user2758113 - 1 year ago 154

Python Question

Lets say I have a list of tuples representing basketball players and their name, position, cost, and their projected points,

`listOfPlayers = [`

("Player1","PG",Cost,projectedPoints),

("Player2","PG",Cost,projectedPoints),

("Player3","SG",Cost,projectedPoints),

("Player4","SG",Cost,projectedPoints),

("Player5","SF",Cost,projectedPoints),

("Player6","SF",Cost,projectedPoints),

("Player7","PF",Cost,projectedPoints),

("Player8","PF",Cost,projectedPoints),

("Player9","C",Cost,projectedPoints),

("Player10","C",Cost,projectedPoints)

]

Assume all of the names, costs, and projected points are variable.

I have a traditional knapsack problem working, they can sort and pack a knapsack based on a given weight. But this does not account for the positions.

I was wondering if there is a way to edit the knapsack code to only include one of every position, i.e., (pg, sg, sf, pf, c).

Can a traditional 0/1 knapsack do this or do i need to switch to something else?

Answer Source

This is called the "multiple-choice knapsack problem".

You can use an algorithm similar to the dynamic programming solution for the 0/1 knapsack problem.

The 0/1 knapsack problem's solution is as follows: (from Wikipedia)

Define

`m[i, w]`

to be the maximum value that can be attained with weight less than or equal to`w`

using items up to`i`

.

We can define`m[i, w]`

recursively as follows:`m[i, w] = m[i-1, w] if w_i > w (new item is more than current weight limit) m[i, w] = max(m[i-1, w], m[i-1, w-w_i] + v_i) if w_i <= w.`

The solution can then be found by calculating

`m[n,W]`

. To do this efficiently we can use a table to store previous computations.

Now the extension is just to find the maximum of all choices instead.

For `n`

players available as choices for some position `i`

(with `c_i_j`

being the cost of choice `j`

and `p_i_j`

being the points), we'd have:

```
m[i, c] = max(m[i-1, c],
m[i-1, c-c_i_1] + p_i_1 if c_i_1 <= c, otherwise 0,
m[i-1, c-c_i_2] + p_i_2 if c_i_2 <= c, otherwise 0,
...
m[i-1, c-c_i_n] + p_i_n if c_i_n <= c, otherwise 0)
```

So, say we have:

```
Name Position Cost Points
Player1 PG 15 5
Player2 PG 20 10
Player3 SG 9 7
Player4 SG 8 6
```

Then we'd have 2 positions "PG" and "SG" and each position will have 2 choices.

Thus, for position "PG" (at `i=1`

), we'll have:

```
m[i, c] = max(m[i-1, c],
m[i-1, c-15] + 5 if 15 <= c, otherwise 0,
m[i-1, c-20] + 10 if 20 <= c, otherwise 0)
```

And for position "SG" (at `i=2`

), we'll have:

```
m[i, c] = max(m[i-1, c],
m[i-1, c-9] + 7 if 9 <= c, otherwise 0,
m[i-1, c-8] + 6 if 8 <= c, otherwise 0)
```