Vertex 1 as the common root

Throughout this subsection, let T0, T1, . . . , Tn−1 be the output of Algorithm 1 when the input is the F of LT Qn and the root is r = 1. The purpose of this subsection is to prove that T₀, T₁, . . . , T_n−1 are n ISTs rooted at r = 1 for LT Q_n. For S ⊆ V (LT Q_n), define f_i(S) to be

f_i(S) = {f_i(v) | for all v ∈ S}.

This definition will be used in the following proofs.

Lemma 12. T₀, T₁, . . . , T_n−1 are n spanning trees rooted at r for LT Q_n when r = 1.

Proof. The proof of this lemma is similar to that of Lemma 9 except that r = 0 is replaced by r = 1 and the proof of the claim is modified as follows.

Proof of the claim. This claim is true when k = 1 since line 3 sets S = {son} and hence

|S| = 1 = 2⁰. We now prove that if this claim is true before the k-th iteration of the outer for-loop, then it remains true before the next iteration. According to which Ti is considered, there are three possibilities.

1. Suppose T₀ is considered. Then i = 0 and there are two cases.

Case 1: k ∈ {1, 2, . . . , n−1}. The proof of this case is the same as Case 1 in Lemma 9.

Case 2: k = n. The proof of this case is the same as Case 2 in Lemma 9 except that:

the i-th bit of each vertex v ∈ S is 0 and the i-th bit of each vertex in S⁰ is 1.

2. Suppose T_n−1 is considered. Then i = n − 1 and there are two cases.

Case 1: k ∈ {1, 2, . . . , n − 1}. The proof of this case is the same as Case 1 in Lemma 9 except that: when k = n − 1, the (n − 2)-th bit of each vertex v ∈ S is 1 and the (n − 2)-th bit of each vertex in S⁰ is 0.

Case 2: k = n. The proof of this case is the same as Case 2 in Lemma 9.

3. Suppose Ti is considered, where i ∈ {1, 2, . . . , n − 2}. Then there are two cases.

Case 1: k ∈ {1, 2, . . . , n − 1}. The proof of this case is the same as Case 1 in Lemma 9 except that: when k = n − 1, the (n − 2)-th bit of each vertex v ∈ S is 1 and the (n − 2)-th bit of each vertex in S⁰ is 0.

Case 2: k = n. This is the last (the n-th) iteration of the outer for-loop of Algorithm 1.

Before the n-th iteration of the outer for-loop, |S| = 2ⁿ⁻¹and a total of 2⁰+2¹+· · ·+2ⁿ⁻²= 2ⁿ⁻¹−1 edges have been put into T_i; these edges form a connected subgraph since each newly generated edge in Algorithm 1 is incident to an edge that is already generated.

Thus S induces a tree. Partition S into S⁰ and S¹ such that

S⁰ = {all the vertices in the subtree rooted at f_i+1(f_i(1))} and S¹ = S \ S⁰.

Figure 5: An illustration for the proof of Lemma 12.

By (2) and by Lemma 3, we have: (i) the i-th bit of all the vertices in S⁰ is 0 and hence

Suppose i = n−2. Then the (n−1)-bit of all the vertices in S⁰and f_n−2(S⁰) is 1; however, the (n−1)-bit of all the vertices in S¹ and fn−2(S¹) is 0. Thus when i = n−2, S⁰∩fn−2(S¹) = ∅ and S¹ ∩ f_n−2(S⁰) = ∅. Now suppose i ∈ {1, 2, . . . , n − 3}. Partition S⁰ into S₀⁰ and S₁⁰ such that

S₀⁰ = {all the vertices in the subtree rooted at fi+2(fi+1(fi(1)))} and S₁⁰ = S⁰ \ S₀⁰. Partition S¹ into S₀¹ and S₁¹ such that

