**2. Basic Properties**

In this section, we study the basic properties of the ID-stable graph, and we construct new ID-stable graphs from an old one.

**Proposition 2.** *If G is an ID-stable graph, then every support vertex in G is adjacent to exactly one leaf.*

**Proof.** Let *G* be an ID-stable graph. Suppose, to the contrary, that *G* has a support vertex *x* with |*L*(*x*)| ≥ 2, and let *y*, *z* ∈ *<sup>L</sup>*(*x*). If *G* has an *<sup>i</sup>*(*G*)-set *S* such that *x* ∈ *S*, then *y*, *z* ∈ *S*, and clearly, *S* − {*y*} is an independent dominating set of *G* − *y* yielding *i*(*G* − *y*) < *<sup>i</sup>*(*G*), which is a contradiction. Hence, we assume that every *<sup>i</sup>*(*G*)-set contains *x*. Now, consider the graph *G* − *x*, and let *D* be an *i*(*G* − *x*)-set. Since each vertex in *<sup>L</sup>*(*x*) is isolated in *G* − *x*, *D* contains all vertices in *<sup>L</sup>*(*x*). Clearly, *D* is an independent dominating set of *G* such that *x* ∈ *D*. It follows from the assumption that *i*(*G* − *x*) = |*D*| > *<sup>i</sup>*(*G*), a contradiction again. This completes the proof.

**Proposition 3.** *If G is an ID-stable graph, then G does not have two adjacent support vertices.*

**Proof.** Let *G* be an ID-stable graph. Suppose, to the contrary, that there exist two adjacent support vertices *x*, *y* in *G*. Assume that *<sup>L</sup>*(*x*) = {*x*}, *<sup>L</sup>*(*y*) = {*y*}, and let *S* be an *<sup>i</sup>*(*G*)-set. Then, *x* ∈ *S* or *y* ∈ *S*. Assume, without loss of generality, that *y* ∈ *S*. Then, *y* ∈/ *S*. If (*NG*(*y*) − {*y*}) ∩ *S* = ∅, then *S* − {*y*} is an independent dominating set of *G* − *y*, which leads to a contradiction. Hence, (*NG*(*y*) − {*y*}) ∩ *S* = ∅. In particular, *x* ∈ *S*, and so, *x* ∈ *S*. Now, (*S* − {*y*, *x*}) ∪ {*y*} is an independent dominating set of *G* − *<sup>x</sup>*, which leads to a contradiction.

The spider *Sq* is the graph obtained from the star *<sup>K</sup>*1,*q* by subdividing its edges once. Clearly, *i*(*Sq*) = *q*. Assume that *V*(*Sq*) = {*s*}∪{*ai*, *bi*|*i* = 1, 2, ... , *q*} and *E*(*Sq*) = {*sai*, *aibi*|*i* = 1, 2, ... , *q*}. The vertex *s* is called the *head*; the vertices *ai* are called the *knees*; and the vertices *bi* are called the *feet* of the spider for 1 ≤ *i* ≤ *q*.

**Proposition 4.** *Let G be a graph and v* ∈ *<sup>V</sup>*(*G*)*. Let G be the graph obtained from G by adding a spider Sq* (*q* ≥ 1) *and possibly joining the head s to v. Then, i*(*G*) = *i*(*G*) + *q.*

**Proof.** Clearly, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* by adding *a*1,..., *aq*, and so, *i*(*G*) ≤ *i*(*G*) + *q*.

Now, we show that *i*(*G*) ≥ *i*(*G*) + *q*. Let *S* be an *<sup>i</sup>*(*G*)-set. To dominate *bi*, we must have |*S* ∩ {*ai*, *bi*}| ≥ 1 for each *i*. If *s* ∈ *S* or *sv* ∈ *<sup>E</sup>*(*G*), then the set *S* − *V*(*Sq*) is an independent dominating set of *G*, and this implies that *i*(*G*) ≥ *i*(*G*) + *q*. Suppose that *s* ∈ *S* and *sv* ∈ *<sup>E</sup>*(*G*). It follows that {*b*1, ... , *bq*} ⊆ *S* and *S* ∩ *NG*[*v*] = ∅. Then, the set (*S* − {*<sup>s</sup>*, *b*1, ... , *bq*}) ∪ {*v*} is an independent dominating set of *G* yielding *i*(*G*) ≥ *i*(*G*) + *q*. Thus, *i*(*G*) = *i*(*G*) + *q*, and the proof is complete.

**Proposition 5.** *Let G be an ID-stable graph. Then:*


**Proof.** Our arguments apply equally well to both parts, so we prove them simultaneously. Let *v* ∈ *V*(*G*) be an arbitrary vertex. If *v* ∈ *<sup>V</sup>*(*G*), then we have *i*(*G* − *v*) = *i*(*G*) because *G* is an ID-stable graph, and by Proposition 4, we have

$$i(G'-v) = i(G-v) + q = i(G) + q = i(G').$$

Assume that *v* ∈ *<sup>V</sup>*(*Sq*). We consider three cases.

### **Case 1.** *v* = *s*.

Then, clearly, *G* − *v* is the union of *G* with *qK*2 (*q* ≥ <sup>1</sup>), and so, *i*(*G* − *v*) = *i*(*G*) + *q*. It follows from Proposition 4 that *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

### **Case 2.** *v* = *ai* for some *i* ∈ {1, 2, . . . , *q*}.

Assume, without loss of generality, that *v* = *aq*. First, we prove (1). Clearly, we have *G* − *v* = (*G* + *us*) ∪ *K*1. Obviously, any *<sup>i</sup>*(*G*)-set containing *u* can be extended to an independent dominating set of *G* − *v* by adding *b*1, and so, *i*(*G* − *v*) ≤ *i*(*G*) + 1 = *<sup>i</sup>*(*G*). On the other hand, any *i*(*G* − *v*)-set is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*), yielding *i*(*G* − *v*) = *<sup>i</sup>*(*G*). Now, we prove (2). Clearly, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding {*bq*, *a*1, ... , *aq*−<sup>1</sup>}, and so, *i*(*G* − *v*) ≤ *i*(*G*) + *q* = *<sup>i</sup>*(*G*). Furthermore, any *i*(*G* − *v*)-set is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Thus, *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

**Case 3.** *v* = *bi* for some *i* ∈ {1, 2, . . . , *q*}.

Assume, without loss of generality, that *v* = *b*1. Obviously, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding {*<sup>a</sup>*1, ... , *aq*}, and so, *i*(*G* − *v*) ≤ *i*(*G*) + *q* = *<sup>i</sup>*(*G*). Now, let *S* be an *i*(*G* − *v*)-set. If *a*1 ∈ *S*, then *S* is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Assume that *a*1 ∈/ *S*. Then, *s* ∈ *S*. If (*NG*(*u*) − {*s*}) ∩ *S* = ∅, then (*S* − {*s*}) ∪ {*<sup>a</sup>*1} is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Suppose that (*NG*(*u*) − {*s*})∩*S* = ∅. Then, *S* − {*<sup>s</sup>*, *b*2, ... , *bq*} is an independent dominating set of *G* − *u*, and since *G* is an ID-stable graph, we deduce that *i*(*G* − *v*)=(|*S*| − *q*) + *q* ≥ *i*(*G*) + *q* = *<sup>i</sup>*(*G*). Hence, *i*(*G* − *v*) = *<sup>i</sup>*(*G*). Therefore, *G* is an ID-stable, and the proof is complete.

Let *k*1 and *k*2 be non-negative integers, and let *Hk*1,*k*2 be the graph obtained from the star *<sup>K</sup>*1,3 centered at *s* with *<sup>V</sup>*(*<sup>K</sup>*1,3) = {*<sup>s</sup>*, *a*, *b*, *c*} by attaching *k*1 pendent paths *P*2 to *a* and *k*2 pendent paths *P*2 to *b* (see, e.g., Figure 1). For each 1 ≤ *i* ≤ *k*1, the vertex set of *i*th *P*2 is {*ri*, *ti*} with *ari* ∈ *<sup>E</sup>*(*Hk*1,*k*2 ), and for each 1 ≤ *j* ≤ *k*2, the vertex set of *j*th *P*2 is {*pj*, *qj*} with *bpj* ∈ *<sup>E</sup>*(*Hk*1,*k*2 ).

**Figure 1.** The operations O3, or O4, or O5.

**Proposition 6.** *Let G be a graph and x*, *y* ∈ *V*(*G*) *(possibly x* = *y). Let G be the graph obtained from G by adding a graph Hk*1,*k*2*and adding possibly the edges xa or yb. Then, i*(*G*) = *i*(*G*) + *k*1 + *k*2 + 1*.*

**Proof.** Clearly, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* by adding *<sup>s</sup>*,*r*1,...,*rk*1 , *p*1,..., *pk*2 , and so, *i*(*G*) ≤ *i*(*G*) + *k*1 + *k*2 + 1.

