1 1 c (a) 1 (b) 1 Figure 1: (a) First ath followed by salesman in the stris method. (b) Alternative ath. 4. D = distance travelled closing the loo. Th

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1 18.415/6.854 Advanced Algorithms ovember 7, 1996 Euclidean TSP (art I) Lecturer: Michel X. Goemans MIT These notes are based on scribe notes by Marios Paaefthymiou and Mike Klugerman. 1 Euclidean TSP Consider the travelling salesman roblem in the lane. Given n oints in the lane, we would like to nd a tour that visits all of them and that minimizes the distance travelled, where the distance between two oints is given by the Euclidean distance. This roblem will be denoted by ETSP (Euclidean TSP). In a comanion set of notes, we will resent very recent algorithms of Arora and Mitchell that roduce in olynomial time a tour which is within 1 + of the otimum, for any xed >. In this set of notes, we rst resent some reliminaries and older related results. Before we continue, we should oint out that it is not known whether the Euclidean TSP is in P. Even if we are resented with a candidate tour for the \yes-no" version of the roblem, we do not know how to avoid comuting a ossibly exonential number of decimal digits, in order to calculate the square root required for the Euclidean distance. First, we resent an algorithm which generates a ath of length no more than k (k a constant to be determined) given that the oints all lie within a unit square. Here we use the \Stris Method". First, break the square down into horizontal c stris of equal height. The \Salesman's" strategy will then be the following: He will begin at the left side of the tomost stri and travel to the right along the horizontal line slitting the stri in half. If at any oint the salesman reaches a sot where a oint,, in the roblem is located in the stri directly above or below him, he travels directly to, and back to the center line. When he reaches the end of the line, he travels along the edge of the square to the middle of the next stri and then travels left across the middle line. The salesman goes back and forth in this way until he has assed through all stris. At the end, the salesman travels from the lower right corner to the uer left to nish the loo (see Figure 1, art a). Analysis: Let 1. A = length travelled horizontally across each stri = 1.. B = distance travelled vertically along edge of square. 3. C = worst case distance travelled getting to each oint and back. TSP-1

2 1 1 c (a) 1 (b) 1 Figure 1: (a) First ath followed by salesman in the stris method. (b) Alternative ath. 4. D = distance travelled closing the loo. The total distance, Z SM c A + B + C + D c c + = ( 1 c + c) = for c = 1 ow, we can imrove this method in the following way. We consider another ath, as well. The saleman travels along the boundary lines of the stris rather than along the middle of the stri, and treats what were the middle lines as the new boundary lines (see Figure 1, art b). The salesman then takes the shorter of this ath and the ath described above. We can see why this imroves our bound by looking at the sum of the lengths of the two aths. early each argument in the sum above will double in size. The dierence is that the sum of the distance travelled to each oint and back will remain the same ( c ) (see Figure ). TSP-

3 Figure : Distance travelled to each oint and back. So the total will be: Z SM A + B + C + D c c c + = ( c + c) + + = + + for c = Thus, one of the two aths has length The rst roof of this bound is due to Few []. This has been imroved to :984 by Karlo [3]. Kar's artitioning scheme We now resent Kar's artitioning algorithm [4] for the Euclidean TSP roblem on n oints. A deterministic analysis of the algorithm shows that the length of the tour given by this algorithm does not exceed the length of the otimum tour by more than a factor of o( n). A robabilistic analysis of the algorithm shows that if the oints are uniformly distributed in the unit square, then the length of the tour yielded by the algorithm aroaches the otimum length with robability 1, as the number of oints in the lane becomes arbitrarily large. Given n oints in [; 1], the algorithm roceeds as follows. Algorithm Partition 1. Choose s such that s! n (s = lg n= lg lg n is a good choice, for examle ). In fact, any s such that s! = O(n k ) for some k will do.. Divide the unit square into q n=s vertical stris (see Figure 3). Each stri must contain exactly ns oints. 3. Divide each vertical stri into q n=s horizontal stris, so that every one of the n=s rectangles contains s oints. TSP-3

4 4. For each rectangle Q i solve the TSP otimally for the s oints inside Q i. Even brute-force solutions will cost s! = O(n k ) stes. 5. For each Q i select a vertex v i. Then construct a tour on fv i j i = 1; : : : ; n=sg of length L no greater than n This can be done using double STRIPS, for instance. In fact, any tour of length less than c n, for a constant c, will do. 6. Shortcut the resulting grah to get a tour of length Z KARP. Q 1 Q i v i Figure 3: Kar's artitioning algorithm. 3 Determinisitc Analysis of Kar's Algorithm In this section we give a deterministic analysis of Kar's artitioning algorithm. We show that the length Z KARP of the tour yielded by the artitioning algorithm does not q exceed the length Z T SP of the otimal tour by more than an additive factor of O( n=s) = o( n). First, we rove the following claim, that bounds the length Zi of the locally otimum tour in Q i, constructed in Ste 4 of the algorithm, in terms of the length Z i of the edges of the otimum tour in Q i and the erimeter of Q i. TSP-4