S₀¹ = {all the vertices in the subtree rooted at f_i+2(f_i(1))} and S₁¹ = S⁰\ S₀¹. By (2) and by Lemma 3, the pair of the (i+1)-th and the i-th bit of all the vertices in S₀⁰ and f_i(S₁¹) is (0,0); in f_i(S₀⁰) and S₁¹ is (0,1); in S₁⁰ and f_i(S₀¹) is (1,0) and in f_i(S₁⁰) and S₀¹ is (1,1). Thus to prove (10), it suffices to prove that

S₀⁰∩ fi(S₁¹) = ∅, S₁¹∩ fi(S₀⁰) = ∅, S₀¹∩ fi(S₁⁰) = ∅ and S₁⁰∩ fi(S₀¹) = ∅. (11) For each v = (v_n−1, v_n−1, . . . , v₀)₂ ∈ V (LT Q_n) such that v 6= 0, define q to be the index so that v_q is the leftmost nonzero bit, i.e., v_n−1= v_n−2 = · · · = v_q+1 = 0 and v_q= 1 (since v 6= 0, q exists). For v = 0, define q to be −1. By (2) and by Lemma 3, we have Table 2.

We now use two claims to prove (11).

Table 2: The value of q for every vertex in the given set.

S₀⁰∪ f_i(S₀⁰) S₁¹∪ f_i(S₁¹) S₀¹∪ f_i(S₀¹) S₁⁰∪ f_i(S₁⁰) q ≥ i + 2 q ≤ i + 1 or q ≥ i + 3 q ≥ i + 3 q = i + 1 or q ≥ i + 3

Claim A: S₀⁰∩ f_i(S₁¹) = ∅ and S₁¹∩ f_i(S₀⁰) = ∅. This claim holds since:

By Table 2, each vertex in S₁¹∩ f_i(S₁¹) with q ≤ i + 1 does not belong to S₀⁰∪ f_i(S₀⁰) since every vertex in S₀⁰∪ f_i(S₀⁰) has q ≥ i + 2. By Table 2, each vertex in S₀⁰ ∪ f_i(S₀⁰) with q = i + 2 does not belong to S₁¹ ∩ f_i(S₁¹) since each vertex in S₁¹ ∩ f_i(S₁¹) has q 6= i + 2.

From the above, we may focus on vertices with q = i + 3 or q > i + 3. Note that each vertex in S₀⁰ ∪ fi(S₀⁰) with q = i + 3 will have its (i + 2)-th bit to be 0; however, from Table 2, we know that each vertex in f_i(S₁¹) ∪ S₁¹ with q ≥ i + 3 will have its (i + 2)-th bit to be 1. Therefore, each vertex in S₀⁰∪f_i(S₀⁰) with q = i+3 does not belong to S₁¹∪f_i(S₁¹).

It remains to consider vertices with q > i + 3. Note that the bit string of those bits from

However, the bit string of those bits from the q-th bit to the (i + 2)-th bit of all the vertices in S₁¹∪ f_i(S₁¹) is one of the strings in the execution of line 11. Thus at the start of the next iteration of the outer for-loop,

|S| = 2^k. which can be determined by the ordered set

C_i(v, f_i(1)) = {c_m−1, c_m−2, . . . , c₀}

as follows. Let ce−1 be the first (from left to right) member in Ci(v, fi(1)) that is larger than i. Suppose v = (vn−1vn−2· · · v0)2. When i = 0, since r = 1, we have

P_i(v, f_i(1)) = C_i(v, f_i(1)). (13)

When i 6= 0 and v₀ = 0, since r = 1, we have c_e= 0 and and define C_i¹ to be the ordered sequence

C_i¹ =

For v = (11101)₂ ∈ T₃, we have C₃(v, 13) = {4}, C₃² = {4}, C₃¹ = {3}, ζ_j(v, f_j(1)) = {3, 0, 4, 3} and P3(v, 13) = {I(3, 0), I(4, 3)} = {3, 2, 1, 4}; so the v, 13-path in T3 is

(11101)2 f₃⁻¹=f3

→ (10001)2 f₂⁻¹=f2

→ (10111)2 f₁⁻¹=f1

→ (10101)2 f₄⁻¹=f4

→ (01101)2.

