Constructing independent spanning trees for locally twisted cubes

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Contents lists available atScienceDirect

Theoretical Computer Science

journal homepage:www.elsevier.com/locate/tcs

Constructing independent spanning trees for locally twisted cubes

✩

Yi-Jiun Liu, James K. Lan, Well Y. Chou, Chiuyuan Chen

∗ Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan

a r t i c l e i n f o Article history: Received 11 December 2009 Accepted 27 December 2010 Communicated by D.-Z. Du Keywords:

Independent spanning trees Data broadcasting

Design and analysis of algorithms Locally twisted cubes

Hypercubes Hypercube variants Parallel algorithm

a b s t r a c t

The independent spanning trees (ISTs) problem attempts to construct a set of pairwise independent spanning trees and it has numerous applications in networks such as data broadcasting, scattering and reliable communication protocols. The well-known ISTs conjecture, Vertex/Edge Conjecture, states that any n-connected/n-edge-connected graph has n vertex-ISTs/edge-ISTs rooted at an arbitrary vertex r. It has been shown that the Vertex Conjecture implies the Edge Conjecture. In this paper, we consider the independent spanning trees problem on the n-dimensional locally twisted cube LTQn. The very recent

algorithm proposed by Hsieh and Tu (2009) [12] is designed to construct n edge-ISTs rooted at vertex 0 for LTQn. However, we find out that LTQnis not vertex-transitive when n≥4;

therefore Hsieh and Tu’s result does not solve the Edge Conjecture for LTQn. In this paper,

we propose an algorithm for constructing n vertex-ISTs for LTQn; consequently, we confirm

the Vertex Conjecture (and hence also the Edge Conjecture) for LTQn.

1. Introduction

Two spanning trees in a graph G are said to be vertex/edge independent if they are rooted at the same vertex r and for each vertex

v

of G,

v ̸=

r, the paths from r to

v

in two trees are vertex/edge disjoint except the two end vertices. A set of spanning trees of G are said to be vertex/edge independent if they are pairwise vertex/edge independent. The vertex/edge independent spanning trees (ISTs) problem attempts to construct a set of pairwise vertex/edge independent spanning trees and it has has applications such as data broadcasting, scattering and reliable communication protocols. For example, a rooted spanning tree in the underlying graph of a network can be viewed as a broadcasting scheme for data communication and fault-tolerance can be achieved by sending n copies of the message along the n independent spanning trees rooted at the source node [1]. For other applications, see [3] for the multi-node broadcasting problem, [21] for one-to-all broadcasting, and [2] for n-channel graphs, reliable broadcasting and secure message distribution.

The independent spanning trees problem has been widely studied in the last two decades. Two well-known conjectures on this problem are raised by Zehavi and Itai [27]: (refer to [4] or [23] for graph terminologies)

Conjecture 1.1 (Vertex Conjecture). Any n-connected graph has n vertex-ISTs rooted at an arbitrary vertex r. Conjecture 1.2 (Edge Conjecture). Any n-edge-connected graph has n edge-ISTs rooted at an arbitrary vertex r.

Zehavi and Itai [27] also raised the question: It would be interesting to show that either the Vertex Conjecture implies the Edge Conjecture, or vice versa. Later, Khuller and Schieber [16] successfully proved that the Vertex Conjecture implies the Edge Conjecture, i.e., if any n-connected graph has n vertex-ISTs, then any n-edge-connected graph has n edge-ISTs. Khuller

✩ _{This research was partially supported by the National Science Council of the Republic of China under the grants grant NSC97-2628-M-009-006-MY3.}

∗_{Corresponding author. Tel.: +886 3 5731767.}

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Table 1

The connectivity, edge-connectivity and diameters of Qn and its variants. Topology κ(G) λ(G) Diameter Qn n n n LTQn n n ⌈(n+1)/2⌉ if n<5 ⌈(n+3)/2⌉ if n≥5 TQn n n ⌈(n+1)/2⌉ MQn n n ⌈(n+2)/2⌉ in 0-MQnfor n≥4 ⌈(n+1)/2⌉ in 1-MQnfor n≥1

and Schieber’s proof also works for the directed graphs. For the directed case, Edmonds [7] solved the Edge Conjecture. Khuller and Schieber [16] pointed out that the Vertex Conjecture for directed graphs is the strongest conjecture since it implies all the other conjectures.

The vertex and the edge conjectures have been confirmed only for n

≤

4. In particular, in [15], Itai and Rodeh proposed a linear-time algorithm for constructing two edge-ISTs for a 2-edge-connected graph; they also solved the Vertex Conjecture for n

=

2. In [27], Zehavi and Itai solved the Vertex Conjecture for n

=

3, but they did not proposed an algorithm for constructing three vertex-ISTs. In [6], Cheriyan and Maheshwari proposed an O

(|

V

(

G

)|

2

)

-time algorithm for constructing three vertex-ISTs in a 3-connected graph. In [5], Curran et al. proposed an O

(|

V

(

G

)|

3

₎

_{-time algorithm for constructing four}

vertex-ISTs in a 4-connected graph. When n

≥

5, both the vertex and the edge conjectures are still open. It has been proven that the Vertex/Edge Conjecture holds for several restricted classes of graphs or digraphs, such as planar graphs [9,10,17,18], maximal planar graphs [19], product graphs [20], chordal rings [14,24], de Bruijn and Kautz digraphs [8,11], and hypercubes [22,26]. Note that the development of algorithms for constructing vertex-ISTs tends toward pursuing two research goals: One is to design efficient construction schemes (for example, [14,17,19,24] proposed linear-time algorithms) and the other is to reduce the heights of vertex-ISTs (for example, [11,22,24] proposed the idea of height improvements).

The hypercube (Qn

)

is one of the most popular interconnection network topologies due to its simple structure and ease of implementation. Several commercial machines with hypercube topology have been built and a huge amount of research work, both theoretical and practical, has been done on various aspects of the hypercube. However, it has been shown that the hypercube does not achieve the smallest possible diameter for its resources. Therefore, many variants of the hypercube have been proposed. The most well-known variants are locally twisted cubes (LTQn), twisted cubes (TQn), crossed cubes (CQn) and Möbius cubes (MQn). A concise comparison including the connectivity, edge-connectivity and diameters of Qnand its variants is shown inTable 1. Clearly, one advantage of LTQnover Qnis that the diameter of LTQnis only about half of that of Qn.

Before going further, we now briefly review results of the vertex-ISTs problem for Qn. It is well known that Qn is n-connected. Since Qnis a product graph, the algorithm proposed by Obokata et al. [20] can be used to construct n vertex-ISTs for Qn. As to the construction of the height-reduced vertex-ISTs on Qn, Tang et al. [22] modified the algorithm in [20] and proposed an O

(

n2n

₎

_{-time algorithm for constructing an optimal set (in the sense of smallest average path lengths) of n} vertex-ISTs for Qn. It was pointed out by Yang et al. [26] that the algorithms in [20,22] are designed by a recursive fashion and such a construction forbids the possibility that the algorithm could be parallelized; Yang et al. [26] therefore proposed a parallel construction for an optimal set of n vertex-ISTs for Qn.

