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 LTQ_ninAlgorithm 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 LTQn is 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 T_i(i.e., the parent of u is set as

v

^{in T}i). After n iterations of the outer for-loop, T_iis 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 n spanning trees of LTQ_nin Section4.2. The vertex-independency of T₀

,

^T1

, . . . ,

^Tn−1is shown in Section4.3.

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

4.1. The notations

Definition 3. For V^′

⊆

V

(

^LTQn

)

^{, define f}i

(

^V′

)

^{to be} f_i

(

^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, O⁴₀

= {

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

|

x_i

̸=

y_i

,

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 H_i

(

^x

,

^y

) = {

^cm−1

,

^cm−2

, . . . ,

^c0

}

with m

≥

2 and H_i

(

^x

,

^y

) ̸=

^Oni. We say that j is between c_uand c_u−1for some 0

≤

u

≤

m

−

1 with respect to Oⁿ_i if j

̸∈

H_i

(

^x

,

^y

)

^{and when 0}

,

¹

, . . . ,

ⁿ

−

1 are arranged on a circle, the location of j on the circle is between c_uand c_u−1.

For example, consider LTQ4. Suppose

v =

^{12. Then H}0

(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 cu

=

3

,

^cu−1

=

2; 0

̸∈

H0

(v,

⁰

)

, 0 is between cu

=

2

,

^cu−1

=

3.

Definition 7. For two vertices x

,

^y

∈

V

(

^LTQn

)

^{, define}

Π

i

(

^x

,

^y

)

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

,

y-path in T_i, 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. Then}

Π

²

(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

,

a if a

=

b

.

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 Ti}in 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

)

^{, T}iremains 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

+

2^t−1

=

2^t and

|

E

(

^Ti

)| =

^2t−1

−

1

+

2^t−1

=

2^t

−

1

= |

V

(

^Ti

)| −

1. When the outer for-loop terminates, t

=

n

+

1. Therefore, T_iis 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

,

^fi

(

^r

))

^{to T}i. Therefore T_iis a spanning tree rooted at r of LTQ_n. 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 T_i, 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 V}1

=

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 f_i

(

^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 f_i

(

^V1

)

is 0. Notice that

S

=

_fi

(

^V0

) ∪

^fi

(

^V1

).

Therefore, to proveClaim 4.1, it suffices to prove that

V₀

∩

f_i

(

^V1

) = ∅

^{and V}1

∩

f_i

(

^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 f_n−2

(

^V1

)

is 0. Thus when i

=

n

−

2, V₀

∩

f_n−2

(

^V1

) = ∅

^{and V}1

∩

f_n−2

(

^V0

) = ∅

. Now suppose i

∈ {

1

,

²

, . . . ,

ⁿ

−

3

}

. Partition V₀into V_0,0and V_0,1such that

V_0,0

= {

all the vertices in the subtree rooted at f_i+2

(

^fi+1

(

^fi

(

¹

)))}

^{and V}0,¹

=

V₀

\

V_0,0

.

Partition V1into V1,⁰and V1,¹such that

V1,⁰

= {

all the vertices in the subtree rooted at fi+2

(

^fi

(

¹

))}

^{and V}1,¹

=

V1

\

V1,⁰

.

By (1) andLemma 4.6, the pair of the

(

ⁱ

+

1

)

th and the ith bit of all the vertices in V0,⁰and fi

(

^V1,¹

)

is (0, 0); in fi

(

^V0,⁰

)

^and V_1,1is (0, 1); in V_0,1and f_i

(

^V1,0

)

is (1, 0) and in f_i

(

^V0,1

)

^{and V}1,0is (1, 1). Thus to prove (3), it suffices to prove that

V0,⁰

∩

fi

(

^V1,¹

) = ∅,

^V1,¹

∩

fi

(

^V0,⁰

) = ∅,

^V1,⁰

∩

fi

(

^V0,¹

) = ∅

^{and V}0,¹

∩

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.

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)

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 LTQ_n. 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 T}iand the r

, v

^{-path in T}jare internally vertex-disjoint. Recall that the child of the root in T_iand T_jare f_i

(

^r

)

^and f_j

(

^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 T}i

and the r

, v

^{-path in T}jare clearly internally vertex-disjoint. Let parent_i

(v)

(resp., parent_j

(v)

) be the parent of vertex

v

^{in T}i

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 f}i

(

^r

)

^{in T}iuses each perfect matching in

{

M0

,

^M1

, . . . ,

^Mn−1

}

at most once.

Proof. It follows from the fact that f₍i+t)^{mod n} used 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 LTQ_n.

