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POImprovement of the Greedy Algorithm
 

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    <p>I've been working on an abstract chess algorithm using Haskell (trying to expand my understanding of different paradigms), and I've hit a challenge that I've been pondering about for weeks.</p> <p>Here's the problem:</p> <blockquote> <p>Given a board (represented by a list of lists of integers; each integer represents a subsequent point value), with dimensions n x n, determine the path that provides the most points. If there is a tie for best path, return either of them.</p> </blockquote> <p>Here are the specifics:</p> <pre><code>A = [[5,4,3,1],[10,2,1,0],[0,1,2,0],[2,3,4,20]] </code></pre> <p>which renders as: </p> <pre><code>R1: 5 4 3 1, R2: 10 2 1 0, R3: 0 1 2 0, R4: 2 3 4 20. </code></pre> <p>The rules are: </p> <ol> <li><p>You may start anywhere on the top row </p></li> <li><p>You may move one square at a time, either straight down, down-left (diagonal) , or down-right (diagonal). </p></li> <li><p>The output must be a tuple of integers. </p></li> </ol> <p>First element is a list representing the columns vs. row, and the second element is the total number of points. Eg. for the above board, the best solution is to travel from top-left (5) and go diagonally for the remaining steps (until the 20 point square). This would result in the tuple <code>([1,2,3,4], 29)</code>. </p> <p>Remember, this is all in Haskell so it is a functional-paradigm recursive problem. At first, I was thinking about using the greedy algorithm, that is, choosing the highest value in r1, and recursing through comparing the next 3 possibilities; choosing the highest of the 3. However, the downfall is that the greedy algorithm doesn't have the ability to see potential ahead of the next row. </p> <p>How would I go about this? I'm not looking for code per se, since I enjoy solving things on my own. However, pseudocode or some algorithmic guidance would be much appreciated!</p>
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    1. COCould this maybe be done using an adaptation of Dijkstras algorithm? The numbers are nodes of the graph, the value of each node will be the weight of each edge pointing to it, and instead of comparing against shorter paths, we could compare against longer paths.
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    2. COApparently that intuition is false for the general case, since the longest path problem is NP-hard in general graphs (see for example http://stackoverflow.com/questions/10462736/graph-dijkstra-for-the-single-source-longest-path). However (!), your problem would be equivalent to find the longest path in a directed acyclic(!) graph, both the longest and the shortest path can be calculated by finding the longest/shortest path in the topological order of the nodes.
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    3. COIs this a duplicate of http://stackoverflow.com/questions/14666835/recursion-using-lists-haskell ?
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