The purpose of this paper is to confirm the Vertex Conjecture for the n-dimensional locally twisted cube LTQn. The very recent algorithm proposed by Hsieh and Tu [12] is designed to construct n edge-ISTs rooted at vertex 0 for LTQn. However, we find out that LTQnis not vertex-transitive whenever n

≥

4 (see Section2). Therefore, Hsieh and Tu did not solve the Edge Conjecture for LTQn. In this paper, we will propose an algorithm for constructing n vertex-ISTs rooted at an arbitrary vertex of LTQn. Therefore, we will confirm the Vertex Conjecture for LTQn. Since vertex-ISTs are edge-ISTs, we also confirm the Edge Conjecture for LTQn.

In the remaining discussion, we will simply use ISTs to denote vertex-ISTs unless otherwise specified. This paper is organized as follows. In Section2, we give definitions and notations used in the paper. In Section3, we present an algorithm to construct n ISTs rooted at an arbitrary vertex of LTQn. In Section4, we prove the correctness of our algorithm. Concluding remarks are given in the last section.

2. Preliminaries

All graphs in this paper are simple undirected graphs. Let G be a graph with vertex set V

(

G

)

and the edge set E

(

G

)

. Let

x

,

y

∈

V

(

G

)

. A path from x to y is denoted as x

,

y-path. The distance between two vertices x and y, denoted by d

(

x

,

y

)

, is the length of a shortest x

,

y-path. Two x

,

y-paths P and Q are edge-disjoint if E

(

P

) ∩

E

(

Q

) = ∅

. Two x

,

y-paths P and Q are internally vertex-disjoint if V

(

P

) ∩

V

(

Q

) = {

x

,

y

}

. A subgraph T of G is a spanning tree if T is a tree and V

(

T

) =

V

(

G

)

. Two spanning trees T and T′of G are vertex-independent/edge-independent if T and T′are rooted at the same vertex, say r, and

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a

b

Fig. 1. (a) LTQ3. (b) A symmetric drawing of LTQ3.

Fig. 2. LTQ4and its perfect matchings{M0,M1,M2,M3}.

for each

v ∈

V

(

G

)

,

v ̸=

r, the r

, v

-path in T and the r

, v

-path in T′are (internally) vertex-disjoint/edge-disjoint. A set of spanning trees of G are vertex-independent/edge-independent if they are pairwise vertex-independent/edge-independent.

2.1. The locally twisted cube

The n-dimensional locally twisted cube LTQn(n

≥

2), proposed first by Yang et al. [25], has 2nvertices. Each vertex is an

n-string on

{

0

,

1

}

, i.e., a binary string of length n. The LTQnis defined recursively as follows.

Definition 1 ([25]). 1. LTQ2 is the graph consisting of four vertices labeled with 00, 01, 10, and 11, respectively, and

connected by the four edges (00, 01) (00, 10), (01, 11), and (10, 11).

2. LTQn (n

≥

3) is built from two disjoint copies of LTQn−1’s as follows: Let 0LTQn−1(respectively, 1LTQn−1) denote the

graph obtained by prefixing the label of each vertex in one copy of LTQn−1with 0 (respectively, 1). Connect each vertex

0xn−2xn−3

. . .

x0of 0LTQn−1to the vertex 1

(

xn−2

⊕

x0

)

xn−3

. . .

x0of 1LTQn−1with an edge, where ‘‘

⊕

’’ represents the XOR

operation, or equivalently, the modulo 2 addition.

Figs. 1and2illustrate LTQ3and LTQ4, respectively. Yang et al. [25] also mentioned that the locally twisted cube can be

equivalently defined by the following non-recursive fashion.

Definition 2 ([25]). Let x

=

xn−1xn−2

. . .

x0and y

=

yn−1yn−2

. . .

y0be two vertices of LTQn

(

n

≥

2

)

. Then vertices x and y are adjacent if and only if one of the following conditions are satisfied.

1. There is an integer 2

≤

k

≤

n

−

1 such that (a) xk

= ¯

yk(y

¯

kis the complement of ykin

{

0

,

1

}

) (b) xk−1

=

yk−1

⊕

x0

(c) all the remaining bits of x and y are identical.

2. There is an integer 0

≤

k

≤

1 such that x and y only differ in the kth bit.

FromDefinition 2, LTQnis obviously an n-regular graph, and the labels of any two adjacent vertices of LTQndiffer in at most two consecutive bits. Note that in the remaining part of this paper, the label of a vertex in LTQnis presented in binary representation and decimal representation interchangeably when there is no ambiguity.

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Fig. 3. The in-vertex-transitivity of LTQ4.

2.2. The neighbor information and the perfect matchings of the locally twisted cube

FromDefinition 2, the n neighbors of an arbitrary vertex x

=

xn−1xn−2

. . .

x0of LTQnis given by

f₀

(

x

) =

x_n−1xn−2xn−3

. . .

x2x1x0

,

f1

(

x

) =

xn−1xn−2xn−3

. . .

x2x1x0

,

f2

(

x

) =

xn−1xn−2xn−3

. . .

x2

(

x1

⊕

x0

)

x0

,

...

=

...

fn−2

(

x

) =

xn−1xn−2

(

xn−3

⊕

x0

)

xn−4

. . .

x1x0

,

fn−1

(

x

) =

xn−1

(

xn−2

⊕

x0

)

xn−3

. . .

x2x1x0

,

(1)

where fk

(

x

),

0

≤

k

≤

n

−

1, is called the kth dimensional neighbor of x; see also Lemma 4 in [13]. By (1), the n neighbors of vertices 0 and 1 can be determined as follows.

Lemma 2.1. The n neighbors of vertex 0 in LTQnis given by

fk

(

0

) =

2k

,

for k

=

0

,

1

, . . . ,

n

−

1. The n neighbors of vertex 1 in LTQnis given by

fk

(

1

) =











0 if k

=

0

,

3 if k

=

1

,

2k

₊

₂k−1

₊

₁ _{if 2}

_≤

_k

_≤

_n

₋

₁

_.

Given a graph G

=

(

V

,

E

)

, a matching M of G is a set of pairwise non-adjacent edges of G. A perfect matching is a matching that saturates all the vertices; in other words, every vertex in the graph is incident to exactly one edge in the matching. From Eq. (1), for all vertices x of LTQnand for all 0

≤

k

≤

n

−

1, we have

fk

(

fk

(

x

)) =

x

.

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Therefore, for a fixed k, the set of edges connecting a vertex and its k-th dimensional neighbor forms a perfect matching of

LTQn. More precisely,

Mk

= {

(

x

,

fk

(

x

) |

x

∈

V

(

LTQn

)}

is a perfect matching of LTQn. SeeFig. 2for an illustration.

2.3. The even–odd-vertex-transitivity of the locally twisted cube

A graph is vertex-transitive if for every pair of vertices u and

v

, there is an automorphism that maps u to

v

. Intuitively, a vertex-transitive network looks the same from every node. The vertex-transitive property is advantageous to the design and simulation of some algorithms. It is not difficult to see that LTQ2and LTQ3are vertex-transitive; seeFig. 1. However, in

the following, we will show that LTQnis not vertex-transitive when n

≥

4.

Theorem 2.2. The locally twisted cube LTQnis not vertex-transitive for n

≥

4.