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

v

to the child of the root in T_i when r

=

0. For any

v ∈

V

(

^Ti

) \ {

⁰

,

^fi

(

⁰

)}

^{, the}

v,

^fi

(

⁰

)

^{-path in T}ican be determined by

Π

i

(v,

^fi

(

⁰

))

. In addition,

Π

i

(v,

^fi

(

⁰

))

can be determined by H_i

(v,

^fi

(

⁰

))

as follows. Suppose

v = v

n−1

v

n−2

. . . v

0and H_i

(v,

^fi

(

⁰

)) = {

^cm−1

,

^cm−2

, . . . ,

^c0

}

. If

v

0

=

0, then

Π

i

(v,

^fi

(

⁰

))

^can be determined by

Π

ⁱ

(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,

Π

ⁱ

(v,

^fi

(

⁰

))

can be determined by

Π

i

(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} H_i

(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

H_j

(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 in

Π

ⁱ

(v,

²ⁱ

)

^{. Let x}

∈

V

(

^P1

)

^{. Then the}

(

^cm−1

)

th bit of x is

v

c_m−1only when (i)

(

^cm−1

+

1

) ∈ Π

ⁱ

(v,

²ⁱ

)

, and (ii) cm−1

+

1

≥

2, and (iii) there exists q

=

qn−1qn−2

. . .

^q0

∈

V

(

^P1

)

such that x

=

fc_m−1+1

(

^q

)

^{and q}0

=

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

c_m−1. If

|

Hi

(v,

²ⁱ

)| =

^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

=

fc_m−1+1

(

^q

)

^{and q0}

=

1. By (7), cm−1

+

1 is the last element in

Π

ⁱ

(v,

²ⁱ

)

^{. Since q0}

=

1, I

(

^c0

,

⁰

) ⊆ Π

ⁱ

(v,

²ⁱ

)

^. ByLemma 4.6and by the fact that I

(

^c0

,

⁰

) = {

^c0

,

^c0

−

1

, . . . ,

¹

}

, we have c_m−1

+

1

=

1; thus (ii) does not occur and consequently the

(

^cm−1

)

th bit of all the vertices in V

(

^P1

)

^is

v

c_m−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

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

Case 2:

v

i

=

0 and

v

j

=

0. Then c_m−1

=

i. If

|

H_i

(v,

²ⁱ

)| =

^{1, then H}i

(v,

²ⁱ

) = {

ⁱ

}

, which implies that

v =

0; this contradicts to the assumption that

v ̸=

^{0. Thus}

|

H_i

(v,

²ⁱ

)| ≥

2 and there must exist u such that j is between c_uand c_u−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

)

*

Π

ⁱ

(v,

²ⁱ

)

, each vertex in V

(

^P1

)

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

̸=

_i

+

_{1 and} I

(

^c0

,

^cm−2

)

*

Π

^j

(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 a implies that I

(

^cu

,

^cu−1

) ⊆ Π

i

(v,

²ⁱ

)

^{and I}

(

^c0

,

^cm−2

) ⊆ Π

j

(v,

^2j

)

. Note that I

(

^cu

,

^cu−1

) ⊆ Π

i

(v,

²ⁱ

)

implies that i

=

0 and

Case 3:

v

i

=

0 and

v

j

=

1. Then c_m−1

=

i and there must exist u such that c_u

=

j. If

|

H_i

(v,

²ⁱ

)| =

^{1, then H}j

(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

)

*

Π

j

(v,

^2j

)

, each vertex in V

(

^P2

)

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

(

^c0

,

^cm−2

) ⊆ Π

j

(v,

^2j

)

, which further implies

v

0

=

1. Since I

(

^c0

,

^cm−2

) ⊆ Π

j

(v,

^2j

)

^{, V}

(

^P2

)

has only one vertex x

=

x_n−1x_n−2

. . .

^x0such that x_i

=

1 and x

=

f_i+1

(

^q

)

^{for some q}

∈

V

(

^P2

)

. The existence of a implies that x

=

a. Since

v

0

=

1,

Π

i

(v,

²ⁱ

)

starts with I

(

ⁱ

,

^cm−2

)

^{, i.e.,}

Π

i

(v,

²ⁱ

)

is of the form

{

I

(

ⁱ

,

^cm−2

), . . .}

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

(

ⁱ

,

^cm−2

)

ⁱⁿ

Π

ⁱ

(v,

²ⁱ

)

. Recall that

Π

ⁱ

(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 in

Π

ⁱ

(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

Π

i

(v,

²ⁱ

)

to the first vertex in V

(

^P1

)

, and so on. Thus we can partition V

(

^P1

)

^{into V}1,¹and V_1,2such that V_1,1consists of those vertices in V

(

^P1

)

^{before f}c_m−3is applied and V_1,2

=

V

(

^P1

) −

^V1,¹. Let y

=

y_n−1y_n−2

. . .

^y0be an arbitrary vertex in V_1,1. Then Ham

(

^yiy_i−1

. . .

^yc_m−2

, v

i

v

i−1

. . . v

c_m−2

) =

2. However, Ham

(

^xix_i−1

. . .

