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TSP cost cost of this path MST cost
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Little Rock
Dallas El Paso
El Paso
Dallas
15 16 2
Houston San Antonio
Houston
San Antonio
Therefore, this tour has a length at most twice the MST cost, which as we ve already seen is at most twice the TSP cost This is the result we wanted, but we aren t quite done because our tour visits some cities multiple times and is therefore not legal To x the problem, the tour should simply skip any city it is about to revisit, and instead move directly to the next new city in its list:
Wichita Tulsa Amarillo Little Rock Dallas El Paso
Albuquerque
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By the triangle inequality, these bypasses can only make the overall tour shorter General TSP But what if we are interested in instances of TSP that do not satisfy the triangle inequality It turns out that this is a much harder problem to approximate Here is why: Recall that on page 255 we gave a polynomial-time reduction which given any graph G and integer any C > 0 produces an instance I(G, C) of the TSP such that: (i) If G has a Rudrata path, then
OPT(I(G, C))
= n, the number of vertices in G
(ii) If G has no Rudrata path, then OPT(I(G, C)) n + C This means that even an approximate solution to TSP would enable us to solve R UDRATA PATH! Let s work out the details Consider an approximation algorithm A for TSP and let A denote its approximation ratio From any instance G of R UDRATA PATH, we will create an instance I(G, C) of TSP using the speci c constant C = n A What happens when algorithm A is run on this TSP instance In case (i), it must output a tour of length at most A OPT(I(G, C)) = n A , whereas in case (ii) it must output a tour of length at least OPT(I(G, C)) > n A Thus we can gure out whether G has a Rudrata path! Here is the resulting procedure: Given any graph G: compute I(G, C) (with C = n A ) and run algorithm A on it if the resulting tour has length n A : conclude that G has a Rudrata path else: conclude that G has no Rudrata path This tells us whether or not G has a Rudrata path; by calling the procedure a polynomial number of times, we can nd the actual path (Exercise 82) We ve shown that if TSP has a polynomial-time approximation algorithm, then there is a polynomial algorithm for the NP-complete R UDRATA PATH problem So, unless P = NP, there cannot exist an ef cient approximation algorithm for the TSP
Knapsack
Our last approximation algorithm is for a maximization problem and has a very impressive guarantee: given any > 0, it will return a solution of value at least (1 ) times the optimal value, in time that scales only polynomially in the input size and in 1/ The problem is KNAPSACK, which we rst encountered in 6 There are n items, with weights w1 , , wn and values v1 , , vn (all positive integers), and the goal is to pick the most valuable combination of items subject to the constraint that their total weight is at most W Earlier we saw a dynamic programming solution to this problem with running time O(nW ) Using a similar technique, a running time of O(nV ) can also be achieved, where V is the sum of the values Neither of these running times is polynomial, because W and V can be very large, exponential in the size of the input 276
Let s consider the O(nV ) algorithm In the bad case when V is large, what if we simply scale down all the values in some way For instance, if v1 = 117,586,003, v2 = 738,493,291, v3 = 238,827,453, we could simply knock off some precision and instead use 117, 738, and 238 This doesn t change the problem all that much and will make the algorithm much, much faster! Now for the details Along with the input, the user is assumed to have speci ed some approximation factor > 0 Discard any item with weight > W Let vmax = maxi vi Rescale values vi = vi vn max Run the dynamic programming algorithm with values {v i } Output the resulting choice of items Let s see why this works First of all, since the rescaled values v i are all at most n/ , the dynamic program is ef cient, running in time O(n 3 / ) Now suppose the optimal solution to the original problem is to pick some subset of items S, with total value K The rescaled value of this same assignment is vi =
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