Proof. For n

=

4, let Nk

(

r

)

denote the set Nk

(

r

) = {

x

∈

V

(

LTQn

) |

d

(

x

,

r

) =

k

}

. Consider the setΩ

(

r

) = {

x

∈

N2

(

r

) |

N1

(

x

) ∩

N1

(

r

) =

1 and N1

(

x

) ∩

N3

(

r

) =

1

}

. ThenΩ

(

0

) = {

7

}

, butΩ

(

1

) = {

6

,

12

}

; seeFig. 3for an illustration. Therefore

LTQ4is not vertex-transitive.

Now consider LTQn with n

≥

5. It is well-known that vertices 0 and 2n

−

2 are at the farthest distance of LTQnand

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Claim 2.3. For an arbitrary vertex x

∈

V

(

LTQn

)

, n

≥

5, the distance d

(

x

,

1

) ≤ 

n+1₂



.

Proof ofClaim 2.3. Before showing the claim, some notations are introduced first. Let x

=

xn−1xn−2

. . .

x0. Scanning the

bits of x from xn−1to x1(notice that we ignore the bit x0). Suppose there are a total of m bits equal to 1 and a total of k

disjoint pairs of consecutive bits equal to ‘‘11’’, we denoted it by ‘‘11’’-bits. A bit xi, 1

≤

i

≤

n

−

1, is said to be isolated if after removing the k disjoint pairs of ‘‘11’’-bits of x, we have xi

=

1. For example, consider x

=

111011 in LTQ6. Then m

=

4

,

k

=

1

and x1

,

x3are isolated. Clearly, 0

≤

k

≤



_m

2



holds.

It should be noticed that if m

< 

n−1₂



, then there exists a trivial path from x to 1: (i) If x0

=

0, then corrects all xi

=

1 bits, 1

≤

i

≤

n

−

1, to 0, and then corrects x0to 1; (ii) If x0

=

1, then corrects x0to 0. Then corrects all xi

=

1 bits, 1

≤

i

≤

n

−

1, to 0, and then correct x0to 1. Clearly, both paths have length at most m

+

2

≤



_n₊₁

2



. In the following, we assume m

≥



_n₋₁

2



. Therefore, m

−



n

−

1 2



≤

k

≤



m 2



holds. There are two cases.

Case 1: x0

=

0. A path from x to 1 can be found as follows: Step 1: Remove all the isolated bits of x. Step 2: Correct x0to

1. Step 3: Match all ‘‘11’’-bits. Clearly, Steps 1, 2 and 3 take m

−

2k, 1 and k steps, respectively. The total number of steps is

m

−

k

+

1

≤

m

−



m

−



n

−

1 2



+

1

=



n

+

1 2



.

For example, consider x

=

11101010 in LTQ8. We have m

=

5, k

=

1 and x1, x3, x5are isolated bits. A path from x to 1 is

built as follows: 11101010Step 1

−→

11001010Step 1

−→

11000010Step 1

−→

11000000Step 2

−→

11000001Step 3

−→

00000001.

Case 2: x0

=

1. We further divide this case into two subcases:

Subcase 2.1: m

+

1

−



n−1

2

 ≤

k

≤



_m

2



. Then a path from x to 1 can be found as follows: Step 1: Correct x0to 0. Step 2:

Remove all the isolated bits of x. Step 3: Correct x0to 1. Step 4: Match all ‘‘11’’-bits. Clearly, Steps 1, 2, 3 and 4 take 1, m

−

2k,

1 and k steps, respectively. Thus the total number of steps is

m

−

k

+

2

≤



n

+

1 2



.

=

5, k

=

2 and x1is a isolated bit. A path from x to 1 is built as

follows: 11011011Step 1

−→

11011010Step 2

−→

11011000Step 3

−→

11011001Step 4

−→

00011001Step 4

−→

00000001.

Subcase 2.2: k

=

m

−



n−1

2



. In this case, all bits xn−1

,

xn−3

, . . . ,

x1 must equal to 1 if n is even; either all bits

xn−2

,

xn−3

, . . . ,

x1or all bits xn−1

,

xn−3

, . . . ,

x2must equal to 1 if n is odd. Thus a path from x to 1 can be found by bitwise

correcting the bits to 0 (by scanning the bits from xn−1to x1). Since it takes one step to correct an isolated bit and one step

to correct a ‘‘11’’-bits, the total step is

(

m

−

2k

) +

k

=



n

−

1 2



.

=

5, k

=

1. A path from x to 1 is built as follows: 10111011 isolated

−→

01111011‘‘11’’-bits

−→

00011011‘‘11’’-bits

−→

00000011isolated

−→

00000001.

From the above discussion, we have d

(

x

,

1

) ≤ 

n+1₂



. As a result, LTQnis not vertex-transitive for n

≥

4.

Although LTQnfails to be vertex-transitive for n

≥

4, it does satisfy the even–odd-vertex-transitive property: for every pair of vertices x

=

xn−1xn−2

. . .

x0, y

=

yn−1yn−2

. . .

y0with the same parity, i.e., x0

=

y0, there is an automorphism

ψ

that maps

x to y. In other words, in LTQn, all even-numbered vertices are symmetric and all odd-numbered vertices are symmetric. By using this property, we may pay our attention of constructing ISTs to use vertex 0 and vertex 1 as the common root without loss of generality.

Theorem 2.4. The locally twisted cube LTQnsatisfies the even–odd-vertex-transitive property.

Proof. It suffices to prove that there exists an automorphism which maps

v (̸=

0

)

to 0 (resp.,

v (̸=

1

)

to 1), whenever

v

is an even-numbered (resp., odd-numbered) vertex. For two n-bits binary strings x and y, let x

⊕

y denote the bitwise XOR

(modulo 2) of x and y. Let

v = v

n−1

v

n−2

. . . v

0

∈

V

(

LTQn

)

.

Suppose

v

is an even-numbered vertex. For x

=

xn−1xn−2

. . .

x0

∈

V

(

LTQn

)

, define a function

ψ

0as follows:

ψ

0

(

x

) = v ⊕

x

.

It is not difficult to see that

ψ

0is a bijection from V

(

LTQn

)

to V

(

LTQn

)

. Now we verify that

ψ

0preserves the adjacency.

Consider any edge

(

x

,

fk

(

x

)) ∈

E

(

LTQn

)

. Since

v

0

=

0, we have

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Algorithm 1 Construct_IST

Input: All vertices of LTQnand root r.

Output: n ISTs T0

,

T1

, . . . ,

Tn−1rooted at r.

