All 0-1 Polytopes are. Abstract. We study the facial structure of two important permutation polytopes
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- Caroline Nash
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1 All 0-1 Polytopes are Traveling Salesman Polytopes L.J. Billera and A. Sarangarajan y Abstract We study the facial structure of two important permutation polytopes in R n2, the Birkho or assignment polytope B n, dened as the convex hull of all n n permutation matrices, and the asymmetric traveling salesman polytope T n, dened as the convex hull of those n n permutation matrices corresponding to n-cycles. Using an isomorphism between the face lattice of B n and the lattice of elementary bipartite graphs, we show, for example, that every pair of vertices of is contained in a cubical face, showing faces of B n to be fairly B n special 0-1 polytopes. On the other hand, we show that T n has every 0-1 d-polytope as a face, for d = log n, by showing that every 0-1 d-polytope is the asymmetric traveling salesman polytope of some directed graph with n nodes. The latter class of polytopes is shown to have maximum diameter n 3. 1 Introduction The (asymmetric) traveling salesman problem, to nd the shortest (directed) Hamiltonian tour in a complete (directed) graph, is one of the widely studied problems in combinatorial optimization, both for its utility and for the fact that it represents, in a well-dened sense, all hard combinatorial problems. A standard approach to solving this problem is to consider it as a linear Department of Mathematics and School of Operations Research, Cornell University, Ithaca, New York, Partially supported by NSF grant DMS y Center for Applied Mathematics, Cornell University, Ithaca, New York,
2 programming problem over the (asymmetric) traveling salesman polytope, dened as the convex hull of all (directed) Hamiltonian tours, and to use known classes of bounding inequalities (facets) to try to nd an optimal tour. (See [4] for a discussion of this and other approaches to this problem.) The diculty with this approach is that the facets of these polytopes are not all known, and, it seems, not knowable. We give explicit evidence for this assertion for the asymmetric traveling salesman polytope (ATSP) by showing that, up to a certain dimension, all 0-1 polytopes are among its faces. In particular, we show that if P is a 0-1 polytope, then P appears as a face of an ATSP of suciently large dimension. The dimension of this ATSP is in general exponential in the dimension of P, and we show that it is not possible to get all 0-1 polytopes in R d as faces of an ATSP of a dimension that is polynomial in d. Another way of viewing the main result of this paper is that every 0-1 polytope is the AT SP of some directed graph. In section 2 we introduce the Birkho polytope and associate the faces of this polytope with elementary bipartite graphs. This establishes an isomorphism between the face lattice of the Birkho polytope and the lattice of elementary bipartite graphs. We use this isomorphism to show that every pair of vertices of the Birkho polytope is contained in a cubical face. Section 3 is devoted to proving