**5. Bounds**

In this section, we provide sharp bounds on ID-stable trees. First, we present a lower bound and characterize all extremal trees. Let T1 be the family of trees *T* that can be obtained from a sequence *T*1, *T*2, ..., *Tk* of trees for some *k* ≥ 1, where *T*1 is *P*2 and *T* = *Tk*. If *k* ≥ 2, then all but at most one of *Ti*+<sup>1</sup> can be obtained from *Ti* by operation T1, and that one (if any) can be obtained from *Ti* by operation T2 for *q* = 2.

**Theorem 6.** *Let T be an ID-stable tree of order n* ≥ 2*. Then:*

$$i(T) \ge \left\lceil \frac{n}{3} \right\rceil$$

*with equality if and only if T* ∈ T1*.*

**Proof.** By Theorem 2, we have *T* ∈ T . Thus, there exists a sequence of trees *T*1, *T*2, ... , *Tk* (*k* ≥ 1) such that *T*1 is *P*2, and if *k* ≥ 2, then *Ti*+<sup>1</sup> can be obtained from *Ti* by one of the operations T1 or T2. We proceed by induction on the number of operations used to construct *T*. If *k* = 1, the result is trivial. Assume the result holds for each tree *T* ∈ T , which can be obtained from a sequence of operations of length *k* − 1, and let *T* = *Tk*−1. By Proposition 4 and the induction hypothesis, we obtain:

$$i(T) = i(T') + i(\mathbb{S}\_q) \ge \left\lceil \frac{n - 2q - 1}{3} \right\rceil + \left\lceil \frac{2q + 1}{3} \right\rceil \ge \left\lceil \frac{n}{3} \right\rceil. \tag{1}$$

The equality holds if and only if *i*(*T*) = # *<sup>n</sup>*−2*q*−1 3 \$ and *i*(*Sq*) = # 2*q*+1 3 \$. It follows from the induction hypothesis that *T* ∈ T1. Furthermore, we deduce from *i*(*Sq*) = # 2*q*+1 3 \$ that *q* ≤ 3. First, let *q* = 3. It follows from Equation (1) that:

$$
\left\lceil \frac{n-7}{3} \right\rceil + \left\lceil \frac{7}{3} \right\rceil = \left\lceil \frac{n}{3} \right\rceil
$$

yielding *n* ≡ 1(mod <sup>3</sup>). This implies that <sup>3</sup>|*n*(*T*), which is a contradiction by construction of trees in T1. Hence, *q* ≤ 2. If *Ti*+<sup>1</sup> is obtained from *Ti* by operation T1 for each 2 ≤ *i* ≤ *k* − 1, then clearly, *T* ∈ T1. Assume that one of the *Ti*+1's is obtained from *Ti* by operation T2 for *q* = 2. Then, clearly, *n*(*T*) = *n* − 2*q* − 1 = 3(*k* − 1) + 1. If *q* = 2, then *n*(*T*) = 3(*k* − 1) + 6, and we have # *<sup>n</sup>*−2*q*−1 3 \$ + # 2*q*+1 3 \$ = *k* + 2 > *k* + 1 = # *n*(*T*) 3 \$, which is a contradiction. Thus, *q* = 1, and this implies that *T* ∈ T1.

Let F1 be the family of all spiders *Sq* for *q* ≥ 2, F2 be the family of trees obtained from two spiders *Sp* and *Sq* by joining their heads, F3 be the family of trees obtained from two spiders *Sp* and *Sq* by joining the head of *Sp* to a knee of *Sq*, and F4 be the family of trees obtained from two spiders *Sp* and *Sq* by joining the head of *Sp* to a foot of *Sq* where *p* ≥ *q* = 2 or *p*, *q* ≥ 3. For example, the trees obtained by F2, F3, and F4 when *p* = *q* = 3 are illustrated in Figure 5.

**Figure 5.** (**a**) Tree F2; (**b**) tree F3; (**c**) tree F4.

The next result is an immediate consequence of Proposition 4.

**Observation 1.** *If T* ∈ <sup>∪</sup>4*i*=1F*i, then i*(*T*) = *n*−2 2 *.*

**Theorem 7.** *Let T be an ID-stable tree of order n* ≥ 5*. Then:*

$$i(T) \le \left\lceil \frac{n-2}{2} \right\rceil$$

*with equality if and only if T* ∈ <sup>∪</sup>4*i*=1F*i.*

**Proof.** The proof is by induction on *n*. If *n* = 5, then by Propositions 2 and 3, we have *T* = *P*5, and the result holds. Let *n* ≥ 6, and let the statement hold for all ID-stable trees of order less than *n*. Assume that *T* is an ID-stable tree of order *n*. By Propositions 2 and 3, we deduce that diam(*T*) ≥ 4. If diam(*T*) = 4, then by Propositions 2 and 3, *T* is the healthy spider *Spider*(*dT*(*<sup>v</sup>*3)), and so, *i*(*T*) = *dT*(*<sup>v</sup>*3) = # *<sup>n</sup>*(*T*)−<sup>2</sup> 2 \$ and *T* ∈ F1. Suppose that diam(*T*) ≥ 5. Let *v*1*v*2 ... *vk* (*k* ≥ 5) be a diametrical path in *T* such that *dT*(*<sup>v</sup>*3) is as large as possible and root *T* at *vk*. By Propositions 2 and 3, we have *dT*(*<sup>v</sup>*2) = 2 and that *v*3 is not a support vertex. Hence, *Tv*3 = *SdT*(*<sup>v</sup>*3)−1. Assume that *p* = *dT*(*<sup>v</sup>*3) − 1. Let *T* = *T* − *Tv*3 . Since *T* is an ID-stable tree, we deduce from Proposition 4 that for any vertex *v* ∈ *<sup>V</sup>*(*T*),

$$i(T'-\upsilon) + p = i(T-\upsilon) = i(T) = i(T') + p$$

and this implies that *i*(*T* − *v*) = *<sup>i</sup>*(*T*). Hence, *T* is an ID-stable tree. It follows from the induction hypothesis that *i*(*T*) ≤ # *<sup>n</sup>*−2*p*−3 2 \$, and hence,

The equality holds if and only if diam(*T*) = 4 or diam(*T*) ≥ 5 and *i*(*T*) = # *<sup>n</sup>*(*T*)−<sup>2</sup> 2 \$ = # *<sup>n</sup>*−2*p*−3 2 \$ and *n* is even, and this if and only if *T* ∈ F1 or diam(*T*) ≥ 5 and *T* ∈ F1 by the induction hypothesis. Thus, the equality holds if and only if *T* ∈ <sup>∪</sup>4*i*=1F*i*, and the proof is complete.