^xc_m−2

, v

i

v

i−1

. . . v

c_m−2

) =

^{0. Thus x}

̸∈

V_1,1. On the other hand, xc_m−3

= v

c_m−3but the

(

^cm−3

)

th bit of all the vertices in V1,²is

v

c_m−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 Oⁿ_i. Thus Hj

(v,

^2j

) =



{

_j

,

ⁱ

,

^cm−1

,

^cm−2

, . . . ,

^cu=0

}

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

,

{

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 c_m−1with respect to Oⁿ_i. Then V

(

^P2

)

has only one vertex f_j

(v)

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

Π

ⁱ

(v,

²ⁱ

)

^{. Thus the}

(

^cm−1

)

th bit of those vertices in V

(

^P1

)

with the jth bit being 1 is

v

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

(

^cm−1

)

th bit of fj

(v)

^is

v

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

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

(

^P1

)

is 1. If

|

Hi

(v,

²ⁱ

)| =

^{1, then H}i

(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

)

*

Π

ⁱ

(v,

²ⁱ

)

, each vertex in V

(

^P1

)

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

(

^cu

,

^cu−1

) ⊆ Π

i

(v,

²ⁱ

)

^. Since I

(

^cu

,

^cu−1

) ⊆ Π

i

(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

)

th bit of those vertices in V

(

^P2

)

^{with the} ith bit being 1 is

v

^cm−1. However, the xc_m−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 LTQ_n. 

Now we consider the case of r

=

1.

Lemma 4.8. T₀

,

^T1

, . . . ,

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

=

1 for LTQ_n.

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 T}i can be determined by

Π

ⁱ

(v,

^fi

(

¹

))

. Furthermore,

Π

ⁱ

(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 c_m−₁to c₀) member in H_i

(v,

^fi

(

¹

))

that is larger than i. If i

=

0,

Π

i

(v,

^fi

(

¹

))

can be determined by

Π

⁰

(v,

^f0

(

¹

)) =

^H0

(v,

^f0

(

¹

)).

⁽¹²⁾

If i

̸=

0 and

v

0

=

0, we have c_e

=

0 for some e. Thus

Π

i

(v,

^fi

(

¹

))

can be determined by

Π

ⁱ

(v,

^fi

(

¹

))=



 

 

{

c_m−1

,

^cm−2

, . . . ,

^ce

,

^I

(

^ce−1

,

^ce−2

),

^I

(

^ce−3

,

^ce−4

), . . . ,

^I

(

^c1

,

^c0

)}

if e is even

,

{

c_m−2

,

^cm−3

, . . . ,

^ce

,

^I

(

^ce−1

,

^ce−2

),

^I

(

^ce−3

,

^ce−4

), . . . ,

^I

(

^c0

,

ⁱ

)}

if e is odd and c_m−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 c_m−1

̸=

i

.

(13)

H_i¹

=



 

 

cm−1

,

^cm−2

, . . . ,

^ce if

|

H_i²

|

is even

,

i

,

^cm−1

,

^cm−2

, . . . ,

^ce if

|

H_i²

|

is odd and cm−1

̸=

i cm−2

,

^cm−3

, . . . ,

^ce if

|

H_i²

|

is odd and cm−1

=

i

,

and define H_i²to be the sequence

H_i²

=

ce−1

,

^ce−2

, . . . ,

^c0

.

Define

ζ

i

(v,

^fi

(

¹

))

to be the sequence

ζ

i

(v,

^fi

(

¹

)) =



 

 

 

 

H_i¹

,

^Hi² if

|

H_i¹

|

is even and

|

H_i²

|

is even

,

H_i¹

,

^Hi²

,

^{i if}

|

H_i¹

|

is even and

|

H_i²

|

is odd

,

H_i¹

,

⁰

,

^Hi² if

|

H_i¹

|

is odd and

|

H_i²

|

is even

,

H_i¹

,

⁰

,

^H_i²

,

^{i if}

|

H_i¹

|

is odd and

|

H_i²

|

is odd

.

(14)

Suppose

ζ

i

(v,

^fi

(

¹

)) = ς

u

, ς

u−1

, . . . , ς

0

.

Thus if i

̸=

0 and

v

⁰

=

1,

Π

ⁱ

(v,

^fi

(

¹

))

can be determined by

Π

ⁱ

(v,

^fi

(

¹

)) = {

^I

(ς

^u

, ς

^u−1

),

^I

(ς

^u−2

, ς

^u−3

), . . . ,

^I

(ς

¹

, ς

⁰

), }.

⁽¹⁵⁾ Now we show thatClaim 4.5is true for r

=

1. Suppose not, then there exists a vertex a

(̸=v) ∈

^V

(

^P1

) ∩

^V

(

^P2

)

^{. Suppose}

H_i

(v,

^fi

(

¹

)) = {

^cm−1

,

^cm−2

, . . . ,

^c0

} .