Lemma 13. T0, T1, . . . , Tn−1 are n vertex-independent trees rooted at r for LT Qn when r = 1.

Proof. It suffices to prove that any two T_i and T_j with 0 ≤ i < j ≤ n − 1 are vertex-independent. Let v = (v_n−1v_n−2· · · v₀)₂ be an arbitrary vertex in LT Q_n. We assume v 6∈ {r, f_i(r), f_j(r)} since if v ∈ {r, f_i(r), f_j(r)}, then the r, v-path in T_i and the r, v-path in Tj are clearly internally vertex-disjoint. By the same arguments used in the proof of Lemma 10, it suffices to prove that the v, f_i(r)-path in T_i and the v, f_j(r)-path in T_j are internally vertex-disjoint. Let V₁ and V₂ be defined as in Lemma 10. We now claim that:

Claim: V₁∩ V₂ = ∅.

Proof of the claim. Suppose this claim is not true and there exists a vertex a ∈ V₁ ∩ V₂. Let

C_i(v, f_i(1)) = {c_m−1, c_m−2, . . . , c₀}. (17)

There are four cases.

Case 1: 0 = i < j ≤ n − 1. The proof of this case is divided into two parts, depending on v₀ = 1 or v₀ = 0. Suppose v₀ = 1. Then 0 6∈ C_j(v, f_j(1)). Thus the 0-th bit of all the vertices in V₂ is 1. By (13) and (17), 0 is the first element in C₀(v, f₀(1)); this implies that the 0-th bit of all the vertices in V₁ is 0. Thus V₁ ∩ V₂ = ∅. Suppose v₀ = 0. Then 0 6∈ C₀(v, f₀(1)). Thus the 0-th bit of all the vertices in V₁ is 0; this implies that the 0-th bit of a is 0. There are two possibilities: j = 1 or j > 1.

1. j = 1. Note that either 1 ∈ P1(v, f1(1)) or 1 6∈ P1(v, f1(1)). Suppose 1 6∈ P1(v, f1(1)).

Then 0 is the first element in P1(v, f1(1)); this implies that the 0-th bit of all the vertices in V₂ is 1. Thus no such a exists and V₁ ∩ V₂ = ∅. Suppose 1 ∈ P₁(v, f₁(1)). Then 1 and 0 are the first element and the second element in P₁(v, f₁(1)), respectively. Thus the

0-th bit of all the vertices in V₂\ {f₁(v)} is 1. The existence of a implies that f₁(v) = a.

Suppose v1 = 0. Then 1 6∈ C0(v, f0(1)); this implies that the 1-st bit of all the vertices in V₁ is 0. However, it is obvious that the 1-st bit of f₁(v) is 1. Therefore f₁(v) 6∈ V₁. Thus no such a exists and V₁ ∩ V₂ = ∅. Suppose v₁ = 1. Since 1 ∈ P₁(v, f₁(1)), there must exist some k > 1 such that v_k = 1; this implies that c_m−1 6= 1. By (13) and (17), the (c_m−1)-th bit of all the vertices in V₁ is v_cm−1. However, the (c_m−1)-th bit of f₁(v) is v_cm−1. Therefore f₁(v) 6∈ V₁. Thus no such a exists and V₁∩ V₂ = ∅.

2. j > 1. By (13), (14), (15), (16) and (17), we have: cm−1 is the first element in Ci(v, fi(1)), cm−1 ∈ Cj(v, fj(1)), 0 ∈ Cj(v, fj(1)), and cm−1 appears after 0 in the ordered set C_j(v, f_j(1)). Thus the (c_m−1)-th bit of all the vertices in V₁ is v_cm−1. However, the (c_m−1)-th bit of those vertices with the 0-th bit being 0 in V₂ is v_cm−1. Thus no such a exists and V₁∩ V₂ = ∅.

Case 2: 1 = i < j ≤ n − 1. The proof of this case is divided into two parts, depending on v0 = 0 or v0 = 1.