our main result on the AT SP. In section 4 we study the asymmetric TSP of an arbitrary directed graph and give a tight bound on its diameter. We dene some terms that will be used for the rest of this paper. We denote the set f1; 2; : : : ; ng by [n]. The symmetric group of degree n is the set of all permutations of [n]. Permutations that are cycles of length k will be called k-cycles. If v = (v 1 ; v 2 ; : : : ; v n ) 2 R n then Supp(v) := fi : v i 6= 0g is the support of v. A 0-1 polytope is a polytope whose vertices have coordinates 0 or 1. K n;n is the complete bipartite graph with bipartition ([n]; [n]) and edge set f(i; j) : 1 i; j ng. If C = fc 1 ; : : : ; C d g is an ordered collection of permutations C 1 ; : : : ; C d and if S [d] then C(S) := Q s2s C s ; we write C[d] instead of C([d]). Throughout this paper, C will denote an ordered collection of disjoint cycles. The graph ~ G(P ) of a polytope P is the graph whose nodes are the vertices of P, and which has an edge joining two nodes if and only 2
3 if the corresponding vertices in P are adjacent on P. The diameter of P is dened as diam(p ) := diam( ~ G(P )) = maxfd(u; v) : u; v are nodes of ~ G(P )g where d(u; v) is the length of the shortest path between u and v in ~ G(P ). 2 The Birkho Polytope Let S n denote the symmetric group of degree n. Given 2 S n, we dene the corresponding n n permutation matrix X() 2 R n2 by X() ij := 1 if (i) = j 0 otherwise. We denote by B n the Birkho polytope (or the assignment polytope) of order n, given by B n := convfx() : 2 S n g: It is well known that B n is an (n?1) 2 dimensional polytope with the following inequality description: B n = fx 2 R n2 : x ij 0 ; 1 i ; j n; and nx i=1 x ij = 1 for j = 1; : : : ; ng: nx j=1 x ij = 1 for i = 1; : : : ; n A detailed study of this polytope is given in [2] (see also [3]). For convenience, we shall often denote a vertex X() by. With each vertex 2 B n, we associate the bipartite graph G() which is the matching on K n;n with the edge set f(i; (i)) : i = 1; : : : ; ng. If F is a face of B n, then G(F ) is the subgraph of K n;n which is the union of G() over all the vertices 2 F. G(F ) has the property that every edge of G(F ) is in some matching. Such graphs are called elementary graphs (see [5]; the denition of elementary graphs given there also requires them to be connected, but we will not require that here). Also if G K n;n is an elementary graph, dene F (G) := convf : G() Gg = B n \ fx 2 R n2 : x ij = 0 if (i; j) 62 Gg: 3
4 We see that F (G) is an intersection of facets of B n and therefore is a face of B n. Clearly, if F 1 ; F 2 are faces of B n, then F 1 F 2 if and only if G(F 1 ) G(F 2 ). This yields the following (see [5] for details). Theorem 2.1 The face lattice of B n is isomorphic to the lattice of all elementary subgraphs of K n;n ordered by inclusion. 