⁽¹⁶⁾

There are four cases.

Case 1: 0

=

i

<

^j

≤

n

−

1. The proof of this case is divided into two parts, depending on

v

0

=

1 or

v

0

=

0. Suppose

v

0

=

1. Then 0

̸∈

H_j

(v,

^fj

(

¹

))

. Thus the 0th bit of all the vertices in V

(

^P2

)

is 1. By (12) and (16), 0 is the first element in H₀

(v,

^f0

(

¹

))

; this implies that the 0th bit of all the vertices in V

(

^P1

)

is 0. Thus no such a exists. In the following, we assume

v

0

=

0. Then 0

̸∈

H0

(v,

^f0

(

¹

))

. The 0th bit of all the vertices in V

(

^P1

)

is 0; this implies that the 0th bit of a is 0. There are two possibilities: j

=

1 or j

>

^1.

Subcase 1.1: j

=

1. Note that either 1

∈ Π

¹

(v,

^f1

(

¹

))

^{or 1}

̸∈ Π

¹

(v,

^f1

(

¹

))

^{. If 1}

̸∈ Π

¹

(v,

^f1

(

¹

))

, then 0 is the first element in

Π

1

(v,

^f1

(

¹

))

. This implies that the 0th bit of all the vertices in V

(

^P2

)

is 1. Thus no such a exists. If 1

∈ Π

1

(v,

^f1

(

¹

))

^{, then} 1 and 0 are the first element and the second element in

Π

1

(v,

^f1

(

¹

))

, respectively. Thus the 0th bit of all the vertices in V

(

^P2

) \ {

^f1

(v)}

is 1. The existence of a implies that f₁

(v) =

^a.

If

v

1

=

0, then 1

̸∈

H0

(v,

^f0

(

¹

))

. This implies that the 1st bit of all the vertices in V

(

^P1

)

is 0. However, it is obvious that the 1st bit of f1

(v)

is 1. Therefore f1

(v) ̸∈

^V

(

^P1

)

. Thus no such a exists. Now suppose

v

1

=

1. Since 1

∈ Π

¹

(v,

^f1

(

¹

))

, there must exist some k

>

1 such that

v

k

=

1; this implies that cm−1

>

1. By (12) and (16), the

(

^cm−1

)

th bit of all the vertices in V

(

^P1

)

is

v

c_m−1. However, the

(

^cm−1

)

th bit of f₁

(v)

^is

v

c_m−1. Therefore f₁

(v) ̸∈

^V

(

^P1

)

. Thus no such a exists and V

(

^P1

) ∩

^V

(

^P2

) = ∅

^. Subcase 1.2: j

>

1. By (12), (13) and (16), we have: c_m−1is the first element in H_i

(v,

^fi

(

¹

))

^{, c}m−1

∈

H_j

(v,

^fj

(

¹

))

^{, 0}

∈

H_j

(v,

^fj

(

¹

))

^, and c_m−1appears after 0 in the ordered set H_j

(v,

^fj

(

¹

))

. Thus the (c_m−1)th bit of all the vertices in V

(

^P1

)

^is

v

c_m−1. However, the (cm−1)th bit of those vertices with the 0th bit being 0 in V

(

^P2

)

^is

v

c_m−1. Thus no such a exists.

From the above discussion,Claim 4.5is true for Case 1.

Case 2: 1

=

i

<

^j

≤

n

−

1. The proof of this case is divided into two parts, depending on

v

0

=

0 or

v

0

=

1.

Subcase 2.1:

v

0

=

0. Then it is not difficult to see (by comparing the jth and the 0th bits of fj

(v)

and all the vertices in V

(

^P1

)

⁾ that f_j

(v) ̸∈

^V

(

^P1

)

. Thus a can not be f_j

(v)

. It remains to consider those vertices in V

(

^P2

)\

^fj

(v)

. The remaining proof is further divided into two parts, depending on

v

j−1

=

0 or

v

j−1

=

1.

Subcase 2.1.1:

v

j−1

=

0. Since

v

0

=

0 and

v

j−1

=

0, j

−

1

∈ Π

j

(v,

^fj

(v))

^{. Since}

v

0

=

0 and j

−

1

∈ Π

j

(v,

^fj

(v))

^{, the}

(

^j

−

1

)

^th bit of all the vertices in V

(

^P2

) \

^fj

(v)

is 1. However, the

(

^j

−

₁

)

th bit of all the vertices in V

(

^P1

)

is 0. Thus no such a exists and Claim 4.5is true.

