Just when you thought the cap is gone for good.. Here are some thinking cap questions you can think/comment on:

1. If I have n propositions, how many possible clauses are there? (Assume that clauses of the form A V A are automatically simplified to A). Does this help in convincing you that the resolution theoremproving search procedure is *complete* in that it will terminate whether or not the theorem you are trying to prove actually holds?

2. Suppose you have a propositional knowledge base KB, and you just proved a theorem alpha. Now more facts are added to the KB to give KB'. Will the theorem you proved in KB still hold in KB'? What if the facts I added made KB' *inconsistent*? [The property we are talking of here is called "Monotonicity"--propositional logic is monotonic.]

2.1. Question 2 points out that propositional logic is no John Kerry (or Mitt Romney--for you dems)--in that it doesn't "flip-flop". But is not "flip-flopping" really all that great a property? Consider this. I tell you Tweety is a bird. Do you think Tweety flies?

Now I tell you Tweety is an ostrich. Do you think it flies? Now I tell you Tweety is a magical ostrich. Do you think it flies? Now I tell you Tweety just got cursed by the local witch/druid. Do you think it flies?

Do you think it would be good to not "flip-flop" in the example above?

3. The "informedness" of a heuristic is not as great an idea as it is touted to be. You will likely found, during your project 1, that the search with manhattan distance heuristic sometimes does *more* expansions than the search with misplaced tiles

heuristic! In this thinking-cap question you will try to understand why this happens.

Consider a search problem, for which you have two *admissible* heuristics. h1(n) is h*(n)/5 ; h2(n) is a constant C (except of course at goal nodes, where it has to be zero), such that h2(n) is always higher than h1(n). Which is more informed? Which do

In particular, given a heuristics h2 which is more informed than h2 (in that forall nodes n, h2(n) >= h1(n)), we can only show that the "interior f-contour" nodes expanded by search with h1 are also expanded by h2.

Specifically, consider a node n' such that f2(n') = g(n') + h2(n') < f* (where f*, the optimal cost of the goal node, is the same for both heuristics). So, n' will clearly be expanded by the search with h2. Since h1 <= h2, clearly f1(n') is also less than f*, and so search with h1 also expands node n'. This shows that the number of nodes on the interior f-contours for h1 can not be fewer than the nodes on the interior f-contours for h1. (An interior f-contour is the set of equi-f-value nodes with f-values less than f*)

The tricky part is that this subsumption doesn't necessrily hold for nodes that are on the "f* contour"--i.e. nodes that have

f1 or f2 equal to f*. Based on the tie-breaking strategy, a subset of these nodes will likely be expanded before the search terminates. If the subset expanded by h2 is more than that expanded by h1, search with h2 can do more expansions even though it is more informed.

See if you can think of possible reasons why this could happen. [hint: there is a slide in the informed search lecture--with annotation "advanced"--that gives the insights needed, if you are having trouble..]

*you*think is a better heuristic?In particular, given a heuristics h2 which is more informed than h2 (in that forall nodes n, h2(n) >= h1(n)), we can only show that the "interior f-contour" nodes expanded by search with h1 are also expanded by h2.

Specifically, consider a node n' such that f2(n') = g(n') + h2(n') < f* (where f*, the optimal cost of the goal node, is the same for both heuristics). So, n' will clearly be expanded by the search with h2. Since h1 <= h2, clearly f1(n') is also less than f*, and so search with h1 also expands node n'. This shows that the number of nodes on the interior f-contours for h1 can not be fewer than the nodes on the interior f-contours for h1. (An interior f-contour is the set of equi-f-value nodes with f-values less than f*)

The tricky part is that this subsumption doesn't necessrily hold for nodes that are on the "f* contour"--i.e. nodes that have

f1 or f2 equal to f*. Based on the tie-breaking strategy, a subset of these nodes will likely be expanded before the search terminates. If the subset expanded by h2 is more than that expanded by h1, search with h2 can do more expansions even though it is more informed.

See if you can think of possible reasons why this could happen. [hint: there is a slide in the informed search lecture--with annotation "advanced"--that gives the insights needed, if you are having trouble..]

Rao

1. n propositions * n - 1 propositions, since using the same one just gets simplified to a proposition, / 2 since ordering doesn't matter. So, n(n - 1)/2.

ReplyDelete2. Yes. All the original facts you had are still there, so if KB' is consistent you'll still be able to prove it. On the other hand, if it's inconsistent, you'll still be able to prove it but it won't mean anything, since you can prove anything from an inconsistent KB.

2.1. I think that the same proposition is just changing over and over. In other words, the proposition "if Tweety is a bird, Tweety flies" would have to be removed in order to have "if Tweety is an ostrich, Tweety doesn't fly". If we were really adding propositions and not doing this, our knowledge base would be inconsistent.

3. h2 is more informed, but I think h1 is a better heuristic. If we're using h2 it seems we'd just be doing Djikstra's algorithm, since the order in which we consider paths won't even change from if h = 0.

I'm not quite sure why h2 might expand more nodes than h1 though.

Re: qn 1, aren't you assuming that you are considering only clauses of length 2?

Delete(If there are only n^2 clauses, then you can't possibly need exponential time to do all resolutions...)

Rao

Oh oops, I got confused because I saw A v A and assumed there were only two. If there's more than two, then I'd say would be 2^n including the empty clause, since each proposition can either be in or out.

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