2 Suppose F is a face of B n and G(F ) has components G 1 ; : : : ; G k. Suppose G i is bipartite with bipartition fi i ; J i g, so that the sets I 1 ; : : : ; I k (and J 1 ; : : : ; J k ) form a partition of [n]. Let B i be the Birkho polytope dened on the coordinates fx kl : k 2 I i ; l 2 J i g. Since G(F ) is elementary, each G i is elementary bipartite on fi i ; J i g. Hence, F i := F (G i ) is a face of B i. Then F = F 1 F 2 F k, embedded as block diagonal matrices in R n2 (up to permutation of rows and columns). This follows from the fact that v is a vertex of F if and only if v = v 1 v k where, for each i, v i is a vertex of F i for each i. If ; 2 S n, then G(; ) := G() [ G() is a union of two matchings which is a set of disjoint cycles and edges. Hence, by the above remark, F ; := F (G(; )) is a k dimensional cube where k is the number of cycles in the graph. If?1 = Q k C i=1 i 2 S n where C 1 ; : : : ; C k are disjoint cycles in S n, and if C = fc 1 ; : : : ; C k g then the vertices of F ; are given j by C(S) n over all subsets S [k]. We note that k can be at most 2k. We shall associate F ; with the unit k-cube with C(S) corresponding to the vertex with support S (so corresponds to the origin and to (1; 1; : : : ; 1)). For convenience, we shall often denote F ; by F (; C). This proves the following. Theorem 2.2 B n (or more generally, any face of B n ) has the property that any two of its vertices are contained in a cubical face of dimension at most b n 2 c. 2 We note here that the k-cube F ; is actually a zonotope (Minkowski sum of line segments) generated by k mutually orthogonal segments z i 2 R n2. Each z i is supported on a cycle of the graph G(; ) and has coordinates +1 or?1 alternately for edges of G() and G(). 4
5 3 The Asymmetric TSP Let T n := f 2 S n : is a cycle of length ng S n The asymmetric TSP of order n is dened by T n := convfx() : 2 T n g so that T n B n R n2 This means that if F is a face of B n, then F \ T n is a face of T n induced by F. We exploit this relationship between the faces of B n and T n to derive some results about the faces of T n. We rst describe the following procedure to generate some lower dimensional faces of T n : Let 2 S n and C = fc 1 ; : : : ; C k g where C 1 ; : : : ; C k 2 S n are disjoint cycles. Then F (; C) B n is a k-cube and if V is the set of AT SP vertices of this cube then conv(v ) = F (; C) \ T n is a face of T n. Call such a face F 0 (; C). We shall identify F 0 (; C) with a 0-1 polytope embedded in the k-cube. We note that if = C[k] = C 1 C k, then F (; C) = B n \ fx 2 R n2 : x ij = 0 if (i; j) 62 G(; )g and hence F 0 (; C) = T n \ fx 2 R n2 : x ij = 0 if (i; j) 62 G(; )g: Example 1: (Rao [6]) Let n = 9; = (1; 2; 3; 4; 5; 6; 7; 8; 9) and C = fc 1 ; C 2 ; C 3 g where C 1 = (1; 7; 4); C 2 = (2; 8; 5) and C 3 = (3; 9; 6). Then F (; C) B 9 is a 3-cube and the other vertices of this cube are given by 1 = C 1 = (1; 8; 9)(2; 3; 4)(5; 6; 7) (1; 0; 0) 2 = C 2 = (1; 2; 9)(3; 4; 5)(6; 7; 8) (0; 1; 0) 5
6 3 = C 3 = (1; 2; 3)(4; 5; 6)(7; 8; 9) (0; 0; 1) 12 = C 1 C 2 = (1; 8; 6; 7; 5; 3; 4; 2; 9) (1; 1; 0) 23 = C 2 C 3 = (1; 2; 9; 7; 8; 6; 4; 5; 3) (0; 1; 1) 13 = C 1 C 3 = (1; 8; 9; 7; 5; 6; 4; 2; 3) (1; 0; 1) 123 = C 1 C 2 C 3 = (1; 8; 6; 4; 2; 9; 7; 5; 3) (1; 1; 1) We note that 5 of the vertices of this cube are AT SP vertices and thus F 0 (; C) is a bipyramid over a triangle as shown in Figure Figure 1: F (; C) and F 0 (; C) for Example 1 It is natural to ask whether any 0-1 polytope can appear as a face of the AT SP