1: for i

=

0 to n

−

1 do in parallel ◃construct Tisimultaneously

2: child_of _the_root

←

fi

(

r

)

3: V

(

Ti

) ← {

child_of _the_root

}

4: for t

=

1 to n do ◃outer for-loop 5: S

← ∅

;

6: for each vertex

v ∈

V

(

Ti

)

do ◃inner for-loop 7: u

←

f₍i+t)mod n

(v)

8: E

(

Ti

) ←

E

(

Ti

) ∪ {(v,

u

)}

◃set the parent of vertex u asvin Ti

9: S

←

S

∪ {

u

}

10: end for 11: V

(

Ti

) ←

V

(

Ti

) ∪

S 12: end for 13: end for Also,

ψ

0

(

fk

(

x

)) =

(v

n−1

⊕

xn−1

) (v

n−2

⊕

xn−2

) . . . (v

1

⊕

u1

)

x0 if k

=

0

,

(v

n−1

⊕

xn−1

) (v

n−2

⊕

xn−2

) . . . (v

2

⊕

u2

) (v

1

⊕

x1

)

x0 if k

=

1

,

and for 2

≤

k

≤

n

−

1,

ψ

0

(

fk

(

x

)) = (v

n−1

⊕

xn−1

) (v

n−2

⊕

xn−2

) . . . (v

k+1

⊕

xk+1

) (v

k

⊕

xk

) (v

k−1

⊕

xk−1

⊕

x0

) (v

k−2

⊕

xk−2

) . . . (v

1

⊕

x1

)

x0

.

Since

v

k

⊕

xk

=

v

k

⊕

xkno matter

v

k

=

xkor

v

k

̸=

xk, we have

ψ

0

(

fk

(

x

)) =

fk

(ψ

0

(

x

))

and hence

(ψ

0

(

x

), ψ

0

(

fk

(

x

))) ∈

E

(

LTQn

)

.

Similar arguments can be applied to the case of

v

being an odd-numbered vertex, except that the bijection function from

V

(

LTQn

)

to V

(

LTQn

)

is replaced by

ψ

1

(

x

) = v ⊕

x

⊕

1

.

3. The algorithm

We now present an algorithm, called Construct_IST, for constructing n ISTs T0

,

T1

, . . . ,

Tn−1rooted at an arbitrary vertex

r for the locally twisted cube LTQninAlgorithm 1. For convenience, call the for-loop in lines4–12of this algorithm the ‘‘outer for-loop’’ and call the for-loop in lines6–10the ‘‘inner for-loop’’. This algorithm constructs T0

,

T1

, . . . ,

Tn−1simultaneously

and it works as follows. Since LTQnis n-regular, the n neighbors of the root r must be the unique child of the root r in

T0

,

T1

, . . . ,

Tn−1, respectively. In this algorithm, the unique child of the root r in Tiis set as fi

(

r

)

. Thus, initially V

(

Ti

) = {

fi

(

r

)}

. At the tth iteration of the outer for-loop, each vertex

v

in V

(

Ti

)

is connected to a new vertex u

=

f(i+t)mod n

(v)

by using the

edges in perfect matching M₍i+t)mod n, and the edge

(v,

u

)

is added to Ti(i.e., the parent of u is set as

v

in Ti). After n iterations of the outer for-loop, Tiis constructed.

Example 1. We now demonstrate how Algorithm Construct_IST constructs T2rooted at vertex 1 in LTQ4. In line 2 of the

algorithm, the unique child of the root 1 is set as f2

(

1

) =

7. Thus V

(

T2

) = {

7

}

. Now consider the outer for-loop. For t

=

1,

each vertex in V

(

T2

)

is connected to a new vertex by using the edges in M3; thus the edge

(

7

,

11

)

is added to T2; so S becomes

{

11

}

and V

(

T2

)

becomes

{

7

,

11

}

. For t

=

2, each vertex in V

(

T2

)

is connected to a new vertex by using the edges in M0; thus

the edges

(

7

,

6

)

and

(

11

,

10

)

are added to T2; so S becomes

{

6

,

10

}

and V

(

T2

)

becomes

{

7

,

11

,

6

,

10

}

. For t

=

3, each vertex

in V

(

T2

)

is connected to a new vertex by using the edges in M1; thus the edges

(

7

,

5

)

,

(

11

,

9

)

,

(

6

,

4

)

and

(

10

,

8

)

are added to

T2; so S becomes

{

5

,

9

,

4

,

8

}

and V

(

T2

)

becomes

{

7

,

11

,

6

,

10

,

5

,

9

,

4

,

8

}

. Finally, for t

=

4, each vertex in V

(

T2

)

is connected

to a new vertex by using the edges in M2; thus the edges

(

7

,

1

)

,

(

11

,

13

)

,

(

6

,

2

)

,

(

10

,

14

)

,

(

5

,

3

)

,

(

9

,

15

)

,

(

4

,

0

)

and

(

8

,

12

)

are

added to T₂; so S becomes

{

1

,

13

,

2

,

14

,

3

,

15

,

0

,

12

}

and V

(

T₂

)

becomes

{

7

,

11

,

6

,

10

,

5

,

9

,

4

,

8

,

1

,

13

,

2

,

14

,

3

,

15

,

0

,

12

}

. SeeFig. 4for an illustration.

4. Correctness

The purpose of this section is to prove that T0

,

T1

, . . . ,

Tn−1generated by Algorithm Construct_IST are n ISTs rooted at an

arbitrary vertex r for LTQn. To this end, some notations are first introduced in Section4.1. We show that T0

,

T1

, . . . ,

Tn−1are

(7)

Fig. 4. Four ISTs rooted at vertex 1 in LTQ4constructed by Algorithm Construct_IST.

4.1. The notations

Definition 3. For V′

_⊆

_V

₍

_LTQ

n

)

, define fi

(

V′

)

to be

fi

(

V′

) = {

fi

(v) | v ∈

V′

}

.

Definition 4. For a fixed integer i, 0

≤

i

≤

n

−

1, define On

i to be the ordered set

On_i

= {

i

, (

i

−

1

)

mod n

, (

i

−

2

)

mod n

, . . . , (

i

−

n

+

1

)

mod n

}

.

Notice that On_i can be obtained by arranging 0

,

1

, . . . ,

n

−

1 around a circle, starting from the number i and picking up these n numbers counterclockwise. For example, O4

0

= {

0

,

3

,

2

,

1

}

, O41

= {

1

,

0

,

3

,

2

}

and O43

= {

3

,

2

,

1

,

0

}

.

Definition 5. The Hamming distance between two vertices x

,

y

∈

V

(

LTQn

)

, denoted by Ham

(

x

,

y

)

, is the number of positions at which the corresponding symbols are different. More precisely, Ham

(

x

,

y

) = |{

i

|

xi

̸=

yi

,

0

≤

i

≤

n

−

1

}|

. For two fixed vertices x

,

y

∈

V

(

LTQn

)

, suppose Ham

(

x

,

y

) =

m. Define Hi

(

x

,

y

)

to be an ordered set consisting of the indices of the m different bits, listed according to the order given by On_i.

Definition 6. For two fixed vertices x

,

y

∈

V

(

LTQn

)

, suppose Hi

(

x

,

y

) = {

cm−1

,

cm−2

, . . . ,

c0

}

with m

≥

2 and Hi

(

x

,

y

) ̸=

Oni. We say that j is between cuand cu−1for some 0

≤

u

≤

m

−

1 with respect to Oni if j

̸∈

Hi

(

x

,

y

)

and when 0

,

1

, . . . ,

n

−

1 are arranged on a circle, the location of j on the circle is between cuand cu−1.

For example, consider LTQ4. Suppose

v =

12. Then H0

(v,

0

) = {

3

,

2

}

, H1

(v,

3

) = {

1

,

0

,

3

,

2

}

, H2

(v,

7

) = {

1

,

0

,

3

}

and

H₃

(v,

13

) = {

0

}

. Since 1

̸∈

H₀

(v,

0

)

, 1 is between c_u

=

3

,

c_u−1

=

2; 0

̸∈

H0

(v,

0

)

, 0 is between cu

=

2

,

cu−1

=

3.

