Returned Midterms - Artificial Intelligence | CSE 571, Study notes of Computer Science

Material Type: Notes; Class: Artificial Intelligence; Subject: Computer Science and Engineering; University: Arizona State University - Tempe; Term: Unknown 1989;

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CSE571 Artificial Intelligence
Nov 12th Notes
William Cushing
0) Returned midterms
1) Went over questions
a) Question 5. Liberal grading, one answer in book – so not going over in class
b) Question 4. Also liberal grading, as there are many ways of answering it.
As long as it works, it works – so good grading.
There was supposed to be an abnormal predicate – for the discussed way Of
handling normative statements. So some points could be lost there.
c) Question 3. The separators in the representation were periods, not commas.
i) For the correct stratification, the answer set is the empty set.
For the incorrect version of the program (reading the program wrong) the
answer set is {r,p}.
ii) For the correct reading, the answer set is {q}
For the incorrect reading, the answer set is {r,p}
iii) For the correct reading, the answer set is {r}
For the incorrect reading, the answer set is also {r}
d) Question 2. Almost everyone got this right – typical example was (a if not b, b
if not a).
e) Question 1. Herbrand Universe is all of the objects; including those produced
by functions. The Herbrand Base is just all of the predicates applied to all of the
objects (ie: Base = Pred(Universe))
i)
HB={p(a),p(b),p(c),q(a),q(b),q(c)}
LM={q(b),q(c),p(a)}
AM={q(b),p(a),p(b),q(c)}
ii)
Add f(a),f(b),f(c),f(f(a)),f(f(b)),f(f(c)), ... to the HB above
Add q(f(a)) to LM above
f) Question 0 – take-home. Graded liberally, despite the fact that it was take-
home and people could have read the referenced sections (5.1, 5.2).
The basic argument is that the answer sets do not encode a plan because the plan
does not work in every possible initial state – there are multiple possible initial
states (uncertain information), and a plan needs to work in all of them.
About 30% of people had a good argument and example, and got 20 – others had
some thoughts, but not enough to get more than a 10. The last portion had a good
example, but didn’t abstract out well enough, and got about 15.
2) Bayes nets
Covering 2 chapters out of a different textbook:
Artificial Intelligence: a new synthesis. Nilsson. Chapter 19, 20.1.
Chapter 19 is available in the library (Noble) on hold. (by Thursday).
“see how much you remember from last class” ==> bigger example:
pf3
pf4
pf5

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CSE571 Artificial Intelligence Nov 12th^ Notes William Cushing

  1. Returned midterms
  2. Went over questions a) Question 5. Liberal grading, one answer in book – so not going over in class b) Question 4. Also liberal grading, as there are many ways of answering it. As long as it works, it works – so good grading. There was supposed to be an abnormal predicate – for the discussed way Of handling normative statements. So some points could be lost there. c) Question 3. The separators in the representation were periods, not commas. i) For the correct stratification, the answer set is the empty set. For the incorrect version of the program (reading the program wrong) the answer set is {r,p}. ii) For the correct reading, the answer set is {q} For the incorrect reading, the answer set is {r,p} iii) For the correct reading, the answer set is {r} For the incorrect reading, the answer set is also {r} d) Question 2. Almost everyone got this right – typical example was (a if not b, b if not a). e) Question 1. Herbrand Universe is all of the objects; including those produced by functions. The Herbrand Base is just all of the predicates applied to all of the objects (ie: Base = Pred(Universe)) i) HB={p(a),p(b),p(c),q(a),q(b),q(c)} LM={q(b),q(c),p(a)} AM={q(b),p(a),p(b),q(c)} ii) Add f(a),f(b),f(c),f(f(a)),f(f(b)),f(f(c)), ... to the HB above Add q(f(a)) to LM above f) Question 0 – take-home. Graded liberally, despite the fact that it was take- home and people could have read the referenced sections (5.1, 5.2). The basic argument is that the answer sets do not encode a plan because the plan does not work in every possible initial state – there are multiple possible initial states (uncertain information), and a plan needs to work in all of them. About 30% of people had a good argument and example, and got 20 – others had some thoughts, but not enough to get more than a 10. The last portion had a good example, but didn’t abstract out well enough, and got about 15.
  3. Bayes nets Covering 2 chapters out of a different textbook: Artificial Intelligence: a new synthesis. Nilsson. Chapter 19, 20.1. Chapter 19 is available in the library (Noble) on hold. (by Thursday). “see how much you remember from last class” ==> bigger example:

P(Q|P4,P5) =

Sum over P6,P7 of P(Q,P6,P7|P4,P5) How do we do: P(Q,P6,P7|P4,P5)? = P(Q|P6,P7,P4,P5) P(P6,P7|P4,P5) = P(Q|P6,P7)P(P6,P7|P4,P5) = P(Q|P6,P7)P(P6|P7,P4,P5)P(P7|P4,P5) But, better is to do: = P(Q|P6,P7) P(P6|P4,P5) P(P7|P4,P5) Because P6 and P7 are separable (due to Q) However, that is a deeper result that depends on the structure of P6 and P An example involving a sprinkler, rain, and wet grass – the sprinkler and rain are independent, but both give wet grass. If the wet grass is unknown, then we can use the independence of the sprinkler and rain. If, however, wet grass is known, then that can’t be used. Regardless of the value that is known – true or false. For the other two relations between 3 variables, (that remain trees), the opposite is true – given information on the ‘third’ node, the first two become independent. In the above, given information, they became dependent. The generalization is known as de-separation: Given Vi and Vj (nodes of a Bayesian network) The question is: Are the two nodes conditionally independent given a set of nodes E? If, for all undirected paths between Vi and Vj There is a node X such that:

Also, P11 is independent of P14:

P(Q) P(P11|Q) P(P14|Q) P(P12,P13|Q) % P(P11,P12,P13,P14) No more independence, so insert parents. = P(Q) P(P11|Q) P(P14|Q) sum over P9 with P(P12,P13,P9|Q) % P(P11,P12,P13,P14) Just considering the summation P(P12,P13,P9|Q) = P(P12,P13|P9,Q) P(P9|Q) = P(P12,P13|P9) P(P9|Q) Going back... considering how to evaluate P(Q) = sum over {P6, P7} of P(Q,P6,P7) If we are evaluating one term of the sum, then we can write: P(Q|P6,P7) P(P6,P7) Where we can repeat the process to evaluate P(P6,P7) – P(Q|P6,P7) can be evaluated from the table. Also, parents introduced in this fashion are always independent, because of their child, so the above can be simplified to P(P6) P(P7) Which is one step closer to the beginning of the tree – so recursion will work.

  1. Going back to the bayes net programming idea: Some people didn’t do well on the test, so doing the program is worth an extra 20 points. Definitely don’t copy from any source on the internet: Prof. Baral expressed, repetitively, an admonition against it – besides the obvious academic integrity policy that is university wide.
  2. Generalized procedure: P(Q|E-, E+) Where the first is a set of evidence below, and the second is a set of evidence above. = P(E-|Q, E+) P(Q|E+)

P(E-|E+)

So, all remaining are in evidence above form – which can be evaluated by recursively introducing parents. Lastly, some comment about the homework being due next week – Wednesday, I think. Should be on the web page.