in the above manner. Surprisingly, this turns out to be true; the remainder of this section is devoted to proving this assertion. Let I d denote the unit d-cube. We will usually refer to vertices of I d by their corresponding supports which are subsets of [d]. If V is a subset of vertices of I d, then by I d? V we mean the convex hull of the vertices of I d that are not in V. Proposition 3.1 I d is a face of T n for n = 3d. Proof: Let = (1; : : : ; n) and C = f(1; 2; 3); (4; 5; 6); : : : ; (n? 2; n? 1; n)g. Then for any S [d], C(S) = (a 1 ; : : : ; a n ) 2 T n where 3i? 2; 3i; 3i? 1 if i 2 S a 3i?2 ; a 3i?1 ; a 3i = 3i? 2; 3i? 1; 3i otherwise 6
7 This shows that F 0 (; C) = F (; C) = I d. 2 Next we show that I d? S is also a face of the T 3d for any vertex S 2 I d. We begin with S = [d]. Proposition 3.2 Let d be a positive integer, n = 3d; = (1; 2; : : : ; n) and let C = fc 1 ; C 2 ; : : : ; C d g = f(1; 3; 5); (4; 6; 8); : : : ; (n? 2; n; 2)g ( i.e. C i = (3i? 2; 3i; 3i + 2) for i < d and C d = (n? 2; n; 2)). Then F 0 (; C) = I d? [d] Proof: Since C[d] = (1; 4; 7; : : : ; n? 2)(3; 6; 9; : : : ; n)(n? 1; n? 4; : : : ; 2) 62 T n this implies that [d] 62 F 0 (; C). Now we need to show that all the other vertices of the cube F (; C) are also vertices of T n. For the rest of the proof, the numbers we indicate are modulo n. Let C[i; j] = C i C i+1 C j ( if i > j, then set C[i; j] = C i C d C 1 C j ). Let ij denote the sequence 3i?2; 3i+1; 3i+4; : : : ; 3j+1; 3j+2; 3j?1; 3j?4; : : : ; 3i?1; 3i; 3i+3; : : : ; 3j+3 that is ij increases from 3i? 2 to 3j + 1 in steps of 3 then decreases from 3j + 2 to 3i? 1 in steps of 3 and nally increases from 3i to 3j + 3 in steps of 3. Then C[i; j] = ( ij ; 3j + 4; 3j + 5; : : : ; 3i? 3) 2 T n ; that is C[i; j] is a cyclic permutation that diers from by inserting ij in the interval [3i? 2; 3j + 3]. Let S [d] be a proper subset of [d]. We write C(S) = C[i 1 ; j 1 ]C[i 2 ; j 2 ] C[i k ; j k ] 7
8 where i 1 < i 2 < < i k and the above representation for C(S) is minimal (i.e., C[i r ; j r ]C[i r+1 ; j r+1 ] 6= C[i r j r+1 ] for r = 1; : : : ; k ). Then, C(S) is a cyclic permutation that diers from by inserting the sequences irjr in the interval [3i r?2; 3j r +3] for r = 1; : : : ; k, hence C(S) 2 T n and we are done. 2 Proposition 3.2 generalizes to the following result. Proposition 3.3 If S [d]; S 6= ;; n = 3d and = (1; : : : ; n) then there is an ordered collection of disjoint cycles C = f C1 ; : : : ; Cd g S n such that F 0 (; C) = I d? S. Proof: Let C = fc 1 ; C 2 : : : ; C d g be the ordered set as dened in the last proposition. Let S = [d]? S. We show that there exists a 2 S n such that C i = (?1 C i i 2 S?1 C?1 i i 2 S: Since S is non empty, S 6= [d] and hence ^ := C( S) 2 Tn. Let ^C = f ^C 1 ; : : : ; ^C d g where Ci i 2 S ^C i = C?1 i i 2 S: Then F 0 (^; ^C) = I d? S. Since and ^ are both cycles of length n they are conjugate, that is, there is 2 S n such that =?1^ =?1 C( S). Conjugating the cycles ^C i by to get C i, we see that for any subset R [d] C(R) =?1^(?1^C(R)) =?1^^C(R) which has the same cycle structure as ^^C(R). Thus we must have F 0 (; C) = I d? S. 2 Example 2: Let v = (1; 0; 0; 0), = (1; : : : ; 12). We nd C = fc 1 ; C 2 ; C 3 ; C 4 g such that F 0 (; C) = I 4? v. By Proposition 3.2, if C = fc 1 ; C 2 ; C 3 ; C 4 g = f(1; 3; 5); (4; 6; 8); (7; 9; 11); (10; 12; 2)g 8 (1)
