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5020631
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2011-02-16T18:36:02.463
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The problem is NP-complete for the general case. This can be shown via a reduction of 3-partition (which is a still strong NP-complete version of bin packing). Let w1, ..., wn be the weights of objects of the 3-partition instance, let b be the bin size, and k = n/3 the number of bins that are allowed to be filled. Hence, there is a 3-partition if objects can be partitioned such that there are exactly 3 objects per bin. For the reduction, we set N=kb and each bin is represented by b price labels of the same price (think of Pi increasing every bth label). Let ti, 1≤i≤k, be the price of the labels corresponding to the ith bin. For each wi we have one product Sj of quantity wi + 1 (lets call this the root product of wi) and another wi - 1 products of quantity 1 which are required to be cheaper than Sj (call these the leave products). For ti = (2b + 1)i, 1≤i≤k, there is a 3-partition if and only if Bob can sell for 2bΣ1≤i≤k ti:<ul> <li>If there is a solution for 3-partition, then all the b products corresponding to objects wi, wj, wl that are assigned to the same bin can be labeled with the same price without violating the restrictions. Thus, the solution has cost 2bΣ1≤i≤k ti (since the total quantity of products with price ti is 2b). <li>Consider an optimal solution of Bob's Sale. First observe that in any solution were more than 3 root products share the same price label, for each such root product that is "too much" there is a cheaper price tag which sticks on less than 3 root products. This is worse than any solution were there are exactly 3 root products per price label (if existent). Now there can still be a solution of Bob's Sale with 3 root labels per price, but their leave products do not wear the same price labels (the bins sort of flow over). Say the most expensive price label tags a root product of wi which has a cheaper tagged leave product. This implies that the 3 root labels wi, wj, wl tagged with the most expensive price do not add up to b. Hence, the total cost of products tagged with this price is at least 2b+1. Hence, such a solution has cost tk(2b+1) + some other assignment cost. Since the optimal cost for an existent 3-partition is 2bΣ1≤i≤k ti , we have to show that the just considered case is worse. This is the case if tk > 2b Σ1≤i≤k-1 ti (note that it's k-1 in the sum now). Setting ti = (2b + 1)i, 1≤i≤k, this is the case. This also holds if not the most expensive price tag is the "bad" one, but any other. </ul> So, this is the destructive part ;-) However, if the number of different price tags is a constant, you can use dynamic programming to solve it in polynomial time.
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1. POProblem: Bob's Sale
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1. USGero Greiner
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