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CHAPTER 9 MAGIC METHODS, PROPERTIES, AND ITERATORS
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If you have never heard of graphs and trees before, you should learn about them as soon as possible, because they are very important concepts in programming and computer science. To find out more, you should probably get a book about computer science, discrete mathematics, data structures, or algorithms. For some concise definitions, you can check out the following web pages: http://mathworld.wolfram.com/Graph.html http://mathworld.wolfram.com/Tree.html http://www.nist.gov/dads/HTML/tree.html http://www.nist.gov/dads/HTML/graph.html A quick web search or some browsing in Wikipedia (http://wikipedia.org) will turn up a lot of material.
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This strategy of backtracking is useful for solving problems that require you to try every combination until you find a solution. Such problems are solved like this: # Pseudocode for each possibility at level 1: for each possibility at level 2: ... for each possibility at level n: is it viable To implement this directly with for loops, you need to know how many levels you ll encounter. If that is not possible, you use recursion.
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The Problem
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This is a much loved computer science puzzle: you have a chessboard and eight queen pieces to place on it. The only requirement is that none of the queens threatens any of the others; that is, you must place them so that no two queens can capture each other. How do you do this Where should the queens be placed This is a typical backtracking problem: you try one position for the first queen (in the first row), advance to the second, and so on. If you find that you are unable to place a queen, you backtrack to the previous one and try another position. Finally, you either exhaust all possibilities or find a solution.
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CHAPTER 9 MAGIC METHODS, PROPERTIES, AND ITERATORS
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In the problem as stated, you are provided with information that there will be only eight queens, but let s assume that there can be any number of queens. (This is more similar to realworld backtracking problems.) How do you solve that If you want to try to solve it yourself, you should stop reading now, because I m about to give you the solution.
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Note You can find much more efficient solutions for this problem. If you want more details, a web search should turn up a wealth of information. A brief history of various solutions may be found at http:// www.cit.gu.edu.au/~sosic/nqueens.html.
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State Representation
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To represent a possible solution (or part of it), you can simply use a tuple (or a list, for that matter). Each element of the tuple indicates the position (that is, column) of the queen of the corresponding row. So if state[0] == 3, you know that the queen in row one is positioned in column four (we are counting from zero, remember ). When working at one level of recursion (one specific row), you know only which positions the queens above have, so you may have a state tuple whose length is less than eight (or whatever the number of queens is).
Note I could well have used a list instead of a tuple to represent the state. It s mostly a matter of taste in
this case. In general, if the sequence is small and static, tuples may be a good choice.
Finding Conflicts
Let s start by doing some simple abstraction. To find a configuration in which there are no conflicts (where no queen may capture another), you first must define what a conflict is. And why not define it as a function while you re at it The conflict function is given the positions of the queens so far (in the form of a state tuple) and determines if a position for the next queen generates any new conflicts: def conflict(state, nextX): nextY = len(state) for i in range(nextY): if abs(state[i]-nextX) in (0, nextY-i): return True return False The nextX parameter is the suggested horizontal position (x coordinate, or column) of the next queen, and nextY is the vertical position (y coordinate, or row) of the next queen. This function does a simple check for each of the previous queens. If the next queen has the same x coordinate, or is on the same diagonal as (nextX, nextY), a conflict has occurred, and True is returned. If no such conflicts arise, False is returned. The tricky part is the following expression: abs(state[i]-nextX) in (0, nextY-i)