In this section we determine the domination number in 4-regular Knödel graphs *W*_{4,n}. Note that *n* ≥ 16 by the definition. For this purpose, we prove the following lemmas namely Lemma 3.1, 3.2, 3.3, 3.4 and 3.5.

#### Lemma 3.1

*For each even integer n* ≥ 16, *we have*
$\begin{array}{}\gamma ({W}_{4,n})=2\lfloor \frac{n}{10}\rfloor +\left\{\begin{array}{cc}0& n\equiv 0\text{\hspace{0.17em}(mod\hspace{0.17em}}10)\\ 2& n\equiv 2,4\text{\hspace{0.17em}(mod\hspace{0.17em}}10)\end{array}\right..\end{array}$

#### Proof

First assume that *n* ≡ 0 (mod 10). Let *n* = 10*t*, where *t* ≥ 2. By Theorem 1.2, *γ*(*W*_{4,n}) ≥
$\begin{array}{}\frac{n}{5}\end{array}$ = 2*t*. On the other hand, we can see that the set *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t−4}} ∪ {*v*_{5}, *v*_{10}}, ⋯, *v*_{5t}} is a dominating set with 2*t* elements, and we have *γ* (*W*_{4,n}) = 2*t* = 2⌊
$\begin{array}{}\frac{n}{10}\end{array}$⌋, as desired.

Next assume that *n* ≡ 2 (mod 10). Let *n* = 10*t*+2, where *t* ≥ 2. By Theorem 1.2, we have *γ* (*W*_{4,n}) ≥
$\begin{array}{}\frac{n}{5}\end{array}$ > 2*t*. Suppose that *γ*(*W*_{4,n}) = 2*t*+1. Let *D* be a minimum dominating set of *W*_{4,n}. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ *t* or |*D* ∩ *V*| ≤ *t*. Without loss of generality, assume that |*D* ∩ *U*| ≤ *t*. Let |*D* ∩ *U*| = *t* − *a*, where *a* ≥ 0. Then |*D* ∩ *V*| = *t*+1+*a*. Observe that *D* ∩ *U* dominates at most 4*t* − 4*a* vertices of *V* and therefore *D* dominates at most (4*t* − 4*a*) + (*t*+1+*a*) = 5*t* − 3*a*+1 vertices of *V*. Since *D* dominates all vertices of *V*, we have 5*t* − 3*a*+1 ≥ 5*t*+1 and so *a* = 0, |*D* ∩ *U*| = *t* and |*D* ∩ *V*| = *t*+1. Let *D* ∩ *U* = {*u*_{i1}, *u*_{i2}, ⋯, *u*_{it}} and *n*_{1}, *n*_{2}, ⋯, *n*_{t} be the cyclic-sequence of *D* ∩ *U*. By Observation 2.2, we have
$\begin{array}{}\sum _{k=1}^{t}{n}_{k}=5t+1\end{array}$ and, therefore, there exists some *k* such that *n*_{k} ∈ ℳ_{4} = {1, 2, 3, 4, 6, 7}. Then by Lemma 2.3, |*N*(*u*_{k}) ∩ *N*(*u*_{k+1}) ≥ 1. Hence, *D* ∩ *U* dominates at most 4*t* − 1 vertices of *V*, that is, *D* dominates at most (4*t* − 1)+(*t*+1) = 5*t* vertices of *V*, a contradiction. Now we deduce that *γ*(*W*_{4,n}) ≥ 2*t*+2. On the other hand the set *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}} ∪ {*v*_{5}, *v*_{10}}, ⋯, *v*_{5t}} ∪ {*v*_{5t+1}} is a dominating set for *W*_{4,n} with 2*t*+2 elements. Consequently, *γ*(*W*_{4,n}) = 2*t*+2.

It remains to assume that *n* ≡ 4 (mod 10). Let *n* = 10*t*+4, where *t* ≥ 2. By Theorem 1.2, we have *γ* (*W*_{4,n}) ≥
$\begin{array}{}\frac{n}{5}\end{array}$ > 2*t*. Suppose that *γ*(*W*_{4,n}) = 2*t*+1. Let *D* be a minimum dominating set of *γ* (*W*_{4,n}). Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ *t* or |*D* ∩ *V*| ≤ *t*. Without loss of generality, assume that |*D* ∩ *U*| = *t* − *a* and *a* ≥ 0. Then |*D* ∩ *V*| = *t*+1+*a*. Observe that *D* ∩ *U* dominates at most 4(*t* − *a*) elements of *V* and therefore *D* dominates at most 4(*t* − *a*)+(*t*+1+*a*) = 5*t* − 3*a*+1 vertices of *V*. Since *D* dominates all vertices of *V*, we have 5*t* − 3*a*+1 ≥ |*V*| = 5*t*+2 and so −3*a* ≥ 1, a contradiction. Thus *γ* (*W*_{4,n}) > 2*t*+1. On the other hand, the set {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}} ∪ {*v*_{5}, *v*_{10}, ⋯, *v*_{5t}} ∪ {*v*_{3}} is a dominating set with 2*t*+2 elements. Consequently, *γ* (*W*_{4,n}) = 2*t*+2. □

#### Lemma 3.2

*For each even integer n* ≥ 46 *with n* ≡6 (mod 10), *we have γ* (*W*_{4,n}) = 2⌊
$\begin{array}{}\frac{n}{10}\end{array}$⌋+3.

