- As human cognition is studied more and more, we're steadily learningjust how far human reasoning is from the "logic" that some of us know and love.I suspect that in the future, AI will depend more and more on Bayesian and mental modelapproaches to intelligence.A lot of this has to do with the fact that humans and machines are not and cannotbe reasonably endowed with enough concise knowledge for hard-core logicto function. The first few pages of Chap. 13 (Uncertainty) in AIMA make this point.- Bob FutrelleRobert P. FutrelleAssociate ProfessorBiological Knowledge LaboratoryCollege of Computer and Information ScienceNortheastern University MS WVH202360 Huntington Ave.Boston, MA 02115Office: 617-373-4239Fax: 617-373-5121On Sep 23, 2007, at 8:09 AM, Lalit Gidwani wrote:
Your e-mail is one of the best e-mails I have seen on this site. Unfortunately,

I cannot help you. As and when I get the time I will study your email; it has been on my mind for a long time.

However, I have a question. I hope you can answer it. Why is it that the summation of P(other) over 'other' equal to 1. I think I am missing something very simple here.

Finally, I wonder if you have looked at the chapters on logic. Given the importanc e of logic I am surprised that there are very few good books on the use of logic for computer science. I could not trace very many on the web. I have found some instructors' s website that seem to be promising. I will provide you with that information in case you want it.

Sincerely,

LG.*"spam@b-event.*wrote:*de" <spam@b-event. de>*Hi,

I have some questions regarding chapter 13.7 "The Wumpus World

Revisited" and the derivation of the distribution of P1,3 given known

and breeze evidence in particular:

1.) The probability that a particular cell contains a pit is 0.2 = 3/15,

so that pits are distributed over every cell except for a starting cell,

here 1,1. Now, when calculating the probability of a configuration

containing n pits in Equation 13.15 the product of each cell's

probability distribution is used because of the absolute independence

relationships. The resulting term is 0.2^n x 0.8^(16-n), although the

starting cell is not allowed to contain a pit, shouldn't it be 0.2^n x

0.8^(15-n) x 1 for configurations with no pit in the starting cell and 0

for configurations with a pit in the starting cell because of a

distribution of <0,1> for the starting cell to contain a pit?

2.) P(fringe) is the prior probability of a fringe configuration.

However, when showing the five possible models of the fringe in Figure

13.7 those three possibilities that do not hold the evidence of

perceived breezes in 1,2 and 2,1 are not shown. Does this fact mean that

P(fringe) is in any kind conditioned on the evidence or is it still the

prior distribution with those three combinations not shown?

3.) Later, P(known) is taken out of the summation and put into the

normalizing constant, the sum over the possible configurations of other

is said to be 1. But, that the sum over other of P(other) is 1 is

unclear to me. Of course, having 12 cells in unknown I have 2^12 = 4096

configurations of pits in these cells and a probability of 1/(2^12) to

get one particular configuration. Summing over 2^12 terms results of

course in 1. Same for fringe with 2 cells (2^2 configurations with a

probability of 1/(2^2) each) and other with 10 cells (2^10

configurations with a probability of 1/(2^10) each). But how is this

related to the prior distribution of <0.2,0.8> of each cell and the

product in Eq. 13.15 for the prior of one configuration?

Using combinatorics I have n!/( k!(n-k)! ) combinations to distribute k

holes over n cells, e.g. for distributing 3 pits over 15 cells

15!/(3!x12!) = 455 combinations.

Considering the 5 possible models in Figure 13.7, we have three cases

for other:

3 pits in the fringe: no pit in other - 10!/(0!x10!) * 0.2^0 * 0.8^10

2 pits in the fringe: one pit in other - 10!/(1!x9!) * 0.2^1 * 0.8^9

1 pit in the fringe: two pits in other - 10!/(2!x8!) * 0.2^2 * 0.8^8

However, these values do, of course, not sum up to 1. Even adding the

case, that is not shown in Figure 13.7 with three pits in other does not

correct that, neither does the summing over all 8 possibilities of pits

in the fringe. Actually the above equations do only fulfill our goal of

having the sum of other equal to 1, if we extend those three cases to

distribute every possible number of holes (with k ranging from 0 to 10

and n being 10) over the cells in other. Am I missing something out

here, e.g. a normalization, or am I on a completely misleading way?

So the fundamental question is:

"Is sum_other P(other) = 1?", "Why is it one?"

and, furthermore, is the result of the summation relevant at all or

could it just be put in the normalizing constant just as we did with

P(known)?

Despite my lousy explanations, I hope that my concerns are

understandable and I hope to receive some comprehensible answers.

Best regards

Dirk

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