9 then F 0 (; C) = I 4? (1; 1; 1; 1). Let S = f1g = Supp(v), S = f2; 3; 4g, ^ = C( S) = C 2 C 3 C 4 = (1; 2; 11; 8; 5; 6; 9; 12; 3; 4; 7; 10) =: (a 1 ; a 2 ; : : : ; a 12 ) with a 1 = 1. Setting (i) = a i for i = 1; : : : ; 12 we get = (3; 11; 7; 9)(4; 8; 12; 10) and =?1^. Then C 1 =?1 C 1 = (1; 9; 5) Hence C 2 =?1 C?1 2 = (4; 6; 10) C 3 =?1 C?1 3 = (3; 7; 11) C 4 =?1 C?1 4 = (2; 8; 12) C = f(1; 9; 5); (4; 6; 10); (3; 7; 11); (2; 8; 12)g Using the same method, one can show that if C = f(2; 4; 12); (1; 5; 7); (6; 8; 10); (3; 9; 11)g; then F 0 (; C) = I 4? (0; 1; 1; 1). 2 Example 2 shows how we can nd C of Proposition 3.3. If ^ := C( S) = (a 1 ; : : : ; a n ) then setting (i) = a i for i = 1; : : : ; n we would get =?1^. This denes C i (by (1)) and hence C. Now we tackle the general case of removing any set of vertices. P R d be a 0-1 polytope. Assume without loss of generality that 0 2 P and let V := fs 1 ; : : : ; S k g be the set of vertices of I d that are not in P ( i.e. P = I d? V ). By Proposition 3.3, we can nd C i = fc i1 ; : : : ; C id g such that F 0 (; C i ) = I d? S i for i = 1; : : : ; k, where n = 3d; = (1; : : : ; n) and each C ij is a 3-cycle. The idea is to concatenate the cycles C 1j ; : : : ; C kj for j = 1; : : : ; d to eliminate S 1 ; : : : ; S k. For this, the cycles must be dened on distinct sets. So, assume that C i is dened on f(i? 1)n + 1; : : : ; ing and let i = ((i? 1)n + 1; : : : ; in) so that F 0 ( i ; C i ) = I d? S i for i = 1; : : : ; k (i.e. i C i (S) is an n-cycle for all S [d] except when S = S i ). Proposition 3.4 There is an integer N and an ordered set of disjoint cycles Y 0 = fy 0 ; : : : ; Y 0 1 dg S N such that if 0 = (1; : : : ; N) then F 0 ( 0 ; Y 0 ) = P. 9 Let
10 Proof: Dene the permutations Y j = C 1j C 2j C kj 2 S kn for j = 1; : : : ; d and let Y = fy 1 ; : : : ; Y d g, and N = kn + (k + 1)d. The following array illustrates how Y j is dened. C 1 C 11 C 1j C 1d F 0 ( 1 ; C 1 ) = I d? S C k C k1 C kj C kd F 0 ( k ; C k ) = I d? S k Y j We rst show that if S [d] then 0 Y(S) 2 T N, r C r (S) is an n-cycle for r = 1; : : : ; k (2) To see this, note that 0 Y(S) = 0 Y s2s Y s = 0 Y s2s(c 1s C ks ) = 0 C 1 (S) C k (S): Suppose r C r (S) = (x (r?1)n+1 ; : : : ; x rn ) is an n-cycle for r = 1; : : : ; k with x (r?1)n+1 = (r? 1)n + 1. Then it follows that 0 Y(S) = (x 1 ; x 2 ; : : : ; x kn ; kn + 1; kn + 2; : : : ; N) 2 T N since 0 Y(S)(l) = 0 C r (S)(l) = r C r (S)(l) if (r? 1)n < l rn; l 6= x rn and 0 Y(S)(x rn ) = 0 (rn) = rn + 1 = x rn+1 for r = 1; : : : ; k. Conversely, suppose that for some r, r C r (S) = C 0 is not an n-cycle where C 0 is a cycle not involving (r? 1)n + 1. Then, as above 0 Y(S) has C 0 as a cycle, so is not in T N proving (2). It follows from (2) that 0 Y(S) 2 T N if and only if S 62 V, i.e., if and only if S is a vertex of P. We now modify the permutations Y 1 ; : : : ; Y d to dene cycles Y 0 ; : : : ; Y 0 1 d such that 0 Y(S) 2 T N, 0 Y 0 (S) 2 T N for all S [d]. Suppose the 3-cycle C ij = (a ij ; b ij ; c ij ) and let ij denote the sequence a ij ; b ij ; c ij. For j = 1; : : : ; d dene Y 0 j = (r j + k + 1; kj ; r j + k; (k?1)j ; : : : ; r j + 2; 1j ; r j + 1) = (r j + k + 1; a kj ; b kj ; c kj ; r j + k; : : : ; r j + 2; a 1j ; b 1j ; c 1j ; r j + 1): 10