5 Claim 1 Let C be the otimal tour on all n oints. Let Z i be the length of the edges of C in Q i. Let Zi be the length of the otimal tour in Q i. Then where P (Q i ) denotes the erimeter of Q i. Z i Z i + 3 P (Q i); Proof: Let's focus on some Q i (see Figure 4). For each edge crossing the boundary of Q i we dene a new boundary oint (cfr. A through F on the gure). There is an even number of boundary oints. We connect all boundary oints in their order of aearance on the boundary with a cycle. This cycle is of length less than or equal to P (Q i ). We also add edges which constitute a short matching, i.e. one of length no greater than (1=)P (Q i ), on the boundary nodes. One of the two ways of selecting every other edge of the cycle constructed above constitutes such a short matching. The urose of these additions is to create an eulerian grah. ow, we can short-cut to get a tour on the inside nodes and the boundary nodes. This tour can be further short-cut to obtain a tour on the inside nodes only. Clearly, the length l of this tour satises Z i l Z i + 3 P (Q i); since the total length of the edges in the eulerian grah is at most Z i + P (Q i ) + (1=)P (Q i ). F E A D C Figure 4: A subrectangle Q i. Its boundary intersections are denoted by A through F. ow, we can rove the following theorem, which rovides a erformance guarantee for the artitioning algorithm. B TSP-5

6 Theorem Let Z T SP be the length of the otimal tour on the given n oints and let Z KARP be the length of the tour yielded by Kar's artitioning algorithm. Then where s! n. Z T SP Z KARP Z T SP + O( q n=s); Proof: By the denition of Z KARP and the revious claim we have: Z KARP n=s X n=s X Z i + L Z i + 3 = Z T SP + 3 n=s X = Z T SP + O( q P (Q i ) + q n=s + O(1) q n=s + n=s): q n=s + q n=s + O(1) 4 Probabilistic Analysis of Kar's Algorithm In this section we resent a robabilistic analysis of Kar's artitioning algorithm. We state three theorems, one of them along with its roof, and we conclude with two imortant corollaries about the roerties of the solutions yielded by the algorithm. Let X 1 ; : : : ; X n be n indeendent random variables uniformly distributed in [; 1]. The following theorem gives an uer and a lower bound on the exected length of the otimum tour on these n oints. Theorem 3 E[T SP (X 1 ; : : : ; X n )] = ( n): Proof: For convenience let us denote E[T SP (X 1 ; : : : ; X n )] by E n. Alying the double STRIPS method we deduce an uer bound for E n : E n n : We furthermore claim that E n n=. First, since any edge incident to X i has length at least d(x i ; fx 1 ; : : : ; X i 1 ; X i+1 ; : : : ; X n g) = min j6=i jjx j X i jj, we observe that " n # X E n E d(x i ; fx 1 ; : : : ; X i 1 ; X i+1 ; : : : ; X n g) = nx E [d(x i ; fx 1 ; : : : ; X i 1 ; X i+1 ; : : : ; X n g)] = ne [d(x n ; fx 1 ; : : : ; X n 1 g)] : TSP-6

7 Since E[Y ] = R 1 P r[y y]dy for any nonnegative random variable Y, we can bound the exectation in the following way: E [d(x n ; fx 1 ; : : : ; X n 1 g)] = where we had used the fact that: Z 1 Z 1= P r [d(x n ; fx 1 ; : : : ; X n 1 g) r] dr (1 r ) n 1 dr: P r[d(x n ; fx 1 ; : : : ; X n 1 g) r j X n ] = P r[min i<n jjx i X n jj r j X n ] = Y i<n P r[jjx i X n jj r j X n ] (1 r ) n 1 ; for all X n. The last integral can be seen to be equal to which is greater than or equal to 1 (n) (n + 1); 1 1 : n Here is a quick way to show that this integral is ( 1 n ): Z 1= (1 r ) n 1 dr Z 1= n Z 1= n e 1 n : (1 r ) n 1 dr (1 1 n )n 1 dr It follows that E n n 1 n = n ; thus comleting the roof of the theorem. We can rove a stronger result than that of Theorem 3. Secically, we can show that the ratio E n = n is not only bounded but it also aroaches a limit, as the number of oints n aroaches innity. Theorem 4 There exists a number such that E[T SP (X 1 ; : : : ; X n )] lim = : n!1 n TSP-7

8 The limit is known to be in the range :65 :94. An even stronger theorem was roven by Beardwood, Halton and Hammersley [1]: Theorem 5 There exists a number such that P r " lim n!1 T SP (X 1 ; : : : ; X n ) n # = = 1: The roofs of Theorems 4 and 5 are omitted. Based on Theorem we can show that Kar's algorithm is otimal in exected value. Corollary 6 1 Z KARP Z T SP Proof: 1 + o(1): From Theorem we have Z KARP Z T SP + c n s : Therefore E[Z KARP ] E[Z T SP ] + c n s : From Theorem 3 we have 1 E[Z KARP ] E[Z T SP ] 1 + c n s n= = 1 + c s = 1 + o(1); since 1= s! as n! 1. Based on Theorem 5 we can show that the length of the tour yielded by Kar's algorithm aroaches the otimal length almost surely, as the number of oints aroaches innity. Corollary 7 P r h lim n!1 Z KARP Z T SP = 1 i = 1: References [1] J. Beardwood, J. Halton, J. Hammersley, \The shortest ath through many oints", Proc. Cambridge Philos. Soc., 55, 99{37, [] L. Few, \The shortest ath and shortest road through n oints", Mathematika,, 141{144, [3] H.J. Karlo, \How long can a Euclidean Traveling Salesman Tour be", SIAM J. Disc. Math.,, 91{99, [4] R. Kar, \Probabilistic Analysis of Partitioning Algorithms for the Traveling Salesman Problem in the lane", Math. of Oerations Research,, TSP-8