**4. The Crossing Number of** *G<sup>∗</sup>* **+** *Dn*

A drawing *D* of *G*<sup>∗</sup> + *Dn* is said to be antipode-free if cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) ≥ 1 for any two different vertices *ti* and *tj*. In the proof of Theorem 2, the following statements related to some restricted subdrawings of the graph *G*<sup>∗</sup> + *Dn* are required.

**Lemma 1.** *Let D be a good and antipode-free drawing of G*<sup>∗</sup> + *Dn, n* > 1*, with the vertex notation of the graph G*<sup>∗</sup> *in such a way as shown in Figure 1a. For T<sup>i</sup>* ∈ *RD, let* A*<sup>k</sup>* ∈ M*<sup>D</sup> be a configuration of the corresponding subgraph F<sup>i</sup>* = *G*<sup>∗</sup> ∪ *T<sup>i</sup> for some k* ∈ {1, ... , 6}*. If there is a T<sup>j</sup>* ∈ *SD such that* cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1*, then all possible* rot*D*(*tj*) *are given in Table 2.*

**Table 2.** The corresponding rotations of *tj*, for *T<sup>i</sup>* ∈ *RD*, *F<sup>i</sup>* = *G*<sup>∗</sup> ∪ *T<sup>i</sup>* and *T<sup>j</sup>* ∈ *SD* satisfying the restriction cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1.


**Proof.** Assume the configuration A<sup>1</sup> of the subgraph *F<sup>i</sup>* = *G*<sup>∗</sup> ∪ *T<sup>i</sup>* for some *T<sup>i</sup>* ∈ *RD*, i.e., rot*D*(*ti*)=(13245). The subdrawing of *F<sup>i</sup>* induced by *D* contains just five regions with *ti* on their boundaries, see Figure 2. If there is a *T<sup>j</sup>* ∈ *SD* such that cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1, then the vertex *tj* must be placed in the region with the four vertices *v*1, *v*2, *v*4, and *v*<sup>5</sup> of *G*<sup>∗</sup> on its boundary. Besides that only the edge *v*1*v*<sup>2</sup> of *G*<sup>∗</sup> can be crossed by *tjv*3, and cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1 is fulfilling for *T<sup>j</sup>* with rot*D*(*tj*)=(14523) if *tjv*<sup>4</sup> crosses *tiv*5. The same idea also force that the rotations of the vertex *tj* are (15423), (12345), (12354), (12453), and (12543) for the remaining configurations A2, A3, A4, A5, and A<sup>6</sup> of *F<sup>i</sup>* , respectively.

**Corollary 1.** *Let D be a good and antipode-free drawing of G*<sup>∗</sup> + *Dn, for n* > 3*, with the vertex notation of the graph G*<sup>∗</sup> *in such a way as shown in Figure 1a. If T<sup>i</sup>* , *T<sup>j</sup> , and T<sup>k</sup>* ∈ *RD are three different subgraphs with* cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1*,* cr*D*(*T<sup>i</sup>* , *Tk*) = 1 *and such that F<sup>i</sup> , F<sup>j</sup> , and F<sup>k</sup> have three mutually different configurations from any of the sets* {A1, A4, A6}*,* {A2, A3, A5}*,* {A3, A2, A6}*,* {A1, A4, A5}*,* {A2, A4, A5}*, and* {A1, A3, A6}*, then*

$$\mathsf{crr}\_{D}(T^{i}\cup T^{j}\cup T^{k},T^{l})\geq\mathsf{6} \qquad\qquad\qquad\qquad\text{for any }T^{l}\in\mathbb{S}\_{D}.$$

*i.e.,*

$$\operatorname{crr}\_D(G^\* \cup T^i \cup T^j \cup T^k, T^l) \ge \mathcal{T} \qquad\qquad\qquad\text{for any } T^l \in \mathcal{S}\_D.$$

