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769Questions to "The Wumpus World Revisited"

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  • spam@b-event.de
    Aug 1, 2007

      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

      Despite my lousy explanations, I hope that my concerns are
      understandable and I hope to receive some comprehensible answers.

      Best regards
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