#### Proof

Let *n* = 10*t*+6 and *t* ≥ 4. By Theorem 1.2, we have *γ* (*W*_{4,n}) ≥
$\begin{array}{}\frac{n}{5}\end{array}$ > 2*t*+1. The set *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}} ∪ {*v*_{5}, *v*_{10}, ⋯, *v*_{5t}} ∪ {*v*_{2}, *v*_{3}} is a dominating set with 2*t*+3 elements. Thus, 2*t*+2 ≤ *γ* (*W*_{4,n}) ≤ 2*t*+3. We show that *γ*(*W*_{4,n}) = 2*t*+3. Suppose to the contrary that *γ*(*W*_{4,n}) = 2*t*+2. Let *D* be a minimum dominating set of *W*_{4,n}. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ *t*+1 or |*D* ∩ *V*| ≤ *t*+1. Without loss of generality, assume that |*D* ∩ *U*| = *t*+1 − *a*, where *a* ≥ 0. Then |*D* ∩ *V*| = *t*+1+*a*. Note that *D* ∩ *U* dominates at most 4*t*+4 − 4*a* vertices and therefore *D* dominates at most (4*t*+4 − 4*a*)+(*t*+1+*a*) = 5*t* − 3*a*+5 vertices of *V*. Since *D* dominates all vertices of *V*, we have 5*t* − 3*a*+5 ≥ 5*t*+3 and so *a* = 0 and |*D* ∩ *U*| = |*D* ∩ *V*| = *t*+1. Also we have |*V* − *D*| = 4*t*+2 ≤ |*N*(*D* ∩ *U*)| ≤ 4*t*+4, and similarly, |*D* − *U*| = 4*t*+2 ≤ |*N*(*D* ∩ *V*)| ≤ 4*t*+4.

Let *D* ∩ *U* = {*u*_{i1}, *u*_{i2}, ⋯, *u*_{it+1}} and *n*_{1}, *n*_{2}, ⋯, *n*_{t+1} be the cyclic-sequence of *D* ∩ *U*. By Observation 2.2, we have
$\begin{array}{}\sum _{k=1}^{t+1}{n}_{k}\end{array}$ = 5*t*+3 and, therefore, there exists *k*′ such that *n*_{k′} < 5. Then *n*_{k′} ∈ ℳ_{4} and by Lemma 2.3, |*N*(*u*_{ik′}) ∩ *N*(*u*_{ik′+1})| ≥ 1. Hence, *D* ∩ *U* dominates at most 4*t*+3 vertices from *V* and therefore 4*t*+2 ≤ *N*(*D* ∩ *U*) ≤ 4*t*+3.

If |*N*(*D* ∩ *U*)| = 4*t*+3, then for each *k* ≠ *k*′ we have *n*_{k} ∉ ℳ_{4}. If there exists *k*″ ≠ *k*′ such that *n*_{k″} ≥ 8, then 5*t*+3 =
$\begin{array}{}\stackrel{t+1}{\underset{k=1}{\mathit{\Sigma}}}{n}_{k}\end{array}$ ≥ 1+8+5(*t* − 1) = 5*t*+4, a contradiction. By symmetry we have *n*_{1} = *n*_{2} = ⋯ = *n*_{t} = 5, *n*_{t+1} = 3 and *D* ∩ *U* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}}. Observe that *D* ∩ *U* doesn’t dominate vertices *v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t} and so {*v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t}} ⊆ *D*. Thus *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}, *v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t}}. But the vertices *u*_{4}, *u*_{5}, *u*_{5t−2}, *u*_{5t+2} are not dominated by *D*, a contradiction.

Thus, |*N*(*D* ∩ *U*)| = 4*t* + 2. Then there exists precisely two pairs of vertices in *D* ∩ *U* with index-distances belonging to ℳ_{4}. If there exists an integer 1 ≤ *i*′ ≤ *t*+1 such that *n*_{i′}+*n*_{i′+1} ∈ ℳ_{4}, then min{*n*_{i′}, *n*_{i′+1}} ≤ 3. Then min{*n*_{i′}, *n*_{i′+1}} ∈ ℳ_{4} and \max{*n*_{i′}, *n*_{i′+1}} ∉ ℳ_{4}. Now we have max{*n*_{i′}, *n*_{i′+1}} = 5, min{*n*_{i′}, *n*_{i′+1}} ∈ {1, 2}, and *n*_{i} ∉ ℳ_{4}, for each *i*∉{*i*′, *i*′+1}. Now a simple calculation shows that the equality 5*t*+3 =
$\begin{array}{}\sum _{k=1}^{t+1}{n}_{k}\end{array}$ does not hold. (Note that if each *n*_{i} is less than 8, then we have
$\begin{array}{}\sum _{k=1}^{t+1}{n}_{k}\end{array}$ ≤ 2+5+5(*t* − 1) = 5*t*+2; otherwise we have
$\begin{array}{}\sum _{k=1}^{t+1}{n}_{k}\end{array}$ ≥ 1+5+8+5(*t* − 2) = 5*t*+4.) Thus there exist exactly two indices *j* and *k* such that *n*_{j}, *n*_{k} ∈ ℳ_{4} and {*n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, ⋯, *n*_{t}+*n*_{t+1}, *n*_{t+1}+*n*_{1}} ∩ ℳ_{4} = ∅. By this hypothesis, the only possible cases for the cyclic-sequence of *D* ∩ *U* are those demonstrated in .

Note that each column of shows the cyclic sequence of *D* ∩ *U*. We show that each case is impossible. For this purpose, we show that the cyclic-sequence of *D* ∩ *U* posed in the column *i*, for *i* ≥ 1 is impossible.

*i* = 1). If *n*_{1} = 8, *n*_{2} = 1, *n*_{3} = 4, *n*_{4} = ⋯ = *n*_{t+1} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{9}, *u*_{10}, *u*_{14}, *u*_{19}, ⋯, *u*_{5t − 1}}, then *D* ∩ *U* does not dominate the vertices *v*_{5}, *v*_{6}, *v*_{7}, *v*_{18}, *v*_{23}, *v*_{28}, …, *v*_{5t+3}.