**Proof.** Let us assume the configurations A<sup>1</sup> of *F<sup>i</sup>* , A<sup>4</sup> of *F<sup>j</sup>* , and A<sup>6</sup> of *F<sup>k</sup>* with respect to the restrictions cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = cr*D*(*T<sup>i</sup>* , *Tk*) = 1 and recall that they are represented by the cyclic permutations rot*D*(*ti*)=(13245), rot*D*(*tj*)=(15432), and rot*D*(*tk*)=(13542). If there is a subgraph *T<sup>l</sup>* ∈ *SD* with cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) = 1, then the subgraph *F<sup>l</sup>* can be represented only by rot*D*(*tl*)=(14523), where the edge *tlv*<sup>3</sup> crosses *v*1*v*<sup>2</sup> of *G*<sup>∗</sup> and either *tlv*<sup>4</sup> or *tlv*<sup>5</sup> crosses corresponding edge of *T<sup>i</sup>* . Any such subgraph *T<sup>l</sup>* must cross edges of both subgraphs *T<sup>j</sup>* and *T<sup>k</sup>* at least twice because the minimum number of interchanges of adjacent elements of (14523) required to produce (15432)=(12345) and (13542)=(12453) is two. Clearly, if cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) > 2 or cr*D*(*Tk*, *T<sup>l</sup>* ) > 2, we obtain the desired result cr*D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *Tk*, *T<sup>l</sup>* ) ≥ 1 + 3 + 2 = 6. Further, if cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) = 2 and cr*D*(*Tk*, *T<sup>l</sup>* ) = 2, then the edge *tiv*<sup>5</sup> is crossed by *tlv*<sup>4</sup> in *D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *T<sup>l</sup>* ) and also *tiv*<sup>4</sup> by *tlv*<sup>5</sup> in *D*(*T<sup>i</sup>* ∪ *T<sup>k</sup>* ∪ *T<sup>l</sup>* ), respectively. However, then cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) ≥ 2, which contradicts the fact that cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) = 1 in *D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *T<sup>k</sup>* ∪ *T<sup>l</sup>* ).

If there is a *T<sup>l</sup>* ∈ *SD* with cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) = 1, then the subgraph *F<sup>l</sup>* is represented only by the cyclic permutation (12354). Using same properties as in the previous subcase, we have cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) ≥ 2 and cr*D*(*Tk*, *T<sup>l</sup>* ) ≥ 3. This in turn implies that cr*D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *Tk*, *T<sup>l</sup>* ) ≥ 2 + 1 + 3 = 6. Of course, we can apply the same idea for the case of cr*D*(*Tk*, *T<sup>l</sup>* ) = 1.

To finish the proof, let us consider a subgraph *T<sup>l</sup>* ∈ *SD* with cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) = 2, cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) = 2, and cr*D*(*Tk*, *T<sup>l</sup>* ) = 2. This enforces that the minimum number of interchanges of adjacent elements of rot*D*(*tl*) required to produce (13245)=(15423), (15432)=(12345), and (13542)=(12453) must be exactly two. However, it is not difficult to show that such cyclic permutation does not exist. Similar arguments can be applied for remaining five cases (or using the transformations Π<sup>1</sup> and Π2), and the proof is complete.

**Corollary 2.** *Let D be a good and antipode-free drawing of G*<sup>∗</sup> + *Dn, for n* > 3*, with the vertex notation of the graph G*<sup>∗</sup> *in such a way as shown in Figure 1a. If T<sup>i</sup>* , *T<sup>j</sup> , and T<sup>k</sup>* ∈ *RD are three different subgraphs such that F<sup>i</sup> , F<sup>j</sup> , and F<sup>k</sup> have three mutually different configurations from any of the sets* {A1, A3, A5} *and* {A2, A4, A6}*, then*

$$\mathsf{crr}\_{D}(T^{i}\cup T^{j}\cup T^{k},T^{l}) \geq 5\tag{10.43} \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad for \ any \ T^{l} \in \mathcal{S}\_{D,l}$$

*i.e.,*

$$\operatorname{crr}\_D(G^\* \cup T^i \cup T^j \cup T^k, T^l) \ge 6 \qquad \qquad \qquad \text{for any } T^l \in S\_D.$$