11 where r j = kn + (j? 1)(k + 1). Let S [d]. Let = 0 Y(S) and 0 = 0 Y 0 (S) and write and 0 as products of disjoint cycles. Noting that Y j = (a kj ; b kj ; c kj ) (a 1j ; b 1j ; c 1j ), we see that for i = 1; : : : ; k and j 2 S, the sequence c ij ; a ij +1 in is replaced by the sequence c ij ; r j +i+1; a ij +1 in 0 and the sequence r j +1; : : : ; r j +k+2 in is replaced by r j + 1; r j + k + 2 in 0. Thus if is an N-cycle then 0 is a rearranged N-cycle and conversely. Therefore, 0 = 0 Y(S) 2 T N, = 0 Y 0 (S) 2 T N (3) It follows from (2) and (3) that F 0 ( 0 ; Y 0 ) = P and N = kn + (k + 1)d = 3kd + (k + 1)d = (4k + 1)d. 2 The following is a direct consequence of Proposition 3.4. Theorem 3.1 If P R d is a 0-1 polytope with 2 d? k vertices, k > 1, then P appears as a face of T n for n (4k + 1)d. 2 Example 3: Let P = I 4? fv 1 ; v 2 g where v 1 = (1; 0; 0; 0) ; v 2 = (0; 1; 1; 1). In Example 2, we found C 1 and C 2 such that if = (1; : : : ; 12) then F 0 (; C 1 ) = I 4? v 1 and F 0 (; C 2 ) = I 4? v 2. Now we add 12 to every number in C 2 to get the permutations in dierent sets. Let 1 = (1; : : : ; 12) ; C 1 = f(1; 9; 5); (4; 6; 10); (3; 7; 11); (2; 8; 12)g 2 = (13; : : : ; 24) ; C 2 = f(14; 16; 24); (13; 17; 19); (18; 20; 22); (15; 21; 23)g so that F 0 ( 1 ; C 1 ) = I 4? v 1 and F 0 ( 2 ; C 2 ) = I 4? v 2 : Now dene Y 1 = (1; 9; 5)(14; 16; 24) ; Y 0 1 = (27; 14; 16; 24; 26; 1; 9; 5; 25) Y 2 = (4; 6; 10)(13; 17; 19) ; Y 0 2 = (30; 13; 17; 19; 29; 4; 6; 10; 28) Y 3 = (3; 7; 11)(18; 20; 22) ; Y 0 3 = (33; 18; 20; 22; 32; 3; 7; 11; 31) Y 4 = (2; 8; 12)(15; 21; 23) ; Y 0 4 = (36; 15; 21; 23; 35; 2; 8; 12; 34) 11
12 where we have underlined the new elements in each Yi 0. Let 0 = (1; : : : ; 36) and Y 0 = fy 0 ; Y 0 ; Y 0 ; Y g. Then for example while and 0 Y 3 = (1; 2; 3; 8; 9; 10; 11; 4; 5; 6; 7; 12; 13; : : : ; 18; 21; 22; 19; 20; 23; 24; : : : ; 31; 32; 33; 34; 35; 36) 2 T 36 0 Y 0 3 = (1; 2; 3; 8; 9; 10; 11; 32; 4; 5; 6; 7; 12; 13; : : : ; 18; 21; 22; 33; 19; 20; while 23; 24; : : : ; 31; 34; 35; 36) 2 T 36 0 Y 2 Y 3 Y 4 = (14; 15; 22; 19) 62 T 36 0 Y 0 2 Y 0 3 Y 0 4 = (14; 15; 22; 33; 19; 30) 62 T 36 where we have underlined the places where the permutations dier. Similarly multiplying 0 with all subsets of cycles of Y 0 we get F 0 ( 0 ; Y 0 ) = P. 2 Remarks: (1) The above result is not valid for B n since B n has the property that every pair of its vertices is contained in a cubical face. For instance, the bipyramid over a triangle (of Example 1), cannot be a face of B n. (2) A natural question to ask is whether the bound for n can be improved. We ask this question in two dierent forms. Given d, is there n d k such that (i) All 0-1 polytopes in R d appear as faces of T n? The answer is no for the following reason: The number of 0-1 polytopes in R d is 2 2d, since I d has 2 d vertices and the convex hull of any subset of these vertices is a 0-1 polytope. If f i is the number of i-dimensional faces of T n, then f i (n? 1)! i + 1! (n? 1)! i+1 n (i+1)n n n3 since i < n 2? 1 12
