Question

So I have to solve the knapsack problem for class. So far, I’ve come up with the following. My comparators are functions that determine which of two subjects will be the better option (by looking at the corresponding (value,work) tuples).

I decided to iterate over the possible subjects with work less than maxWork, and in order to find which subject is the best option at any given turn, I compared my most recent subject to all the other subjects that we have not used yet.

def greedyAdvisor(subjects, maxWork, comparator):
    """
    Returns a dictionary mapping subject name to (value, work) which includes
    subjects selected by the algorithm, such that the total work of subjects in
    the dictionary is not greater than maxWork.  The subjects are chosen using
    a greedy algorithm.  The subjects dictionary should not be mutated.

    subjects: dictionary mapping subject name to (value, work)
    maxWork: int >= 0
    comparator: function taking two tuples and returning a bool
    returns: dictionary mapping subject name to (value, work)
    """

    optimal = {}
    while maxWork > 0:
        new_subjects = dict((k,v) for k,v in subjects.items() if v[1] < maxWork)
        key_list = new_subjects.keys()
        for name in new_subjects:
            #create a truncated dictionary
            new_subjects = dict((name, new_subjects.get(name)) for name in key_list)
            key_list.remove(name)
            #compare over the entire dictionary
            if reduce(comparator,new_subjects.values())==True:
                #insert this name into the optimal dictionary
                optimal[name] = new_subjects[name]
                #update maxWork
                maxWork = maxWork - subjects[name][1]
                #and restart the while loop with maxWork updated
                break
    return optimal  

The problem is I don’t know why this is wrong. I’m getting errors and I have no idea where my code is wrong (even after throwing in print statements). Help would be much appreciated, thanks!

Answer 1

Using a simple greedy algorithm will not provide any bounds on the quality of the solution in comparison to OPT.

Here is a fully polynomial time (1 – epsilon) * OPT approximation psuedocode for knapsack:

items = [...]  # items
profit = {...} # this needs to be the profit for each item
sizes = {...}  # this needs to be the sizes of each item
epsilon = 0.1  # you can adjust this to be arbitrarily small
P = max(items) # maximum profit of the list of items
K = (epsilon * P) / float(len(items))
for item in items:
    profit[item] = math.floor(profit[item] / K)
return _most_prof_set(items, sizes, profit, P)

We need to define the most profitable set algorithm now. We can do this with some dynamic programming. But first lets go over some definitions.

If P is the most profitable item in the set, and n is the amount of items we have, then nP is clearly a trivial upper bound on the profit allowed. For each i in {1,…,n} and p in {1,…,nP} we let Sip denote a subset of items whose total profit is exactly p and whose total size is minimized. We then let A(i,p) denote the size of set Sip (infinity if it doesn’t exist). We can easily show that A(1,p) is known for all values of p in {1,…,nP}. We will define a recurrance to compute A(i,p) which we will use as a dynamic programming problem, to return the approximate solution.

A(i + 1, p) = min {A(i,p), size(item at i + 1 position) + A(i, p - profit(item at i + 1 position))} if profit(item at i + 1) < p otherwise A(i,p)

Finally we give _most_prof_set

def _most_prof_set(items, sizes, profit, P):
    A = {...}
    for i in range(len(items) - 1):
        item = items[i+1]
        oitem = items[i]
        for p in [P * k for k in range(1,i+1)]:
            if profit[item] < p:
                A[(item,p)] = min([A[(oitem,p)], \
                                     sizes[item] + A[(item, p - profit[item])]])
            else:
                A[(item,p)] = A[(oitem,p)] if (oitem,p) in A else sys.maxint
    return max(A) 

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