1. v₀ = 0. Then it is not difficult to see (by comparing the j-th and the 0-th bits of f_j(v) and all the vertices in V₁) that f_j(v) 6∈ V₁. Thus a can not be f_j(v). It remains to consider those vertices in V₂\ f_j(v). The remaining proof is further divided into two parts, depending on v_j−1 = 0 or v_j−1 = 1.

1.1. vj−1 = 0. Since v0 = 0 and vj−1 = 0, j − 1 ∈ Pj(v, fj(v)). Since v0 = 0 and j − 1 ∈ Pj(v, fj(v)), the (j − 1)-th bit of all the vertices in V2\ fj(v) is 1. However, the (j − 1)-th bit of all the vertices in V₁ is 0. Thus no such a exists and V₁∩ V₂ = ∅.

1.2. v_j−1= 1. We claim that: the bits from v_j−2 to v₂ are all 0, i.e., v_j−2 = v_j−3 = · · · = v₂ = 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 (17) and (14), the (j − 1)-th and the k-th bits of all the vertices in V2\ fj(v) is 1 and 0, respectively. However, the (j − 1)-th bit of those vertices in V1 with k-th bit being 0 is 0. Thus vj−2 = vj−3 = · · · = v2 = 0. So the 1-st bit of all the vertices in V₁ is 1 and the 1-st bit of all the vertices in V₂\ f_j(v) is 0. Thus no such a exists and V₁∩ V₂ = ∅.

2. v₀ = 1. The proof of this part is further divided into six parts as follows.

2.1. j = 2, v1 = 1 and v2 = 1. Since v0 = 1 and v1 = 1 and v2 = 1, C_j(v, f_j(1)) = (c_m−1, c_m−2, . . . , c₁).

Suppose m is even. Then

P_i(v, f_i(1)) = {I(c_m−1, c_m−2), . . . , I(c₁, c₀= 2)}

and

P_j(v, f_j(1)) = {I(2, 0), I(c_m−1, c_m−2), . . . , I(c₁, 2)}.

By (15) and (16), the 2-nd bit of all the vertices in V₁ are 1. However, the 2-nd bit of all the vertices in V₂ are 0. Thus no such a exists and V₁∩ V₂ = ∅. Suppose m is odd. Then

P_i(v, f_i(1)) = {1, I(c_m−1, c_m−2), . . . , I(c₀, 1)}

and

P_j(v, f_j(1)) = {I(c_m−1, c_m−2), . . . , I(c₂, c₁)}.

By (15) and (16), the 1-st bit of all the vertices in V₁ is 0. However, the 1-st bit of all the vertices in V₂ is 1. Thus no such a exists and V₁ ∩ V₂ = ∅.

2.2. j = 2, v₁ = 0 and v₂ = 1. Since v₀ = 1 and v₁ = 0 and v₂ = 1, we have c_m−1 = 1, c₀ = 2 and

C_j(v, f_j(1)) = {c_m−1, c_m−2, . . . , c₁}.

Suppose m − 1 is odd. Then

P_i(v, f_i(1)) = {I(c_m−2, c_m−3), . . . , I(c₀, 1)}

and

P_j(v, f_j(1)) = {1, I(c_m−2, c_m−3), . . . , I(c₂, c₁)}.

By (15) and (16), the 1-st bit of all vertices in V₁ are 0. However, the 1-st bit of all vertices in V₂ is 1. Thus no such a exists and V₁ ∩ V₂ = ∅. Suppose m − 1 is even. Then

P_i(v, f_i(1)) = {1, I(c_m−2, c_m−3), . . . , I(c₁, c₀)}

and

Pj(v, fj(1)) = {2, 1, I(cm−2, cm−3), . . . , I(c1, 2)}.

By (15) and (16), the 2-nd bit of all vertices in V₁ are 1. However, the 2-nd bit of all vertices in V₂ are 0. Thus no such a exists and V₁∩ V₂ = ∅.

2.3. j = 2, v₁ = 1 and v₂ = 0 (resp., v₁ = 0 and v₂ = 0). Then

Cj(v, fj(1)) = {2, cm−1, cm−2, . . . , c0}.

