’ bits, while keeping the most ‘latenty’ bits.

  So in this case, we have started from a stochastic natural latent and used it to construct a deterministic natural latent which is almost as good. I haven’t done the calculation, but hopefully we could say something like ‘if $\Lambda$ satisfies the natural latent conditions within $ϵ$ then the first bit of $\Lambda$ satisfies the natural latent conditions within $2ϵ$ (or $3ϵ$ or something else)’. Would an explicit proof of a statement like this for this case be a special case of the general problem?

  The problem question could be framed as something like: “Is there some standard process we can do for every stochastic natural latent, in order to obtain a deterministic natural latent which is almost as good (in terms of \epsilon)”. This process will be analogous to the ‘throwing away the less useful/​more random bits of \lambda’ which we did in the example above. Does this sound right?

  Also, can all stochastic natural latents can be thought of as ‘more approximate’ deterministic latents? If a latent satisfies the the three natural latents conditions within ${ϵ}_1$, we can always find a (potentially much bigger) ${ϵ}_2$ such that this latent also satisfies the deterministic latent condition, right? This is why you need to specify that the problem is showing that a deterministic natural latent exists with ‘almost the same’ $ϵ$. Does this sound right?

  1. ^

     I’m going to talk about the ‘first bit’ but an equivalent argument might also hold for the ‘first two bits’ or something. I haven’t actually checked the maths.

  What links here?

  • Ap­ply for the 2025 Dove­tail fellowship by Alex_Altair (17 Aug 2025 19:09 UTC; 42 points)

The In­ter­nal Model Prin­ci­ple: A Straight­for­ward Ex­pla­na­tion

• Alfred Harwood 21 Feb 2025 16:42 UTC

  3 points

  2

  in reply to: Cole Wyeth’s comment on: Rul­ing Out Lookup Tables

  It seems that the relevant thing is not so much how many values you have tested as the domain size of the function. A function with a large domain cannot be explicitly represented with a small lookup table. But this means you also have to consider how the black box behaves when you feed it something outside of its domain, right?

  This sounds right. Implicitly, I was assuming that when the black box was fed an x outside of its domain it would return an error message or at least, something which is not equal to f(x). I realise that I didn’t make this clear in the post.

  But what if it is a lookup table, but the index is taken mod the length of the table? Or more generally, what if a hash function is applied to the index first?

  In the post I was considering programs which are ‘just’ a lookup tables. But I agree that there are other programs that use lookup tables but are not ‘just’ tables. The motivation is that if you want simulate f(x) over a large domain using a function smaller than the bound given in the post, then the program needs to contain some feature which captures something nontrivial about the nature of f.

  Like (simple example just to be concrete) suppose f(x)=[x(mod10)]^2 defined on all real numbers. Then the program in the black box might have two steps: 1) compute x(mod10) using a non-lookup table method then 2) use a lookup table containing x^2 for x=1 to 10. This (by my slightly arbitrary criteria) would count as a program which ‘actually computes the function’ since it has fixed length and its input/​output behaviour coincides with f(x) for all possible inputs. It seems to me that if a blackbox program contains a subsection which computes x(mod10) then it has captured something important about f(x), more so than if the program only contains a big lookup table.

  It seems that counting the number of unique output values taken by the function is more robustly lower-bounds its size as a (generalized) lookup table?

  I think this is right. If you wanted to rule out lookup tables being used at any point in the program then this is bottlenecked by the number of unique outputs.

Towards build­ing blocks of ontologies

• Alfred Harwood 7 Feb 2025 11:55 UTC

  1 point

  0

  in reply to: Dagon’s comment on: Rul­ing Out Lookup Tables

  Cool, that all sounds fair to me. I don’t think we have any substantive disagreements.