Constructing Independent Spanning Trees on Locally Twisted Cubes in Parallel

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1 Constructing Independent Spanning Trees on Locally Twisted Cubes in Parallel Yu-Huei Chang Jinn-Shyong Yang, Jou-Ming Chang Department of Information Management, National Taipei College of Business, Taipei, Taiwan, ROC Institute of Information and Decision Sciences, National Taipei College of Business, Taipei, Taiwan, ROC Abstract Let LT Q n denote the n-dimensional locally twisted cube. Hsieh and Tu (009) [3] presented an algorithm to construct n edge-disjoint spanning trees rooted at vertex 0 in LT Q n. Later on, Lin et al. (00) [3] proved that Hsieh and Tu s spanning trees are indeed independent spanning trees (ISTs for short), i.e., all spanning trees are rooted at the same vertex r and for any other vertex v( r), the paths from v to r in any two trees are vertex-disjoint except the two end vertices v and r. Shortly afterwards, Liu et al. (0) [4] pointed out that LT Q n fails to be vertex-transitive for n 4 and proposed an algorithm for constructing n ISTs rooted at an arbitrary vertex of LT Q n. Although this algorithm can simultaneously construct n ISTs in parallel, it is not fully parallelized for the construction of each spanning tree. In this paper, we revisit the problem of constructing n ISTs rooted at an arbitrary vertex of LT Q n. As a consequence, we present a fully parallelized approach that is obtained from Hsieh and Tu s algorithm with a slight modification. Keyword: independent spanning trees; edgedisjoint spanning trees; locally twisted cubes; interconnection networks; fault-tolerant broadcasting; Introduction Interconnection networks are usually modeled as undirected simple graphs G = (V, E), where the vertex set V (= V (G)) represents the set of processing elements and the edge set E(= E(G)) represents the set of communication channels, respectively. A tree is a connected graph without cycle. This research was partially supported by National Science Council under the Grants NSC0--E-4-00 and NSC0--E-4-00-MY3. Corresponding author : shyong@mail.ntcb.edu.tw A rooted tree is a tree with a distinguished vertex called the root. A subgraph T in a graph G is called a spanning tree if T is a tree such that V (T ) = V (G). Let T be a set of k spanning trees of G rooted at a vertex r. We say that T is edgedisjoint if the paths from any vertex v( r) to r on the k trees share no common directed edges. By contrast, T is said to be independent if the paths from any vertex v( r) to r on the k trees have no common vertex except x and y (i.e., the paths are internally vertex-disjoint). Constructing independent spanning trees (ISTs for short) in networks have been studied from not only the theoretical point of view but also some practical applications such as fault-tolerant broadcasting [, 9] and secure message distribution [, 3, 40]. Let G be a graph and denote G F the graph obtained from G by removing a set of vertices F. A graph G is k-connected if V (G) > k and G F is connected for every subset F V (G) with F < k. A conjecture proposed by Zehavi and Itai [49] says that any k-connected graph has k ISTs rooted at an arbitrary vertex r. Henceforth, we refer the conjecture as the IST- Conjecture. From then on, the IST-Conjecture has been confirmed only for k-connected graphs with k 4 (see [9] for k =, [8,49] for k = 3, and [9] for k = 4), and it is still open for k-connected graphs with k 5. In addition, by providing construction schemes of ISTs, the IST-Conjecture has been agreed for several restricted classes of graphs or digraphs. For example, the graph classes related to planarity [6, 7, 7, 8], graph classes defined by Cartesian product [3, 9, 3, 33, 36, 4, 46], special classes of digraphs [0,, 8, 37], variations of hypercubes [4 7, 4, 34, 35, 40], subclasses of Cayley graphs [,,3,4,44,45], and chordal ring [0,43]. The family of locally twisted cubes was first introduced by Yang et al. [47] as a variation of 56