Subcase 2.1.2:

v

j−1

=

1. We claim that: the bits from

v

j−2to

v

2are all 0, i.e.,

v

j−2

= v

j−3

= · · · = v

2

=

0. Suppose this claim is not true and let k be the largest number between j

−

2 and 2 (inclusive) such that

v

k

=

1. By (13) and (16), the

(

^j

−

1

)

^th and the kth bits of all the vertices in V

(

^P2

) \

^fj

(v)

is 1 and 0, respectively. However, the

(

^j

−

1

)

th bit of those vertices in V

(

^P1

)

with kth bit being 0 is 0. Thus

v

j−2

= v

j−3

= · · · = v

2

=

0. So the 1st bit of all the vertices in V

(

^P1

)

is 1 and the 1st bit of all the vertices in V

(

^P2

) \

^fj

(v)

is 0. Thus no such a exists andClaim 4.5is true.

Hj

(v,

^fj

(

¹

)) = (

^cm−1

,

^cm−2

, . . . ,

^c1

).

Suppose m is even. Then by (14) and (15),

Π

i

(v,

^fi

(

¹

))={

^I

(

^cm−1

,

^cm−2

), . . . ,

^I

(

^c1

,

^c0

=

2

)}

and

Π

j

(v,

^fj

(

¹

))={

^I

(

²

,

⁰

),

^I

(

^cm−1

,

^cm−2

), . . . ,

^I

(

^c1

,

²

)}.

Thus, the 2nd bit of all the vertices in V

(

^P1

)

are 1. However, the 2nd bit of all the vertices in V

(

^P2

)

are 0. Thus no such a exists. Suppose m is odd. Then by (14) and (15),

Π

ⁱ

(v,

^fi

(

¹

)) = {

¹

,

^I

(

^cm−1

,

^cm−2

), . . . ,

^I

(

^c0

,

¹

)}

and

Π

^j

(v,

^fj

(

¹

)) = {

^I

(

^cm−1

,

^cm−2

), . . . ,

^I

(

^c2

,

^c1

)}.

Hence the 1st bit of all the vertices in V

(

^P1

)

is 0. However, the 1st bit of all the vertices in V

(

^P2

)

is 1. Thus no such a exists.

Subcase 2.2.2: j

=

2,

v

1

=

0 and

v

2

=

1. Since

v

0

=

1 and

v

1

=

0 and

v

2

=

1, we have c_m−1

=

1, c₀

=

2 and H_j

(v,

^fj

(

¹

)) = {

^cm−1

,

^cm−2

, . . . ,

^c1

} .

Suppose m

−

1 is odd. Then by (14) and (15),

Π

i

(v,

^fi

(

¹

)) = {

^I

(

^cm−2

,

^cm−3

), . . . ,

^I

(

^c0

,

¹

)}

and

Π

^j

(v,

^fj

(

¹

)) = {

¹

,

^I

(

^cm−2

,

^cm−3

), . . . ,

^I

(

^c2

,

^c1

)}.

Thus, the 1st bit of all vertices in V

(

^P1

)

are 0. However, the 1st bit of all vertices in V

(

^P2

)

is 1. Thus no such a exists. Suppose m

−

1 is even. Then by (14) and (15),

Π

ⁱ

(v,

^fi

(

¹

)) = {

¹

,

^I

(

^cm−2

,

^cm−3

), . . . ,

^I

(

^c1

,

^c0

)}

and

Π

j

(v,

^fj

(

¹

)) = {

²

,

^I

(

^cm−2

,

^cm−3

), . . . ,

^I

(

^c1

,

²

)}.

Thus, the 2nd bit of all vertices in V

(

^P1

)

are 1. However, the 2nd bit of all vertices in V

(

^P2

)

is 0. Thus no such a exists.

Subcase 2.2.3: j

=

2,

v

1

=

1 and

v

2

=

0 (resp.,

v

1

=

0 and

v

2

=

0). Then H_j

(v,

^fj

(

¹

)) = {

²

,

^cm−1

,

^cm−2

, . . . ,

^c0

} .

Suppose m (resp., m

−

1) is even. Then by (14) and (15), the 2nd bit of all vertices in V

(

^P1

)

is 0. However, the 2nd bit of all vertices in V

(

^P2

)

is 1. Suppose m (resp., m

−

1) is odd. Then by (14) and (15), the 1st bit of all vertices in V

(

^P1

)

is 0. However, the 1st bit of all vertices in V

(

^P2

)

is 1. Thus no such a exists.

Subcase 2.2.4: j

̸=

2 and

v

j−1

=

0. Then the

(

^j

−

1

)

th bit of all the vertices in V

(

^P1

)

are 0. However, the

(

^j

−

1

)

th bit of all the vertices in V

(

^P2

)

are 1. Thus no such a exists.

Subcase 2.2.5: j

̸=

2,

v

j−1

=

1 and at least one of the bits in

v

j−2

v

j−3

. . . v

2is 1. Then there exists q such that

v

q

=

1 and q is the largest number between j

−

2 and 2 (inclusive).