Definition 7. For two vertices x

,

y

∈

V

(

LTQn

)

, defineΠi

(

x

,

y

)

to be the ordered set consisting of all the indices of perfect matchings used in the x

,

y-path in Ti, 0

≤

i

≤

n

−

1, listed according to the order from x to y.

For example, consider T2rooted at vertex 1 of LTQ4inFig. 4. Suppose

v =

12. ThenΠ2

(v,

7

) = {

2

,

1

,

0

,

3

}

. Moreover,

the path from

v

to 7 is

1100

−→

M2 1000

−→

M1 1010

−→

M0 1011

−→

M3 0111

.

Definition 8. Define I

(

a

,

b

)

, where a

≥

b, to be the sequence such that I

(

a

,

b

) =



a

,

a

−

1

, . . . ,

b

+

1 if a

>

b

,

(8)

4.2. The spanning trees

Throughout this subsection, let T0

,

T1

, . . . ,

Tn−1be the output of Algorithm Construct_IST. The purpose of this subsection

is to prove that T0

,

T1

, . . . ,

Tn−1are n spanning trees rooted at r. ByTheorem 2.4, we assume r

=

0 and r

=

1 as the common

roots without loss of generality. To prove that Ti, 0

≤

i

≤

n

−

1, is a spanning tree rooted at r, we prove the following loop

invariant:

Loop invariant: At the start of the tth iteration of the outer for-loop, Tiis connected,

|

V

(

Ti

)| =

2t−1and

|

E

(

Ti

)| = |

V

(

Ti

)| −

1.

The loop invariant is trivial true prior to the first loop iteration since in line 3, Algorithm Construct_IST sets V

(

Ti

) = {

fi

(

r

)}

. Hence Ti is connected,

|

V

(

Ti

)| =

20 and

|

E

(

Ti

)| = |

V

(

Ti

)| −

1. We now prove that if the loop invariant is true before the tth iteration of the outer for-loop, then it remains true before the next iteration. Algorithm Construct_IST first resets

S to be empty in line 5. For each vertex

v

in V

(

Ti

)

, Algorithm Construct_IST adds the edge

(v,

u

)

to Tiin line 8, where

u

=

f₍i+t)mod n

(v)

, by using the edges in M(i+t)mod n, and adds u to S in line 9. Since each newly generated edge is incident

to a vertex in V

(

Ti

)

, Tiremains to be connected. Now we claim that

Claim 4.1. V

(

Ti

) ∩

S

= ∅

.

IfClaim 4.1is true, then at the end of the inner for-loop, the newly generated edges between V

(

Ti

)

and S clearly form a matching that saturates V

(

Ti

)

and S. Thus

|

V

(

Ti

)| = |

S

|

. Consequently, after the tth iteration of the outer for-loop, Tiis connected,

|

V

(

Ti

)| =

2t−1

+

2t−1

=

2t and

|

E

(

Ti

)| =

2t−1

−

1

+

2t−1

=

2t

−

1

= |

V

(

Ti

)| −

1. When the outer for-loop terminates, t

=

n

+

1. Therefore, Tiis connected,

|

V

(

Ti

)| =

2nand

|

E

(

Ti

)| = |

V

(

Ti

)| −

1. Also, at the end of the

(

t

=

n

)

th iteration of the outer for-loop, Algorithm Construct_IST adds the edge

(

r

,

f_i

(

r

))

to T_i. Therefore T_iis a spanning tree rooted at r of LTQn. In the following, we prove thatClaim 4.1is true for r

=

0 and r

=

1. We first consider the case of r

=

0.

Lemma 4.2. Claim 4.1is true for r

=

0.

Proof. Consider the tth iteration of the outer for-loop. Set k

=

(

i

+

t

)

mod n for easy writing. Let

v ∈

V

(

Ti

)

and u

∈

S. If

t

∈ {

1

,

2

, . . . ,

n

−

1

}

, then

(v

k

,

uk

) = (

0

,

1

)

. If t

=

n, then we have

(v

i

,

ui

) = (

1

,

0

)

. Therefore V

(

Ti

) ∩

S

= ∅

.

Lemma 4.3. Claim 4.1is true for r

=

1.

Proof. Consider Ti, 0

≤

i

≤

n

−

1. Set k

=

(

i

+

t

)

mod n for easy writing. Let

v ∈

V

(

Ti

)

and u

∈

S.

Case 1: i

=

0. If t

∈ {

1

,

2

, . . . ,

n

−

1

}

, then

(v

k

,

uk

) = (

0

,

1

)

. If t

=

n, then

(v

i

,

ui

) = (

1

,

0

)

. Therefore V

(

Ti

) ∩

S

= ∅

.

Case 2: i

=

n

−

1. If t

∈ {

1

,

2

, . . . ,

n

−

2

}

, then

(v

k

,

uk

) = (

0

,

1

)

. If t

=

n

−

1, then we have

(v

n−2

,

un−2

) = (

1

,

0

)

. If

t

=

n, then we have

(v

i

,

ui

) = (

1

,

0

)

. Therefore V

(

Ti

) ∩

S

= ∅

.

Case 3: i

∈ {

1

,

2

, . . . ,

n

−

2

}

. We further divide this case into two subcases.

Subcase 3.1: t

∈ {

1

,

2

, . . . ,

n

−

2

}

. The proof of this case is the same as Case 2.

Subcase 3.2: t

=

n. By the loop invariant, Tiinduces a tree before the tth iteration of the outer for-loop. Partition V

(

Ti

)

into

V0and V1as follows:

V0

= {

all the vertices in the subtree rooted at fi+1

(

fi

(

1

))}

and V1

=

V

(

Ti

) \

V0

.

SeeFig. 5for an illustration.

By (1) and byLemma 4.6, we have: (i) the ith bit of all the vertices in V0is 0 and hence the ith bit of all the vertices in fi

(

V0

)

is 1, and (ii) the ith bit of all the vertices in V1is 1 and hence the ith bit of all the vertices in fi

(

V1

)

is 0. Notice that

S

=

fi

(

V0

) ∪

fi

(

V1

).

Therefore, to proveClaim 4.1, it suffices to prove that

V0

∩

fi

(

V1

) = ∅

and V1

∩

fi

(

V0

) = ∅.

(3)

If i

=

n

−

2, then the

(

n

−

1

)

-bit of all the vertices in V0and fn−2

(

V0

)

is 1; however, the

(

n

−

1

)

-bit of all the vertices in

V1and fn−2

(

V1

)

is 0. Thus when i

=

n

−

2, V0

∩

fn−2

(

V1

) = ∅

and V1

∩

fn−2

(

V0

) = ∅

. Now suppose i

∈ {

1

,

2

, . . . ,

n

−

3

}

.

Partition V0into V0,0and V0,1such that

V_0,0

= {

(

fi+1

(

fi

(

1

)))}

and V0,1

=

V0

\

V0,0

.

Partition V1into V1,0and V1,1such that

V_1,0

= {

(

fi

(

1

))}

and V1,1

=

V1

\

V1,0

.