2 hypercube architecture to achieving the improvement in diameter. Hsieh and Tu [3] studied the construction of edge-disjoint spanning trees on locally twisted cubes. Since the n-dimensional locally twisted cube LT Q n is n-connected, they presented an algorithm to construct n edge-disjoint spanning trees rooted at vertex 0. At a later time, Lin et al. [3] proved that Hsieh and Tu s spanning trees are indeed independent. Liu et al. [4] pointed out that LT Q n fails to be vertex-transitive for n 4 and it does satisfy the even-odd-vertex-transitive property. Thus, the proof of [3], together with Hsieh and Tu s algorithm [3], does not solve the IST-Conjecture on LT Q n. Furthermore, Liu et al. [4] proposed an algorithm for constructing n ISTs rooted at an arbitrary vertex of LT Q n, and thus confirmed the IST-Conjecture for LT Q n. Although the algorithm in [4] can simultaneously construct n ISTs in parallel, it is not fully parallelized for the construction of each spanning tree (in fact, it looks like a constructing scheme of binomial tree in a recursive fashion). In this paper, with a slight modification from Hsieh and Tu s algorithm, we present a fully parallelized approach for constructing n ISTs rooted at an arbitrary vertex of LT Q n. The rest of this paper is organized as follows. Section formally gives the definition of locally twisted cubes and introduces our constructing scheme of ISTs for LT Q n. Section 3 shows the correctness by proving the independency of the constructed spanning trees. The final section contains our concluding remarks. () LT Q is a graph consisting of four vertices labeled with 00, 0, 0, and, respectively, connected by four edges (00,0), (00,0), (0,), and (0,). () For n 3, LT Q n is constructed from two disjoint copies of LT Q n according to the following steps: Denote 0LT Q n (respectively, LT Q n ) the graph obtained by prefixing the label of each vertex in one copy of LT Q n with 0 (respectively, ). Each vertex x = 0x n x n 3 x 0 in 0LT Q n is connected with the vertex (x n x 0 )x n 3 x 0 in LT Q n by an edge LT Q LT Q 4 Figure : Locally twisted cubes LT Q 3 and LT Q 4. Constructing ISTs on LT Q n in parallel Let denote the modulo addition. For n, the n-dimensional locally twisted cube LT Q n is a graph with {0, } n as its vertex set, and two vertices x = x n x n x 0 and y = y n y n y 0 are adjacent in LT Q n if and only if either () there is an integer i {, 3,..., n } such that x i = ȳ i and x i = y i x 0, and x j = y j for all remaining bits, or () there is an integer i {0, } such that x i = ȳ i, and x j = y j for all remaining bits. If one of the above conditions is fulfilled, then y is called the i-neighbor of x and is denoted by y = N i (x). Figure shows the graphs LT Q 3 and LT Q 4, respectively. The locally twisted cube can be equivalently defined by the following recursive fashion: From the above definition, it is clear that LT Q n is an n-regular graph, and the binary strings of any two adjacent vertices in LT Q n differ in at most two successive bits. Yang et al. [47] showed that LT Q n has a connectivity of n. Also, from [4], we know that LT Q n possesses the property of evenodd-vertex-transitive, i.e., for every pair of vertices x = x n x n x 0 and y = y n y n y 0 with the same parity (i.e., x 0 = y 0 ), there is an automorphism that maps x to y. More previous results on LT Q n can be found in the literature, e.g., the studies of diagnosability [39], mesh embedding [], fault-hamiltonicity [4], panconnectivity [5], pancyclicity and fault-pancyclicity [, 5, 6, 30, 38, 48]. In this paper, we also use the following notation. Two paths P and Q joining two distinct vertices x and y are internally vertex-disjoint, denoted by P Q, if V (P ) V (Q) = {x, y}. Let T be a spanning tree rooted at a vertex r of LT Q n. The parent of a vertex x( r) in T is denoted by parent(t, x). For x, y V (T ), the unique path from x to y is denoted by T [x, y]. Hence, two spanning trees T and T with the same root r are ISTs if and only if T [x, r] T [x, r] for every vertex x V (T ) \ {r}. 57