Subcase 2.2.5.1: Suppose I

(

^j

,

^q

)

*

Π

j

(v,

^fj

(

¹

))

. Then the qth and the

(

^j

−

1

)

th bit of all the vertices in V

(

^P2

)

are 0 and 1, respectively; however, the

(

^j

−

1

)

th bit of those vertices in V

(

^P1

)

with the qth bit being 0 is 0. Thus no such a exists.

Subcase 2.2.5.2: Suppose I

(

^j

,

^q

) ⊆ Π

j

(v,

^fj

(

¹

))

. Then we partition V

(

^P2

)

^{into V}2,¹and V_2,2such that V2,¹

= {

all the vertices in V

(

^P2

)

before the perfect matching Mqis applied

}

and V2,²

=

V

(

^P2

) \

^V2,¹

.

Consider the vertices in V2,¹. Suppose

v

^j

=

0. Since j

∈

I

(

^j

,

^q

)

, we can compare the jth bit of all vertices in V

(

^P1

)

^{and in V2},¹

to see that no such a exists. Suppose

v

j

=

1. Then the number of bits in

v

n−1

v

n−2

. . . v

j+1that are 1 is odd, i.e.,

|

H_j²

|

is odd.

This implies that cm−1

̸=

j. Since cm−1

̸=

j, by comparing the cm−1th bit of all the vertices in V

(

^P1

)

^{and in V}2,¹, we know that V

(

^P1

) ∩

^V2,1

= ∅

. Consider the vertices in V_2,2. Then the qth and the

(

^j

−

1

)

th bit of all the vertices in V_2,2are 0 and 1, respectively. However, the

(

^j

−

1

)

th bit of those vertices in V

(

^P1

)

with the qth bit being 0 is 0. Hence V

(

^P1

) ∩

^V2,²

= ∅

. Since V

(

^P1

) ∩

^V2,¹

= ∅

and V

(

^P1

) ∩

^V2,²

= ∅

, no such a exists.

Subcase 2.2.6.1: Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

is even. Then t

(

ⁿ

−

1

,

^j

)

is odd. Thus the ith bit of all the vertices in V

(

^P2

)

^{is 0.}

However, the ith bit of all the vertices in V

(

^P1

)

is 1. Thus no such a exists.

Subcase 2.2.6.2: Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

^{is odd and}

v

^j

=

0. Then t

(

ⁿ

−

1

,

^j

+

1

)

is even. Thus the jth bit of all the vertices in V

(

^P2

)

is 1. However, the jth bit of all the vertices in V

(

^P1

)

is 0. Thus no such a exists.

Subcase 2.2.6.3: Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

^{is odd and}

v

j

=

1. Then t

(

ⁿ

−

1

,

^j

+

1

)

is odd. Thus the ith bit of all the vertices in V

(

^P1

)

is 0 and the jth bit of all the vertices in V

(

^P2

)

is 0. Then the. ith and the jth bit of a are 0. By (15), the

(

^j

−

1

)

th bit of all the vertices in V

(

^P2

)

with the ith and the jth bit be 0 is 1. However, only the vertex 2^j−1

+

1 in V

(

^P1

)

^{with the}

(

^j

−

1

)

^{th bit is} 1, and the ith and the jth bit are 0. The existence of a implies a

=

2^j−1

+

1. Since t

(

ⁿ

−

1

,

^j

+

1

)

is odd, there exists

v

k

=

1, where k

>

j. Then it is easy to find that a

̸∈

V

(

^P2

)

by comparing the kthBthe jth and the ith bit of a and all vertices in V

(

^P2

)

^. Thus no such a exists.

From the above discussion,Claim 4.5is true for Case 2.

Case 3: 3

≤

i

+

1

=

j

≤

n

−

1. By (12)–(16), we have the following results. Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

is odd. Then the ith bit of all vertices in V

(

^P1

)

^{is 0 and j}

̸∈ Π

^j

(v,

^fj

(

¹

))

; however, the ith bit of all the vertices in V

(

^P2

)

is 1. Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

^is even and

v

j

=

0. Then the jth bit of all the vertices in V

(

^P2

)

is 1; however, the jth bit of all the vertices in V

(

^P1

)

is 0. Suppose t

(

ⁿ

−

1

,

ⁱ

+

1

)

is even and

v

j

=

1. Then the jth bit of all the vertices in V

(

^P2

)

is 0; however, the jth bit of all the vertices in V

(

^P1

)

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

Case 4: 3

≤

i

+

1

<

^j

≤

n

−

1. We divide the proof into three parts, depending on the values of

v

j−1and

v

i−1.