By (1) andLemma 4.6, the pair of the

(

i

+

1

)

th and the ith bit of all the vertices in V_0,0and fi

(

V1,1

)

is (0, 0); in fi

(

V0,0

)

and

V_1,1is (0, 1); in V_0,1and fi

(

V1,0

)

is (1, 0) and in fi

(

V0,1

)

and V1,0is (1, 1). Thus to prove (3), it suffices to prove that

V_0,0

∩

fi

(

V1,1

) = ∅,

V1,1

∩

fi

(

V0,0

) = ∅,

V1,0

∩

fi

(

V0,1

) = ∅

and V0,1

∩

fi

(

V1,0

) = ∅.

(4) For

v = v

n−1

, v

n−1

, . . . , v

0

∈

V

(

LTQn

)

with

v ̸=

0, let q be the largest index of

v

such that

v

q

=

1. If

v =

0, then let q

= −

1. By (1) andLemma 4.6, we haveTable 2.

(9)

Fig. 5. An illustration for the proof ofLemma 4.3.

Table 2

The value of q for every vertex in the given set.

V0,0∪fi(V0,0) V1,1∪fi(V1,1) V1,0∪fi(V1,0) V0,1∪fi(V0,1)

q≥i+2 q≤i+1 or q≥i+3 q≥i+3 q=i+1 or q≥i+3

We first prove that V_0,0

∩

fi

(

V1,1

) = ∅

and V1,1

∩

fi

(

V0,0

) = ∅

. ByTable 2, each vertex in V1,1

∩

fi

(

V1,1

)

with q

≤

i

+

1 does not belong to V_0,0

∪

fi

(

V0,0

)

since every vertex in V0,0

∪

fi

(

V0,0

)

has q

≥

i

+

2. Also, each vertex in V0,0

∪

fi

(

V0,0

)

with

q

=

i

+

2 does not belong to V_1,1

∩

fi

(

V1,1

)

since each vertex in V1,1

∩

fi

(

V1,1

)

has q

̸=

i

+

2. Thus, we may focus on vertices with q

=

i

+

3 or q

>

i

+

3. Note that each vertex in V_0,0

∪

fi

(

V0,0

)

with q

=

i

+

3 has its

(

i

+

2

)

th bit to be 0; however, fromTable 2, we know that each vertex in fi

(

V1,1

) ∪

V1,1with q

≥

i

+

3 has its

(

i

+

2

)

th bit to be 1. Therefore, each vertex in

V_0,0

∪

fi

(

V0,0

)

with q

=

i

+

3 does not belong to V1,1

∪

fi

(

V1,1

)

. It remains to consider the vertices with q

>

i

+

3. For each

x

∈

V0,0

∪

fi

(

V0,0

)

, the bit string of x formed by xqto xi+2is in

L0

= {

1 q−i−2 0’s







00

· · ·

0







q−i−1 bits

,

1 q−i−4 0’s







00

· · ·

0 11







q−i−1 bits

,

1 q−i−5 0’s







00

· · ·

0 101







q−i−1 bits

,

1 q−i−6 0’s







00

· · ·

0 1001







q−i−1 bits

, . . . ,

101 q−i−5 0’s







00

· · ·

0 1







q−i−1 bits

,

11 q−i−4 0’s







00

· · ·

0 1







q−i−1 bits

}

.

However, for each y

∈

V1,1

∪

fi

(

V1,1

)

, the bit string of y formed by yqto yi+2is in

L1

= {

1 q−i−3 0’s







00

· · ·

0 1







q−i−1 bits

,

1 q−i−4 0’s







00

· · ·

0 10







q−i−1 bits

,

1 q−i−5 0’s







00

· · ·

0 100







q−i−1 bits

,

1 q−i−6 0’s







00

· · ·

0 1000







q−i−1 bits

, . . . ,

101 q−i−4 0’s







00

· · ·

0







q−i−1 bits

,

11 q−i−3 0’s







00

· · ·

0







q−i−1 bits

}

.

It is not difficult to check that L0

∩

L1

= ∅

. Hence we have V0,0

∩

fi

(

V1,1

) = ∅

and V1,1

∩

fi

(

V0,0

) = ∅

.

Similar arguments can show that V_0,1

∩

fi

(

V1,0

) = ∅

and V1,0

∩

fi

(

V0,1

) = ∅

, except that V0,0

∪

fi

(

V0,0

)

is replaced by

V_1,0

∪

fi

(

V1,0

)

and V1,1

∪

fi

(

V1,1

)

is replaced by V0,1

∪

fi

(

V0,1

)

. From the above discussion, we have (4) and hence have (3). Therefore V

(

Ti

) ∩

S

= ∅

.

ByTheorem 2.4andLemmas 4.2and4.3, we have the following result.

Lemma 4.4. T0

,

T1

, . . . ,

Tn−1are n spanning trees rooted at r for LTQn.

4.3. The vertex-independency of the n spanning trees

In this subsection, we show that T0

,

T1

, . . . ,

Tn−1generated by Algorithm Construct_IST are vertex-independent trees

rooted at an arbitrary vertex r for LTQn. ByTheorem 2.4, without loss of generality, we may assume r

=

0 and r

=

1 as the common roots. To this end, we need to show that for any i

,

j with 0

≤

i

<

j

≤

n

−

1 and for each

v(̸=

r

) ∈

V

(

LTQn

)

, the

r

, v

-path in Tiand the r

, v

-path in Tjare internally vertex-disjoint. Recall that the child of the root in Tiand Tjare fi

(

r

)

and

fj

(

r

)

, respectively. In the following, we further assume

v ̸∈ {

r

,

fi

(

r

),

fj

(

r

)}

since if

v ∈ {

r

,

fi

(

r

),

fj

(

r

)}

, then the r

, v

-path in Ti and the r

, v

-path in Tjare clearly internally vertex-disjoint. Let parenti

(v)

(resp., parentj

(v)

) be the parent of vertex

v

in Ti

(10)

(resp., Tj). Let P1(resp., P2) be the parenti

(v),

fi

(

r

)

-path (resp., parentj

(v),

fj

(

r

)

-path) in Ti(resp., Tj). Since fi

(

r

) ̸=

fj

(

r

)

, the

r

, v

-path in Tiand the r

, v

-path in Tjare internally vertex-disjoint if and only if V

(

P1

) ∩

V

(

P2

) = ∅

. We prove Tiand Tjare vertex-independent by showing the following claim:

Claim 4.5. V

(

P1

) ∩

V

(

P2

) = ∅

.

Before provingClaim 4.5, we need a lemma.

Lemma 4.6. Ti, 0

≤

i

≤

n

−

1, constructed by Algorithm Construct_IST has the property that for each

v ∈

V

(

LTQn

) \ {

r

,

fi

(

r

)}

,

the path from

v

to fi

(

r

)

in Tiuses each perfect matching in

{

M0

,

M1

, . . . ,

Mn−1

}

at most once.

Proof. It follows from the fact that f₍i+t)mod nused in the for-loop between the inner for-loop are distinct when the outer

for-loop iterates from t

=

1 to t

=

n. We first consider the case of r

=

0.

