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Arbitrary Adventures Analysis (Posted on 2006-11-20) Difficulty: 3 of 5
In the "Your own adventure" books, the reader starts at page 1, and every page either (1) ends the story, or (2) sends him to another page, or (3) offers a choice among two possible pages.

Knowing that:

  • each page can be reached from only one other page -- except for the 1st page, that cannot be reached from any page;
  • that all pages can eventually be reached by picking an appropriate path from page 1;
  • that if a book can be converted into another just by reordering choices and renumbering pages, they are considered to be the same;
  • and that these books are always 100 pages long...
  • How many essentially different books can be published?

    No Solution Yet Submitted by Old Original Oskar!    
    Rating: 4.0000 (1 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Some Thoughts Program needed . . . | Comment 5 of 9 |

    I arrived at the following number of different books for 1 through 12 pages:

    1, 1, 2, 3, 6, 11, 24, 47, 103, 214, 481, 1030

    As given in previous posts f(1) = 1, f(2) = 1, f(3) = 2, f(4) = 3

    When adding another page, add it at the begining.  This new first page can send the reader to the first page of a book one page shorter, or to the first pages of two books whose total is one page shorter.

    Thus f(n) = f(n-1) + f(n-2)*f(1) + f(n-3)*f(2) + . . . f(n-m)*f(m-1) while (n-m) >= (m-1)

    My calculations were getting cumbersome, and I ask those who program to give the calculation a go.

    Edited on November 20, 2006, 1:48 pm
      Posted by Leming on 2006-11-20 13:44:06

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