> > >> I guess my previous question was not clear enough: if the only domain > >> knowledge PLN has is > >> > >> > Ben is an author of a book on AGI <tv1> > >> > This dude is an author of a book on AGI <tv2> > >> > >> and > >> > >> > Ben is odd <tv1> > >> > This dude is odd <tv2> > >> > >> Will the system derives anything? > > > > Yes, via making default assumptions about node probability... > > Then what are the conclusions, with their truth-values, in each of the > two cases? >
Without node probability tv's, PLN actually behaves pretty similarly to NARS in this case... If we have Ben ==> AGI-author <s1> Dude ==> AGI-author <s2> |- Dude ==> Ben <s3> the PLN abduction rule would yield s3 = s1 s2 + w (1-s1)(1-s2) where w is a parameter of the form w = p/ (1-p) and if we set w=1 which is a principle of indifference type assumption then we just have s3 = 1 - s1 - s2 + 2s1s2 In any case, regardless of w, s1=s2=1 implies s3=1 in this formula, which is the same answer NARS gives in this case (of crisp premises) Similar to NARS, PLN also gives a fairly low confidence to this case, but the confidence formula is a pain and I won't write it out here... (i.e., PLN assigns this a beta distribution with 1 in its support, but a pretty high variance...) So, similar to NARS, without node probability info PLN cannot distinguish the two inference examples I gave .. no system could... However, PLN incorporates the node probabilities when available, immediately and easily, without requiring knowledge of math on the part of the system... and it incorporates them according to Bayes rule which I believe the right approach ... What is counterintuitive to me is having an inference engine that does not immediately and automatically use the node probability info when it is available... As evidence about Bayesian neural population coding in the brain suggests, use of Bayes rule is probably MORE cognitively primary than use of these other more complex inference rules... -- ben g p.s. details: In PLN, simple abduction consists of the inference problem: Given P(A), P(B), P(C), P(B|A) and P(B|C), find P(C|A). and the simplest, independence-assumption + Bayes rule based formula for this is abdAC:=(sA,sB,sC,sAB,sCB)->(sAB*sCB*sC/sB+(1-sAB)*(1-sBC)*sC/(1-sB)) [or, more fully including all consistency conditions, abdAC:= (sA,sB,sC,sAB,sCB)->(sAB*sCB*sC/sB+(1-sAB)*(1-sBC)*sC/(1-sB))*(Heaviside(sAB-max(((sA+sB-1)/sA),0))-Heaviside(sAB-min(1,(sB/sA))))*(Heaviside(sCB-max(((sB+sC-1)/sC),0))-Heaviside(sCB-min(1,(sB/sC)))); ] (This is Maple notation...) ------------------------------------------- agi Archives: https://www.listbox.com/member/archive/303/=now RSS Feed: https://www.listbox.com/member/archive/rss/303/ Modify Your Subscription: https://www.listbox.com/member/?member_id=8660244&id_secret=114414975-3c8e69 Powered by Listbox: http://www.listbox.com