Suppose m (resp., m − 1) is even. Then by (15) and (16), the 2-nd bit of all vertices in V₁. However, the 2-nd bit of all vertices in V₂ are 1. Suppose m (resp., m − 1) is odd.

Then by (15) and (16), the 1-st bit of all vertices in V₁ are 0. However, the 1-st bit of all vertices in V₂ are 1. Thus no such a exists and V₁∩ V₂ = ∅.

2.4. j 6= 2 and v_j−1 = 0. Then the (j − 1)-th bit of all the vertices in V₁ are 0. However, the (j − 1)-th bit of all the vertices in V₂ are 1. Thus no such a exists and V₁∩ V₂ = ∅.

2.5. j 6= 2, vj−1 = 1 and at least one of the bits in vj−2vj−3· · · v2 is 1. Then there exist q such that

q = max{ t | t ∈ C_i(v, f_i(1)), 1 < t < j − 1}.

2.5.1. Suppose I(j, q) * P_j(v, f_j(1)). Then the q-th and the (j − 1)-th bit of all the vertices in V₂ are 0 and 1, respectively; however, the (j − 1)-th bit of those vertices in V₁ with the q-th bit being 0 is 0. Thus no such a exists and V1 ∩ V2 = ∅.

2.5.2. Suppose I(j, q) ⊆ P_j(v, f_j(1)). Then we partition V₂ into V_2,1 and V_2,2 such that

V_2,1 = {all the vertices in V₂before the perfect matching f_qis applied} and V_2,2 = V₂\V_2,1.

Consider the vertices in V_2,1. Suppose v_j = 0. Since j ∈ I(j, q), we can compare the j-th bit of all vertices in V₁ and in V_2,1 to see that no such a exists and V₁∩ V₂ = ∅. Suppose v_j = 1. Then the number of bits in v_n−1v_n−2· · · v_j+1 that are 1 is odd. This implies that cm−1 6= j. Since cm−1 6= j, by comparing the cm−1-th bit of all the vertices in V1 and in V2,1, we know that V1 ∩ V2,1 = ∅. Consider the vertices in V2,2. Then the q-th and the

(j − 1)-th bit of all the vertices in V_2,2 are 0 and 1, respectively. However, the (j − 1)-th bit of those vertices in V1 with the q-th bit being 0 is 0. Hence V1 ∩ V2,2 = ∅. Since V₁∩ V_2,1 = ∅ and V₁∩ V_2,2 = ∅, no such a exists and V₁∩ V₂ = ∅.

2.6. j 6= 2, v_j−1 = 1 and all the bits in v_j−2v_j−3· · · v₂ are 0 (i.e., v_j−2= v_j−3= · · · = v₂= 0).

For convenience, let t(w₁, w₂) denote the number of bits in v_w1v_w1−1· · · v_w2 that are 1.

There are three possibilities.

2.6.1. Suppose t(n − 1, i + 1) is odd. Then t(n − 1, j) is even. Thus the i-th bit of all the vertices in V₂ is 0. However, the i-th bit of all the vertices in V₁ is 1. Thus no such a exists and V1∩ V2 = ∅.

2.6.2. Suppose t(n − 1, i + 1) is even and v_j = 0. Then t(n − 1, j + 1) is even. Thus the j-th bit of all the vertices in V₂ is 1. However, the j-th bit of all the vertices in V₁ is 0.

Thus no such a exists and V₁∩ V₂ = ∅.

2.6.3. Suppose t(n − 1, i + 1) is even and v_j = 1. Then t(n − 1, j + 1) is odd. Thus the i-th bit of all the vertices in V₂\ {f_j(v)} is 0. However, the i-th bit of all the vertices in V1 is 1. Since cm−1 6= j, we can find that fj(v) 6∈ V1 by comparing the cm−1-th bit. Thus no such a exists and V1∩ V2 = ∅.