**Proof.** Let us assume the configurations A<sup>1</sup> of *F<sup>i</sup>* , A<sup>3</sup> of *F<sup>j</sup>* , and A<sup>5</sup> of *Fk*. If there is a subgraph *T<sup>l</sup>* ∈ *SD* with cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) = 1, then the subgraph *F<sup>l</sup>* can be represented only by the cyclic permutations (14523). Uniqueness of all rotations in Table 2 confirms that cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) ≥ 2 and cr*D*(*Tk*, *T<sup>l</sup>* ) ≥ 2. Hence, cr*D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *Tk*, *T<sup>l</sup>* ) ≥ 1 + 2 + 2 = 5, and the similar way can be applied for the case if cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) = 1 or cr*D*(*Tk*, *T<sup>l</sup>* ) = 1 with *T<sup>l</sup>* ∈ *SD*. It remains to consider the case where cr*D*(*T<sup>i</sup>* , *T<sup>l</sup>* ) ≥ 2, cr*D*(*T<sup>j</sup>* , *T<sup>l</sup>* ) ≥ 2, and cr*D*(*Tk*, *T<sup>l</sup>* ) ≥ 2, which yields that cr*D*(*T<sup>i</sup>* ∪ *T<sup>j</sup>* ∪ *Tk*, *T<sup>l</sup>* ) ≥ 2 + 2 + 2 = 6 clearly holds for any such *T<sup>l</sup>* , as claimed. The proof proceeds in the similar way for the second triple of configurations {A2, A4, A6}, and this completes the proof.

**Lemma 2.** cr(*G*<sup>∗</sup> + *D*2) = 1*.*

**Proof.** If we consider the configurations A<sup>2</sup> of *F<sup>i</sup>* and A<sup>3</sup> of *F<sup>j</sup>* , then one can easily find a subdrawing of *T<sup>i</sup>* ∪ *T<sup>j</sup>* in which cr*D*(*T<sup>i</sup>* , *T<sup>j</sup>* ) = 1, i.e., cr(*G*<sup>∗</sup> + *D*2) ≤ 1. The graph *G*<sup>∗</sup> + *D*<sup>2</sup> contains a subgraph that is a subdivision of the complete graph *K*<sup>5</sup> and it is well-known by Guy [32] that cr(*K*5) = 1. As cr(*G*<sup>∗</sup> + *D*2) ≥ cr(*K*5) = 1, the proof of Lemma 2 is complete.

**Theorem 2.** cr(*G*<sup>∗</sup> + *D*1) = 0 *and* cr(*G*<sup>∗</sup> + *Dn*) = *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . *for n* ≥ 2*, i.e.,* cr(*G*<sup>∗</sup> + *Dn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . *for n even and* cr(*G*<sup>∗</sup> + *Dn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . − 1 *for n odd at least 3.*

**Proof.** The graph *G*<sup>∗</sup> + *D*<sup>1</sup> is planar, hence cr(*G*<sup>∗</sup> + *D*1) = 0. For *n* ≥ 2, both special drawings in Figures 4 and 5 produce *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . crossings, and so cr(*G*<sup>∗</sup> + *Dn*) ≤ *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . . The opposite inequality can be proved by induction on *n*, and the result holds for *n* = 2 by Lemma 2. For some *n* ≥ 3, suppose a drawing *D* of *G*<sup>∗</sup> + *Dn* with

$$\text{crr}\_D(G^\* + D\_n) < n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor \tag{3}$$

and that

$$\text{crr}(G^\* + D\_m) = m^2 - 2m + \left\lfloor \frac{m}{2} \right\rfloor \quad \text{for any integer } 2 \le m < n. \tag{4}$$

**Figure 4.** The good drawing of *G*<sup>∗</sup> + *Dn* with 4- *<sup>n</sup>* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . crossings for *n* even, *n* ≥ 2.

**Figure 5.** The good drawing of *G*<sup>∗</sup> + *Dn* with 4- *<sup>n</sup>* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . − 1 crossings for *n* odd, *n* ≥ 3, where three subgraphs *T*1, *T*2, and *T<sup>n</sup>* are fixed.