Now, we show that *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + 1. Let *S* be an *<sup>i</sup>*(*G*)-set such that |*S* ∩ {*<sup>a</sup>*, *b*}| is as small as possible. To dominate *c*, *ti* (1 ≤ *i* ≤ *k*1) and *qj* (1 ≤ *j* ≤ *k*2), we must have |*S* ∩ {*<sup>s</sup>*, *c*}| ≥ 1, |*S* ∩ {*ri*, *ti*}| ≥ 1 (1 ≤ *i* ≤ *k*1) and |*S* ∩ {*pj*, *qj*}| ≥ 1 (1 ≤ *j* ≤ *k*2). We claim that |*S* ∩ {*<sup>a</sup>*, *b*}| = 0. Suppose, on the contrary, that |*S* ∩ {*<sup>a</sup>*, *b*}| ≥ 1. We consider the following cases.

**Case 4.** |*S* ∩ {*<sup>a</sup>*, *b*}| = 1.

Assume without loss of generality that *a* ∈ *S* and *b* ∈ *S*. Then, we must have *c* ∈ *S* and *t*1, ... , *tk*1 ∈ *S* if *k*1 ≥ 1. If *xa* ∈ *E*(*G*) or *S* ∩ *NG*(*x*) = ∅, then the set (*S* − {*<sup>a</sup>*, *c*}) ∪ {*s*} is an independent dominating set of *G* of size less that *<sup>i</sup>*(*G*), which is a contradiction. Hence, *xa* ∈ *E*(*G*) or *S* ∩ *NG*(*x*) = ∅, but then the set (*S* − {*a*}) ∪ {*x*} is an *<sup>i</sup>*(*G*)-set, which contradicts the choice of *S*. **Case 5.** |*S* ∩ {*<sup>a</sup>*, *b*}| = 2.

Then, we must have *c* ∈ *S*, {*<sup>t</sup>*1, ... , *tk*1 } ⊆ *S* if *k*1 ≥ 1 and {*q*1, ... , *qk*2 } ⊆ *S* if *k*2 ≥ 1. If *S* ∩ *NG*[*x*] = ∅ and *S* ∩ *NG*[*y*] = ∅, then (*S* − {*<sup>a</sup>*, *b*, *c*}) ∪ {*s*} is an independent dominating set of *G* of size *i*(*G*) − 2, which is a contradiction. Assume without loss of generality that *S* ∩ *NG*[*x*] = ∅. If *x* = *y*, then (*S* − {*<sup>a</sup>*, *b*, *c*}) ∪ {*<sup>x</sup>*,*<sup>s</sup>*} is an independent dominating set of *G* of size *i*(*G*) − 1, a contradiction again. Hence, *x* = *y*. Now, to dominate *x*, we must have *xa* ∈ *<sup>E</sup>*(*G*), but then the set (*S* − {*a*}) ∪ {*x*} is an *<sup>i</sup>*(*G*)-set, contradicting the choice of *S*.

Therefore *a*, *b* ∈ *S*. Now, the set *S* ∩ *V*(*G*) is an independent dominating set of *G*, and this implies that *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + 1. Thus, *i*(*G*) = *i*(*G*) + *k*1 + *k*2 + 1, and the proof is complete.

### **Proposition 7.** *Let G be an ID-stable graph. Then:*


**Proof.** Let *v* be a vertex in *G*. If *v* ∈ *<sup>V</sup>*(*G*), then by Proposition 6, we have *i*(*G* − *v*) = *i*(*G* − *v*) + *k*1 + *k*2 + 1. Since *G* is an ID-stable graph, we have *i*(*G* − *v*) = *<sup>i</sup>*(*G*), and so, *i*(*G* − *v*) = *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Assume that *v* ∈ *<sup>V</sup>*(*G*). We consider the following cases.

### **Case 6.** *v* = *s*.

Clearly, any *i*(*G* − *v*)-set is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). In the case (a), any *<sup>i</sup>*(*G*)-set containing *x*, *y* can be extended to an independent dominating set of *G* − *v* by adding *c*, and so, *i*(*G* − *v*) ≤ *i*(*G*) + 1 = *<sup>i</sup>*(*G*). In the case (b), any *<sup>i</sup>*(*G*)-set containing *x* can be extended to an independent dominating set of *G* − *v* by adding *c*, *p*1, ... , *pk*2 , and so, *i*(*G* − *v*) ≤ *i*(*G*) + *k*2 + 1 = *<sup>i</sup>*(*G*). In the case (c), any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding *c*, *p*1, ... , *pk*2 ,*r*1, ... ,*rk*1 , and so, *i*(*G* − *v*) ≤ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Thus, *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

### **Case 7.** *v* = *a* (the case *v* = *b* is similar).

It is easy to see that there exists an *i*(*G* − *v*)-set containing *s*. On the other hand, any *i*(*G* − *v*)-set containing *s* is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Using an argumen<sup>t</sup> similar to that described in Case 6, we obtain *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

### **Case 8.** *v* = *c*.

Obviously, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding the vertices *s*, *r*1, ... ,*rk*1 if *k*1 ≥ 1 and *p*1, ... , *pk*2 if *k*2 ≥ 1, and so, *i*(*G* − *v*) ≤ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Now, let *S* be an *i*(*G* − *v*)-set. To dominate *s*, *ti* (1 ≤ *i* ≤ *k*1) and *qj* (1 ≤ *j* ≤ *k*2), we must have |*S* ∩ {*<sup>a</sup>*, *b*,*<sup>s</sup>*}| ≥ 1, |*S* ∩ {*ri*, *ti*}| ≥ 1 for 1 ≤ *i* ≤ *k*1 and |*S* ∩ {*pj*, *qj*}| ≥ 1 for 1 ≤ *j* ≤ *k*2. If *s* ∈ *S*, then *S* is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Assume that *s* ∈/ *S*. Then, *a* ∈ *S* or *b* ∈ *S*. Assume, without loss of generality, that *a* ∈ *S*. If *b* ∈/ *S*, then *S* − *<sup>V</sup>*(*Hk*1,*k*2 ) is an independent dominating set of *G* − *x*, and since *G* is an independent domination stable graph, we have *i*(*G* − *v*)=(|*S*| − *k*1 − *k*2 − 1) + *k*1 + *k*2 + 1 ≥ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Let *b* ∈ *S*. This implies that {*q*1, ... , *qk*2 } ⊆ *S* if *k*2 ≥ 1. If (*N*(*y*) − {*b*}) ∩ *S* = ∅, then the set *S* − {*b*} if *k*2 = 0, and the set (*S* − {*b*, *q*1, ... , *qk*2 }) ∪ {*p*1, ... , *pk*2 } if *k*2 ≥ 1 is an independent dominating set of *G* − *v*, which leads to a contradiction. Hence, (*N*(*y*) − {*b*}) ∩ *S* = ∅, and similarly, (*N*(*x*) − {*a*}) ∩ *S* = ∅. Then, (*S* − *<sup>V</sup>*(*Hk*1,*k*2 )) ∪ {*y*} is an independent dominating set of *G* − *x*, and since *G* is an ID-stable graph, we deduce that *i*(*G* − *v*)=(|*S*| − *k*1 − *k*2 − 1) + *k*1 + *k*2 + 1 ≥ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Therefore, *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

### **Case 9.** *v* = *ri* for some *i* ∈ {1, 2, . . . , *k*1} or *v* = *pj* for some *j* ∈ {1, 2, . . . , *k*2}.

Assume, without loss of generality, that *v* = *r*1. Obviously, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding *s*, *t*1,*r*2, ... ,*rk*1 and *p*1, ... , *pk*2 if *k*2 ≥ 1, and so, *i*(*G* − *v*) ≤ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). On the other hand, any *i*(*G* − *v*)-set is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Therefore, *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