Case 3: 3 ≤ i + 1 = j ≤ n − 1. For convenience, let t denote the number of bits in v_n−1v_n−2· · · v_i+1 that are 1. By (13)∼(17), we have the following results for t. Suppose t is odd. Then the i-th bit of all vertices in V₁ is 0 and j 6∈ P_j(v, f_j(1)); however, the i-th bit of all the vertices in V₂ is 1. Suppose t is even and v_j = 0. Then the j-th bit of all the vertices in V₂ is 1; however, the j-th bit of all the vertices in V₁ is 0. Suppose t is even and v_j = 1. Then the j-th bit of all the vertices in V₂ is 0; however, the j-th bit of all the vertices in V1 is 1. Thus no such a exists and V1 ∩ V2 = ∅.

Case 4: 3 ≤ i+1 < j ≤ n−1. The proof of this case is divided into xxx parts, depending on the values of v_j−1 and v_i−1.

4.1. v_j−1 = 0. Then if j ∈ P_i(v, f_i(1)), then V₁ has only one vertex (say, vertex x) with its (j − 1)-th bit being 1. By comparing from the j-th to the (i − 1)-th bits of x with the

j-th to the (i − 1)-th bits of each vertex in V₂, we have x 6∈ V₂. If j ∈ P_j(v, f_j(1)), then fj(v) is the unique vertex in V2 with its (j − 1)-th bit being 0. By comparing from the j-th to the (i − 1)-th bits of f_j(v) with the j-th to the (i − 1)-th bits of each vertex in V₁, we have f_j(v) 6∈ V₁. Then by (13)∼(17), the (j − 1)-th bit of all the vertices in V₁\ {x} is 0; however, the (j − 1)-th bit of all the vertices in V₂\ f_j(v) is 1. Thus no such a exists and V₁∩ V₂ = ∅.

4.2. vi−1= 0. Then we can use similar arguments to prove that no such a exists and V1∩ V2= ∅.

4.3. v_i−1 = 1 and v_j−1 = 1. By (13)∼(16), we have following the results. When i ∈ C_i(v, f_i(1)) and v₀ = 1, V₁ has only one vertex (say, vertex z) with its (i − 1)-th bit being 0. By comparing the (j − 1)-th and the (i − 1)-th bits of z with the (j − 1)-th and the (i − 1)-th bits of each vertex in V₂, we have z 6∈ V₂. Thus the (i − 1)-th bit of all the vertices in V1\ {z} is 1. Hence the existence of a implies that the (i − 1)-th bit of a must be 1. Partition V2 into two V2,1 and V2,2 such that

V2,1 = {all the vertices in V2before the perferct matching fiis applied} and V2,2 = V2\V2,1.

Thus the (i − 1)-th bit of all the vertices in V_2,1 is 1 and if a exist, then a ∈ V_2,1. We claim that:

If a exists, then vj−2= vj−3= · · · = vi+1= 0.

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 = (y_n−1y_n−2· · · y₀)₂ be an arbitrary vertex in V_2,1 \ {f_j(v)}. Note that f_j(v) ∈ V_2,1 only when j ∈ P_j(v, f_j(1)). Also note that q ∈ P_j(v, f_j(1)). Moreover, if j ∈ C_j(v, f_j(1)), then q is the first element after j in C_j(v, f_j(1));

if j 6∈ C_j(v, f_j(1)), then q is the first element in C_j(v, f_j(1)). Since q exists, by (14)∼(16), the bits yj−2yj−3· · · yi+1 will be different from the bits vj−2vj−3· · · vi+1. However, let x = (xn−1xn−2· · · x0)2 be an arbitrary vertex in V1. Then the bits xj−2xj−3· · · xi+1 are identical to the bits v_j−2v_j−3· · · v_i+1. Thus every vertex in V_2,1 \ {f_j(v)} is not in V₁. Although f_j(v) ∈ V_2,1, f_j(v) is not in V₁ (this can be observed by comparing the j-th bit

and the bits from the (j − 2)-th to the (i + 1)-th bits of all the vertices in V₁ with j-th bit and the bits from the (j − 2)-th to the (i + 1)-th bits of fj(v)). Thus V1∩ V2,1 = ∅.