Let us first show that *D* must be antipode-free. Suppose that, without loss of generality, cr*D*(*Tn*<sup>−</sup>1, *Tn*) = 0. If at least one of *Tn*−<sup>1</sup> and *Tn*, say *Tn*, does not cross *G*∗, it is not difficult to verify in Figure 1 that {*Tn*<sup>−</sup>1, *Tn*} ⊆ *RD*, i.e., cr*D*(*G*∗, *Tn*−<sup>1</sup> ∪ *Tn*) ≥ 1. By (1), we already know that cr*D*(*K*5,3) ≥ 4, which yields that edges of the subgraph *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>* must be crossed at least four times by each other *Tk*. So, by fixing the subgraph *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>* in *D*, we have

$$\text{crr}\_D(G^\* + D\_{n-2}) + \text{crr}\_D(T^{n-1} \cup T^n) + \text{crr}\_D(K\_{5,n-2}, T^{n-1} \cup T^n) + \text{crr}\_D(G^\*, T^{n-1} \cup T^n)$$

$$\geq (n-2)^2 - 2(n-2) + \left\lfloor \frac{n-2}{2} \right\rfloor + 0 + 4(n-2) + 1 = n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor$$

The obtained crossing number contradicts the assumption (3) and confirms that the considered drawing *D* is antipode-free. For easier reading, if *r* = |*RD*| and *s* = |*SD*|, then again (3) together with cr*D*(*K*5,*n*) ≥ 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . using (1) imply the following inequality with respect to possible edge crossings of *G*<sup>∗</sup> in *D*:

$$\text{crr}\_D(\mathbf{G}^\*) + s + 2(n - r - s) < \left\lfloor \frac{n}{2} \right\rfloor. \tag{5}$$

The inequality (5) forces more than / *<sup>n</sup>* 2 0 subgraphs *T<sup>i</sup>* by which edges of *G*<sup>∗</sup> are not crossed, that is, *r* ≥ / *<sup>n</sup>* 2 0 + 1 ≥ 3 and *s* < - *<sup>n</sup>* 2 . . Of course, if *n* is odd then previous inequalities could be strengthened, but this is not necessary in the following process of obtaining a contradiction with number of crossings in *D*. Moreover, if *n* = 3 then *r* = 3, and cr*D*(*G*<sup>∗</sup> + *D*3) ≥ cr(*K*5,3) = 4 with the assumption (3) enforce *n* at least four.

Case 1: cr*D*(*G*∗) = 0 and choose the vertex notation of the graph *G*<sup>∗</sup> in such a way as shown in Figure 1a. In this case, we deal with configurations from the nonempty set M*D*. As the set *RD* is nonempty, recall that

$$\sum\_{i \in I\_{\mathcal{S}}} \alpha\_i + \sum\_{i \in I\_{\mathcal{S}}} \alpha\_i = r \ge 3\dots$$

Let us first suppose that either *α*<sup>1</sup> + *α*<sup>3</sup> + *α*<sup>5</sup> = 0 or *α*<sup>2</sup> + *α*<sup>4</sup> + *α*<sup>6</sup> = 0. For the rest of the proof we may therefore assume that *α*<sup>2</sup> + *α*<sup>4</sup> + *α*<sup>6</sup> = 0, that is, *α*<sup>1</sup> + *α*<sup>3</sup> + *α*<sup>5</sup> > 0. Since G*<sup>D</sup>* is the subgraph of G induced by *VD* with respect to weights 2 of all its edges (without possible loops), three possible subcases presented in Figure 6 may occur:

**Figure 6.** Three possible components of the graph G*<sup>D</sup>* if *α*<sup>2</sup> = *α*<sup>4</sup> = *α*<sup>6</sup> = 0. (**a**): *α<sup>i</sup>* > 0 for each *i* ∈ *Io*; (**b**): *α<sup>i</sup>* > 0 and *α<sup>j</sup>* > 0 for exactly two different *i*, *j* ∈ *Io*; (**c**): *α<sup>i</sup>* > 0 for only one *i* ∈ *Io*.

(a) *α<sup>i</sup>* > 0 for each *i* ∈ *Io*. Let us assume three subgraphs *Tn*<sup>−</sup>2, *Tn*<sup>−</sup>1, *T<sup>n</sup>* ∈ *RD* such that *Fn*<sup>−</sup>2, *Fn*−<sup>1</sup> and *F<sup>n</sup>* have three mutually different configurations from the set M*<sup>D</sup>* = {A1, A3, A5}. Then, cr*D*(*Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *Tn*, *T<sup>i</sup>* ) ≥ 4 + 2 + 2 = 8 holds for any other *T<sup>i</sup>* ∈ *RD* by summing values in corresponding three rows of Table 1, and cr*D*(*G*<sup>∗</sup> ∪ *Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *Tn*, *T<sup>i</sup>* ) ≥ 6 is true by Corollary 2 for any *T<sup>i</sup>* ∈ *SD*. Then, by fixing the graph *G*<sup>∗</sup> ∪ *Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>*

$$\text{crr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor + 8(r-3) + 6s + 7(n-r-s) + 6s$$

$$= 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor + 7n + r - s - 18 \ge 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor$$

$$+ 7n + \left( \left\lceil \frac{n}{2} \right\rceil + 1 \right) + \left( 1 - \left\lfloor \frac{n}{2} \right\rfloor \right) - 18 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