**Case 10.** *v* = *ti* for some *i* ∈ {1, 2, . . . , *k*1} or *v* = *qj* for some *j* ∈ {1, 2, . . . , *k*2}.

Assume, without loss of generality that *v* = *t*1. Clearly, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding the vertices *s*,*r*1, ... ,*rk*1 and *p*1, ... , *pk*2 if *k*2 ≥ 1, and so, *i*(*G* − *v*) ≤ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). To prove the inverse inequality, let *S* be an *i*(*G* − *v*)-set. To dominate *c*, *ti* (2 ≤ *i* ≤ *k*1) and *qj* (1 ≤ *j* ≤ *k*2), we must have |*S* ∩ {*<sup>c</sup>*,*<sup>s</sup>*}| ≥ 1, |*S* ∩ {*ri*, *ti*}| ≥ 1 for 2 ≤ *i* ≤ *k*1 and |*S* ∩ {*pj*, *qj*}| ≥ 1 for 1 ≤ *j* ≤ *k*2. If *r*1 ∈ *S*, then *S* is obviously an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Assume that *r*1 ∈/ *S*. It follows that *a* ∈ *S* yielding *c*, *ti* ∈ *S* for 2 ≤ *i* ≤ *k*2. If *b* ∈ *S*, then we may assume that *p*1, ... , *pk*2 ∈ *S*, and clearly, the set *S* − *<sup>V</sup>*(*Hk*1,*k*2 ) is an independent dominating set of *G* − *x*. Since *G* is an ID-stable graph, we obtain *i*(*G* − *v*)=(|*S*| − *k*1 − *k*2 −<sup>1</sup>) + *k*1 + *k*2 +1 ≥ *i*(*G*) + *k*1 + *k*2 +1 = *<sup>i</sup>*(*G*). Let *b* ∈ *S*. Then, *q*1, ... , *qk*2 ∈ *S* if *k*2 ≥ 1. It is easy to see that (*N*(*y*) − {*b*}) ∩ *S* = ∅. If (*N*(*x*) − {*a*}) ∩ *S* = ∅, then (*S* − {*a*}) ∪ {*<sup>r</sup>*1} is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Suppose that (*N*(*x*) − {*a*}) ∩ *S* = ∅. Then, (*S* − *<sup>V</sup>*(*Hk*1,*k*2)) ∪ {*y*} is an independent dominating set of *G* − *x*, and since *G* is an ID-stable

graph, we have *i*(*G* − *v*)=(|*S*| − *k*1 − *k*2 − 1) + *k*1 + *k*2 + 1 ≥ *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*). Hence, *i*(*G* − *v*) = *<sup>i</sup>*(*G*). Thus, *G* is an ID-stable graph, and the proof is complete.

Let N be the set of non-negative integers, *n* ≥ 3 and *Q* ∈ N*n* with *Q* = (*q*1, *q*2, ... , *qn*). Let *D*(*Q*) = {*i* | *qi* > <sup>0</sup>}. For any *i* with *qi* > 0 and *qi*+<sup>1</sup> = 0, if *qj* = 0 for *j* = *i* + 1, *i* + 2, ... , *i* + *k* and *qi*+*k*+<sup>1</sup> > 0 where the subscript is taken modulo *n*, we define *<sup>H</sup>*(*Q*, *i*) = *k*. For example, if *Q* = (0, 2, 0, 1, 3, 0, 1, 2, 0, <sup>0</sup>), then *<sup>H</sup>*(*Q*, 2) = 1, *<sup>H</sup>*(*Q*, 5) = 1 and *<sup>H</sup>*(*Q*, 8) = 3.

The graph *<sup>C</sup>*(*<sup>n</sup>*, *Q*) (resp. *<sup>P</sup>*(*<sup>n</sup>*, *Q*)) is the graph obtained from *Cn* = (*<sup>v</sup>*1*v*2 ... *vn*) (resp. *Pn* = *v*1*v*2 ... *vn*) by attaching *qi* disjoint pendent paths *P*2 to *vi*. If *qi* > 0, then let *Hvi* = {*vi*+1,..., *vi*+*<sup>H</sup>*(*Q*,*<sup>i</sup>*)}, and assume for any 1 ≤ *j* ≤ *qi*, the vertex set of *j*th *P*2 attached to *vi* is {*vi*,*j*,*a*, *vi*,*j*,*<sup>b</sup>*} with leaf *vi*,*j*,*<sup>b</sup>* (see Figure 2).

**Figure 2.** (**a**) The graph *C*(6,(1, 0, 0, 0, 0, <sup>0</sup>)); (**b**) the graph *C*(9,(2, 0, 1, 3, 0, 1, 2, 0, <sup>0</sup>)).

**Proposition 8.** *Let G be a graph, and x*, *y* ∈ *V*(*G*) (*possibly x* = *y*)*. If G is a graph obtained from G by adding H* = *<sup>P</sup>*(*<sup>n</sup>*,(0, 0, *k*1, 0, ... , 0, *k*2, 0, <sup>0</sup>))*, where n* ≡ 0 (mod <sup>6</sup>)*, k*1 ≥ 0, *k*2 ≥ 0*, and adding possibly the edges xv*3 *and yvn*−2*, then i*(*G*) = *i*(*G*) + *k*1 + *k*2 + *n*3 *.*

**Proof.** Clearly, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* by adding *v*3*i*−1 (1 ≤ *i* ≤ *n*3 ), *<sup>v</sup>*3,*j*,*<sup>a</sup>* (1 ≤ *j* ≤ *k*1) and *vn*−2,*j*,*<sup>a</sup>* (1 ≤ *j* ≤ *k*2), and so, *i*(*G*) ≤ *i*(*G*) + *k*1 + *k*2 + *n*3 .

Now, we show that *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + *n*3 . Let *S* be an *<sup>i</sup>*(*G*)-set. To dominate the vertices *v*1, *vn*, *v*3*i*−1 (2 ≤ *i* ≤ *n*3 <sup>−</sup><sup>1</sup>), *<sup>v</sup>*3,*j*,*<sup>b</sup>* (1 ≤ *j* ≤ *k*1) and *vn*−2,*j*,*<sup>b</sup>* (1 ≤ *j* ≤ *k*2), we must have |*S* ∩ {*<sup>v</sup>*1, *<sup>v</sup>*2}| ≥ 1, |*S* ∩ {*vn*−1, *vn*}| ≥ 1, |*S* ∩ {*<sup>v</sup>*3*i*−2, *v*3*i*−1, *<sup>v</sup>*3*i*}| ≥ 1 (2 ≤ *i* ≤ *n*3 − <sup>1</sup>), |*S* ∩ {*<sup>v</sup>*3,*j*,*a*, *<sup>v</sup>*3,*j*,*<sup>b</sup>*}| ≥ 1 (1 ≤ *j* ≤ *k*1), and |*S* ∩ {*vn*−2,*j*,*a*, *vn*−2,*j*,*<sup>b</sup>*}| ≥ 1 (1 ≤ *j* ≤ *k*2). We may assume without loss of generality that {*<sup>v</sup>*3*i*−<sup>1</sup> | 2 ≤ *i* ≤ *n*3 − 1} ⊆ *S*. If *v*3, *vn*−2 ∈ *S*, then the set *S* ∩ *V*(*G*) is an independent dominating set of *G*, and this implies that *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + *n*3 . Assume without loss of generality that *v*3 ∈ *S*. Then, we must have {*<sup>v</sup>*3,1,*b*, ... , *<sup>v</sup>*3,*k*1,*<sup>b</sup>*} ⊆ *S* and *S* ∩ *NG*(*<sup>v</sup>*3) = ∅. If *v*3 is not adjacent to *x* or *NG*(*x*) ∩ *S* = ∅, then the set *S* = (*S* − {*<sup>v</sup>*1, *v*2, *v*3, *<sup>v</sup>*3,1,*b*, ... , *<sup>v</sup>*3,*k*1,*<sup>b</sup>*}) ∪ {*<sup>v</sup>*2, *<sup>v</sup>*3,1,*<sup>a</sup>*, ... , *<sup>v</sup>*3,*k*1,*<sup>a</sup>*} is an independent dominating set of *G* of size *i*(*G*) − 1, a contradiction. Hence, *v*3*x* ∈ *<sup>E</sup>*(*G*), *NG*(*x*) ∩ *S* = ∅, and so, *x* ∈ *S*. If *vn*−2 ∈ *S*, then the set (*S* − *V*(*H*)) ∪ {*x*} is an independent dominating set of *G*, yielding *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + *n*3 . Assume that *vn*−2 ∈ *S*. Then, we have {*vn*−2,1,*b*, ... , *vn*−2,*k*2,*<sup>b</sup>*} ⊆ *S* and *S* ∩ *NG*(*vn*−<sup>2</sup>) = ∅. Using the above arguments, we have *vn*−<sup>2</sup>*y* ∈ *<sup>E</sup>*(*G*), *y* ∈ *S*, and *NG*(*y*) ∩ *S* = ∅. If *x* = *y* or *x* and *y* are adjacent in *G*, then the set (*S* − *V*(*H*)) ∪ {*<sup>x</sup>*, *v*2, *v*5, ... , *vn*−2, *<sup>v</sup>*3,1,*<sup>a</sup>*, ... , *<sup>v</sup>*3,*k*1,*a*, *vn*−2,1,*<sup>a</sup>*, ... , *vn*−2,*k*2,*<sup>a</sup>*} is an independent dominating set of *G* of size *i*(*G*) − 1, which is a contradiction. Hence, *x* = *y* and *x* and *y* are not adjacent in *G*. Now, the set (*S* − *V*(*H*)) ∪ {*<sup>x</sup>*, *y*} is an independent dominating set of *G*, implying that *i*(*G*) ≥ *i*(*G*) + *k*1 + *k*2 + *n*3 . Therefore, *i*(*G*) = *i*(*G*) + *k*1 + *k*2 + *n*3 , and the proof is complete.