Subcase 4.1:

v

j−1

=

0. Then if j

∈ Π

ⁱ

(v,

^fi

(

¹

))

^{, then V}

(

^P1

)

has only one vertex (say, vertex x) with its

(

^j

−

1

)

th bit being 1. By comparing from the jth to the

(

ⁱ

−

1

)

th bits of x with the jth to the

(

ⁱ

−

1

)

th bits of each vertex in V

(

^P2

)

, we have x

̸∈

V

(

^P2

)

^. If j

∈ Π

^j

(v,

^fj

(

¹

))

^{, then f}j

(v)

is the unique vertex in V

(

^P2

)

^{with its}

(

^j

−

1

)

th bit being 0. By comparing from the jth to the

(

ⁱ

−

1

)

th bits of fj

(v)

with the jth to the

(

ⁱ

−

1

)

th bits of each vertex in V

(

^P1

)

, we have fj

(v) ̸∈

^V

(

^P1

)

. Then by (12)–(16), the

(

^j

−

1

)

th bit of all the vertices in V

(

^P1

) \ {

^x

}

is 0; however, the

(

^j

−

1

)

th bit of all the vertices in V

(

^P2

) \

^fj

(v)

is 1. Thus no such a exists.

Subcase 4.2:

v

i−1

=

0. Then we can use similar arguments to prove that no such a exists.

Subcase 4.3:

v

i−1

=

1 and

v

j−1

=

1. By (12)–(15), we have following the results. When i

∈

Hi

(v,

^fi

(

¹

))

^and

v

0

=

1, V

(

^P1

)

has only one vertex fi

(v)

^{with the}

(

ⁱ

−

1

)

th bit being 0. It is easy to find fi

(v) ̸∈

^V

(

^P2

)

by comparing those bits from the

(

^j

−

1

)

^{th to the}

(

ⁱ

−

1

)

^{th of f}i

(v)

with each vertex in V

(

^P2

)

. And since the

(

ⁱ

−

1

)

th bit of all the vertices in V

(

^P1

) \

^fi

(v)

^{is 1,} the existence of a implies that the

(

ⁱ

−

1

)

th bit of a must be 1.

Partition V

(

^P2

)

^{into two V}2,¹and V_2,2such that

V_2,1

= {

all the vertices in V

(

^P2

)

before the perfect matching M_iis applied

}

and V_2,2

=

V

(

^P2

) \

^V2,1

.

Thus the

(

ⁱ

−

1

)

th bit of all the vertices in V2,¹is 1, and if a exists, then a

∈

V2,¹. We now claim that:

Claim 4.9. If a exists, then

v

j−2

= v

j−3

= · · · = v

i+1

=

0.

Proof ofClaim 4.9. Suppose this claim is not true. Then let q be the largest index between j

−

2 and i

+

1 (inclusive) such that

v

q

=

1. Let y

=

yn−1yn−2

. . .

^y0be an arbitrary vertex in V2,¹

\ {

fj

(v)}

. Note that fj

(v) ∈

^V2,¹only when j

∈ Π

^j

(v,

^fj

(

¹

))

^. Also note that q

∈ Π

^j

(v,

^fj

(

¹

))

. Moreover, if j

∈

Hj

(v,

^fj

(

¹

))

, then q is the first element after j in Hj

(v,

^fj

(

¹

))

^{; if j}

̸∈

Hj

(v,

^fj

(

¹

))

^, then q is the first element in H_j

(v,

^fj

(

¹

))

. Since q exists, by (13)–(15), the bits y_j−2y_j−3

. . .

^yi+1will be different from the bits

v

j−2

v

j−3

. . . v

i+1. However, let x

=

x_n−1x_n−2

. . .

^x0be an arbitrary vertex in V

(

^P1

)

. Then the bits x_j−2x_j−3

. . .

^xi+1are identical to the bits

v

j−2

v

j−3

. . . v

i+1. Thus every vertex in V_2,1

\ {

f_j

(v)}

is not in V

(

^P1

)

. Although f_j

(v) ∈

^V2,¹, f_j

(v)

is not in V

(

^P1

)

^(this can be observed by comparing the jth bit and from the

(

^j

−

2

)

^{th to the}

(

ⁱ

+

1

)

th bits of all the vertices in V

(

^P1

)

with jth bit and the bits from the

(

^j

−

2

)

^{th to the}

(

ⁱ

+

1

)

th bits of fj

(v)

^{). Thus V}

(

^P1

) ∩

^V2,¹

= ∅

. Since if a exists, then a

∈

V2,¹. Thus a does not exists and we have this claim. 

ByClaim 4.9, in the remaining proof, we assume

v

i−1

=

1,

v

j−1

=

1 and

v

j−2

= v

j−3

= · · · = v

i+1

=

0. For convenience, let t denote the number of bits in

v

n−1

v

n−2

. . . v

j+1that are 1. We further divided the proof into four subcases.