(b) Assuming that *α<sup>i</sup>* > 0 for exactly two *i* ∈ *Io*, without lost of generality, let us consider two different subgraphs *Tn*<sup>−</sup>1, *T<sup>n</sup>* ∈ *RD* such that *Fn*−<sup>1</sup> and *F<sup>n</sup>* have configurations A<sup>1</sup> and A3, respectively. As M*<sup>D</sup>* = {A1, A3}, we have cr*D*(*Tn*−<sup>1</sup> ∪ *Tn*, *T<sup>i</sup>* ) ≥ 4 + 2 = 6 for any *T<sup>i</sup>* ∈ *RD*, *i* = *n* − 1, *n*. Therewith, the antipode-free property of *D* forces that, cr*D*(*Tn*−<sup>1</sup> ∪ *Tn*, *T<sup>i</sup>* ) ≥ 2 trivially holds for any subgraph *T<sup>i</sup>* with *i* = *n* − 1, *n*. Hence, by fixing the graph *G*<sup>∗</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>*

$$\operatorname{cr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 6(r-2) + 3s + 4(n - r - s) + 2$$

$$= 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 4n + 2r - s - 10 \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor$$

$$+ 4n + 2\left( \left\lceil \frac{n}{2} \right\rceil + 1 \right) + \left( 1 - \left\lfloor \frac{n}{2} \right\rfloor \right) - 10 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

(c) *α<sup>i</sup>* > 0 for only one *i* ∈ *Io*. As M*<sup>D</sup>* = {A*i*}, in the rest of the paper, we may consider *T<sup>n</sup>* ∈ *RD* with the configuration A<sup>1</sup> of *Fn*. Then edges of each other subgraph *T<sup>j</sup>* ∈ *RD* cross at least four times edges of *T<sup>n</sup>* provided by rot*D*(*tn*) = rot*D*(*tj*). Thus, by fixing the graph *G*<sup>∗</sup> ∪ *T<sup>n</sup>*

$$\begin{split} \operatorname{cr}\_D(G^\* + D\_n) &\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4(r-1) + 2s + 3(n - r - s) + 0 \\ &= 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 3n + r - s - 4 \geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor \\ &+ 3n + \left( \left\lceil \frac{n}{2} \right\rceil + 1 \right) + \left( 1 - \left\lfloor \frac{n}{2} \right\rfloor \right) - 4 \geq n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor. \end{split}$$

All three subcases contradict the assumption (3). In addition, let us suppose that *α*<sup>1</sup> + *α*<sup>3</sup> + *α*<sup>5</sup> > 0 and *α*<sup>2</sup> + *α*<sup>4</sup> + *α*<sup>6</sup> > 0. Remark that the subgraph G*<sup>D</sup>* can be either connected (consisting of a single component) or also disconnected with several components. Now, we are able to discuss over remaining possible components of G*<sup>D</sup>* in the following subcases:

1. There are no two adjacent edges with weights 1 in the subgraph G*D*, that is, there are four possibilities presented in Figure 7.

**Figure 7.** Four possible components of the subgraph G*<sup>D</sup>* in which there are no two adjacent edges with weights 1. Green, blue, brown, and black correspond to the values 1, 2, 3, and 4, respectively. (**a**): the complete graph *K*<sup>3</sup> with edge weights 1, 2, and 3; (**b**): the complete graph *K*<sup>4</sup> with edge weights 1, 1, 2, 2, 3, and 3; (**c**): the complete graph *K*<sup>2</sup> with edge weight 1; (**d**): the complete graph *K*<sup>2</sup> with edge weight 3.