Lemma 4.7. T0

,

T1

, . . . ,

Tn−1are n vertex-independent trees rooted at r

=

0 for LTQn.

Proof. To proveClaim 4.5, we first describe the path from

v

to the child of the root in Ti when r

=

0. For any

v ∈

V

(

Ti

) \ {

0

,

fi

(

0

)}

, the

v,

fi

(

0

)

-path in Ti can be determined byΠi

(v,

fi

(

0

))

. In addition,Πi

(v,

fi

(

0

))

can be determined by

Hi

(v,

fi

(

0

))

as follows. Suppose

v = v

n−1

v

n−2

. . . v

0and Hi

(v,

fi

(

0

)) = {

cm−1

,

cm−2

, . . . ,

c0

}

. If

v

0

=

0, thenΠi

(v,

fi

(

0

))

can be determined by Πi

(v,

fi

(

0

)) =











Hi

(v,

fi

(

0

))

if i

̸=

0

,

{

cm−1

=

0

,

I

(

cm−2

,

cm−3

), . . . ,

I

(

c3

,

c2

),

I

(

c1

,

c0

)}

if i

=

0 and m

−

1 is even

,

{

cm−1

=

0

,

I

(

cm−2

,

cm−3

), . . . ,

I

(

c2

,

c1

),

I

(

c0

,

0

)}

if i

=

0 and m

−

1 is odd

.

(5)

If

v

0

=

1 and i

̸=

0, then Hi

(v,

fi

(

0

))

must contain 0; in this case, we assume ce

=

0 for some e. Thus if

v

0

=

1,Πi

(v,

fi

(

0

))

can be determined by Πi

(v,

fi

(

0

))=











{

I

(

cm−1

,

cm−2

),

I

(

cm−3

,

cm−4

), . . . ,

I

(

c1

,

c0

)}

if i

=

0 and m is even

,

{

I

(

cm−1

,

cm−2

),

I

(

cm−3

,

cm−4

), . . . ,

I

(

c2

,

c1

),

I

(

c0

,

0

)}

if i

=

0 and m is odd

,

{

I

(

cm−1

,

cm−2

),

I

(

cm−3

,

cm−4

), . . . ,

I

(

ce+2

,

ce+1

),

ce

,

ce−1

, . . . ,

c0

}

if i

̸=

0 and m

−

e is odd

,

{

I

(

cm−1

,

cm−2

),

I

(

cm−3

,

cm−4

), . . . ,

I

(

ce+1

,

0

),

ce

,

ce−1

, . . . ,

c0

}

if i

̸=

0 and m

−

e is even

.

(6)

Now we show thatClaim 4.5is true for r

=

0. Suppose not, then there exists a vertex a

(̸=v) ∈

V

(

P1

) ∩

V

(

P2

)

. Suppose

Hi

(v,

fi

(

0

)) =

Hi

(v,

2i

) = {

cm−1

,

cm−2

, . . . ,

c0

}

.

(7)

There are four cases.

Case 1:

v

i

=

1 and

v

j

=

1. Then there must exist u such that cu

=

j. Thus

Hj

(v,

fj

(

0

)) =

Hj

(v,

2j

) = {

cu−1

,

cu−2

, . . . ,

c0

,

i

,

cm−1

,

cm−2

, . . . ,

cu+1

}

.

(8)

By (5)–(7), cm−1is the first element inΠi

(v,

2i

)

. Let x

∈

V

(

P1

)

. Then the

(

cm−1

)

th bit of x is

v

cm−1only when (i)

(

cm−1

+

1

) ∈

Πi

(v,

2i

)

, and (ii) cm−1

+

1

≥

2, and (iii) there exists q

=

qn−1qn−2

. . .

q0

∈

V

(

P1

)

such that x

=

fcm−1+1

(

q

)

and q0

=

1. We now prove that (i)–(iii) will not occur simultaneously; hence for all x

∈

V

(

P1

)

, the

(

cm−1

)

th bit of x is

v

cm−1. If

|

Hi

(v,

2

i

_{)| =}

_1, then (i) cannot occur. Suppose

|

Hi

(v,

2i

)| ≥

2 and both (i) and (iii) occur; that is, there exists q

=

qn−1qn−2

. . .

q0

∈

V

(

P1

)

such that x

=

fcm−1+1

(

q

)

and q0

=

1. By (7), cm−1

+

1 is the last element inΠi

(v,

2

i

₎

_{. Since q}

0

=

1, I

(

c0

,

0

) ⊆

Πi

(v,

2i

)

. ByLemma 4.6and by the fact that I

(

c0

,

0

) = {

c0

,

c0

−

1

, . . . ,

1

}

, we have cm−1

+

1

=

1; thus (ii) does not occur and

consequently the

(

cm−1

)

th bit of all the vertices in V

(

P1

)

is

v

cm−1. Since

v

i

=

1, the ith bit of all the vertices in V

(

P1

)

is 1. By (5) and (6) and (8), the

(

cm−1

)

th bit of those vertices in V

(

P2

)

with the ith bit being 1 is

v

cm−1. Thus no such a exists and Claim 4.5is true.

Case 2:

v

i

=

0 and

v

j

=

0. Then cm−1

=

i. If

|

Hi

(v,

2i

)| =

1, then Hi

(v,

2i

) = {

i

}

, which implies that

v =

0; this contradicts to the assumption that

v ̸=

0. Thus

|

Hi

(v,

2i

)| ≥

2 and there must exist u such that j is between cuand cu−1with respect to

On_i. Thus Hj

(v,

2j

) =



{

j

,

cm−2

,

cm−3

, . . . ,

cu+1

,

cu=0

}

if j

=

i

+

1

,

{

j

,

cu−1

,

cu−2

, . . . ,

c0

,

cm−2

,

cm−3

, . . . ,

cu+1

}

if otherwise. (9) By (5)–(7), the ith bit of all vertices in V

(

P1

)

is 1. By (5) and (6) and (9), the jth bit of all the vertices in V

(

P2

)

is 1. The ith

bit and the jth bit of a are both 1. If I

(

cu

,

cu−1

)

*Πi

(v,

2i

)

, each vertex in V

(

P1

)

has its jth bit to be 0. If (i) j

̸=

i

+

1 and

I

(

c0

,

cm−2

)

*Πj

(v,

2j

)

, or if (ii) j

=

i

+

1 and

v

0

̸=

1, then each vertex in V

(

P2

)

has its ith bit to be 0. Thus the existence of

(11)

hence

v

0

=

0 (since case 2 requires

v

i

=

0). However, I

(

c0

,

cm−2

) ⊆

Πj

(v,

2j

)

implies

v

0

=

1, which contradicts to

v

0

=

0.

Thus no such a exists andClaim 4.5is true.

Case 3:

v

i

=

0 and

v

j

=

1. Then cm−1

=

i and there must exist u such that cu

=

j. If

|

Hi

(v,

2i

)| =

1, then Hj

(v,

2i

) = ∅

. This implies that

v =

2j, which contradicts to the assumption that

v ̸=

2j. Thus

Hj

(v,

2j

) = {

cu−1

,

cu−2

, . . . ,

c0

,

cm−2

,

cm−3

, . . . ,

cu+1

}

.