**Proposition 9.** *Let G be an ID-stable graph. If x*, *y* ∈ *V*(*G*) *and G is a graph obtained from G by adding P*(6,(0, 0, *k*1, *k*2, 0, 0)) *and adding the edges xv*3, *yv*<sup>4</sup>*, then G is an ID-stable graph.*

**Proof.** Let *v* be a vertex in *G*. If *v* ∈ *<sup>V</sup>*(*G*), then by Proposition 8 and the fact that *G* is an ID-stable graph, we obtain:

$$i(G'-\upsilon) = i(G-\upsilon) + k\_1 + k\_2 + 2 = i(G) + k\_1 + k\_2 + 2 = i(G').$$

Let *v* ∈ *<sup>V</sup>*(*G*). We consider the following cases.

**Case 11.** *v* ∈ {*<sup>v</sup>*2, *v*5, *<sup>v</sup>*3,*j*,*a*, *<sup>v</sup>*4,*k*,*<sup>a</sup>* | 1 ≤ *j* ≤ *k*1 and 1 ≤ *k* ≤ *k*2}. As in Case 9 in Proposition 7, we have *i*(*G* − *v*) = *<sup>i</sup>*(*G*).

**Case 12.** *v* ∈ {*<sup>v</sup>*1, *v*6, *<sup>v</sup>*3,*j*,*b*, *<sup>v</sup>*4,*k*,*<sup>b</sup>* | 1 ≤ *j* ≤ *k*1 and 1 ≤ *k* ≤ *k*2. *<sup>i</sup>*(*G*).

As in Case 10 in Proposition 7, we have *i*(*G* − *v*) =

**Case 13.** *v* ∈ {*<sup>v</sup>*3, *<sup>v</sup>*4}.

We may assume, without loss of generality, that *v* = *v*3. Clearly, any *i*(*G* − *v*)-set containing *v*2 is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). On the other hand, any *<sup>i</sup>*(*G*)-set can be extended to an independent dominating set of *G* − *v* by adding {*<sup>v</sup>*2, *v*5, *<sup>v</sup>*3,*j*,*a*, *<sup>v</sup>*4,*l*,*<sup>a</sup>* | 1 ≤ *j* ≤ *k*1 and 1 ≤ *l* ≤ *k*2}, and Proposition 8 yields *i*(*G* − *v*) ≤ *i*(*G*) + *k*1 + *k*2 + 2 = *<sup>i</sup>*(*G*).

**Proposition 10.** *Let Q* = (*q*1, *q*2, ... , *qn*) ∈ N*n such that* |*D*(*Q*)| ≥ 2 *and <sup>H</sup>*(*Q*, *i*) ≥ 1 *for each i* ∈ *<sup>D</sup>*(*Q*)*. If <sup>H</sup>*(*Q*, *i*) ≡ 1(*mod* 3) *for some i* ∈ *D*(*Q*) *or <sup>H</sup>*(*<sup>i</sup>*, *Q*) ≡ 2 (*mod* 3) *for each i* ∈ *<sup>D</sup>*(*Q*)*, or <sup>H</sup>*(*Q*, *i*) ≡ 0 (*mod* 3) *and <sup>H</sup>*(*Q*, *j*) ≡ 0 (*mod* 3) *for some i*, *j* ∈ *<sup>D</sup>*(*Q*)*, then the graph <sup>C</sup>*(*<sup>n</sup>*, *Q*) *is not an ID-stable graph.*

**Proof.** Suppose, to the contrary, that *G* = *<sup>C</sup>*(*<sup>n</sup>*, *Q*) is an ID-stable graph. If *G* has an *<sup>i</sup>*(*G*)-set *S* containing *vi* for some *i* ∈ *<sup>D</sup>*(*Q*), then *S* − {*vi*,1,*<sup>b</sup>*} is an independent dominating stable set for *G* − *vi*,1,*b*, which leads to a contradiction. Hence, for any *<sup>i</sup>*(*G*)-set *S* and any *i* ∈ *<sup>D</sup>*(*Q*), we have *vi* ∈ *S*. Assume that *D*(*Q*) = {*<sup>i</sup>*1, *i*2,..., *ir*}. Now, we show that:

$$i(G) = \sum\_{j=1}^{r} q\_{i\_j} + \sum\_{j=1}^{r} \left\lceil \frac{H(i\_{j\prime}, Q)}{3} \right\rceil.$$

For 1 ≤ *j* ≤ *r*, let *Pij* be the path *vij*+<sup>1</sup>*vij*+<sup>2</sup> ... *vij*+*<sup>H</sup>*(*Q*,*ij*), and let *Sj* be an *<sup>i</sup>*(*Pij*)-set. Clearly, the set *I* = <sup>∪</sup>*rj*=<sup>1</sup>(*Sij* ∪ {*vij*,*k*,*<sup>a</sup>* | 1 ≤ *k* ≤ *qij*}) is an independent dominating set of *G*, and we conclude from Proposition 1 that *i*(*G*) ≤ ∑*rj*=<sup>1</sup> *qij* + ∑*rj*=<sup>1</sup> # *<sup>H</sup>*(*ij*,*Q*) 3 \$ . To prove the inverse inequality, let *S* be an *<sup>i</sup>*(*G*)-set. To dominate the vertices *vij*,*k*,*<sup>b</sup>* for 1 ≤ *j* ≤ *r* and 1 ≤ *k* ≤ *qij* , we must have |*S* ∩ {*vij*,*k*,*a*, *vij*,*k*,*<sup>b</sup>*}| ≥ 1, and since *S* ∩ {*vi*1 , ... , *vir*} = ∅, we must have |*S* ∩ *Hvij* | ≥ # *<sup>H</sup>*(*ij*,*Q*) 3 \$ for each *j*, by Proposition 1. This implies that *i*(*G*) = |*S*| ≥ <sup>∑</sup>*rj*=<sup>1</sup>(*dG*(*vij*) − 2) + ∑*rj*=<sup>1</sup> # *<sup>H</sup>*(*ij*,*Q*) 3 \$ . Hence, *i*(*G*) = ∑*rj*=<sup>1</sup> *qij*) + ∑*rj*=<sup>1</sup> # *<sup>H</sup>*(*ij*,*Q*) 3 \$ .

If *<sup>H</sup>*(*Q*, *ij*) ≡ 1 (mod 3) for some *ij* ∈ *<sup>D</sup>*(*Q*), say *j* = 1, then the set (*I* − (*Si*1 ∪ {*vi*1,*k*,*<sup>a</sup>* | 1 ≤ *k* ≤ *qi*1 })) ∪ ({*vi*1,*k*,*<sup>b</sup>* | 1 ≤ *k* ≤ *qi*1 }∪{*vi*1+3*<sup>s</sup>* | 0 ≤ *s* ≤ *qi*13 − 1}) when *qir* ≡ 0 (mod <sup>3</sup>), the set (*I* − (*Si*1 ∪ *Si*2 ∪ {*vi*1,*k*,*<sup>a</sup>* | 1 ≤ *k* ≤ *qi*1 })) ∪ ({*vi*1,*k*,*<sup>b</sup>* | 1 ≤ *k* ≤ *qi*1 }∪{*vi*1+3*<sup>s</sup>* | 0 ≤ *s* ≤ *qi*13 } ∪ {*vir*+3*s*+<sup>1</sup> | 0 ≤ *s* ≤ *qir*3 − 1}) when *qir* ≡ 2 (mod <sup>3</sup>), and the set (*I* − (*Si*1 ∪ *Si*2 ∪ {*vi*1,*k*,*<sup>a</sup>* | 1 ≤ *k* ≤ *qi*1 })) ∪ ({*vi*1,*k*,*<sup>b</sup>* | 1 ≤ *k* ≤ *qi*1 }∪{*vi*1+3*<sup>s</sup>* | 0 ≤ *s* ≤ *qi*13 } ∪ {*vir*<sup>+</sup>*qir*−1, *vir*+3*s*+<sup>1</sup> | 0 ≤ *s* ≤ *qir*3 − 2}) when *qir* ≡ 1 (mod 3) is an *<sup>i</sup>*(*G*)-set, which is a contradiction. Thus, *<sup>H</sup>*(*Q*, *ij*) ≡ 1 (mod 3) for each *ij* ∈ *<sup>D</sup>*(*Q*).

