Yes—at least that’s the assumption I’m working under.
It seems like this should be lower complexity than the intended result, since True has much lower complexity than H_understands?
I agree that the θ1 you’ve described has lower complexity than the intended θ1—but the θ2 in this case has higher complexity, since θ2 is no longer getting any of its complexity for free from conditioning on the f? condition. And in fact what you’ve just described is precisely the unintended model—what I call M−—that I’m trying to compete against, with the hope being that the savings that M+ gives you in θ2 are sufficient to compensate for the loss in having to specify f+ and H_understands in θ1.
If we calculate the complexity of your proposal, we get
complexity(M−)=complexity(θ−1)+complexity(θ−2|M−|f?)=complexity(W−H)+complexity(f−)+complexity(H|True)=complexity(W−H)+complexity(f−)+complexity(H)≈complexity(W)
whereas, if we calculate the complexity of the intended M+, we get
complexity(M+)=complexity(θ+1)+complexity(θ+2|M+|f?)=complexity(W−H)+complexity(f−)+complexity(f+)+complexity(H_understands)+complexity(H|H_understands→f+=f−)≈complexity(W−H)+complexity(f+)+complexity(H_understands)+complexity(H)−minθ2{complexity(θ2)|H_understandsH=θ2→f+H=θ2=f−H=θ2}≈complexity(W)+complexity(f+)+complexity(H_understands)−minθ2{complexity(θ2)|H_understandsH=θ2→f+H=θ2=f−H=θ2}
such that you can see that the question of which one wins is precisely dependent on whether the savings from conditioning on H_understands→f+=f− offsets the cost of having to specify f+ and H_understands.
such that you can see that the question of which one wins is precisely dependent on whether the savings from conditioning on H_understands→f+=f− offsets the cost of having to specify f+ and H_understands.
Yeah, that makes sense. I guess I don’t really see the intuition about why this should be true, but fair enough to leave that as an open question.
Yes—at least that’s the assumption I’m working under.
I agree that the θ1 you’ve described has lower complexity than the intended θ1—but the θ2 in this case has higher complexity, since θ2 is no longer getting any of its complexity for free from conditioning on the f? condition. And in fact what you’ve just described is precisely the unintended model—what I call M−—that I’m trying to compete against, with the hope being that the savings that M+ gives you in θ2 are sufficient to compensate for the loss in having to specify f+ and
H_understandsin θ1.If we calculate the complexity of your proposal, we get complexity(M−)=complexity(θ−1)+complexity(θ−2 | M−|f?)=complexity(W−H)+complexity(f−)+complexity(H | True)=complexity(W−H)+complexity(f−)+complexity(H)≈complexity(W) whereas, if we calculate the complexity of the intended M+, we get complexity(M+)=complexity(θ+1)+complexity(θ+2 | M+|f?)=complexity(W−H)+complexity(f−)+complexity(f+)+complexity(H_understands)+complexity(H | H_understands→f+=f−)≈complexity(W−H)+complexity(f+)+complexity(H_understands)+complexity(H)−minθ2{complexity(θ2) | H_understandsH=θ2→f+H=θ2=f−H=θ2}≈complexity(W)+complexity(f+)+complexity(H_understands)−minθ2{complexity(θ2) | H_understandsH=θ2→f+H=θ2=f−H=θ2} such that you can see that the question of which one wins is precisely dependent on whether the savings from conditioning on H_understands→f+=f− offsets the cost of having to specify f+ and H_understands.
Yeah, that makes sense. I guess I don’t really see the intuition about why this should be true, but fair enough to leave that as an open question.