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Questions to "The Wumpus World Revisited"

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  • 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
    Message 1 of 3 , Aug 1, 2007
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      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
    • Lalit Gidwani
      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
      Message 2 of 3 , Sep 23, 2007
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        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@..." <spam@...> wrote:
        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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      • Bob Futrelle CCIS
        As human cognition is studied more and more, we re steadily learning just how far human reasoning is from the logic that some of us know and love. I suspect
        Message 3 of 3 , Sep 23, 2007
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          As human cognition is studied more and more, we're steadily learning 
          just 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 model
          approaches to intelligence.

          A lot of this has to do with the fact that humans and machines are not and cannot
          be reasonably endowed with enough concise knowledge for hard-core logic
          to function.  The first few pages of Chap. 13 (Uncertainty) in AIMA make this point.

           - Bob Futrelle

          Robert P. Futrelle
            Associate Professor
          Biological Knowledge Laboratory
          College of Computer and Information Science
          Northeastern University MS WVH202
          360 Huntington Ave.
          Boston, MA 02115

          Office: 617-373-4239
          Fax:     617-373-5121


          On 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. de" <spam@b-event. de> wrote:

          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




          Looking for a deal? Find great prices on flights and hotels with Yahoo! FareChase.


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