Subcase 4.3.1:

v

i

=

1 and

v

j

=

1. Suppose t is even. Then the first member in

Π

j

(v,

^fj

(

¹

))

is i. However, i

̸∈ Π

i

(v,

^fi

(

¹

))

^{. Thus} no such a exists and V

(

^P1

) ∩

^V

(

^P2

) = ∅

. Suppose t is odd. Then j

∈ Π

j

(v,

^fj

(

¹

))

^{and I}

(

^j

−

1

,

ⁱ

) ⊂ Π

i

(v,

^fi

(

¹

))

. Thus the jth bit of all the vertices in V

(

^P2

)

is 0. Partition V

(

^P1

)

^{into V}1,¹and V_1,2such that

V_1,1

= {

all the vertices in V

(

^P1

)

before the perfect matching M_(j+1)^{mod n}is applied

}

and V_1,2

=

V

(

^P1

) \

^V1,¹

.

Thus the jth bit of all vertices in V_1,1is 1 and the jth bit of all vertices in V_1,2is 0. By the fact that the jth bit of all the vertices in V

(

^P2

)

is 0, to prove V

(

^P1

) ∩

^V

(

^P2

) = ∅

, it suffices to prove V_1,2

∩

V

(

^P2

) = ∅

^{. If}

v

0

=

1, then the

(

^j

−

1

)

th bit of all the vertices in V

(

^P2

) \

^fj

(v)

is 1; however, the

(

^j

−

1

)

th bit of all the vertices in V_1,2is 0. Since the ith bit of is 1, but the ith bit of

comparing the kth bit of them. Therefore, no such a exists in this case.

Subcase 4.3.2:

v

i

=

0 and

v

j

=

0. Suppose t is even. Then the jth bit of all the vertices in V

(

^P2

)

is 1. However, the jth bit of all the vertices in V

(

^P1

)

is 0. Suppose t is odd. Then the number of bits in

v

n−1

v

n−2

. . . v

i+1that are 1 is even; this implies that i is the first member in

Π

i

(v,

^fi

(

¹

))

. Thus the ith bit of all the vertices in V

(

^P2

)

is 0. However, the ith bit of all the vertices in V

(

^P1

)

is 1. Thus no such a exists.

Subcase 4.3.3:

v

i

=

_{0 and}

v

j

=

1. Suppose t is even. Then the first member in

Π

^j

(v,

^fj

(

¹

))

^{is i}

−

1 and the first member in

Π

ⁱ

(v,

^fi

(

¹

))

is i. So the ith bit of all the vertices in V

(

^P2

)

is 0; however, the ith bit of all the vertices in V

(

^P1

)

is 1. Suppose t is odd. Define q to be the index of the leftmost nonzero bit of

v

^{. Then q}

>

j. Thus the

(

ⁱ

−

1

)

th bit of all the vertices in V

(

^P2

) \ {

^fj

(v)}

is 0; however, the

(

ⁱ

−

1

)

th bit of all the vertices in V

(

^P1

)

is 1. By comparing the jth and the qth bits of f_j

(v)

with the jth and the qth bits of every vertex in V

(

^P1

)

, we have f_j

(v) ̸∈

^V

(

^P1

)

. Thus no such a exists.

Subcase 4.3.4:

v

i

=

1 and

v

j

=

0. If the number of those bits from

v

n−1to

v

j+1being 1 is even, then the jth bit of all the vertices in V

(

^P2

)

is 1; however the jth bit of all the vertices in V

(

^P1

)

is 0. If the number of those bits from

v

n−1to

v

j+1being 1 is odd, then the number of bits in

v

n−1

v

n−2

. . . v

i+1that are 1 is even. Thus i is the first member of

Π

^j

(v,

^fj

(

¹

))

^{; but i}

̸∈ Π

ⁱ

(v,

^fj

(

¹

))

^, which implies that the ith bit of all the vertices in V

(

^P2

)

is 0 but the ith bit of all the vertices in V

(

^P1

)

^{is 1. SoClaim 4.5is} true for this case.

As a result,Claim 4.5is true for Case 4. From the above discussion,Claim 4.5is true for all the cases, and therefore T0

,

^T1

, . . . ,

^Tn−1are vertex-independent rooted at r

=

_{0 of LTQn. }

ByTheorem 2.4andLemmas 4.7and4.8, we have the following result.

Theorem 4.10. T0

,

^T1

, . . . ,

^Tn−1are n vertex-ISTs rooted at r for LTQn.

5. Concluding remarks

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. Both the Vertex and Edge Conjectures are still open on general graphs for n

≥

5.

In this paper, we consider the ISTs problem on 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 that LTQn

is not vertex-transitive when n

≥

4 and therefore Hsieh and Tu’s result does not solve the Edge Conjecture for LTQn. In this paper, we present an algorithm to construct n vertex-independent spanning trees rooted at an arbitrary vertex for LTQ_n. To

is not vertex-transitive when n

≥

4 and therefore Hsieh and Tu’s result does not solve the Edge Conjecture for LTQn. In this paper, we present an algorithm to construct n vertex-independent spanning trees rooted at an arbitrary vertex for LTQ_n. To