3 Since LT Q n has connectivity n, the root in each spanning tree must have a unique child. Let Z n = {0,,..., n }. For i Z n, we denote T i as the tree such that the root r takes its i-neighbor N i (r) as the unique child. Let N i (r) = c n c n c 0. For each vertex x = x n x n x 0 V (T i ) \ {r}, we define I i (x) = {j Z n : x j c j } and α i (x) = {j I i (x): j > i}. Moreover, if i 0, c 0 = and α i (x) is odd, we let H i (x) = (I i (x) {i}) \ (I i (x) {i}); otherwise, let H i (x) = I i (x). Also, we define the following function: next(i, x) = i if H i (x) = ; max H i (x) if H i (x) and i < min H i (x); max{j H i (x): j i} otherwise. That is, we regard H i (x) as a cyclic ordered set in decreasing order. If H i (x) = or i H i (x), the function outputs i; otherwise, the function outputs the next element in the cyclic order of H i (x) with respect to i. In what follows, we present a fully parallelized algorithm for constructing n spanning trees with an arbitrary vertex r = r n r n r 0 as their common root in LT Q n. For each vertex x V (LT Q n )\ {r} with binary string x = x n x n x 0, the construction can be carried out by describing the parent of x in each spanning tree T i. i = and c 0 =, we have H (x) = I (x) \ {} for x {0,,, 3} and H (x) = I (x) {} for x {4, 5, 6, 7}. Also, we have H (x) = I (x) for 8 x 5 and x. Consequently, we can determine j = 0 when x {0, }; j = when x {, 3, 4, 5}; j = 3 when x = ; and j = otherwise. For x {, 5, 7, 9, 3}, since j and x 0 =, we have parent(t, x) = x + ( ) xj j + ( ) xj j according to Line 5 of the algorithm. Otherwise, we have parent(t, x) = x+( ) xj j according to Line 7 of the algorithm. Table summarizes the information for constructing T i for 0 i 3 in LT Q 4. Figure 3 illustrates the construction for LT Q 4 using Algorithm Constructing-ISTs. For convenience, we adopt the notation x ±i y (respectively, y) to mean that x ± i = y (respectively, x ± i ± i = y) and x and y are adjacent in x ±i ± i LT Q n. For instance, we have T [6, ] = in Figure T 0 T Algorithm Constructing-ISTs Input: All vertices of LT Q n and an arbitrary root r = r n r n r 0. Output: n ISTs T 0, T,..., T n root at r. : for i = 0 to n do in parallel /* construct T i simultaneously */ : for each vertex x = x n x n x 0 in LT Q n do in parallel /* generate parent of each vertex x simultaneously */ 3: j = next(i, x) 4: if j and x 0 = then 5: parent(t i, x) = x + ( ) x j j + ( ) x j j 6: else 7: parent(t i, x) = x + ( ) x j j T T Figure 3: Four ISTs of LT Q 4. Figure : Algorithm for constructing n spanning trees in LT Q n. Example. Consider LT Q 4 and suppose we choose r = 0 = as the common root. For conciseness, we represent a vertex in LT Q 4 by decimal. We describe how the Constructing- ISTs algorithm constructs T in LT Q 4 as follows. Clearly, the -neighbor of r is N () = c 3 c c c 0 = 0 = 3. Let x( r) be an arbitrary vertex in LT Q 4 with x = x 3 x x x 0. If 0 x 7, then α (x) = ; otherwise, α (x) = 0. Since 3 Correctness In this section, we will show the validity of the algorithm. Firstly, we prove the reachability between every vertex x( r) and the vertex r in T i, thereby proving the existence of a unique path from x to the root in the tree. Theorem. Let r be an arbitrary vertex of LT Q n. The construction of T i for i Z n are spanning trees with a common root at r. 58