Thus we have *D* ∩ *V* = {*v*_{5}, *v*_{6}, *v*_{7}, *v*_{18}, *v*_{23}, *v*_{28}, …, *v*_{5t+3}}. But *D* does not dominate three vertices *u*_{8}, *u*_{12}, *u*_{13}, a contradiction.

*i* = 2). If *n*_{1} = 4, *n*_{2} = 1, *n*_{3} = 8, *n*_{4} = ⋯ = *n*_{t+1} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{6}, *u*_{14}, *u*_{19}, ⋯, *u*_{5t − 1}}. Then *D* ∩ *U* does not dominate the vertices *v*_{10}, *v*_{11}, *v*_{16}, *v*_{18}, *v*_{23}, *v*_{28}, …, *v*_{5t+3}. Thus we have *D* ∩ *V* = {*v*_{10}, *v*_{11}, *v*_{16}, *v*_{18}, *v*_{23}, *v*_{28}, …, *v*_{5t+3}}. But *D* does not dominate three vertices *u*_{2}, *u*_{12}, *u*_{5t+1}, a contradiction.

*i* ∈ {3, 4}). As before, we obtain that *u*_{8} ∉ *D*. But *N*(*u*_{8}) = {*v*_{8}, *v*_{9}, *v*_{11}, *v*_{15}} ⊆ *N*(*D* ∩ *U*) and therefore *N*(*u*_{8}) ∩ *D* = ∅. Hence, *N*[*u*_{8}] ∩ *D* = ∅ and *D* does not dominate *u*_{8}, a contradiction.

*i* ≥ 5). As before, we obtain that *u*_{3}∉ *D*. But *N*(*u*_{3}) = {*v*_{3}, *v*_{4}, *v*_{6}, *v*_{10}} ⊆ *N*(*D* ∩ *U*) and therefore *N*(*u*_{3}) ∩ *D* = ∅. Hence, *N*[*u*_{3}] ∩ *D* = ∅ and *D* does not dominate *u*_{3}, a contradiction.

Consequently, *γ*(*W*_{4,n}) = 2*t*+3, as desired. □

Lemma 3.2 determines the domination number of *W*_{4,n} when *n* ≡ 6 (mod 10) and *n* ≥ 46. The only values of *n* for *n* ≡ 6 (mod 10) are thus 16, 26 and 36. We study these cases in the following lemma.

#### Lemma 3.3

*For n* ∈ {16, 26, 36}, *we have*:

#### Proof

For *n* = 16, by Theorem 1.2 we have *γ* (*W*_{4,16}) ≥
$\begin{array}{}\frac{16}{5}\end{array}$ > 3. On the other hand, the set *D* = {*u*_{1}, *u*_{2}, *v*_{6}, *v*_{7}} is a dominating set for *W*_{4,16}, and therefore *γ* (*W*_{4,16}) = 4.

For *n* = 26, by Theorem 1.2 we have *γ* (*W*_{4,26}) ≥
$\begin{array}{}\frac{26}{5}\end{array}$ > 5. On the other hand, the set *D* = {*u*_{1}, *u*_{4}, *u*_{9}, *u*_{10}, *v*_{1}, *v*_{2}, *v*_{6}} is a dominating set for *W*_{4,26} and therefore 6 ≤ *γ* (*W*_{4,26}) ≤ 7. We show that *γ* (*W*_{4,26}) = 7. Suppose, on the contrary, that *γ* (*W*_{4,26}) = 6. Let *D* be a minimum dominating set for *W*_{4,26}. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ 3 or |*D* ∩ *V*| ≤ 3. If |*D* ∩ *U*| = 3 − *a*, where *a* ≥ 0, then |*D* ∩ *V*| = 3+*a*. Now, the elements of *D* ∩ *U* dominate at most 4(3 − *a*) elements of *V* and *D* dominates at most 4(3 − *a*) + (3+*a*) = 15 − 3*a* vertices of *V*. Thus 15 − 3*a* ≥ |*V*| = 13, which implies *a* = 0 and |*D* ∩ *U*| = |*D* ∩ *V*| = 3. Let |*D* ∩ *U*| = {*u*_{1}, *u*_{i}, *u*_{j}}, where 1 < *i* < *j* ≤ 13 and *n*_{1} = *i* − 1, *n*_{2} = *j* − *i*, *n*_{3} = 13+1 − *j*. Since *n*_{1}+*n*_{2}+*n*_{3} = 13, we have {*n*_{1}, *n*_{2}, *n*_{3}} ∩ ℳ_{4} ≠ ∅.

If ℳ_{4} includes at least three numbers of *n*_{1}, *n*_{2}, *n*_{3}, *n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{1}, then by Lemma 2.3, *D* ∩ *U* dominates at most 4 × 3 − 3 = 9 vertices of *V* and |*D* ∩ *V*| ≥ 13 − 9 = 4, a contradiction.

If ℳ_{4} includes exactly one number of *n*_{1}, *n*_{2}, *n*_{3}, then we have, by symmetry, *n*_{1} = 3, *n*_{2} = 5, *n*_{3} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{4}, *u*_{9}}, *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, *v*_{3}, *v*_{4}, *v*_{5}, *v*_{7}, *v*_{8}, *v*_{9}, *v*_{10}, *v*_{11}, *v*_{12}}. Then {*v*_{6}, *v*_{13}} ⊆ *D*. But {*u*_{1}, *u*_{4}, *u*_{9}, *v*_{6}, *v*_{13}} does not dominate the vertices {*u*_{2}, *u*_{7}, *u*_{8}, *u*_{11}} and we need at least 2 other vertices to dominate this four vertices, and hence |*D*| ≥ 7, a contradiction.