(10)

By (5)–(7), the ith bit of all vertices in V

(

P1

)

is 1. The ith bit of a is 1. If I

(

c0

,

cm−2

)

* Πj

(v,

2j

)

(

P2

)

has its ith bit to be 0. Thus the existence of a implies that I

(

c0

,

cm−2

) ⊆

Πj

(v,

2j

)

, which further implies

v

0

=

1. Since

I

(

c0

,

cm−2

) ⊆

Πj

(v,

2j

)

, V

(

P2

)

has only one vertex x

=

xn−1xn−2

. . .

x0such that xi

=

1 and x

=

fi+1

(

q

)

for some q

∈

V

(

P2

)

. The existence of a implies that x

=

a. Since

v

0

=

1,Πi

(v,

2i

)

starts with I

(

i

,

cm−2

)

, i.e.,Πi

(v,

2i

)

is of the form

{

I

(

i

,

cm−2

), . . .}

. By (6), cm−3is the first element after I

(

i

,

cm−2

)

inΠi

(v,

2i

)

. Recall thatΠi

(v,

2i

)

is an ordered set of all the indices of perfecting matchings used in the

v,

2i_{-path in T}

ilisted according to the order from

v

to 2i. Thus the first vertex in V

(

P1

)

can be obtained by applying the first perfect matching obtained from the first element inΠi

(v,

2i

)

to

v

, the second vertex in V

(

P1

)

can be obtained by applying the second perfect matching obtained from the second element in Πi

(v,

2i

)

to the first vertex in V

(

P1

)

, and so on. Thus we can partition V

(

P1

)

into V1,1and V1,2such that V1,1consists of those

vertices in V

(

P1

)

before fcm−3is applied and V1,2

=

V

(

P1

) −

V1,1. Let y

=

yn−1yn−2

. . .

y0be an arbitrary vertex in V1,1. Then

Ham

(

yiyi−1

. . .

ycm−2

, v

i

v

i−1

. . . v

cm−2

) =

2. However, Ham

(

xixi−1

. . .

xcm−2

, v

i

v

i−1

. . . v

cm−2

) =

0. Thus x

̸∈

V1,1. On the other hand, xcm−3

=

v

cm−3but the

(

cm−3

)

th bit of all the vertices in V1,2is

v

cm−3; thus x

̸∈

V1,2. Since x

̸∈

V1,1and x

̸∈

V1,2, we have

x

̸∈

V

(

P1

)

. Since x

=

a, it follows that a

̸∈

V

(

P1

)

. Thus no such a exists andClaim 4.5is true.

Case 4:

v

i

=

1 and

v

j

=

0. Then there must exist u such that j is between cuand cu−1with respect to Oni. Thus

Hj

(v,

2j

) =



{

j

,

i

,

cm−1

,

cm−2

, . . . ,

cu=0

}

if i is between c0and cm−1with respect to Oni

,

{

j

,

cu−1

,

cu−2

, . . . ,

c0

,

i

,

cm−1

,

cm−2

, . . . ,

cu

}

if otherwise.

(11)

By (5), (6) and (11), the jth bit of all vertices in V

(

P2

)

is 1. Since

v

i

=

1, the ith bit of all the vertices in V

(

P1

)

is 1. The ith bit

and the jth bit of a are both 1. By (11), we have two subcases.

Subcase 4.1: i is between c0and cm−1with respect to Oni. Then V

(

P2

)

has only one vertex fj

(v)

with its ith bit and jth bit both being 1. By (5)–(7), cm−1is the first element inΠi

(v,

2i

)

. Thus the

(

cm−1

)

(

P1

)

with the jth bit

being 1 is

v

cm−1. However, by (5), (6) and (11), the

(

cm−1

)

th bit of fj

(v)

is

v

cm−1. Thus no such a exists andClaim 4.5is true.

Subcase 4.2: i is not between c₀and c_m−1with respect to Oni. By (5), (6) and (11), the ith bit of all the vertices in V

(

P1

)

is 1. If

|

Hi

(v,

2i

)| =

1, then Hi

(v,

2i

) = {

c0

}

; since

v

j

=

0, we have c0

̸=

j, which implies that each vertex in V

(

P1

)

has its jth bit to be 0 and consequently no such a exists andClaim 4.5is true. Now suppose

|

Hi

(v,

2i

)| ≥

2. Then when

I

(

cu

,

cu−1

)

*Πi

(v,

2i

)

(

P1

)

has its jth bit to be 0. Thus the existence of a implies that I

(

cu

,

cu−1

) ⊆

Πi

(v,

2i

)

. Since I

(

cu

,

cu−1

) ⊆

Πi

(v,

2i

)

, V

(

P1

)

has only one vertex x

=

xn−1xn−2

. . .

x0such that xj

=

1 and x

=

fj+1

(

q

)

for some

q

∈

V

(

P1

)

. The existence of a implies that x

=

a. By (5), (6) and (11), the

(

cm−1

)

(

P2

)

with the

ith bit being 1 is

v

cm−1. However, the xcm−1

=

v

cm−1. So if x

∈

V

(

P1

)

, then x

̸∈

V

(

P2

)

. Thus no such a exists andClaim 4.5is true.

From the above discussion,Claim 4.5is true and therefore T0

,

T1

, . . . ,

Tn−1are vertex-independent rooted at r

=

0 of

LTQn.

Now we consider the case of r

=

1.

Lemma 4.8. T0

,

T1

, . . . ,

Tn−1are n vertex-independent trees rooted at r

=

1 for LTQn.

Proof. To proveClaim 4.5, we first describe the path from

v

to the child of the root in Tiwhen r

=

1. For any

v ∈

V

(

Ti

) \

{

1

,

fi

(

1

)}

, the

v,

fi

(

1

)

-path in Ti can be determined byΠi

(v,

fi

(

1

))

. Furthermore,Πi

(v,

fi

(

1

))

can be determined by the ordered set Hi

(v,

fi

(

1

))

as follows. Suppose

v = v

n−1

v

n−2

. . . v

0and Hi

(v,

fi

(

1

)) = {

cm−1

,

cm−2

, . . . ,

c0

}

. Let ce−1be the

first (from bit cm−1to c0) member in Hi

(v,

fi

(

1

))

that is larger than i. If i

=

0,Πi

(v,

fi

(

1

))

Π0

(v,

f0

(

1

)) =

H0

(v,

f0

(

1

)).

(12)

If i

̸=

0 and

v

0

=

0, we have ce

=

0 for some e. ThusΠi

(v,

fi

(

1

))

Πi

(v,

fi

(

1

))=











{

cm−1

,

cm−2

, . . . ,

ce

,

I

(

ce−1

,

ce−2

),

I

(

ce−3

,

ce−4

), . . . ,

I

(

c1

,

c0

)}

if e is even

,

{

cm−2

,

cm−3

, . . . ,

ce

,

I

(

ce−1

,

ce−2

),

I

(

ce−3

,

ce−4

), . . . ,

I

(

c0

,

i

)}

if e is odd and cm−1

=

i

,

{

i

,

cm−1

,

cm−2

, . . . ,

ce

,

I

(

ce−1

,

ce−2

),

I

(

ce−3

,

ce−4

), . . . ,

I

(

c0

,

i

)}

if e is odd and cm−1

̸=

i

.

(13)