Lemma 3.4. T2∗ and T4∗ are independent spanning trees rooted at 0 of CR(N, d).
Proof. Let P and Q be the unique paths from 0 to x in T2∗ and in T4∗ , respectively. We
+d
−1
first consider 1 ≤ x ≤ d − 1. When d ≤ y + z " , these paths are P = (0 −→ d =⇒ x) and
−d
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−d
Q = (0 =⇒ N − (%(N − d)/d& × d) =⇒ x + d −→ x). Then every internal vertex a ∈ P satisfies
x < a ≤ d. Since [N − d]d )= 0, N − %(N − d)/d& × d > N − (N − d) = d. This shows that
every internal vertex b ∈ Q is larger than d. Hence, P ||Q. When d > y + z " , these paths are
+d
−1
−d
−1
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P = (0 −→ d =⇒ x) and Q = (0 =⇒ N − (%(N − d)/d& × d) =⇒ x + d −→ x). Then every
internal vertex a ∈ P satisfies x < a ≤ d, whereas every internal vertex b ∈ Q satisfies b ≥ x + d.
Thus P ||Q.
Next, we consider d ≤ x ≤ N − d. By Corollary 2.4, we can omit the case z = z " = 0.
We suppose that z = 0 and z " )= 0. When x/d ≤ %N/2d& (i.e., x ∈ class A), these paths are
+d
−d
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P = (0 =⇒ x) and Q = (0 =⇒ N − (%(N − d)/2d& × d) =⇒ N − (%(N − d)/2d& × d + z " ) =⇒ x).
Then every internal vertex a ∈ P is smaller than x, whereas every internal vertex b ∈ Q is larger
+d
than x. Thus P ||Q. When x/d > %N/2d& (i.e., x ∈ class B), these paths are P = (0 =⇒ x) and
−d
−1
Q = (0 =⇒ x + z " =⇒ x). An argument similar to the above case shows that P ||Q.
We suppose that z )= 0 and z " = 0. When (N − x)/d ≤ %N/2d& (i.e., x ∈ class C), these
+d
+1
+d
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paths are P = (0 =⇒ %(N − d)/2d& × d =⇒ %(N − d)/2d& × d + z =⇒ x) and Q = (0 =⇒ x).
Then every internal vertex a ∈ P is smaller than x, whereas every internal vertex b ∈ Q is
larger than x. Thus P ||Q. When (N − x)/d > %N/2d& (i.e., x ∈ class D), these paths are
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+1
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P = (0 =⇒ x − z =⇒ x) and Q = (0 =⇒ x). An argument similar to the above case shows that
P ||Q.
We suppose that z )= 0 and z " )= 0. If z + z " > d, by Corollary 2.4, we have d ≤ y + z
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−d
and d ≤ y + z " (i.e., x ∈ class E). In this case, P = (0 =⇒ x + [−x]d =⇒ x) and Q = (0 =⇒
+1
x − [x − N ]d =⇒ x). Clearly, x + [−x]d = x + d − z and x − [x − N ]d = x − d + z " . Then every
internal vertex a ∈ P satisfies d ≤ a ≤ x − z or x < a ≤ x + d − z, whereas every internal
vertex b ∈ Q satisfies x + z " ≤ b ≤ N − d or x − d + z " ≤ b < x. Since z, z " < d and z + z " > d,
we have x − z < x − d + z " < x < x + d − z < x + z " . We conclude P ||Q. Otherwise, i.e.,
z + z " ≤ d, by Corollary 2.4, we have d > y + z and d > y + z " (i.e., x ∈ class F ). In this case,
−1
+d
+1
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P = (0 =⇒ x − z =⇒ x) and Q = (0 =⇒ x + z " =⇒ x). Then every internal vertex a ∈ P is
smaller than x, whereas every internal vertex b ∈ Q is larger than x. Thus P ||Q.
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Finally, we consider N − d + 1 ≤ x ≤ N − 1. When d ≤ y + z, these paths are P = (0 =⇒
−1
+d
−d
+1
%(N − d)/d& × d =⇒ x − d −→ x) and Q = (0 −→ N − d =⇒ x). Since [N − d]d )= 0, every
internal vertex a ∈ P satisfies a ≤ %(N − d)/d& × d < N − d, whereas every internal vertex
b ∈ Q satisfies N − d ≤ b < x. Thus a < N − d ≤ b and P ||Q. When d > y + z, these paths are
+d
+1
+d
−d
+1
P = (0 =⇒ %(N − d)/d& × d =⇒ x − d −→ x) and Q = (0 −→ N − d =⇒ x). Then every internal
vertex a ∈ P satisfies a ≤ x − d, whereas every internal vertex b ∈ Q satisfies N − d ≤ b < x.
Since x < N , we have a ≤ x − d < N − d ≤ b. Thus P ||Q.
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