Suppose *<sup>H</sup>*(*<sup>i</sup>*, *Q*) ≡ 2 (mod 3) for each *i* ∈ *<sup>D</sup>*(*Q*). Then, clearly, *n* = ∑*rj*=<sup>1</sup> #*<sup>H</sup>*(*ij*,*Q*) 3 \$, and the set {*vi*1+3*<sup>s</sup>* | 0 ≤ *i* ≤ *n*3 − <sup>1</sup>}∪{*vij*,*k*,*<sup>b</sup>* | 1 ≤ *k* ≤ *qij*}) is an independent dominating set of *G*, which leads to a contradiction again.

Finally let, without loss of generality, *<sup>H</sup>*(*Q*, *i*1) ≡ 0 (mod 3) and *<sup>H</sup>*(*Q*, *i*) ≡ 0 (mod 3) for some *i*1, *i* ∈ *D*(*Q*) and *<sup>H</sup>*(*Q*, *ij*) ≡ 2 (mod 3) for each *ij* ∈ *D*(*Q*) − {*ij*, *<sup>i</sup>*}. If |*D*(*Q*)| = 2, then it is not hard to see that *i*(*G* − *vi*1+<sup>2</sup>) > *<sup>i</sup>*(*G*), which is a contradiction. Assume that |*D*(*Q*)| ≥ 3. By symmetry, we may assume that ≥ 3. Let *G* = *G* − *vi*+2, and let *S* be an *<sup>i</sup>*(*G*)-set such that |*S* ∩ {*vi*1 , ... , *vir*}| is as large as possible. Since *G* is an independent domination stable graph, we have *i*(*G*) = *<sup>i</sup>*(*G*). It is not hard to see that the set:

$$D\_1 = \left( \{ v\_{i\_2}, \dots, v\_{i\_\ell} \} \cup \left( \bigcup\_{j=2}^{\ell-1} \{ v\_{i\_j + 3k} \mid 1 \le k \le \left\lfloor \frac{H(i\_j, Q) - 1}{3} \right\rfloor \} \right) \right)$$

is a subset of *S*. It follows that *D*2 = ∪−<sup>1</sup> *j*=2 {*vij*,*k*,*<sup>b</sup>* | 1 ≤ *k* ≤ *qij*} ⊆ *S*. We may also assume that {*vi*1+3*<sup>s</sup>*−<sup>1</sup> | 1 ≤ *s* ≤ ) *<sup>H</sup>*(*<sup>i</sup>*1,*Q*) 3 \*} ⊆ *S*. Let *D*3 = <sup>∪</sup>*j*=<sup>2</sup>{*vij*,*k*,*<sup>a</sup>* | 1 ≤ *k* ≤ *qij*}. Clearly, the set:

$$\left( (S' - (D\_1 \cup D\_2)) \cup \{ v\_{i\_\ell} \} \cup D\_3 \cup \left( \bigcup\_{j=2}^\ell \{ v\_{i\_j + 3k + 1} \mid 0 \le k \le \left\lfloor \frac{H(i\_j, Q) - 1}{3} \right\rfloor \right) \right)$$

is an independent dominating set of *G* of cardinality |*S*| = *i*(*G*) containing *vi* , which is a contradiction. This completes the proof.

### **3. Independent Domination Stable Trees**

In this section, we give a constructive characterization of all ID-stable trees.

In order to present our constructive characterization, we define a family of trees as follows. Let T 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, *Ti*+<sup>1</sup> can be obtained from *Ti* by one of the following operations.

**Operation** T1**:** If *u* ∈ *<sup>W</sup>*(*Ti*), then T1 adds a spider *S*1 with head *s* and an edge *us* to obtain *Ti*+<sup>1</sup> (see Figure 3).

**Operation** T2**:** If *u* ∈ *<sup>V</sup>*(*Ti*), then T2 adds a spider *Sq* (*q* ≥ 2) with head *s* and an edge *us* to obtain *Ti*+<sup>1</sup> (see Figure 3).

**Theorem 1.** *If T* ∈ T *, then T is an ID-stable tree.*

**Proof.** If *T* is *P*2, then obviously *T* is an ID-stable tree. Suppose now that *T* ∈ T . Then 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 the induction hypothesis, *T* is an ID-stable tree. Since *T* = *Tk* is obtained by one of the Operations T1 or T2 from *T*, we conclude from the Proposition 5 that *T* is an ID-stable tree.

Next, we characterize all ID-stable trees.

**Theorem 2.** *Let T be a tree of order n* ≥ 2*. Then, T is an ID-stable tree if and only if T* ∈ T *.*

**Figure 3.** The operations: T1 or T2.

**Proof.** According to Theorem 1, we need only to prove necessity. Let *T* be an ID-stable tree of order *n* ≥ 2. The proof is by induction on *n*. If *n* = 2, then *T* = *P*2 ∈ T . Let *n* ≥ 3, 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. Let *v*1*v*2 ... *vk* (*k* ≥ 5) be a diametrical path in *T* and root *T* at *vk*. By Proposition 2, any support vertex adjacent to *v*3 has degree two. In particular *dT*(*<sup>v</sup>*2) = 2. By Proposition 3, *v*3 is not a support vertex, and so, *Tv*3 = *SdT*(*<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*),

$$i(T'-\upsilon) + d\_T(\upsilon\_3) - 1 = i(T-\upsilon) = i(T) = i(T') + d\_T(\upsilon\_3) - 1$$

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 *T* ∈ T . If *dT*(*<sup>v</sup>*3) ≥ 3, then *T* ∈ T since *T* can be obtained from *T* by operation T2.

Assume that *dT*(*<sup>v</sup>*3) = 2. By Proposition 4, we have *i*(*T*) + 1 = *<sup>i</sup>*(*T*). Since *T* is an ID-stable tree, we have *i*(*T* − *<sup>v</sup>*2) = *<sup>i</sup>*(*T*). Let *S* be an *i*(*T* − *<sup>v</sup>*2)-set. Clearly, *v*1 ∈ *S*. If *v*3 ∈ *S*, then *S* − {*<sup>v</sup>*1} is an independent dominating set of *T* − *v*1, which is a contradiction. Hence, *v*3 ∈/ *S*, and this implies that *v*4 ∈ *S*. Now, *S* − {*<sup>v</sup>*1} is an independent dominating set of *T*, and we deduce from *i*(*T*) + 1 = *i*(*T*) that *S* − {*<sup>v</sup>*1} is an *<sup>i</sup>*(*T*)-set. Thus, *v*4 ∈ *<sup>W</sup>*(*T*). Now, *T* can be obtained from *T* by operation T1, and so, *T* ∈ T . This completes the proof.

### **4. Independent Domination Stable Unicyclic Graphs**

In this section, we give a constructive characterization of all ID-stable unicyclic graphs. We start with introducing the following families of graphs.


Next, we show that each graph in J is an ID-stable graph. By Corollary 1 and Theorem 1, any graph in the family T ∪J1 is an independent domination stable graph.

### **Proposition 11.** *If G* ∈ J<sup>2</sup>*, then G is an ID-stable graph.*

**Proof.** Let *G* ∈ J2. First, we show that *i*(*G*) = *k* + *q*1. Clearly, the set {*<sup>v</sup>*3*i* | 1 ≤ *i* ≤ *<sup>k</sup>*}∪{*<sup>v</sup>*1,*j*,*<sup>a</sup>* | 1 ≤ *j* ≤ *q*1} is an independent dominating set of *G* yielding *i*(*G*) ≤ *k* + *q*1. To prove the inverse inequality, let *S* be an *<sup>i</sup>*(*G*)-set. To dominate *<sup>v</sup>*1,*j*,*b*, we must have |*S* ∩ {*<sup>v</sup>*1,*j*,*a*, *<sup>v</sup>*1,*j*,*<sup>b</sup>*}| ≥ 1 for each *j* ∈ {1, ... , *q*1}. On the other hand, to dominate the vertices *v*3*i* (1 ≤ *i* ≤ *k*), we must have |*S* ∩ {*<sup>v</sup>*3*i*−1, *v*3*i*, *<sup>v</sup>*3*i*+<sup>1</sup>}| ≥ 1 for each *i* ∈ {1, . . . , *k*}, and this implies that *i*(*G*) ≥ *k* + *q*1. Hence, *i*(*G*) = *k* + *q*1.