• *wD*(*aiaj*) = 1 for some *i* ∈ *Io*, *j* ∈ *Ie*, i.e., there are three cases mentioned in Figure 7a–c. Let us consider two subgraphs *Tn*<sup>−</sup>1, *T<sup>n</sup>* ∈ *RD* such that *Fn*<sup>−</sup>1, *F<sup>n</sup>* have different configurations from {A*i*, A*j*}, where *i*, *j* are associated indexes. Using weights of edges in the considered component of G*D*, one can easily verify that edges of the graph *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>* are crossed at least five times by edges of any another subgraph *T<sup>k</sup>* ∈ *RD*. Moreover, since the minimum number of interchanges of adjacent elements of rot*D*(*tn*) required to produce rot*D*(*tn*−1) is three, any subgraph *T<sup>k</sup>* with *k* = *n* − 1, *n* crosses edges of *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>* at least thrice. Thus, by fixing the graph *G*<sup>∗</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>*

$$\operatorname{cr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 5(r-2) + 4s + 5(n-r-s) + 1$$

$$= 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 5n - s - 9 \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor$$

$$+ 5n + \left(1 - \left\lfloor \frac{n}{2} \right\rfloor \right) - 9 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

• *wD*(*aiaj*) > 1 for all *i* ∈ *Io*, *j* ∈ *Ie*, i.e., there is only one case mentioned in Figure 7d. Let us again consider two subgraphs *Tn*<sup>−</sup>1, *T<sup>n</sup>* ∈ *RD* such that *Fn*<sup>−</sup>1, *F<sup>n</sup>* have different configurations from {A*i*, A*j*}, where *i*, *j* are associated indexes. Then, cr*D*(*Tn*−<sup>1</sup> ∪ *Tn*, *Tk*) ≥ 7 holds by summing edge-weights 4 and 3 for any other *T<sup>k</sup>* ∈ *RD*. Hence, by fixing the graph *G*<sup>∗</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>*

$$\operatorname{cr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 7(r-2) + 3s + 4(n - r - s) + 3$$

$$= 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor + 4n + 3r - s - 11 \ge 4\left\lfloor \frac{n-2}{2} \right\rfloor \left\lfloor \frac{n-3}{2} \right\rfloor$$

$$+ 4n + 3\left(\left\lceil \frac{n}{2} \right\rceil + 1\right) + 1 - \left\lfloor \frac{n}{2} \right\rfloor - 11 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

Both discussed cases again confirm a contradiction with (3) in *D*, and so, suppose that there are two adjacent edges with weights 1 in the subgraph G*D*. Further, only in the case if the number *β<sup>j</sup>* is defined, we claim that the following two properties (6) and (7) must be also fulfilled in *D*:

$$
\beta\_j + \sum\_{i \in I\_\theta} a\_i > \left\lfloor \frac{n}{2} \right\rfloor \qquad \text{for some } j \in I\_{o\prime} \tag{6}
$$

$$
\beta\_j + \sum\_{i \in I\_\varepsilon} a\_i > \left\lfloor \frac{n}{2} \right\rfloor \qquad \text{for some } j \in I\_\varepsilon. \tag{7}
$$

For a contradiction, suppose, without loss of generality, that *β*<sup>1</sup> + *α*<sup>1</sup> + *α*<sup>3</sup> + *α*<sup>5</sup> ≤ - *<sup>n</sup>* 2 . , that is, −*α*<sup>1</sup> − *α*<sup>3</sup> − *α*<sup>5</sup> − *β*<sup>1</sup> ≥ −- *<sup>n</sup>* 2 . . In this case, from the definition of *β*1, we have *α*<sup>1</sup> > 0, *α*<sup>4</sup> > 0, and *α*<sup>6</sup> > 0. Thus, in the rest of the paper, let us consider three subgraphs *Tn*<sup>−</sup>2, *Tn*<sup>−</sup>1, *T<sup>n</sup>* ∈ *RD* such that *Fn*<sup>−</sup>2, *Fn*<sup>−</sup>1, and *F<sup>n</sup>* have configurations A1, A4, and A6, respectively. Using values in Table 1, one can easily verify that edges of the graph *Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>* are crossed at least six times and seven times by edges of any another subgraph *T<sup>i</sup>* ∈ *RD* with the configuration A1, A3, A<sup>5</sup> and A2, A4, A<sup>6</sup> of *F<sup>i</sup>* (of course, if A*<sup>k</sup>* ∈ M*<sup>D</sup>* for some *k* ∈ *Io* ∪ *Ie* in *D*), respectively. However, from Corollary 1 we get that cr*D*(*G*<sup>∗</sup> ∪ *Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *Tn*, *T<sup>i</sup>* ) ≥ 7 holds for any *T<sup>i</sup>* ∈ *SD* provided by we can also assume that cr*D*(*Tn*<sup>−</sup>2, *Tn*−1) = 1 and cr*D*(*Tn*<sup>−</sup>2, *Tn*) = 1 due to the congruence property (If rot*D*(*tx*) and rot*D*(*ty*) are two cyclic permutations of odd length, and *Q*(rot*D*(*tx*),rot*D*(*ty*)) denotes the minimum number of interchanges of adjacent elements of rot*D*(*tx*) required to produce the inverse cyclic permutation of rot*D*(*ty*), then cr*D*(*Tx*, *Ty*) = *Q*(rot*D*(*tx*),rot*D*(*ty*)) + 2*z* for some nonnegative integer *z*, for more see Woodall [31]). Hence, by fixing the graph *G*<sup>∗</sup> ∪ *Tn*−<sup>2</sup> ∪ *Tn*−<sup>1</sup> ∪ *T<sup>n</sup>*