Thus, we assume that ℳ_{4} includes two numbers of *n*_{1}, *n*_{2}, *n*_{3} and {*n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{1}} ∩ ℳ_{4} = ∅. We thus have five possibilities for the cyclic-sequence of *D* ∩ *U* that are demonstrated in . Note that each column of shows the cyclic sequence of *D* ∩ *U*. We show that each case is impossible. For this purpose, we show that the cyclic sequence of *D* ∩ *U* posed in the column *i*, for *i* ≥ 1 is impossible.

*i* = 1) If *n*_{1} = 1, *n*_{2} = 4, *n*_{3} = 8, then *D* ∩ *U* = {*u*_{1}, *u*_{2}, *u*_{6}} and *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, ⋯, *v*_{9}, *v*_{13}}. Thus {*v*_{10}, *v*_{11}, *v*_{12}} ⊆ *D* and *D* = {*u*_{1}, *u*_{2}, *u*_{6}, *v*_{10}, *v*_{11}, *v*_{12}}. But *D* does not dominate the vertex *u*_{13}, a contradiction.

*i* = 2) If *n*_{1} = 1, *n*_{2} = 8, *n*_{3} = 4, then *D* ∩ *U* = {*u*_{1}, *u*_{2}, *u*_{10}} and *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, ⋯, *v*_{11}, *v*_{13}}. Thus {*v*_{6}, *v*_{7}, *v*_{12}} ⊆ *D* and *D* = {*u*_{1}, *u*_{2}, *u*_{10}, *v*_{6}, *v*_{7}, *v*_{12}}. But *D* does not dominate the vertex *u*_{8}, a contradiction.

*i* = 3) If *n*_{1} = 2, *n*_{2} = 3, *n*_{3} = 8, then *D* ∩ *U* = {*u*_{1}, *u*_{3}, *u*_{6}}, *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, ⋯, *v*_{10}, *v*_{13}}.Thus {*v*_{5}, *v*_{11}, *v*_{12}} ⊆ *D* and *D* = {*u*_{1}, *u*_{3}, *u*_{6}, *v*_{5}, *v*_{11}, *v*_{12}}. But *D* does not dominate the vertices *u*_{7} and *u*_{13}.

*i* = 4) If *n*_{1} = 2, *n*_{2} = 8, *n*_{3} = 3, then *D* ∩ *U* = {*u*_{1}, *u*_{3}, *u*_{11}} and *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, ⋯, *v*_{6}, *v*_{8}, *v*_{10}, *v*_{11}, *v*_{12}}. Thus {*v*_{7}, *v*_{9}, *v*_{13}} ⊆ *D* and *D* = {*u*_{1}, *u*_{3}, *u*_{11}, *v*_{7}, *v*_{9}, *v*_{13}}. But *D* does not dominate the vertex *u*_{5}, a contradiction.

*i* = 5) If *n*_{1} = 4, *n*_{2} = 4, *n*_{3} = 5, then *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{9}} and *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, ⋯, *v*_{10}, *v*_{12}}. Thus {*v*_{7}, *v*_{11}, *v*_{13}} ⊆ *D* and *D* = {*u*_{1}, *u*_{3}, *u*_{11}, *v*_{7}, *v*_{9}, *v*_{13}}. But *D* does not dominate the vertices *u*_{2} and *u*_{3}.

Consequently, *γ*(*W*_{4,26}) = 7.

We now consider the case *n* = 36. By Theorem 1.2, *γ* (*W*_{4,36}) ≥
$\begin{array}{}\frac{36}{5}\end{array}$ > 7. On the other hand, the set *D* = {*u*_{1}, *u*_{2}, *u*_{10}, *u*_{11}, *v*_{6}, *v*_{7}, *v*_{15}, *v*_{16}} is a dominating set for the graph *W*_{4,36} and, therefore, *γ* (*W*_{4,36}) = 8. □

We now consider the case *n* ≡ 8 (mod 10). For *n* = 18, 28 and 38 we have the following lemma.

#### Lemma 3.4

*For n* ∈ {18, 28, 38}, *we have*:

#### Proof

For *n* = 18, by Theorem 1.2 we have *γ* (*W*_{4,18}) ≥
$\begin{array}{}\frac{18}{5}\end{array}$ > 3. But the set *D* = {*u*_{1}, *u*_{2}, *v*_{6}, *v*_{7}} is a dominating set for *W*_{4,18} and, therefore, *γ* (*W*_{4,18}) = 4.

For *n* = 28, by Theorem 1.2 we have *γ* (*W*_{4,28}) ≥
$\begin{array}{}\frac{28}{5}\end{array}$ > 5 and the set *D* = {*u*_{1}, *u*_{6}, *u*_{11}, *u*_{13}, *v*_{3}, *v*_{5}, *v*_{9}} is a dominating set for *W*_{4,28} and, therefore, 6 ≤ *γ* (*W*_{4,28}) ≤ 7. Suppose on the contradiction that *γ* (*W*_{4,28}) = 6 and *D* is a minimum dominating set for *γ* (*W*_{4,28}). By the Pigeonhole Principal, either |*D* ∩ *U*| ≤ 3 or |*D* ∩ *V*| ≤ 3. If |*D* ∩ *U*| = 3 − *a* and *a* ≥ 0, then |*D* ∩ *V*| = 3+*a*. Now, the elements of *D* ∩ *U* dominate at most 4(3 − *a*) elements of *V* and *D* dominates at most 4(3 − *a*)+(3+*a*) = 15 − 3*a* vertices of *V*. Thus 15 − 3*a* ≥ |*V*| = 14. Therefore, *a* = 0 and so |*D* ∩ *U*| = |*D* ∩ *V*| = 3. Let |*D* ∩ *U*| = {*u*_{1}, *u*_{i}, *u*_{j}} and 1 < *i* < *j* ≤ 14 and *n*_{1} = *i* − 1, *n*_{2} = *j* − *i*, *n*_{3} = 14 + 1 − *j*. Since *n*_{1}+*n*_{2}+*n*_{3} = 14, we have {*n*_{1}, *n*_{2}, *n*_{3}} ∩ ℳ_{4} ≠ ∅. If ℳ_{4} includes at least two numbers in {*n*_{1}, *n*_{2}, *n*_{3}, *n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{1}}, then *D* ∩ *U* dominates at most 4 × 3 − 2 = 10 vertices of *V* and |*D* ∩ *V*| ≥ 14 − 10 = 4, a contradiction.

