Optimal MLB lineup using Knapsack variant

Question

I am writing a program to find the best possible MLB lineup using a knapsack solution. For this I pass in player data which has a players calculated value and salary. The salary will be my "weight" in terms of being a knapsack problem.

My problem is not be able to select players, but rather select the most optimal lineup. I am choosing a pitcher, a center, first baseman, second baseman, third baseman, short stop, and three outfielders. I can do this all successfully. I want my "weight" to be 36,000, but I am currently only choosing a lineup with a total of 21,000.

Here is my knapsack code:

CalculateLineUp.prototype.findOptimalLineUp = function(data, capacity) {
  var items = data.data;
  var idxItem   = 0,
      idxCapSpace = 0,
      idxPosition = 0,
      oldMax    = 0,
      newMax    = 0,
      numItems  = items.length,
      weightMatrix  = new Array(numItems+1),
      keepMatrix    = new Array(numItems+1),
      positionArray = new Array("P", "C", "1B", "2B", "3B", "SS", "OF", "OF", "OF"),
      solutionSet   = [];

  // Setup matrices
  for(idxItem = 0; idxItem < numItems + 1; idxItem++){
    weightMatrix[idxItem] = new Array(capacity+1);
    keepMatrix[idxItem]   = new Array(capacity+1);
  }

  // Build weightMatrix from [0][0] -> [numItems-1][capacity-1]
  for (idxItem = 0; idxItem <= numItems; idxItem++){
    for (idxCapSpace = 0; idxCapSpace <= capacity; idxCapSpace++){  

          // Fill top row and left column with zeros
          if (idxItem === 0 || idxCapSpace === 0){
            weightMatrix[idxItem][idxCapSpace] = 0;
          }

          // If item will fit, decide if there's greater value in keeping it,
          // or leaving it
          else if (items[idxItem-1]["Salary"] <= idxCapSpace){
            newMax = items[idxItem-1]["Value"] + weightMatrix[idxItem-1][idxCapSpace-items[idxItem-1]["Salary"]];
            oldMax = weightMatrix[idxItem-1][idxCapSpace];

            // Update the matrices
            if(newMax > oldMax ){ 
              weightMatrix[idxItem][idxCapSpace]  = newMax;
              keepMatrix[idxItem][idxCapSpace]    = 1;

            }
            else{
              weightMatrix[idxItem][idxCapSpace]  = oldMax; 
              keepMatrix[idxItem][idxCapSpace]    = 0;
            }
          }

          //Else, item can't fit; value and weight are the same as before
           //else
             //weightMatrix[idxItem][idxCapSpace] = weightMatrix[idxItem-1][idxCapSpace];
    }
  }

  // Traverse through keepMatrix ([numItems][capacity] -> [1][?])
  // to create solutionSet
  idxCapSpace = capacity;
  idxItem   = numItems;
  for(idxItem; idxItem < capacity; idxItem--){
    if(keepMatrix[idxItem][idxCapSpace] === 1 && !this.filledAllPositions(items[idxItem - 1]["Position"])){
      solutionSet.push(items[idxItem - 1]);
      idxCapSpace = idxCapSpace - items[idxItem - 1]["Salary"];
    }
  }
  return {"maxValue": weightMatrix[numItems][capacity], "set": solutionSet};
};

Am I making a blatant mistake that I am just not seeing, or is my logic completely off?

Can you share some sample data? And also a fiddle perhaps? – Bhargav Ponnapalli Oct 09 '15 at 09:15 — Bhargav Ponnapalli, Oct 09 '15 at 09:15

score 1 · Accepted Answer · answered Oct 10 '15 at 08:20

1

You are checking the solutionSet right? The logic to accept positions is not a included in the knapsack logic, which means that the solutionSet is a filter on top of the knapsack solution. You did arrive at the right knapsack solution, but because, on top of the solution, you are checking if the position was already filled, a few items were eliminated from the solutionSet (items which were fighting for the same position) and the total weight decreased.

answered Oct 10 '15 at 08:20

Bhargav Ponnapalli

9,224
7
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Thanks, I will check this in a few hours when I get home and see if the solution set before the filter has more optimal values. This would make sense since I am traversing backwards through the keepMatrix and I expect the end of the matrix to hold the lowest weight based on how the data is sorted. – urnotsam Oct 12 '15 at 13:38
That seems to be the path to take for the solution so I will accept your answer. thanks – urnotsam Oct 13 '15 at 12:57

Optimal MLB lineup using Knapsack variant

1 Answers1