$$\begin{split} \text{tr}\_D(G^\* + D\_n) &\geq 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor + 6(a\_1 + a\_3 + a\_5 - 1) + 7(a\_2 + a\_4 + a\_6 - 2) + 7s \\ &+ 6\beta\_1 + 7(n - r - s - \beta\_1) + 4 = 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor + 7n - a\_1 - a\_3 - a\_5 - \beta\_1 - 16 \\ &\geq 4\left\lfloor \frac{n-3}{2} \right\rfloor \left\lfloor \frac{n-4}{2} \right\rfloor + 7n - \left\lfloor \frac{n}{2} \right\rfloor - 16 \geq n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor. \end{split}$$

The obtained crossing number also contradicts the assumption (3) of *D* and confirms that both parity properties (6) and (7) must be fulfilled in *D*.

	- (a) Let the graph G*<sup>D</sup>* consist of one component in such a way as shown in Figure 8a. Without lost of generality, let us assume that *a*2, *a*3, *a*<sup>6</sup> are vertices of the considered path on three vertices with weight 1 of both edges. In this case, it is obvious that *α*<sup>2</sup> + *α*<sup>3</sup> + *α*<sup>6</sup> = *r*. Since the number *β*<sup>3</sup> can be defined, the property (6) forces *β*<sup>3</sup> + *α*<sup>3</sup> > - *<sup>n</sup>* 2 . . Further, let us also assume that *T<sup>n</sup>* ∈ *RD* with the configuration A<sup>3</sup> of *Fn*. Then, by fixing the graph *G*<sup>∗</sup> ∪ *T<sup>n</sup>*

$$\begin{split} \text{cr}\_D(G^\* + D\_n) &\geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4(a\_3 - 1) + 1(a\_2 + a\_6) + 4\beta\_3 + 2s \\ &+ 3(n - r - s - \beta\_3) = 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 3n + (a\_3 + \beta\_3 - a\_2 - a\_6) \\ &- (s + a\_2 + a\_6) - 4 \geq 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 3n + 0 - \left\lfloor \frac{n}{2} \right\rfloor - 4 \geq n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor. \end{split}$$

(b) Let the graph G*<sup>D</sup>* consist of one component in such a way as shown in Figure 8b. Without lost of generality, let us assume that *a*2, *a*3, *a*<sup>6</sup> are vertices of the considered path on three vertices with weight 1 of both edges and let *a*2, *a*4, *a*<sup>6</sup> be vertices of the 3-cycle with respect to weight 2 of all its edges. In this case, it is obvious that *α*<sup>2</sup> + *α*<sup>3</sup> + *α*<sup>4</sup> + *α*<sup>6</sup> = *r*. The property (6) enforces again *β*<sup>3</sup> + *α*<sup>3</sup> > - *<sup>n</sup>* 2 . because the number *β*<sup>3</sup> can be defined. Further, if *T<sup>n</sup>* ∈ *RD* is assumed with the configuration A<sup>4</sup> of *Fn*, then by fixing the graph *G*<sup>∗</sup> ∪ *T<sup>n</sup>*

$$\text{crr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 3(\alpha\_3 + \beta\_3) + 2(n - \alpha\_3 - \beta\_3 - 1).$$

$$\ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 2n + \left( \left\lfloor \frac{n}{2} \right\rfloor + 1 \right) - 2 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