Since if a exists, then a ∈ V_2,1. Thus a does not exists and we have this claim.

By this claim, 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−1v_n−2· · · v_j+1 that are 1.

The remaining proof is further divided into four subcases.

4.3.1. vi = 1 and vj = 1. Suppose t is even. Then the first member in Pj(v, fj(1)) is i.

However, i 6∈ P_i(v, f_i(1)). Thus no such a exists and V₁∩ V₂ = ∅. Suppose t is odd. Then j ∈ P_j(v, f_j(1)) and I(j − 1, i) ⊂ P_i(v, f_i(1)). Thus the j-th bit of all the vertices in V₂ is 0. Partition V₁ into V_1,1 and V_1,2 such that

V_1,1= {all the vertices in V₁before the perfect matching f_j+1is applied} and V_1,2= V₁\V_1,1.

Thus the j-th bit of all vertices in V1,1 is 1 and the j-th bit of all vertices in V1,2is 0. By the fact that the j-th bit of all the vertices in V2 is 0, to prove V1∩ V2 = ∅, it suffices to prove V_1,2∩ V₂ = ∅. If v₀ = 1, then the (j − 1)-th bit of all the vertices in V₂ is 1; however, the (j −1)-th bit of all the vertices in V_1,2 is 0. If v₀ = 0, then V₂ has only one vertex f_j(v) with its (j − 1)-th bit being 0. Obviously, either f_j(v) = (v_n−1v_n−2· · · v_j+10v_j−1v_j−2v_j−3· · · v₀)₂ or f_j(v) = (v_n−1v_n−2· · · v_j+10v_j−1v_j−2v_j−3· · · v₀)₂; the former case occurs when v₀= 0 and the latter, v₀= 1. In either case, we have f_j(v) 6∈ V₁. Thus no such a exists and V₁∩V₂= ∅.

4.3.2. vi = 0 and vj = 0. Suppose t is even. Then the j-th bit of all the vertices in V2 is 1. However, the j-th bit of all the vertices in V₁ is 0. Suppose t is odd. Then the number of bits in v_n−1v_n−2· · · v_i+1 that are 1 is even; this implies that i is the first member in P_i(v, f_i(1)). Thus the i-th bit of all the vertices in V₂ is 0. However, the i-th bit of all the vertices in V₁ is 1. Thus no such a exists and V₁ ∩ V₂ = ∅.

4.3.3. vi = 0 and vj = 1. Suppose t is even. Then the first member in Pj(v, fj(1)) is i − 1 and the first member in Pi(v, fi(1)) is i. So the i-th bit of all the vertices in V2 is 0; however, the i-th bit of all the vertices in V₁ 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 V₂\ {f_j(v)} is 0; however, the (i − 1)-th bit of all the vertices in V₁ is 1.

By comparing the j-th and the q-th bits of fj(v) with the j-th and the q-th bits of every vertex in V₁, we have f_j(v) 6∈ V₁. Thus no such a exists and V₁∩ V₂ = ∅.

4.3.4. v_i = 1 and v_j = 0. If the number of those bits from v_n−1 to v_j+1 being 1 is even, then the j-th bit of all the vertices in V₂ is 1, but the j-th bit of all the vertices in V₁ is 0. If the number of those bits from v_n−1 to v_j+1 being 1 is odd, then the number of bits in v_n−1v_n−2· · · v_i+1 that are 1 is even. Thus i is the first member of P_j(v, f_j(1)) but i 6∈ P_i(v, f_j(1)) which implies that the i-th bit of all the vertices in V₂ is 0 but the i-th bit of all the vertices in V1 is 1. So V1∩ V2 = ∅ in this case.

Since V₁∩ V₂ = ∅, we have this lemma.

Theorem 14. T₀, T₁, . . . , T_n−1 are n ISTs rooted at r for LT Q_n when r = 1.

Proof. This theorem follows from Lemmas 12 and 13.