The only remaining case is that ℳ_{4} includes exactly one of the three numbers *n*_{1}, *n*_{2} and *n*_{3} and also {*n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{1}} ∩ ℳ_{4} = ∅. By symmetry we have *n*_{1} = 4, *n*_{2} = 5, *n*_{3} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{10}}, *N*(*D* ∩ *U*) = {*v*_{1}, *v*_{2}, *v*_{3}, *v*_{4}, *v*_{5}, *v*_{6}, *v*_{8}, *v*_{10}, *v*_{11}, *v*_{12}, *v*_{13}} thus {*v*_{7}, *v*_{9}, *v*_{14}} ⊆ *D* and *D* = {*u*_{1}, *u*_{5}, *u*_{10}, *v*_{7}, *v*_{9}, *v*_{14}}. But *D* does not dominate the vertex *u*_{3}. That is a contradiction and therefore *γ* (*W*_{4,28}) = 7. For *n* = 38, by Theorem 1.2 we have *γ* (*W*_{4,38}) ≥
$\begin{array}{}\frac{38}{5}\end{array}$ > 7 and the set *D* = {*u*_{1}, *u*_{6}, *u*_{11}, *u*_{16}, *u*_{18}, *v*_{3}, *v*_{5}, *v*_{10}, *v*_{13}, *v*_{15}} is a dominating set for *W*_{4,38} and therefore 8 ≤ *γ* (*W*_{4,38}) ≤ 10. Let *γ* (*W*_{4,38}) < 10 and *D* is a dominating set for *γ* (*W*_{4,38}) with |*D*| = 9. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ 4 or |*D* ∩ *V*| ≤ 4. If |*D* ∩ *U*| = 4 − *a* and *a* ≥ 0, then |*D* ∩ *V*| = 5+*a*. Now, the elements of *D* ∩ *U* dominate at most 4(4 − *a*) elements of *V* and *D* dominates at most 4(4 − *a*)+(5+*a*) = 21 − 3*a* vertices of *V*. Thus 21 − 3*a* ≥ |*V*| = 19 that results *a* = 0 and we have |*D* ∩ *U*| = 4 and |*D* ∩ *V*| = 5. Let |*D* ∩ *U*| = {*u*_{1}, *u*_{i}, *u*_{j}, *u*_{k}}, where 1 < *i* < *j* < *k* ≤ 19 and *n*_{1}, *n*_{2}, *n*_{3}, *n*_{4} be the cyclic-sequence of *D* ∩ *U*. Since *n*_{1}+*n*_{2}+*n*_{3}+*n*_{4} = 19, we have {*n*_{1}, *n*_{2}, *n*_{3}, *n*_{4}} ∩ ℳ_{4} ≠ ∅. If ℳ_{4} includes at least three numbers of {*n*_{1}, *n*_{2}, *n*_{3}, *n*_{4}, *n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{4}, *n*_{4}+*n*_{1}}, then *D* ∩ *U* dominates at most 4 × 4 − 3 = 13 vertices of *V* and |*D* ∩ *V*| ≥ 19 − 13 = 6, a contradiction.

If we wish that ℳ_{4} includes exactly one number out of *n*_{1}, *n*_{2}, *n*_{3}, *n*_{4}, we have three cases:

*i* = 1) If *n*_{1} = 4, *n*_{2} = 5, *n*_{3} = 5, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{10}, *u*_{15}}, then {*v*_{7}, *v*_{9}, *v*_{14}, *v*_{19}} ⊆ *D* but {*u*_{1}, *u*_{5}, *u*_{10}, *u*_{15}, *v*_{7}, *v*_{9}, *v*_{14}, *v*_{19}} does not dominate the vertices *u*_{3} and *u*_{17}. For dominating *u*_{3} and *u*_{17}, we need two vertices and therefore |*D*| ≥ 10, a contradiction.

*i* = 2) If *n*_{1} = 1, *n*_{2} = 8, *n*_{3} = 5, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{2}, *u*_{10}, *u*_{15}}, then {*v*_{6}, *v*_{7}, *v*_{12}, *v*_{14}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{2}, *u*_{10}, *u*_{15}, *v*_{6}, *v*_{7}, *v*_{12}, *v*_{14}, *v*_{19}} but *D* does not dominate the vertices *u*_{8} and *u*_{17}.

*i* = 3) If *n*_{1} = 8, *n*_{2} = 1, *n*_{3} = 5, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{9}, *u*_{10}, *u*_{15}}, then {*v*_{5}, *v*_{6}, *v*_{7}, *v*_{14}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{9}, *u*_{10}, *u*_{15}, *v*_{5}, *v*_{6}, *v*_{7}, *v*_{14}, *v*_{19}} but *D* does not dominate the vertex *u*_{8}.

