Hockey pool Algorithm

Question

This is a little fun project that I have started to try and maximize my chances of winning our office hockey pool. I'm trying to find the best way to select 20 players that will give me the most points within a maximum salary cap.

For example, imagine that the raw data is made of

1. The Player Name
2. Position (Forward, Defensemen, Goalie)
3. Predicted amount of points for this season
4. Salary.

Now I want the 20 players that will give me the most point within X salary cap. Later on, as a phase 2, I would like to do the same thing however in this case, I only want 12 forwards, 6 defensemen and 2 goalie.

Now the obvious way is to simply go every possible combination, however while this will work it is not a valid option as with 500 players, this would have too many possible combination. I could add some smart filters to reduce the 500 players to the top 50 forwards, top 30 defensemen and top 15 goalies but still, this would be a very slow process.

I'm wondering if there are other algorithms out there to implement this. This is just for fun and not an important business request. But if you have some thoughts on how to proceed, please let me know.

My first try was using the knapsack algorith with some help from other sources. It seems to work with just the Salary as a parameter. I'm struggling at figuring out how to add the 20 player team parameter. Its in .Net but should be easy to convert to Java.

I was thinking of doing a seperate loop to figure out the best teams with 20 players reguardless of salary and then compare the two list until I find the team that is highest on the two list. Not sure.

namespace HockeyPoolCalculator
{
public class ZeroOneKnapsack
{

protected List<Item> itemList = new List<Item>();
protected int maxSalary = 0;
protected int teamSize = 0;
protected int teamSalary = 0;
protected int points = 0;
protected bool calculated = false;

public ZeroOneKnapsack() { }

public ZeroOneKnapsack(int _maxSalary)
{
setMaxSalary(_maxSalary);
}

public ZeroOneKnapsack(List<Item> _itemList)
{
setItemList(_itemList);
}

public ZeroOneKnapsack(List<Item> _itemList, int _maxSalary)
{
setItemList(_itemList);
setMaxSalary(_maxSalary);
}

// calculte the solution of 0-1 knapsack problem with dynamic method:
public virtual List<Item> calcSolution()
{
int n = itemList.Count;

setInitialStateForCalculation();
if (n > 0 && maxSalary > 0)
{
List<List<int>> playerList = new List<List<int>>();
List<int> salaryList = new List<int>();

//initialise list
playerList.Add(salaryList);
for (int j = 0; j <= maxSalary; j++)
salaryList.Add(0);
// Loop through players
for (int i = 1; i <= n; i++)
{
List<int> prev = salaryList;
playerList.Add(salaryList = new List<int>());
for (int j = 0; j <= maxSalary; j++)
{
if (j > 0)
{
int wH = itemList.ElementAt(i - 1).getSalary();
// Is the players salary more than the current calculated salary? If yes, then keep current max points, else get the highest amount between previous max points at that salary and new max points.
salaryList.Add((wH > j)?prev.ElementAt(j): Math.Max(prev.ElementAt(j),itemList.ElementAt(i - 1).getPoints() + prev.ElementAt(j - wH)));
}
else
{
salaryList.Add(0);
}
} // for (j...)
} // for (i...)
points = salaryList.ElementAt(maxSalary);

for (int i = n, j = maxSalary; i > 0 && j >= 0; i--)
{
int tempI = playerList.ElementAt(i).ElementAt(j);
int tempI_1 = playerList.ElementAt(i - 1).ElementAt(j);
if ((i == 0 && tempI > 0)||(i > 0 && tempI != tempI_1))
{
Item iH = itemList.ElementAt(i - 1);
int wH = iH.getSalary();
iH.setInKnapsack(1);
j -= wH;
teamSalary += wH;
}
} // for()
calculated = true;
} // if()
return itemList;
}

// add an item to the item list
public void add(String name, int Salary, int value)
{
if (name.Equals(""))
name = "" + (itemList.Count() + 1);
itemList.Add(new Item(name, Salary, value));
setInitialStateForCalculation();
}

// add an item to the item list
public void add(int Salary, int value)
{
add("", Salary, value); // the name will be "itemList.size() + 1"!
}

// remove an item from the item list
public void remove(String name)
{
for (int pointer = 0; pointer <= itemList.Count-1; pointer++)

{
itemList[pointer].getName().Equals("");

if (itemList.ElementAt(pointer).getName().Equals(itemList.ElementAt(pointer).getName()))
{
itemList.Remove(itemList.ElementAt(pointer));
}
}
setInitialStateForCalculation();
}

// remove all items from the item list
public void removeAllItems()
{
itemList.Clear();
setInitialStateForCalculation();
}

public int getPoints()
{
if (!calculated)
calcSolution();
return points;
}

public int getSolutionSalary() { return teamSalary; }
public bool isCalculated() { return calculated; }
public int getMaxSalary() { return maxSalary; }

public void setTeamSize(int _teamSize)
{
teamSize = _teamSize;
}

public int getTeamSize()
{
return teamSize;
}

public void setMaxSalary(int _maxSalary)
{
maxSalary = Math.Max(_maxSalary, 0);
}

public void setItemList(List<Item> _itemList) {
if (_itemList != null) {
itemList = _itemList;
foreach (Item item in _itemList) {
item.checkMembers();
}
}
}

// set the member with name "inKnapsack" by all items:
private void setInKnapsackByAll(int inKnapsack) {
foreach (Item item in itemList)
if (inKnapsack > 0)
item.setInKnapsack(1);
else
item.setInKnapsack(0);
}

// set the data members of class in the state of starting the calculation:
protected void setInitialStateForCalculation()
{
setInKnapsackByAll(0);
calculated = false;
points = 0;
teamSalary = 0;
teamSize = 0;
}

} 
}

Thanks for your help!

Answer 1

Unfortunately, you shouldn't expect to find a good solution to this problem as it is NP-hard. Unless P = NP, there aren't any polynomial-time algorithms, and exhaustive search is likely to be one of the best algorithms (though you might use some heuristics to speed it up).

To see that this problem is NP-hard, we'll show how to reduce the knapsack problem to it in polynomial time. Given any instance of the subset sum problem consisting of a set S = {(weight₁, value₁), (weight₂, value₂), ... , (weight_n, value_n)} and weight limit k, we can construct an instance of your hockey problem by creating a set of players whose salary is the weight and whose expected points is the value. We then try to find the maximum-weight combination of players whose salary doesn't exceed k, which is then identical to the maximum sum you can make in the original knapsack problem that doesn't exceed the target weight.

As the code you've posted shows, though, there is a pseudopolynomial time algorithm for solving the knapsack problem. Assuming salaries are low (or that you can normalize them to small numbers), you might be able to use this to good effect.

While it's doubtful that there is a polynomial-time algorithm to get the exact answer, if you're okay with an approximately optimal solution, there is a polynomial-time algorithm for approximating solutions to the knapsack problem. For details, check out these notes, which detail two algorithms. Interestingly, they rely on the pseudopolynomial time algorithm that you seem to be using, so perhaps they'll be easy to implement?

Sorry to dash your hopes for a nice solution with math... NP-hard problems tend to do that. :-(