4 Table : The parent of vertices x V (LT Q 4) \ {} in T i with root r = 0 =. i = 3, N 3() = 0 = 7 x binary string I 3(x) α 3(x) H 3(x) j = next(3, x) x 0 x j x j parent(t 3, x) {0,, } even {0,, } = 0 + = {, } even {, } 0 0 = + + = {0, } even {0, } 0 0 = + = {} even {} 0 = 3 + = {0, } even {0, } = 4 + = {} even {} 0 = 5 + = {0} even {0} = = even 3 0 = = {0,,, 3} even {0,,, 3} = 8 3 = {,, 3} even {,, 3} 3 0 = = {0,, 3} even {0,, 3} = 0 3 = 00 {0,, 3} even {0,, 3} 3 0 = 3 = {, 3} even {, 3} 3 = 3 3 = 4 0 {0, 3} even {0, 3} 3 0 = 4 3 = 6 5 {3} even {3} 3 = 5 3 = 3 i =, N () = 0 = 3 x binary string I (x) α (x) H (x) j = next(, x) x 0 x j x j parent(t, x) {0,, 3} odd {0, 3} = = 000 {, 3} odd {3} = = {0,,, 3} odd {0,, 3} 0 0 = = {,, 3} odd {, 3} = 3 = {0, 3} odd {0,, 3} 0 0 = 4 = {3} odd {, 3} 0 = 5 + = {0,, 3} odd {0,,, 3} 0 = 6 = 7 0 {, 3} odd {,, 3} = 7 = {0, } even {0, } = 8 + = 9 00 {} even {} 0 0 = = {0,, } even {0,, } 0 0 = 0 + = 4 00 {0} even {0} = + 0 = even 0 = 3 + = 4 0 {0, } even {0, } 0 0 = 4 = 5 {} even {} = 5 = 3 i =, N () = 00 = 9 x binary string I (x) α (x) H (x) j = next(, x) x 0 x j x j parent(t, x) {0, 3} odd {0,, 3} = 0 + = 000 {3} odd {, 3} 0 = + = {0,, 3} odd {0, 3} = + 0 = {, 3} odd {3} = = {0,, 3} even {0,, 3} = = {, 3} even {, 3} 3 0 = = {0,,, 3} even {0,,, 3} 0 0 = 6 = {,, 3} even {,, 3} = 7 = {0} even {0} = = even 0 = 9 + = 0 00 {0, } even {0, } 0 0 = 0 = 8 00 {0, } odd {0,, } = + = {} odd {, } 0 = 3 + = {0,, } odd {0, } = = 5 5 {, } odd {} = 5 = 9 i = 0, N 0() = 00 = 0 x binary string I 0(x) α 0(x) H 0(x) j = next(0, x) x 0 x j x j parent(t 0, x) {, 3} even {, 3} = = {0,, 3} even {0,, 3} 0 - = 0 = {3} odd {3} = + 3 = {0, 3} odd {0, 3} 0 - = 3 0 = {,, 3} odd {,, 3} = = 5 00 {0,,, 3} odd {0,,, 3} 0 - = 5 0 = {, 3} even {, 3} = = {0,, 3} even {0,, 3} 0 - = 7 0 = {} odd {} = 8 + = {0, } odd {0, } 0 - = 9 0 = even = = 00 {, } even {, } 0 0 = = {0,, } even {0,, } 0 - = 3 0 = 4 0 {} odd {} 0 = 4 = 0 5 {0, } odd {0, } 0 - = 5 0 = 4 59

5 Proof. From Constructing-ISTs, we know that every vertex v V (LT Q n ) implies v T i. It follows that T i is a spanning subgraph of LT Q n. Hereafter, for every vertex v V (LT Q n ) \ {r}, all indices of elements of H i (v) are taken modulo H i (v). Let x = x n x n x 0 be any vertex of LT Q n. We claim that T i [x, r] is the unique path connecting x and r in T i. Clearly, if H i (x) =, then x = N i (r) and next(i, x) = i. If i and x 0 =, let J = ( ) xi i +( ) xi i ; otherwise, let J = ( ) xi i. Thus, T i [x, r] = x J r is the desired path that connects x and r in T i. Next, we suppose that H i (x) = {j 0, j,..., j p } is nonempty and j 0 < j < < j p. Consider the following two cases: Case : i / H i (x). We assume that j k = next(i, x) is the next element in the cyclic order of H i (x) with respect to i, where 0 k p. If j k and x 0 =, let J = ( ) xj k j k + ( ) xj k jk ; otherwise, let J = ( ) xj k j k. Thus, we have parent(t i, x) = x + J. Let y = y n y n y 0 = x + J be such a vertex and consider the following scenarios for H i (y). For j k and x 0 =, if j k = j k, then H i (y) = H i (x) \ {j k, j k }; otherwise, H i (y) = (H i (x) \ {j k }) {j k }. On the other hand (i.e., j k {0, } or x 0 = 0), we have H i (y) = H i (x) \ {j k }. By a similar argument, we let j l = next(i, y) and z = z n z n z 0 = parent(t i, y) = y + J, where J = ( ) xj l j l + ( ) xj l jl for j l and y 0 = ; or J = ( ) xj l j l otherwise. Again, we can determine H i (z) according to j l, y 0 and the condition j l = j l or j l j l. By this way, we find a sequence of vertices y, z,..., w in T i such that H i (w) =, and thus w = N i (r). Recall that we have constructed T i [w, r] = w J r for connecting N i (r) and r in T i. Therefore, we obtain the following unique path that connects x and r in T i : T i [x, r] = x J y J z J3 J d w J r. Case : i H i (x). Suppose i = j k for some k {0,,..., p }. In this case, we have next(i, x) = i by definition. If i and x 0 =, let Ĵ = ( )xi i + ( ) xi i ; otherwise, let Ĵ = ( )xi i. Moreover, we let y = parent(t i, x) = x + Ĵ. Clearly, j k / H(y), and thus y is in the situation of Case. Let P = T i [y, r] be the path connecting y and r in T i. Therefore, we obtain the unique path T i [x, r] by concatenating x Ĵ y and P. To show the independency of T 0, T,..., T n, we assume that p, q Z n are any