Next we show that *G* is an ID-stable graph. Let *v* ∈ *G*. If *v* = *v*1, then *G* = *P*3*k* ∪ *q*1*K*2, and by Proposition 1, we have *i*(*G* − *v*) = *<sup>i</sup>*(*<sup>P</sup>*3*k*) + *<sup>i</sup>*(*q*1*K*2) = *k* + *q*1 = *<sup>i</sup>*(*G*). If *v* = *<sup>v</sup>*1,*j*,*<sup>a</sup>* for some 1 ≤ *j* ≤ *q*1, then *G* = *K*1 ∪ *C*(3*k* + 1,(*q*1 − 1, 0, 0, ... , <sup>0</sup>)), and as above, we have *i*(*G* − *v*) = *k* + *q*1 = *<sup>i</sup>*(*G*). Suppose that *v* = *<sup>v</sup>*1,*j*,*<sup>b</sup>* for some 1 ≤ *j* ≤ *q*1, say *j* = 1. Clearly, the set {*<sup>v</sup>*3*i* | 1 ≤ *i* ≤ *<sup>k</sup>*}∪{*<sup>v</sup>*1,*j*,*<sup>a</sup>* | 1 ≤ *j* ≤ *q*1} is an independent dominating set of *G* yielding *i*(*G* − *v*) ≤ *k* + *q*1 = *<sup>i</sup>*(*G*). To prove *i*(*G* − *v*) ≥ *k* + *q*1, let *S* be an *i*(*G* − *v*)-set. To dominate *<sup>v</sup>*1,*j*,*a*, we must have |*S* ∩ {*<sup>v</sup>*1,*j*,*a*, *<sup>v</sup>*1}| ≥ 1, and to dominate the vertex *<sup>v</sup>*1,*j*,*b*, we must have |*S* ∩ {*<sup>v</sup>*1,*j*,*a*, *<sup>v</sup>*1,*j*,*<sup>b</sup>*}| ≥ 1 for each *j* ∈ {2, ... , *q*1}. On the other hand, to dominate the vertices in *V*(*Cn*) − {*<sup>v</sup>*1, *v*2, *<sup>v</sup>*3}, *S* must contain at least *k* − 1 vertices in {*<sup>v</sup>*2, ... , *vn*}, and so, *i*(*G* − *v*) ≥ *k* + *q*1. Hence, *i*(*G* − *v*) = *i*(*G*) in this case. Let now *v* = *vi* (*i* = 1). Clearly, any *<sup>i</sup>*(*<sup>P</sup>*3*k*)-set of *P*3*k* = *vi*−1 ... *v*1*vn* ... *vi*+1 can be extended to an independent dominating set of *G* − *v* by adding *<sup>v</sup>*1,*j*,*<sup>a</sup>* for *j* = 1, 2, ... , *q*1, and so, *i*(*G* − *v*) ≤ *<sup>i</sup>*(*<sup>P</sup>*3*k*) + *q*1 = *k* + *q*1 = *<sup>i</sup>*(*G*). On the other hand, if *S* is an *i*(*G* − *v*)-set, then to dominate the vertices in {*<sup>v</sup>*1,*j*,*<sup>b</sup>* | 1 ≤ *j* ≤ *q*1}, we must have |*S* ∩ {*<sup>v</sup>*1,*j*,*a*, *<sup>v</sup>*1,*j*,*<sup>b</sup>* | 1 ≤ *j* ≤ *q*1}| ≥ *q*1, and to dominate the vertices in *V*(*Cn*) − {*vi*, *<sup>v</sup>*1}, we must have |*S* ∩ (*V*(*Cn*) − {*vi*})| ≥ *k*. Thus, *i*(*G* − *v*) = |*S*| ≥ *k* + *q*1 = *<sup>i</sup>*(*G*), and hence, *G* is an ID-stable graph.

**Theorem 3.** *Let G* = *<sup>C</sup>*(*<sup>n</sup>*, *Q*) *where n* ≥ 3 *and D*(*Q*) = 1*. Then, G is an ID-stable graph if and only if G* ∈ J<sup>2</sup>*.*

**Proof.** According to Proposition 11, we only need to prove necessity. Let *G* be an independent domination stable graph. Assume, without loss of generality, that *Q* = (*q*1, 0 ... , 0) where *q*1 ≥ 1. As Proposition 11, we can see that *i*(*G*) = *<sup>n</sup>*−1 3 + *q*1. If *n* ≡ 1 (mod <sup>3</sup>), then the set {*<sup>v</sup>*3*i*+<sup>1</sup> | 0 ≤ *i* ≤ *<sup>n</sup>*−1 3 − <sup>1</sup>}∪{*<sup>v</sup>*1,*j*,*<sup>a</sup>* | 2 ≤ *j* ≤ *q*1} is an independent dominating set of *G* − *<sup>v</sup>*1,1,*b* of size *i*(*G*) − 1, which is a contradiction. Assume that *n* ≡ 1 (mod <sup>3</sup>). If *q*1 = 1, then clearly *G* − *<sup>v</sup>*1,1,*a* = *K*1 ∪ *Cn*, and by Proposition 1, we have *i*(*G*) = *i*(*Cn*) + 1 = *<sup>n</sup>*−1 3 + 2, which is a contradiction. Therefore, *q*1 ≥ 2, and so, *G* ∈ J2.

**Proposition 12.** *If G* ∈ J<sup>3</sup>*, then G is an independent domination stable graph.*

**Proof.** Let *G* = *<sup>C</sup>*(*<sup>n</sup>*, *Q*) ∈ J3, and let *ω* = ∑*rj*=<sup>1</sup> *qij* + ∑*rj*=<sup>1</sup> # *<sup>H</sup>*(*Q*,*ij*) 3 \$. Assume that *D*(*Q*) = {*<sup>i</sup>*1, ... , *ir*}, and suppose, without loss of generality, that *<sup>H</sup>*(*Q*, *i*1) ≡ 0 (mod <sup>3</sup>). Let *S* = <sup>∪</sup>*rj*=<sup>1</sup>{*vij*,*s*,*<sup>a</sup>* | 1 ≤ *s* ≤ *qij*}, *S* = <sup>∪</sup>*rj*=<sup>2</sup>{*vij*,*s*,*<sup>b</sup>* | 1 ≤ *s* ≤ *qij*}, *Spj* = {*vij*+3*k*+*p* | 0 ≤ *k* ≤ *<sup>H</sup>*(*Q*,*ij*) 3 − 1} for *j* ∈ {1, ... ,*<sup>r</sup>*} and *p* ∈ {1, 2, <sup>3</sup>}.

First, we show that *i*(*G*) = *ω*. Clearly, the set *S* = (∪*rj*=1*S*2*<sup>j</sup>* ) ∪ *S* is an *<sup>i</sup>*(*G*)-set, and so, *i*(*G*) ≤ *ω*. To prove the inverse inequality, let *T* be an *<sup>i</sup>*(*G*)-set. To dominate the vertices *vij*,*s*,*b*, we must have |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 1 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* . Now, to dominate the vertices *vi*1+3*k*+2, we must have |*T* ∩ {*vi*1+3*k*+1, *vi*1+3*k*+2, *vi*1+3*k*+<sup>3</sup>}| ≥ 1 for 0 ≤ *k* ≤ # *<sup>H</sup>*(*Q*,*ij*) 3 \$ − 1, and to dominate the vertices *vij*+2, ... , *vij*+1−1, we must have |*T* ∩ {*vij*+1, *vij*+2, ... , *vij*+1−1, *vij*+<sup>1</sup> }| ≥ 1 for each *j* ∈ {2, ... ,*<sup>r</sup>*} yielding *i*(*G*) ≥ *ω*. Thus, *i*(*G*) = *ω* as desired.

Now, we show that *G* is an independent domination stable graph. Let *v* ∈ *<sup>V</sup>*(*G*). Consider the following cases.

### **Case 14.** *v* ∈ *S*.

Clearly, any *i*(*G* − *v*)-set is an independent dominating set of *G*, and so, *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). On the other hand, (∪*rj*=1*S*2*<sup>j</sup>* ) ∪ *S* ∪ {*vi*1,*s*,*<sup>b</sup>* | 1 ≤ *s* ≤ *qi*1 } is an independent dominating set of *G* − *v*, and hence, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). Thus, *i*(*G*) = *i*(*G* − *v*) in this case.

**Case 15.** *v* ∈ {*vi*1 ,..., *vir*}.

Suppose, without loss of generality, that *v* = *vi*1 . Obviously, *S* is an independent dominating set of *G* − *v*, and hence, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). Let *D* be an *i*(*G* − *v*)-set such that |*D* ∩ {*vi*1,*j*,*<sup>a</sup>* | 1 ≤ *j* ≤ *q*1}| is

as large as possible. Then, {*vi*1,*j*,*<sup>a</sup>* | 1 ≤ *j* ≤ *q*1} ⊆ *D*. As above, we can see that |*D*| = *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Therefore, *i*(*G*) = *i*(*G* − *v*) in this case.