Now we consider the cases that ℳ_{4} includes exactly two numbers of the cyclic-sequence *n*_{1}, *n*_{2}, *n*_{3}, *n*_{4} and {*n*_{1}+*n*_{2}, *n*_{2}+*n*_{3}, *n*_{3}+*n*_{4}, *n*_{4}+*n*_{1}} ∩ ℳ_{4} = ∅. By symmetry we have ten cases:

*i* = 1) If *n*_{1} = 1, *n*_{2} = 9, *n*_{3} = 1, *n*_{4} = 8 and *D* ∩ *U* = {*u*_{1}, *u*_{2}, *u*_{11}, *u*_{12}}, then {*v*_{6}, *v*_{7}, *v*_{10}, *v*_{16}, *v*_{17}} ⊆ *D* and *D* = {*u*_{1}, *u*_{2}, *u*_{11}, *u*_{12}, *v*_{6}, *v*_{7}, *v*_{10}, *v*_{16}, *v*_{17}} but *D* does not dominate the vertex *u*_{8}.

*i* = 2) If *n*_{1} = 2, *n*_{2} = 8, *n*_{3} = 1, *n*_{4} = 8 and *D* ∩ *U* = {*u*_{1}, *u*_{3}, *u*_{11}, *u*_{12}}, then {*v*_{5}, *v*_{7}, *v*_{9}, *v*_{16}, *v*_{17}} ⊆ *D* and *D* = {*u*_{1}, *u*_{3}, *u*_{11}, *u*_{12}, *v*_{5}, *v*_{7}, *v*_{9}, *v*_{16}, *v*_{17}} but *D* does not dominate the vertex *u*_{18}.

*i* = 3) If *n*_{1} = 4, *n*_{2} = 1, *n*_{3} = 9, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{6}, *u*_{15}}, then {*v*_{10}, *v*_{11}, *v*_{14}, *v*_{17}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{5}, *u*_{6}, *u*_{15}, *v*_{10}, *v*_{11}, *v*_{14}, *v*_{17}, *v*_{19}} but *D* does not dominate the vertex *u*_{2}.

*i* = 4) If *n*_{1} = 9, *n*_{2} = 1, *n*_{3} = 4, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{10}, *u*_{11}, *u*_{15}}, then {*v*_{5}, *v*_{6}, *v*_{7}, *v*_{9}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{10}, *u*_{11}, *u*_{15}, *v*_{5}, *v*_{6}, *v*_{7}, *v*_{9}, *v*_{19}} but *D* does not dominate the vertices *u*_{13} and *u*_{14}.

*i* = 5) If *n*_{1} = 3, *n*_{2} = 2, *n*_{3} = 9, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{4}, *u*_{6}, *u*_{15}}, then {*v*_{10}, *v*_{12}, *v*_{14}, *v*_{17}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{4}, *u*_{6}, *u*_{15}, *v*_{10}, *v*_{12}, *v*_{14}, *v*_{17}, *v*_{19}} but *D* does not dominate the vertices *u*_{2} and *u*_{8}.

*i* = 6) If *n*_{1} = 9, *n*_{2} = 2, *n*_{3} = 3, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{10}, *u*_{12}, *u*_{15}}, then {*v*_{5}, *v*_{6}, *v*_{7}, *v*_{9}, *v*_{14}} ⊆ *D* and *D* = {*u*_{1}, *u*_{10}, *u*_{12}, *u*_{15}, *v*_{5}, *v*_{6}, *v*_{7}, *v*_{9}, *v*_{14}} but *D* does not dominate the vertex *u*_{16}.

*i* = 7) If *n*_{1} = 8, *n*_{2} = 3, *n*_{3} = 5, *n*_{4} = 3 and *D* ∩ *U* = {*u*_{1}, *u*_{9}, *u*_{12}, *u*_{17}}, then {*v*_{3}, *v*_{6}, *v*_{7}, *v*_{11}, *v*_{14}} ⊆ *D* and *D* = {*u*_{1}, *u*_{9}, *u*_{12}, *u*_{17}, *v*_{3}, *v*_{6}, *v*_{7}, *v*_{11}, *v*_{14}} but *D* does not dominate the vertex *u*_{16}.

*i* = 8) If *n*_{1} = 3, *n*_{2} = 6, *n*_{3} = 5, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{4}, *u*_{10}, *u*_{15}}, then {*v*_{6}, *v*_{9}, *v*_{12}, *v*_{14}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{4}, *u*_{10}, *u*_{15}, *v*_{6}, *v*_{9}, *v*_{12}, *v*_{14}, *v*_{19}} but *D* does not dominate the vertex *u*_{17}.

*i* = 9) If *n*_{1} = 3, *n*_{2} = 5, *n*_{3} = 6, *n*_{4} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{4}, *u*_{9}, *u*_{15}}, then {*v*_{6}, *v*_{13}, *v*_{14}, *v*_{17}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{4}, *u*_{9}, *u*_{15}, *v*_{6}, *v*_{13}, *v*_{14}, *v*_{17}, *v*_{19}} but *D* does not dominate the vertices *u*_{2} and *u*_{8}.

*i* = 10) If *n*_{1} = 3, *n*_{2} = 5, *n*_{3} = 5, *n*_{4} = 6 and *D* ∩ *U* = {*u*_{1}, *u*_{4}, *u*_{9}, *u*_{14}}, then {*v*_{3}, *v*_{6}, *v*_{13}, *v*_{18}, *v*_{19}} ⊆ *D* and *D* = {*u*_{1}, *u*_{4}, *u*_{9}, *u*_{14}, *v*_{3}, *v*_{6}, *v*_{13}, *v*_{18}, *v*_{19}} but *D* does not dominate the vertices *u*_{7} and *u*_{8}.