two distinct integers. Let x = x n x n x 0 V (LT Q n ) \ {r} be any vertex and two paths P = T p [x, r] and Q = T q [x, r] be constructed in Theorem. Without loss of generality, we assume p > q. Let c = N p (r) = c n c n c 0 and d = N q (r) = d n d n d 0. For notational convenience, the subpath of P between two vertices u, v V (P ) is denoted by P (u, v). Moreover, we write y i P (u, v) = b for i Z n and b {0, } to mean that y i is assigned to b for every vertex y = y n y n y 0 in the path P (u, v). Similarly, we can define Q(u, v) and y i Q(u, v) = b by the same way. In what follows, we always assume that y = y n y n y 0 is any vertex in P \ {x, r} and z = z n z n z 0 is any vertex in Q \ {x, r}. In particular, let ŷ = ŷ n ŷ n ŷ 0 (respectively, ẑ = ẑ n ẑ n ẑ 0 ) be the vertex adjacent to x in P (respectively, in Q). To show that P Q, it suffices to prove that P (ŷ, c) Q(ẑ, d) =. For P and Q, we also use underscore to mark the different bits between the two paths. Note that we omit the proof if one of P (ŷ, c) and Q(ẑ, d) is a null path (i.e., H p (x) = or H q (x) = ). Moreover, due to the space limitation, we only prove Lemmas, 3, 4 and 5, and omit the following four cases: (i) c 0 = d 0 = and x 0 = 0; (ii) c 0 = and d 0 = x 0 = 0; (iii) c 0 = d 0 = x 0 = ; and (iv) c 0 = x 0 = and d 0 = 0. Lemma. If c 0 = d 0 = x 0 = 0, then P Q. Proof. Since p > q and c 0 = d 0 = 0, it implies r 0 = 0 and p > q 0. Moreover, since c 0 = d 0 = 0, it follows that H p (x) = I p (x) and H q (x) = I q (x). In addition, we have c p d p, c q d q and c i = d i for i Z n \ {p, q}. Let p = next(p, x) and q = next(q, x). We consider the following three scenarios. Case : x p c p (i.e., p H p (x)). In this case, p = p and q q. Since x 0 = 0, by Line 7 of the algorithm, we have ŷ = x + ( ) xp p. Thus, ŷ p = c p. Moreover, since T p takes ( ) cp p as the last link to connect the root, it implies y p P (ŷ, c) = c p. On the other hand, since x p c p and c p d p, we have x p = d p. Thus, Q never changes the bit z p in the path and z p Q(ẑ, d) = d p c p. As a result, P Q. (For example, consider P = T 3 [0, 4] : and Q = T [0, 4] : in LT Q 4.) Case : x q d q (i.e., q H q (x)). In this case, p p and q = q. Since x q d q and c q d q, we have x q = c q. Thus, P never changes the bit y q in the path and y q P (ŷ, c) = c q. On the other hand, since x 0 = 0, by Line 7 of the algorithm, we have ẑ = x + ( ) xq q. Thus, ẑ q = d q. Moreover, since T q takes ( ) dq q as the last link to connect the root, it implies z q Q(ẑ, d) = d q c q. As a result, P Q. (For example, consider P = T [, 4] : and Q = T [, 4] : in LT Q 4.) 60

6 Case 3: x p = c p and x q = d q (i.e., p / H p (x) and q / H q (x)). In this case, p > p and q > q. Clearly, x p c p and x q d q. There are three subcases as follows. Case 3.: p = q. Since x 0 = 0, by Line 7 of the algorithm, we have ŷ = x + ( ) x p p. Thus, ŷ p = c p. Moreover, since it remains unchanged the bit y p for every vertex y P (ŷ, c), we have y p P (ŷ, c) = c p (i.e., y q P (ŷ, c) = c q ). On the other hand, since x q = d q and T q takes ( ) dq q as the last link to connect the root, it implies z q Q(ẑ, d) = d q c q. Thus, P Q. (For example, consider P = T 3 [0, 4] : and Q = T [0, 4] : in LT Q 4.) Case 3.: q = p. Since x p = c p and T p takes ( ) cp p as the last link to connect the root, it implies y p P (ŷ, c) = c p. On the other hand, since x 0 = 0, by Line 7 of the algorithm, we have ẑ = x+( ) x q q. Thus, ẑ q = d q. Moreover, since it remains unchanged the bit z q for every vertex z Q(ẑ, d), we have z q Q(ẑ, d) = d q (i.e., z p Q(ẑ, d) = d p c p ). Thus, P Q. (For example, consider P = T 3 [0, 4] : and Q = T [0, 4] : in LT Q 4.) Case 3.3: p q and q p. Clearly, c p = d p and c q = d q. In this case, an argument similar to Case 3. shows that y p P (ŷ, c) = c p. Also, an argument similar to Case 3. shows that y p Q(ŷ, c) = c p. On the other hand, since x p = c p d p, x p c p = d p