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During a robbery, a burglar nds much more loot than he had expected and has to decide what to take His bag (or knapsack ) will hold a total weight of at most W pounds There are n items to pick from, of weight w1 , , wn and dollar value v1 , , vn What s the most valuable combination of items he can t into his bag 1 For instance, take W = 10 and Item 1 2 3 4 Weight 6 3 4 2 Value $30 $14 $16 $9
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There are two versions of this problem If there are unlimited quantities of each item available, the optimal choice is to pick item 1 and two of item 4 (total: $48) On the other hand, if there is one of each item (the burglar has broken into an art gallery, say), then the optimal knapsack contains items 1 and 3 (total: $46) As we shall see in 8, neither version of this problem is likely to have a polynomialtime algorithm However, using dynamic programming they can both be solved in O(nW ) time, which is reasonable when W is small, but is not polynomial since the input size is proportional to log W rather than W
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Let s start with the version that allows repetition As always, the main question in dynamic programming is, what are the subproblems In this case we can shrink the original problem in two ways: we can either look at smaller knapsack capacities w W , or we can look at fewer items (for instance, items 1, 2, , j, for j n) It usually takes a little experimentation to gure out exactly what works The rst restriction calls for smaller capacities Accordingly, de ne K(w) = maximum value achievable with a knapsack of capacity w Can we express this in terms of smaller subproblems Well, if the optimal solution to K(w) includes item i, then removing this item from the knapsack leaves an optimal solution to K(w wi ) In other words, K(w) is simply K(w w i ) + vi , for some i We don t know which i, so we need to try all possibilities K(w) = max {K(w wi ) + vi },
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where as usual our convention is that the maximum over an empty set is 0 We re done! The algorithm now writes itself, and it is characteristically simple and elegant
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If this application seems frivolous, replace weight with CPU time and only W pounds can be taken with only W units of CPU time are available Or use bandwidth in place of CPU time, etc The knapsack problem generalizes a wide variety of resource-constrained selection tasks
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K(0) = 0 for w = 1 to W : K(w) = max{K(w wi ) + vi : wi w} return K(W ) This algorithm lls in a one-dimensional table of length W + 1, in left-to-right order Each entry can take up to O(n) time to compute, so the overall running time is O(nW ) As always, there is an underlying dag Try constructing it, and you will be rewarded with a startling insight: this particular variant of knapsack boils down to nding the longest path in a dag!
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On to the second variant: what if repetitions are not allowed Our earlier subproblems now become completely useless For instance, knowing that the value K(w w n ) is very high doesn t help us, because we don t know whether or not item n already got used up in this partial solution We must therefore re ne our concept of a subproblem to carry additional information about the items being used We add a second parameter, 0 j n: K(w, j) = maximum value achievable using a knapsack of capacity w and items 1, , j The answer we seek is K(W, n) How can we express a subproblem K(w, j) in terms of smaller subproblems Quite simple: either item j is needed to achieve the optimal value, or it isn t needed: K(w, j) = max{K(w wj , j 1) + vj , K(w, j 1)} (The rst case is invoked only if wj w) In other words, we can express K(w, j) in terms of subproblems K( , j 1) The algorithm then consists of lling out a two-dimensional table, with W + 1 rows and n + 1 columns Each table entry takes just constant time, so even though the table is much larger than in the previous case, the running time remains the same, O(nW ) Here s the code Initialize all K(0, j) = 0 and all K(w, 0) = 0 for j = 1 to n: for w = 1 to W : if wj > w: K(w, j) = K(w, j 1) else: K(w, j) = max{K(w, j 1), K(w w j , j 1) + vj } return K(W, n)
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