(c) Let the graph G*<sup>D</sup>* consist of one component in such a way as shown in Figure 8c–e. Let us take a maximal path *Pk* on *k* vertices as the subgraph of G*<sup>D</sup>* with weights 1 on all its edges. If *ai* and *aj* are two inner vertices of *Pk* with *i* + 1 ≡ *j* (mod 2) for which the numbers *β<sup>i</sup>* and *β<sup>j</sup>* satisfy the parity properties (6) and (7), then addition of both inequalities thus obtained enforces a contradiction

$$m \ge \beta\_i + \beta\_j + r \ge 2\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right).$$

The obtained contradictions in all three cases complete the proof for the planar subdrawing of *G*<sup>∗</sup> induced by *D* given in Figure 1a.

Case 2. cr*D*(*G*∗) = 2 and choose the vertex notation of the graph *G*<sup>∗</sup> presented as in Figure 1b. Since the set *RD* is nonempty and there is only one subdrawing of a subgraph *F<sup>i</sup>* = *G*<sup>∗</sup> ∪ *T<sup>i</sup>* for all *T<sup>i</sup>* ∈ *RD* represented by the rotation (13524), the subgraph *T<sup>i</sup>* is crossed at least four times by edges of each subgraph *T<sup>j</sup>* ∈ *RD* with *j* = *i*. Hence, by fixing the graph *G*<sup>∗</sup> ∪ *T<sup>i</sup>*

$$\operatorname{crr}\_D(G^\* + D\_n) \ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 4(r-1) + 2(n-r) + 2 = 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor$$

$$+ 2n + 2r - 2 \ge 4\left\lfloor \frac{n-1}{2} \right\rfloor \left\lfloor \frac{n-2}{2} \right\rfloor + 2n + 2\left(\left\lceil \frac{n}{2} \right\rceil + 1\right) - 2 \ge n^2 - 2n + \left\lfloor \frac{n}{2} \right\rfloor.$$

For all these mentioned cases, it turned out that there is no drawing of the graph *G*<sup>∗</sup> + *Dn* with fewer than *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . crossings, and the proof of Theorem 2 is complete.

**Figure 8.** Five possible components of the subgraph G*<sup>D</sup>* in which there are two adjacent edges with weights 1. Green, blue, brown, and black correspond to the values 1, 2, 3, and 4, respectively. (**a**): the complete graph *K*<sup>3</sup> with edge weights 1, 1, and 2; (**b**): the complete graph *K*<sup>4</sup> with edge weights 1, 1, 2, 2, 2, and 3; (**c**): the complete graph *K*<sup>4</sup> with edge weights 1, 1, 1, 2, 2, and 3; (**d**): the complete graph *K*5; (**e**): the complete graph *K*6.

#### **5. Conclusions**

Into both drawings in Figures 4 and 5, we could add *n* − 1 or *n* edges forming paths *Pn*, *n* ≥ 2 or cycles *Cn*, *n* ≥ 3 on vertices of *Dn* with no crossing, respectively. Thus, the following surprising results are obvious.

**Corollary 3.** cr(*G*<sup>∗</sup> + *Pn*) = *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . *for n* ≥ 2*, i.e.,* cr(*G*<sup>∗</sup> + *Pn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . *for n even and* cr(*G*<sup>∗</sup> + *Pn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . − 1 *for n odd.*

**Corollary 4.** cr(*G*<sup>∗</sup> + *Cn*) = *n*<sup>2</sup> − 2*n* + - *<sup>n</sup>* 2 . *for n* ≥ 3*, i.e.,* cr(*G*<sup>∗</sup> + *Cn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . *for n even and* cr(*G*<sup>∗</sup> + *Cn*) = 4 - *n* 2 .- *<sup>n</sup>*−<sup>1</sup> 2 . + - *<sup>n</sup>* 2 . − 1 *for n odd.*

These results extend already known results of join products of graphs on at most six vertices with paths and cycles, see [2,5,18,20,26,33–41].

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The author declares no conflict of interest.

#### **References**