and d 0 = 0, we have p, p H q (x). Let w(= w n w n w 0 ) and w be vertices on Q such that w = w + ( ) wp p. Since Q(ẑ, w) has not dealt with the bit z p for every vertex z Q(ẑ, w), we have z p Q(ẑ, w) = x p c p. Also, since Q(w, d) has dealt with the bit z p, we have z p Q(w, d) = d p c p. Thus, P (ŷ, c) (Q(ẑ, w) Q(w, d)) = and P Q. (For example, consider P = T 4 [6, 4] : and Q = T [6, 4] : in LT Q 5.) Lemma 3. If c 0 = x 0 = 0 and d 0 =, then P Q. Proof. Since p > q, c 0 = x 0 = 0 and d 0 =, it implies r 0 = 0 and p > q = q = 0, where q = next(q, x). Moreover, since x 0 d 0, it follows that 0 I 0 (x)(= H 0 (x)). Since x 0 = c 0, P never changes the bit y 0 in the path, and thus y 0 P (ŷ, c) = c 0 = 0. On the other hand, by Line 7 of the algorithm, we have ẑ = x + ( ) x0 0 = x +. Thus, ẑ 0 =. Moreover, since it remains unchanged the bit z 0 for every vertex z Q(ẑ, d), we have z 0 Q(ẑ, d) =. Thus, P Q. (For example, consider P = T 3 [0, 4] : and Q = T 0 [0, 4] : in LT Q 4.) Lemma 4. If c 0 = d 0 = 0 and x 0 =, then P Q. Proof. Since p > q and c 0 = d 0 = 0, it implies r 0 = 0 and p > q 0. Moreover, since c 0 = d 0 = 0, it follows that H p (x) = I p (x) and H q (x) = I q (x). In addition, we have c p d p, c q d q and c i = d i for i Z n \ {p, q}. Let p = next(p, x) and q = next(q, x). We consider the following three scenarios. Case : x p c p (i.e., p H p (x)). In this case, p = p. Since x 0 =, by Line 5 of the algorithm, we have ŷ = x+( ) xp p +( ) xp p. Thus, ŷ p = c p. Moreover, since T p takes ( ) cp p as the last link to connect the root, it implies y p P (ŷ, c) = c p. On the other hand, since x p c p and c p d p, we have x p = d p. Thus, Q never changes the bit z p in the path and z p Q(ẑ, d) = d p c p. As a result, P Q. (For example, consider P = T 3 [, 4] : and Q = T [, 4] : in LT Q 4.) Case : x p = c p and x q d q (i.e., p / H p (x) and q H q (x)). In this case, p > p and q = q. Moreover, we have x p c p. Since x q d q and c q d q, it implies x q = c q, and thus p q and c p = d p. There are two subcases as follows. Case.: p > q. In this case, we have p. Since x 0 =, by Line 5 of the algorithm, we have ŷ = x + ( ) x p p + ( ) x p p. Thus, ŷ p = c p. Moreover, since P never changes the bit y p in the succedent path again, it follows that y p P (ŷ, c) = c p. Also, since T p takes ( ) cp p as the last link to connect the root, it implies y p P (ŷ, c) = c p. On the other hand, let w(= w n w n w 0 ) and w be vertices on Q such that w = w + ( ) wp p. Since Q(ẑ, w) has not dealt with the bit z p for every vertex z Q(ẑ, w), we have z p Q(ẑ, w) = x p c p. Also, since Q(w, d) has dealt with the bit z p, we have z p Q(w, d) = d p c p. Thus, P (ŷ, c) (Q(ẑ, w) Q(w, d)) = and P Q. (For example, consider P = T 3 [9, 4] : and Q = T [9, 4] : in LT Q 4.) Case.: q > p. Since x q = c q and p > q = q > p, the path P never changes the bit y q. Thus, y q P (ŷ, c) = x q. On the other hand, since x 0 =, by Line 5 of the algorithm, we have ẑ = x + ( ) xq q + ( ) xq q. Thus, ẑ q = d q. Moreover, T q takes ( ) dq q as the last link to 6

7 connect the root, it follows that z q Q(ẑ, d) = d q x q. As a result, P Q. (For example, consider P = T 3 [5, 4] : and Q = T [5, 4] : in LT Q 4.) Case 3: x p = c p and x q = d q (i.e., p / H p (x) and q / H q (x)). Since x 0 d 0, it follows that p > q > q 0, and thus q p. Moreover, we have x p c p and x q d q. There are two subcases as follows. Case 3.: p = q. By Line 5 or Line 7 of the algorithm, P first changes the bit x p. Thus, ŷ p = c p. Moreover, since P never changes the bit y p in the succedent path again, it follows that y p P (ŷ, c) = c p = c q. On the other hand, since x q = d q and T q takes ( ) dq q as the last link to connect the root, it follows that z q Q(ẑ, d) = d q c q. As a result, P Q. (For example, consider P = T 3 [9, 4] : and Q = T [9, 4] : in LT Q 4.) Case 3.: p q. In this case, the proof is similar to Case.. That is, we can show that y