### **Case 16.** *v* ∈ *S*.

Assume, without loss of generality, that *v* = *vi*1,1,*b*. Clearly, *S* is an independent dominating set of *G* − *v*, and hence, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). To prove the inverse inequality, let *T* be a *i*(*G* − *v*)-set. As above, we have |*T* ∩ {*vi*1,*s*,*a*, *vi*1,*s*,*<sup>b</sup>*}| ≥ 1 for 2 ≤ *s* ≤ *qi*1 , and |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 2 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* . Furthermore, we must have |*T* ∩ {*vi*1 , *vi*1,1,*a*}| ≥ 1. Now, to dominate the vertices *vi*1+2, ... , *vi*<sup>2</sup>−1, we must have |*T* ∩ {*vi*1+1, ... , *vi*<sup>2</sup>−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 , and to dominate the vertices *vir*+1, ... , *vi*<sup>1</sup>−2, we must have |*T* ∩ {*vir*, *vir*+1, ... , *vi*<sup>1</sup>−2, *vi*<sup>1</sup>−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*ir*) 3 . Repeating this process, we must have |*T* ∩ {*vij* , *vij*+1, ... , *vij*+1−2, *vij*+1−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*ij*) 3 for each 2 ≤ *j* ≤ *r* − 1. It follows that |*T*| ≥ *<sup>i</sup>*(*G*), and so, *i*(*G*) = *i*(*G* − *<sup>v</sup>*).

**Case 17.** *v* ∈ *S*1 1 (the case *v* ∈ *S*3 1 is similar).

Assume that *v* = *vi*1+3*k*+1. Clearly, the set (∪*<sup>r</sup> <sup>j</sup>*=2*S*<sup>3</sup> *j* ) ∪ {*vi*1+3*t*+<sup>2</sup> | 0 ≤ *t* ≤ *k* − <sup>1</sup>}∪{*vi*1+3*<sup>t</sup>* | *k* + 1 ≤ *t* ≤ # *<sup>H</sup>*(*Q*,*i*1)−3*k*−<sup>1</sup> 3 \$ } is an independent dominating set of *G* − *v* of size *<sup>i</sup>*(*G*), and so, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). To prove the inverse inequality, let *T* be an *i*(*G* − *v*)-set. As above, we have |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 1 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* , and |*T* ∩ {*vij*+1, ... , *vij*+1−1, *vij*+<sup>1</sup> }| ≥ *<sup>H</sup>*(*Q*,*ij*) 3 for 2 ≤ *j* ≤ *r*. Now to dominate the vertices *vi*1+*j* (1 ≤ *j* ≤ <sup>3</sup>*k*), we must have |*T* ∩ {*vi*1+1, ... , *vi*1+3*<sup>k</sup>*}| ≥ *k*, and to dominate the vertices *vi*1+3*k*+2, ... , *vi*<sup>2</sup>−1, we must have |*T* ∩ {*vi*2+3*k*+2, ... , *vi*<sup>2</sup>−1, *vi*2 }| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 − *k*. This implies that |*T*| ≥ *<sup>i</sup>*(*G*), and so, *i*(*G*) = *i*(*G* − *<sup>v</sup>*).

**Case 18.** *v* ∈ *S*2 1. Assume that *v* = *vi*1+3*k*+2. Clearly, the set (∪*<sup>r</sup> <sup>j</sup>*=2*S*<sup>3</sup> *j* ) ∪ {*vi*1+3*t*+<sup>3</sup> | 0 ≤ *t* ≤ *k* − <sup>1</sup>}∪{*vi*1+3*t*+<sup>1</sup> | *k* + 1 ≤ *t* ≤ # *<sup>H</sup>*(*Q*,*i*1)−3*k*−<sup>2</sup> 3 \$ } is an independent dominating set of *G* − *v* of size *<sup>i</sup>*(*G*), and so, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). To prove the inverse inequality, let *T* be an *i*(*G* − *v*)-set. As above, we have |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 1 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* , and |*T* ∩ {*vij*+2, ... , *vij*+1−1, *vij*+<sup>1</sup> }| ≥ *<sup>H</sup>*(*Q*,*ij*) 3 for 2 ≤ *j* ≤ *r*. If *k* = 0, then to dominate the vertices *vi*1+3, ... , *vi*<sup>2</sup>−1, we must have |*T* ∩ {*vi*2+3, ... , *vi*<sup>2</sup>−1, *vi*2 }| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 yielding |*T*| ≥ *<sup>i</sup>*(*G*). If *k* ≥ 1, then to dominate the vertices *vi*1+1, ... , *vi*1+3*k*+1, we must have |*T* ∩ {*vi*1+2, ... , *vi*1+3*k*+<sup>1</sup>}| ≥ *k*, and to dominate the vertices *vi*1+3*k*+2, ... , *vi*<sup>2</sup>−1, we must have |*T* ∩ {*vi*2+3*k*+2, ... , *vi*<sup>2</sup>−1, *vi*2 }| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 − *k*, so |*T*| ≥ *<sup>i</sup>*(*G*). Therefore, *i*(*G*) = *i*(*G* − *<sup>v</sup>*).

**Case 19.** *v* ∈ *r j*=2 *S*1 *j* (the case *v* ∈ *r j*=2 *S*3 *j* is similar).

Suppose, without loss of generality, that *v* = *vi*2+3*k*+1. Clearly, the set *S*2 1 ∪ (∪*<sup>r</sup> <sup>j</sup>*=3*S*<sup>2</sup> *j* ) ∪ {*vi*2+3*t*+<sup>2</sup> | 0 ≤ *t* ≤ *k* − <sup>1</sup>}∪{*vi*3 , *vi*2+3*<sup>t</sup>* | *k* + 1 ≤ *t* ≤ # *<sup>H</sup>*(*Q*,*i*2)−3*k*−<sup>1</sup> 3 \$ } is an independent dominating set of *G* − *v* of size *<sup>i</sup>*(*G*), and so, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). Now, we show that *i*(*G* − *v*) ≥ *<sup>i</sup>*(*G*). Let *T* be a *i*(*G* − *v*)-set. As above, we have |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 1 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* , and |*T* ∩ {*vi*1+1, *vi*1+2, ... , *vi*<sup>2</sup>−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 . Furthermore, to dominate the vertices *vij*+2, ... , *vij*+1−1, we must have |*T* ∩ {*vij*+2, ... , *vij*+1−1, *vij*+<sup>1</sup> }| ≥ *<sup>H</sup>*(*Q*,*ij*) 3 for 3 ≤ *j* ≤ *r*. Now, to dominate the vertices *vi*2+1, ... , *vi*2+3*k*, we must have |*T* ∩ {*vi*2+1, ... , *vi*2+3*<sup>k</sup>*}| ≥ *k*, and to dominate the vertices *vi*2+3*k*+2, ... , *vi*<sup>3</sup>−1, we must have |*T* ∩ {*vi*2+3*k*+2, ... , *vi*<sup>3</sup>−1, *vi*3 }| ≥ *<sup>H</sup>*(*Q*,*i*2) 3 − *k*. This implies that |*T*| ≥ *<sup>i</sup>*(*G*), yielding *i*(*G*) = *i*(*G* − *<sup>v</sup>*).

### **Case 20.** *v* ∈ *r j*=2 *S*2 *j* .

Suppose, without loss of generality, that *v* = *vi*2+3*k*+2. Clearly, the set *S*2 1 ∪ (∪*<sup>r</sup> <sup>j</sup>*=3*S*<sup>2</sup> *j* ) ∪ {*vi*2+3*<sup>t</sup>* | 0 ≤ *t* ≤ *<sup>k</sup>*}∪{*vi*2+3*t*+<sup>1</sup> | *k* + 1 ≤ *t* ≤ # *<sup>H</sup>*(*Q*,*i*2)−3*k*−<sup>2</sup> 3 \$ } is an independent dominating set of *G* − *v* of size *<sup>i</sup>*(*G*), and so, *i*(*G*) ≥ *i*(*G* − *<sup>v</sup>*). To prove the inverse inequality, let *T* be an *i*(*G* − *v*)-set. As above, we have |*T* ∩ {*vij*,*s*,*a*, *vij*,*s*,*<sup>b</sup>*}| ≥ 1 for each 1 ≤ *j* ≤ *r* and 1 ≤ *s* ≤ *qij* , and |*T* ∩ {*vi*1+1, *vi*1+2, ... , *vi*<sup>2</sup>−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*i*1) 3 . Furthermore, to dominate the vertices *vij*+2, ... , *vij*+1−1, we

must have |*T* ∩ {*vij*+2, ... , *vij*+1−1, *vij*+<sup>1</sup> }| ≥ *<sup>H</sup>*(*Q*,*ij*) 3 for 3 ≤ *j* ≤ *r*. Now, to dominate the vertices *vi*2+1, ... , *vi*2+3*k*+1, we must have |*T* ∩ {*vi*2 , ... , *vi*2+3*k*+<sup>1</sup>}| ≥ *k* + 1, and to dominate the vertices *vi*2+3*k*+3, ... , *vi*<sup>3</sup>−1, we must have |*T* ∩ {*vi*2+3*k*+3, ... , *vi*<sup>3</sup>−<sup>1</sup>}| ≥ *<sup>H</sup>*(*Q*,*i*2) 3 − *k* − 1. This implies that |*T*| ≥ *<sup>i</sup>*(*G*), and so, *i*(*G*) = *i*(*G* − *<sup>v</sup>*).