13 Hence, if f is the total number of faces of T n, then f n 2 n n3 = d 2k d kd3k if n = d k : Then log(f) 2k log(d) + kd 3k log(d) which is a polynomial in d, whereas log(2 2d ) = 2 d log(2) is an exponential in d. Hence, n cannot be a polynomial in d. (ii) All (combinatorially) distinct 0-1 polytopes in R d appear as faces of T n? This question remains open as we do not have a good estimate of the number of distinct 0-1 polytopes in R d. (3) Theorem 3.1 does not hold for all \hard" 0-1 problems: For example, the polytope associated with the quadratic assignment problem cannot have every 0-1 polytope among its faces since it is neighborly, i.e., every pair of its vertices are adjacent. This is so since, choosing a hyperplane containing any pair of vertices of B n, and no others, its normal can be used (by forming a tensor product with itself) to dene a supporting hyperplane to the desired edge of the quadratic assignment polytope. See [1; 2.2.3] for some background. (This fact and its proof was pointed out to us by A.I. Barvinok.) 4 The Asymmetric TSP of a Directed Graph For the rest of this section D will be a directed graph on the node set [n]. A Hamiltonian tour < i 1 ; : : : ; i n > in D (i.e. a tour comprising the edges (i 1 ; i 2 ); (i 2 ; i 3 ); : : : ; (i n ; i 1 )) corresponds to the cyclic permutation = (i 1 ; : : : ; i n ) 2 T n and hence to the permutation matrix X(). The asymmetric TSP of the graph D is given by T D := convfx()j corresponds to a tour in Dg If D is the complete directed graph, then T D section. Hence, for a general directed graph D = T n as dened in the last T D = T n \ fx 2 R n2 jx ij = 0 whenever (i; j) 62 Dg (4) 13
14 Since the nonnegativity constraints dene facets of T n, this shows that T D is a face of T n. Let ; 2 S n be permutations with no xed points (i.e. for each i 2 [n]; (i) 6= i and (i) 6= i). Let D(; ) be the directed graph with the edge set f(i; j) : (i) = j or (i) = jg. Lemma 4.1 Suppose C = fc 1 ; : : : ; C k g where?1 = C 1 C k. F 0 (; C) = T D(;). Then Proof: We have seen earlier that if G(; ) is the bipartite graph which is the union of the matchings corresponding to and then the face F (; C) = B n \ fx 2 R n2 : x ij = 0 if (i; j) 62 G(; )g is a k-cube and F 0 (; C) = F (; C) \ T n : Since the edge (i; j) 62 G(; ) if and only if the directed edge (i; j) 62 D(; ), it follows from (4) that F 0 (; C) = T D(;). 2 We showed that any 0-1 polytope appeared as a face of the AT SP. In proving this result we saw that the face was of the form F 0 (; C) where was our generic cycle (1; : : : ; n). The following theorem is therefore a direct consequence of Lemma 4.1: Theorem 4.1 Every 0-1 polytope in R d is the asymmetric TSP of some directed graph. In fact, if the polytope has a pair of diametrically opposite vertices, (vertices with disjoint supports such that the union of their supports is [d]) then the graph is the union of two tours. Otherwise the graph is the union of a tour and disjoint cycles that cover the graph. 2 Finally, we obtain some bounds on the diameter of T D. following preliminary results: We need the Lemma 4.2 Let ; 2 T n correspond to tours in D. If?1 = C 1 C k and C = fc 1 ; : : : ; C k g, then F 0 (; C) is a face of T D. 14