p P (ŷ, c) = c p and y p P (ŷ, c) = c p. On the other hand, z p Q(ẑ, w) = x p c p and z p Q(w, d) = d p c p, where w(= w n w n w 0 ) and w = w + ( ) wp p are two adjacent vertices on Q. (For example, consider P = T 3 [, 4] : and Q = T [, 4] : in LT Q 4.) Lemma 5. If c 0 = 0 and d 0 = x 0 =, then P Q. Proof. Since p > q and c 0 = 0, we have r 0 = 0. Also, since r 0 d 0 and d 0 = x 0, it follows that q = 0 and q q. Moreover, c 0 = 0 and q = 0 imply H p (x) = I p (x) and H 0 (x) = I 0 (x). In addition, we have c p d p, c 0 d 0 and c i = d i for i Z n \ {p, 0}. Let p = next(p, x) and q = next(q, x). We consider the following two scenarios. Case : x p c p (i.e., p H p (x)). In this case, p = p. Since c p d p and x p c p, it implies x p = d p. Since x q d q, we have q p. There are two subcases as follows. Case.: p > q. In this case, we have p = p > q > q = 0. Since p and x 0 =, by Line 5 of the algorithm, we have ŷ = x + ( ) xp p + ( ) xp p. Thus, ŷ p = c p. Moreover, since T p takes ( ) cp p as the last link to connect the root, it follows that y p P (ŷ, c) = c p. On the other hand, since x p = d p and Q never changes the bit z p in the path, we have z p Q(ẑ, d) = d p c p. This shows that P Q. (For example, consider P = T 3 [, 4] : and Q = T 0 [, 4] : in LT Q 4.) Case.: q > p. In this case, we have q > p = p > q = 0. Clearly, P first changes the bit x p in x and takes ( ) cp p as the last link to connect the root of T p. Let w(= w n w n w 0 ) and w be vertices on P such that w = w+( ) wq q = w+ ( ) w0. Since P (ŷ, w) has not dealt with the bit y q for every vertex y P (ŷ, w), we have y q P (ŷ, w) = x q d q. Also, since P (w, c) has dealt with the bit y q (= y 0 ), we have y 0 P (w, c) = c 0 = 0. On the other hand, since q and x 0 =, by Line 5 of the algorithm, we have ẑ = x + ( ) x q q + ( ) x q q. Thus, ẑ q = d q. Moreover, since Q never changes the bit z q in the succedent path again, it follows that z q P (ẑ, d) = d q. Also, since x 0 = d 0 = and Q takes ( ) dq q (= ) as the last link to connect the root of T 0, it follows that z 0 Q(ẑ, d) = d 0 =. Thus, (P (ŷ, w) P (w, c)) Q(ẑ, d) = and P Q. (For example, consider P = T [5, 4] : and Q = T 0 [5, 4] : in LT Q 4.) Case : x p = c p (i.e., p / H p (x)). In this case, p p. Clearly, x p c p. There are three subcases as follows. Case.: p = q. In this case, we have p > p = q = 0. By Line 7 of the algorithm, we have ŷ = x + ( ) x p p = x. Thus, ŷ 0 = 0. Moreover, since P never changes the bit y 0 in the succedent path again, it follows that y 0 P (ŷ, c) = 0. On the other hand, since x 0 = and T q takes ( ) dq q (= ) as the last link to connect the root, it follows that z 0 Q(ẑ, d) =. This shows that P Q. (For example, consider P = T [5, 4] : and Q = T 0 [5, 4] : in LT Q 4.) Case.: q = p. In this case, we have q = p > q = 0. Since x p = c p and T p takes ( ) cp p as the last link to connect the root, it follows that y p P (ŷ, c) = c p. On the other hand, since Q first changes the bit x q (= x p ) in x and never changes the bit z p in the succedent path again, it follows that z p Q(ẑ, d) = d p c p. This shows that P Q. (For example, consider P = T 3 [, 4] : 0 + T 0 [, 4] : and Q = in LT Q 4.) Case.3: p q and q p. In this case, we have q > p > p > q = 0. Since P first changes the bit x p in x, we have ŷ p = c p. Moreover, since P never changes the bit y p in the succedent path again, it follows that y p P (ŷ, c) = c p. Also, since x p = c p and P takes ( ) cp p as the last link 6