Thus, *G* is an independent domination stable graph, and the proof is complete.

The proof of the next result is straightforward and therefore omitted.

**Proposition 13.** *If G* ∈ J<sup>4</sup>*, then G is an independent domination stable graph.*

In order to present our constructive characterization of independent domination stable unicyclic graphs, we define a family of graphs as follows. Let G be the family of graphs *G* that can be obtained from a sequence *G*1, *G*2, ... , *Gk* of graphs for some *k* ≥ 1, where *G*1 ∈J −T if *k* = 1 and *G*1 ∈ J if *k* ≥ 2, and *G* = *Gk*. If *k* ≥ 2, *Gi*+<sup>1</sup> can be obtained from *Gi* by one of the following operations.

**Operation** O1**:** If *u* ∈ *<sup>W</sup>*(*Gi*), then O1 adds a spider *S*1 with head *s* and an edge *us* to obtain *Gi*+<sup>1</sup> (see Figure 3).

**Operation** O2**:** If *u* ∈ *<sup>V</sup>*(*Gi*), then O2 adds a spider *Sq* (*q* ≥ 2) with head *s* and an edge *us* to obtain *Gi*+<sup>1</sup>(see Figure 3).

**Operation** O3**:** If *Gi* is a tree and (*<sup>x</sup>*, *y*) ∈ *<sup>W</sup>*1,1(*Gi*), then O3 adds a graph *Hk*1,*k*2 (*k*1 = *k*2 = 0) and edges *ax*, *by* to obtain *Gi*+<sup>1</sup> (see Figure 1).

**Operation** O4**:** If *Gi* is a tree, *x* ∈ *W*(*Gi*) and *y* ∈ *<sup>V</sup>*(*Gi*), then O4 adds a graph *Hk*1,*k*2 (*k*1 = 0, *k*2 ≥ 1) and edges *ax*, *by* to obtain *Gi*+<sup>1</sup> (see Figure 1).

**Operation** O5**:** If *Gi* is a tree, *x*, *y* ∈ *<sup>V</sup>*(*Gi*), then O5 adds a graph *Hk*1,*k*2 (*k*1 ≥ 1, *k*2 ≥ 1) and edges *ax*, *by* to obtain *Gi*+<sup>1</sup> (see Figure 1).

**Operation** O6**:** If *Gi* is a tree, *x*, *y* ∈ *<sup>V</sup>*(*Gi*), then O6 adds a graph *P*(6,(0, 0, *k*1, *k*2, 0, 0)) (*k*1 ≥ 0, *k*2 ≥ 0) and edges *v*3*x*, *<sup>v</sup>*4*y* to obtain *Gi*+<sup>1</sup> (see Figure 4).

**Theorem 4.** *Let G* ∈ G *be a graph of order n* ≥ 3*. Then, G is an independent domination stable graph.*

**Proof.** Suppose that *G* ∈ G. Then, there exists a sequence of graphs *G*1, *G*2, ... , *Gk* (*k* ≥ 1) such that *G*1 ∈J −T if *k* = 1 and *G*1 ∈ J if *k* ≥ 2, and if *k* ≥ 2, then *Gi*+<sup>1</sup> can be obtained from *Gi* by one of the operations O1, O2, ··· , O6. We proceed by induction on the number of operations used to construct *G*. If *k* = 1, the result holds by Propositions 11, 12, and 13. Assume that the result holds for each graph *G* ∈ G, which can be obtained from a sequence of operations of length *k* − 1, and let *G* = *Gk*−1. By the induction hypothesis, *G* is an independent domination stable graph. Since *G* = *Gk* is obtained by one of the operations O1, O2, ··· , O6 from *G*, we conclude from Propositions 5, 7, and 9 that *G* is an independent domination stable unicyclic graph.

**Theorem 5.** *Let G be a unicyclic graph of order n* ≥ 3*. Then, G is an ID-stable graph if and only if G* ∈ G*.*

**Proof.** According to Theorem 4, we need only to prove necessity. Let *G* be an ID-stable unicyclic graph of order *n* ≥ 3. The proof is by induction on *n*. Let *n* ≥ 11, and let the statement hold for all ID-stable unicyclic graphs of order less than *n*. Assume that *G* is an ID-stable unicyclic graph of order *n*. Let *C* = (*<sup>v</sup>*1*v*2 ... *vp*) be the unique cycle of *G*. If *G* is a cycle, then *p* = *n*, and Proposition 1 implies that *G* ∈ J3 ⊆ G. Now, we consider the case *p* < *n*. Choose a vertex *u* ∈ *V*(*G*) − *V*(*C*) such that the distance between the vertex *u* and the set *V*(*C*) is as large as possible. Assume that *v*1*u*1*u*2 ··· *uu* is the shortest (*<sup>u</sup>*, *<sup>V</sup>*(*C*))-path. If ≥ 2, then similar to the proof of Theorem 2, *G* can be obtained from *Gk*−<sup>1</sup>by one of the operations O1or O2, and so, *G* ∈ G. Assume that ≤ 1.

First, assume *vi* is not a support vertex for each *i* ∈ {1, ... , *p*}. Then, *G* = *<sup>C</sup>*(*<sup>n</sup>*, *Q*) for some *Q* ∈ N *n*. If *D*(*Q*) = 0, then it follows from Corollary 1 that *G* ∈ J1. If *D*(*Q*) = 1, then it follows from Theorem 3 that *G* ∈ J2. If *D*(*Q*) ≥ 2, then we conclude from Propositions 10 and 12 that *G* ∈ J3.

Now, suppose that *vi* is a support vertex for some *i* ∈ {1, ... , *p*}, say *i* = 2. Assume *c* is a leaf adjacent to *v*2. We conclude from Propositions 2 and 3 that *v*2 is not a strong support vertex and is not adjacent to a support vertex. It follows that *dG*(*<sup>v</sup>*2) = 3 and that *v*1, *v*3 are not support vertices. Let *k*1 be the number of pendant paths of length two beginning at *v*1 and *k*2 be the number of pendant paths of length two beginning at *v*3. Let *G* be the graph obtained from *G* by removing *v*1, *v*2, *v*3 and the vertices of all pendant paths at *v*1, *v*3. By Proposition 6, we have *i*(*G*) = *i*(*G*) + *k*1 + *k*2 + 1. If *G* is not an ID-stable graph, then *i*(*G* − *v*) = *i*(*G*) for some vertex *v* ∈ *<sup>V</sup>*(*G*), and it follows from Proposition 6 that *i*(*G* − *v*) = *i*(*G* − *v*) + *k*1 + *k*2 + 1 = *i*(*G*) + *k*1 + *k*2 + 1 = *<sup>i</sup>*(*G*), which is a contradiction. Hence, *G* is an ID-stable graph, and by the induction hypothesis, we have *G* ∈ G. If *k*1 ≥ 1, *k*2 ≥ 1, then *T* can be obtained from *G* by operation O5, and so, *G* ∈ G. Assume that *k*1 = *k*2 = 0. Then, we have *dG*(*<sup>v</sup>*1) = *dG*(*<sup>v</sup>*3) = 2. Let *S* be a *i*(*G* − *<sup>v</sup>*2)-set. Since *G* is an ID-stable graph, we have *i*(*G*) = *i*(*G* − *<sup>v</sup>*2). To dominate the vertices *c*, *v*1, *v*3, we must have *c* ∈ *S*, |*s* ∩ {*<sup>v</sup>*1, *vn*}| ≥ 1 and |*s* ∩ {*<sup>v</sup>*3, *<sup>v</sup>*4}| ≥ 1. Suppose, without loss of generality, that *v*4, *vn* ∈ *S*. Then, *S* − {*c*} is an *<sup>i</sup>*(*G*)-set containing *v*4, *vn*, and so, (*<sup>v</sup>*3, *vn*) ∈ *W*1,1. Now, *T* can be obtained from *G* by operation O3, and so, *G* ∈ G. Finally, let *k*1 = 0 and *k*2 ≥ 1. As above, we can see that *v*4 ∈ *<sup>W</sup>*(*G*), and since *T* can be obtained from *G* by operation O4, we have *T* ∈ G. This completes the proof.

**Figure 4.** The operation O6.
