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POKnapsack: how to add item type to existing solution
 

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  1. POKnapsack: how to add item type to existing solution
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    <p>I've been working with this variation of dynamic programming to solve a knapsack problem:</p> <pre><code>KnapsackItem = Struct.new(:name, :cost, :value) KnapsackProblem = Struct.new(:items, :max_cost) def dynamic_programming_knapsack(problem) num_items = problem.items.size items = problem.items max_cost = problem.max_cost cost_matrix = zeros(num_items, max_cost+1) num_items.times do |i| (max_cost + 1).times do |j| if(items[i].cost &gt; j) cost_matrix[i][j] = cost_matrix[i-1][j] else cost_matrix[i][j] = [cost_matrix[i-1][j], items[i].value + cost_matrix[i-1][j-items[i].cost]].max end end end cost_matrix end def get_used_items(problem, cost_matrix) i = cost_matrix.size - 1 currentCost = cost_matrix[0].size - 1 marked = Array.new(cost_matrix.size, 0) while(i &gt;= 0 &amp;&amp; currentCost &gt;= 0) if(i == 0 &amp;&amp; cost_matrix[i][currentCost] &gt; 0 ) || (cost_matrix[i][currentCost] != cost_matrix[i-1][currentCost]) marked[i] = 1 currentCost -= problem.items[i].cost end i -= 1 end marked end </code></pre> <p>This has worked great for the structure above where you simply provide a name, cost and value. Items can be created like the following:</p> <pre><code> items = [ KnapsackItem.new('david lee', 8000, 30) , KnapsackItem.new('kevin love', 12000, 50), KnapsackItem.new('kemba walker', 7300, 10), KnapsackItem.new('jrue holiday', 12300, 30), KnapsackItem.new('stephen curry', 10300, 80), KnapsackItem.new('lebron james', 5300, 90), KnapsackItem.new('kevin durant', 2300, 30), KnapsackItem.new('russell westbrook', 9300, 30), KnapsackItem.new('kevin martin', 8300, 15), KnapsackItem.new('steve nash', 4300, 15), KnapsackItem.new('kyle lowry', 6300, 20), KnapsackItem.new('monta ellis', 8300, 30), KnapsackItem.new('dirk nowitzki', 7300, 25), KnapsackItem.new('david lee', 9500, 35), KnapsackItem.new('klay thompson', 6800, 28) ] problem = KnapsackProblem.new(items, 65000) </code></pre> <p>Now, the problem I'm having is that I need to add a position for each of these players and I have to let the knapsack algorithm know that it still needs to maximize value across all players, except there is a new restriction and that restriction is each player has a position and each position can only be selected a certain amount of times. Some positions can be selected twice, others once. Items would ideally become this:</p> <pre><code>KnapsackItem = Struct.new(:name, :cost, :position, :value) </code></pre> <p>Positions would have a restriction such as the following:</p> <pre><code>PositionLimits = Struct.new(:position, :max) </code></pre> <p>Limits would be instantiated perhaps like the following:</p> <pre><code>limits = [Struct.new('PG', 2), Struct.new('C', 1), Struct.new('SF', 2), Struct.new('PF', 2), Struct.new('Util', 2)] </code></pre> <p>What makes this a little more tricky is every player can be in the Util position. If we want to disable the Util position, we will just set the 2 to 0. </p> <p>Our original items array would look something like the following:</p> <pre><code>items = [ KnapsackItem.new('david lee', 'PF', 8000, 30) , KnapsackItem.new('kevin love', 'C', 12000, 50), KnapsackItem.new('kemba walker', 'PG', 7300, 10), ... etc ... ] </code></pre> <p>How can position restrictions be added to the knapsack algorithm in order to still retain max value for the provided player pool provided?</p>
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    1. COIf you are not too much specific on ruby implementation, then it will be simpler to understand if you describe your problem in simple non-code language.
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    2. COI could use some help with getting this implemented in Ruby, so I did add a bounty to the question.
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    3. COUnfortunately it seems to incorporate this additional constraint, you may have to either go through a bit of brute-force or Integer Linear Programming solving. Are you open for these or you specifically want a dynamic programming solution?
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