8 to connect the root of T p, we have y p P (ŷ, c) = c p. On the other hand, let w(= w n w n w 0 ) and w be vertices on Q such that w = w + ( ) wp p. Since Q(ẑ, w) has not dealt with the bit z p for every vertex z Q(ẑ, w), we have z p Q(ẑ, w) = x p c p. Also, since Q(w, d) has dealt with the bit z p, we have z p Q(w, d) = d p c p. This shows that P Q. (For example, consider P = T [, 4] : and Q = T 0 [, 4] : in LT Q 4.) From the above lemmas, we conclude that ISTs constructed in this paper are independent. According to Theorem and the result of independency, we obtain the following main theorem. Theorem 6. Let N = n. For LT Q n and an arbitrary vertex r V (LT Q n ), Algorithm Constructing-ISTs can correctly construct n ISTs rooted at r in O(N log N) time. In particular, the algorithm can be parallelized on LT Q n by using N processors to run in O(log N) time. 4 Concluding remarks This paper provides a parallel construction of ISTs rooted at an arbitrary vertex of locally twisted cubes. Indeed, all ISTs constructed in here are isomorphic to those in [3, 4], which have height n +. As we have mentioned earlier, many algorithms have been proposed for constructing ISTs on some interconnection networks. However, some of these networks are not vertex-transitive, and thus the desired ISTs are in need of an arbitrary vertex as the root [4 7, 4, 34, 35]. Although most of these algorithms can simultaneously construct ISTs in parallel, the construction of each spanning tree relies on a recursive expansion and is hard to be fully parallelized. To the best of our knowledge, for class of networks without vertex-transitivity, the present paper is the first to propose the fully parallelized approach for constructing ISTs. References [] F. Bao, Y. Funyu, Y. Hamada, and Y. Igarashi, Reliable broadcasting and secure distributing in channel networks, in: Proc. of 3rd International Symposium on Parallel Architectures, Algorithms and Networks, ISPAN 97, Taipei, December 997, pp [] Q.-Y. Chang, M.-J. Ma, and J.-M. Xu, Faulttolerant pancyclicity of locally twisted cubes (in Chinese), J. China Univ. Sci. Tech., 36 (006) [3] X.-B. Chen, Parallel construction of optimal independent spanning trees on Cartesian product of complete graphs, Inform. Process. Lett., (0) [4] B. Cheng, J. Fan, X. Jia, and J. Jia, Parallel construction of independent spanning trees and an application in diagnosis on Möbius cubes, J. Supercomput., 65 (03) [5] B. Cheng, J. Fan, X. Jia, and J. Wang, Dimension-adjacent trees and parallel construction of independent spanning trees on crossed cubes, J. Parallel Distrib. Comput., 73 (03) [6] B. Cheng, J. Fan, X. Jia, and S. Zhang, Independent spanning trees in crossed cubes, Inform. Sci., 33 (03) [7] B. Cheng, J. Fan, X. Jia, S. Zhang, and B. Chen, Constructive algorithm of independent spanning trees on Möbius cubes, Comput. J., 56 (03) [8] J. Cheriyan and S.N. Maheshwari, Finding nonseparating induced cycles and independent spanning trees in 3-connected graphs. J. Algorithms, 9 (988) [9] S. Curran, O. Lee, and X. Yu, Finding four independent trees. SIAM J. Comput., 35 (006) [0] Z. Ge and S.L. Hakimi, Disjoint rooted spanning trees with small depths in debruijn and Kautz graphs, SIAM J. Comput., 6 (997) [] Y. Han, J. Fan, S. Zhang, J. Yang, and P. Qjan, Embedding meshes into locally twisted cubes, Inform. Sci., 80 (00) [] T. Hasunuma and H. Nagamochi, Independent spanning trees with small depths in iterated line graphs, Discrete Appl. Math., 0 (00) 89. [3] S.-Y. Hsieh and C.-J. Tu, Constructing edgedisjoint spanning trees in locally twisted cubes, Theoret. Comput. Sci., 40 (009) [4] S.-Y. Hsieh and C.-Y. Wu, Edge-fault-tolerant Hamiltonicity of locally twisted cubes under conditional edge faults, J. Combin. Optim. 9 (00) [5] K.S. Hu, S.-S. Yeoh, C. Chen, and L.-H. Hsu, Node-pancyclicity and edge-pancyclicity of hypercube variants, Inform. Process. Lett., 0 (007) 7. [6] A. Huck, Independent trees in graphs, Graphs Combin., 0 (994) [7] A. Huck, Independent trees in planar graphs, Graphs Combin., 5 (999) [8] A. Huck, Independent branching in acyclic digraphs, Discrete Math, 99 (999) [9] A. Itai and M. Rodeh, The multi-tree approach to reliability in distributed networks, Inform. Comput., 79 (988) [0] Y. Iwasaki, Y. Kajiwara, K. Obokata, and Y. Igarashi, Independent spanning trees of 63

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