15 Proof: Since F 0 (; C) = T D(;) T D T n and F 0 (; C) is a face of T n, it follows that F 0 (; C) is a face of T D, proving the lemma. 2 The following is probably a well known result: Proposition 4.1 Let P R d be a 0-1 polytope and let v 1 ; v 2 be vertices of P such that (i) Supp(v 1 ) Supp(v 2 ) and (ii) If v is a vertex of P such that Supp(v 1 ) Supp(v) Supp(v 2 ), then v = v 1 or v = v 2. Then v 1 and v 2 are adjacent on P. Proof: Consider the face F of I d spanned by the vertices v 1 and v 2. The vertices of F are all the vertices v 2 I d such that Supp(v 1 ) Supp(v) Supp(v 2 ). By our assumption, the only vertices of P that lie on F are v 1 and v 2. Hence F \ P = convfv 1 ; v 2 g, i.e. v 1 is adjacent to v 2 on P. 2 Theorem 4.2k The maximum diameter of T D over all directed graphs D on. n nodes is j n 3 Proof: Dene (n) := maxfdiam(t D ) : D is a directed graph on [n]g: Let d = b n c and let D be any directed graph on [n]. We rst show that 3 diam(t D ) d. Let ; 2 T n be vertices of T D. Let?1 = C 1 C 2 C k and C = fc 1 ; : : : ; C k g. Let i = C 1 C i and 0 =. Choose 0 = i 0 < i 1 < < i m = k so that fi 1 ; : : : ; i m g = fj 2 [k] : j 2 T n g: We assume that the cycles C 1 ; : : : ; C k are arranged in such a way that if S [k] such that [i j ] S [i j+1 ] and C(S) 2 T n then S = [i j ] or S = [i j+1 ]. Then by Proposition 4.1, the vertices ij and ij+1 are adjacent on F 0 (; C) (and hence on T D since F 0 (; C) is a face of T D by Lemma 4.2) for j = 0; 1; : : : ; m? 1. 15
16 Since?1 i j ij+1 = C ij +1 C ij+1 is an even permutation, it follows that for j = 0; : : : ; m? 1 l ij l ij+1 3 where l i is the length of the cycle C i. Adding these m inequalities, we get 3m l 1 + l l k n i.e. m d. We have exhibited a path from to of length at most d which shows that diam(t D ) d. This implies that (n) d. To show that (n) = d, we have to nd a nd a directed graph D such that diam(t D ) = d. Let = (1; : : : ; n); C = fc 1 ; : : : ; C d g where C i = (3i? 2; 3i? 1; 3i) for i = 1; : : : ; d. Let = C 1 C d. Then, T D(;) = F 0 (; C) is a d-cube by Proposition 3.1, and hence has diameter d. 2 REFERENCES [1] A.I. Barvinok, \Combinatorial complexity of orbits in representations of the symmetric group", Advances in Soviet Mathematics, vol. 9, 1992, pp [2] R. Brualdi and P. Gibson, \Convex polyhedra of doubly stochastic matrices, I, II, III", Journal of Combinatorial Theory, A22 (1977), pp ; B22 (1977), pp ; A22 (1977), pp [3] V.A. Emelichev, M.M. Kovalev and M.K. Kravtsov, Polytopes, Graphs and Optimization, Cambridge University Press, New York, 1984, pp [4] E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan and D.B. Shmoys, eds., The Traveling Salesman Problem: A Guided Tour of Combinatorial Optimization, Wiley, New York, [5] L.L. Lovasz and M.D. Plummer, Matching Theory, Elsevier Science Pub. Co., New York, [6] M.R. Rao, \Adjacency of the traveling salesman tours and 0-1 vertices", SIAM J. Appl. Math., vol. 30, No. 2, March 1976, pp
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