Hence, *W*_{4,38} has not any dominating set with 9 vertices and *W*_{4,38} = 10 as desired. □

Now for *n* ≥ 48 with *n* ≡ 8 (mod 10) we determine the domination number of *W*_{4,n} as follows.

#### Lemma 3.5

*For each even integer n* ≥ 48, *n* ≡ 8 (mod 10), *we have γ* (*W*_{4,n}) = 2⌊
$\begin{array}{}\frac{n}{10}\end{array}$⌋+4.

#### Proof

Let *n* = 10*t*+8, where *t* ≥ 4. By Theorem 1.2, we have *γ* (*W*_{4,n}) ≥
$\begin{array}{}\frac{n}{5}\end{array}$ > 2*t*+1. The set *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}} ∪ {*v*_{5}, *v*_{10}, ⋯, *v*_{5t}} ∪ {*v*_{3}, *v*_{5t − 2}, *v*_{5t+3}} is a dominating set with 2*t*+4 elements, and so, 2*t*+2 ≤ *γ* (*W*_{4,n}) ≤ 2*t*+4. We show that *γ* (*W*_{4,n}) = 2*t*+4.

First, assume that *γ*(*W*_{4,n}) = 2*t*+2. Let *D* be a minimum dominating set of *W*_{4,n}. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ *t*+1 or |*D* ∩ *V*| ≤ *t*+1. Without loss of generality assume that |*D* ∩ *U*| = *t*+1 − *a*, where *a* ≥ 0. Then |*D* ∩ *V*| = *t*+1+*a*. Observe that *D* ∩ *U* dominates at most 4*t*+4 − 4*a* vertices of *V*, and therefore, *D* dominates at most (4*t*+4 − 4*a*)+(*t*+1+*a*) = 5*t* − 3*a*+5 vertices of *V*. Since *D* dominates all vertices in *V*, we have 5*t* − 3*a*+5 ≥ 5*t*+4 and so *a* = 0. Then |*D* ∩ *U*| = |*D* ∩ *V*| = *t*+1. Also we have |*V* − *D*| = 4*t*+3 ≤ |*N*(*D* ∩ *U*)| ≤ 4*t*+4 and |*U* − *D*| = 4*t*+3 ≤ |*N*(*D* ∩ *V*)| ≤ 4*t*+4. Let *D* ∩ *U* = {*u*_{i1}, *u*_{i2}, ⋯, *u*_{it+1}} and *n*_{1}, *n*_{2}, ⋯, *n*_{t+1}. By Observation 2.2, we have
$\begin{array}{}\stackrel{t+1}{\underset{k=1}{\mathit{\Sigma}}}{n}_{k}\end{array}$ = 5*t*+4 and therefore there exist *k*′ such that *n*_{k′} ∈ ℳ_{4}. By Lemma 2.3, *D* ∩ *U* dominates at most 4*t*+3 vertices from *V*. Then |*N*(*D* ∩ *U*)| = |*N*(*D* ∩ *V*)| = 4*t*+3 and *k*′ is unique. If there exists 1 ≤ *k*″ ≤ *t*+1 such that *n*_{k″} ≥ 8, then 5*t*+4 =
$\begin{array}{}\stackrel{t+1}{\underset{k=1}{\mathit{\Sigma}}}{n}_{k}\end{array}$ ≥ *n*_{k′}+*n*_{k″}+5(*t* − 1) ≥ 1+8+5(*t* − 1) = 5*t*+4 which implies *n*_{k′} = 1, *n*_{k″} = 8 and for each *k*∉{*k*′, *k*″} we have *n*_{k} = 5. Now in each arrangement of the cyclic sequence of *D* ∩ *U*, we have one adjacency between 1 and 5. Then we have two vertices in *D* ∩ *U* with index-distance equal to 6, a contradiction. Thus for *k* ≠ *k*′ we have *n*_{k} = 5 and *n*_{k′} = 4. We have (by symmetry) *n*_{1} = *n*_{2} = ⋯ = *n*_{t} = 5 and *n*_{t+1} = 4 and *D* ∩ *U* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}}. Now *D* ∩ *U* doesn’t dominate the vertices *v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t} and so {*v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t}} ⊆ *D*. Thus *D* = {*u*_{1}, *u*_{6}, ⋯, *u*_{5t+1}, *v*_{3}, *v*_{5}, *v*_{10}, ⋯, *v*_{5t}}, but *D* does not dominate two vertices *u*_{5t−2} and *u*_{5t+3}, a contradiction.

Now, assume that *γ*(*W*_{4,n}) = 2*t*+3. Let *D* be a minimum dominating set of *W*_{4,n}. Then by the Pigeonhole Principal either |*D* ∩ *U*| ≤ *t*+1 or |*D* ∩ *V*| ≤ *t*+1. Without loss of generality, suppose |*D* ∩ *U*| = *t*+1 − *a*, where *a* ≥ 0. Then |*D* ∩ *V*| = *t*+2+*a*. Observe that *D* ∩ *U* dominates at most 4*t*+4 − 4*a* vertices of *V* and, therefore, *D* dominates at most (4*t*+4 − 4*a*)+(*t*+2+*a*) = 5*t* − 3*a*+6 vertices of *V*. Since *D* dominates all vertices in *V*, we have 5*t* − 3*a*+6 ≥ 5*t*+4 and *a* = 0, |*D* ∩ *U*| = *t*+1 and |*D* ∩ *V*| = *t*+2. Also we have 4*t*+2 ≤ |*N*(*D* ∩ *U*)| ≤ 4*t*+4. Since 4*t*+2 ≤ |*N*(*D* ∩ *U*)| ≤ 4*t*+4, at most two elements of *n*_{1}, *n*_{2}, ⋯, *n*_{t+1} can be in ℳ_{4}. If *x* is the number of 5′s in the cyclic-sequence of *D* ∩ *U*, then by Observation 2.2, we have
$\begin{array}{}\stackrel{t+1}{\underset{k=1}{\mathit{\Sigma}}}{n}_{k}\end{array}$ = 5*t*+4 ≥ 1+1+8(t − *x* − 1)+5*x* and, therefore, 3*x* ≥ 3*t* − 10 which implies *x* ≥ *t* − 3. Thus *t* − 3 elements of the cyclic sequence are equal to 5. The sum of the remaining four values of the cyclic sequence is 19, and at most two of them are in ℳ_{4}. In the last case of Lemma 3.3, for *n* = 38, we identified all such cyclic sequences and placed them in two tables, and . We now continue according to and .

Table 5 *n* = 38 with one *n*_{i} in ℳ_{Δ}

Table 6 *n* = 38 with two *n*_{i} in ℳ_{Δ}

In the case (*i* = 1) in we have *n*_{1} = 4, *n*_{2} = ⋯ = *n*_{t+1} = 5 and *D* ∩ *U* = {*u*_{1}, *u*_{5}, *u*_{10}, ⋯, *u*_{5t}}. Thus {*v*_{7}, *v*_{9}, *v*_{14}, ⋯, *v*_{5t+4}} ⊆ *D*. But {*u*_{1}, *u*_{5}, *u*_{10}, ⋯, *u*_{5t}, *v*_{7}, *v*_{9}, *v*_{14}, ⋯, *v*_{5t+4}} does not dominate the vertices *u*_{3} and *u*_{5t+2}. For dominating *u*_{3} and *u*_{5t+2}, we need two vertices and therefore |*D*| ≥ 2*t*+4, a contradiction. In the cases (i ∈{2, 3}) in we have to add 5′s to the end of the cyclic sequence and construct the corresponding set *D* with 2*t*+3 elements. In both cases we obtain that *N*[*u*_{8}] ∩ *D* = ∅. Then *D* is not a dominating set, a contradiction.

In the case (i = 1) in we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic-sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 and their sum is 6 which belongs to ℳ_{4}, a contradiction.

In the case (i = 2) in we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 or 2, and their sum is 6 or 7, which belongs to ℳ_{4}, a contradiction.

In the cases (i ∈ {3, 4, 5, 6}) in we have to add 5′s to the end of the cyclic sequence and construct the corresponding set *D* with 2*t*+3 elements. In (i = 3), we obtain that *N*[*u*_{2}] ∩ *D* = ∅, in (i = 4), we obtain that *N*[{*u*_{13}, *u*_{14}}] ∩ *D* = ∅, in (i = 5), we obtain that *N*[{*u*_{2}, *u*_{8}}] ∩ *D* = ∅, and in (i = 6), we obtain that *N*[*u*_{16}] ∩ *D* = ∅. In all four cases, *D* is not a dominating set, a contradiction.

In the case (i = 7) in , by adding 5′s to the cyclic-sequence, we obtain some different new cyclic-sequences. We divide them into three categories.

c1)

*n*_{1} = 8, *n*_{2} = 3 and *n*_{3} = 5. In this category, the constructed set, *D*, does not dominate *u*_{16}, a contradiction.

c2)

*n*_{1} = 8, *n*_{2} = 5, *n*_{3} = 3 and *n*_{4} = 5. In this category, the constructed set, *D*, does not dominate *u*_{15} a contradiction.

c3)

*n*_{1} = 8, *n*_{2} = *n*_{3} = 5 and if *n*_{i} = *n*_{j} = 3, then |*i* − *j*| ≥ 2. In this category, the constructed set, *D*, does not dominate *u*_{5i+1}, a contradiction. (Notice that this category does not appear for *n* ≤ 5.) In the cases (i ∈{8, 9, 10}) in , by adding 5’s to the cyclic-sequences, we obtain some different new cyclic-sequences. we divide them into two categories.

c1)

*n*_{1} = 3 and *n*_{t+1} = 5. In this category, the constructed set, *D*, does not dominate *u*_{2}, a contradiction.

c2)

*n*_{1} = 3 and *n*_{t+1} = 6. In this category, the constructed set, *D*, does not dominate *u*_{7} and *u*_{8}.
Hence, *D* is not a dominating set, a contradiction.

Hence, *γ* (*W*_{4,n}) = 2*t*+4 = 2⌊
$\begin{array}{}\frac{n}{10}\end{array}$⌋+4 □

Now a consequence of Lemmas 3.1, 3.2, 3.33, 3.4 and 3.5 implies the following theorem which is the main result of this section.

#### Theorem 3.6

*For each integer n* ≥ 16, *we have*:
$$\begin{array}{}{\displaystyle \gamma ({W}_{4,n})=2\lfloor \frac{n}{10}\rfloor +\left\{\begin{array}{cc}0& n\equiv 0\text{\hspace{0.17em}(mod 10)}\\ 2& n=16,18,36\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}n\equiv 2,4\text{\hspace{0.17em}(mod 10)}\\ 3& n=28\phantom{\rule{thinmathspace}{0ex}};\phantom{\rule{thinmathspace}{0ex}}n\equiv 6\text{\hspace{0.17em}(mod 10)},n\ne 16,36\\ 4& n\equiv 8\text{\hspace{0.17em}(mod 10)},n\ne 18,28\end{array}\right..}\end{array}$$

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