On Critical Unicyclic Graphs with Cutwidth Four

: The cutwidth minimization problem consists of ﬁnding an arrangement of the vertices of a graph G on a line P n with n = | V ( G ) | vertices in such a way that the maximum number of overlapping edges (i.e., the congestion) is minimized. A graph G with a cutwidth of k is k -cutwidth critical if every proper subgraph of G has a cutwidth less than k and G is homeomorphically minimal. In this paper, we ﬁrst veriﬁed some structural properties of k -cutwidth critical unicyclic graphs with k > 1. We then mainly investigated the critical unicyclic graph set T with a cutwidth of four that contains ﬁfty elements, and obtained a forbidden subgraph characterization of 3-cutwidth unicyclic graphs.


Introduction
All graphs in this paper are finite, simple, and connected, with undefined notation following [1].The cutwidth minimization problem consists of finding an arrangement of the vertices of a graph G on a path P n with n = |V(G)| vertices in such a way that the maximum number of overlapping edges (i.e., the congestion) is minimized.As one of the most well-known optimization problems, it is also known as the minimum cut linear arrangement (or linear layout, optimal embedding, optimal labeling, etc.) problem [2].Cutwidth has been extensively examined [2].Computing cutwidth for general graphs is an NP-complete problem except for trees [3][4][5][6], and it remains NP-complete even if the input graph G is restricted to planar graphs with a maximum degree of three [7].Hence, a number of studies have focused on polynomial-time approximation algorithms for general graphs and polynomial-time algorithms for some special graphs to solve their cutwidth [2,8].Relatively little work has been conducted on detecting special graph classes whose cutwidths can be computed polynomially [2] and critical graph classes with cutwidths of k ≥ 1.Let T k ( * ) be the set of critical graphs with the graph parameter * = k.From [9], |T 1 (c(G))| = 1, |T 2 (c(G))| = 2, |T 3 (c(G))| = 5 (see Figure 1).For critical graphs with cutwidth k ≥ 4, |T k (c(G))| has been unknown except that |T 4 (c(T))| = 18, as reported by [10], where T is a tree (see Figure 2).Similar studies have been conducted for the treewidth, pathwidth, and branchwidth of a graph G (abbreviated by tw(G), pw(G), and bw(G), respectively).A graph G is said to be k-treewidth (pathwidth, branchwidth) critical if tw(G) (pw(G), bw(G)) = k but tw(G ) (pw(G ), bw(G )) < k for any minor G of G. From [11][12][13] [14], the critical graphs for parameters with a similar nature are worthy of further study, and the number of these critical graphs for a given value of the parameter would be finite and have yet to be characterized.The cutwidth problem for graphs and a class of optimal embedding (or layout) problems have significant applications in VLSI layouts [15,16], network reliability [17], automatic graph drawing [18], information retrieval [19], urban drainage network design [20], and other domains.In particular, the cutwidth is related to a basic parameter, called the congestion, in designing microchip circuits [2,21,22].Herein, a graph G may be viewed as a model of the wiring diagram of an electronic circuit with the vertices representing components and the edges representing wires connecting them.When a circuit is laid out in a certain architecture (i.e., a path P n ), the maximum number of overlap wires is the congestion, which is one of major parameters in the determination of electronic performance.This motivates the enthusiasm for studying the cutwidth problem in graph theory practically.Theoretically, it appears to be closely related with other graph parameters such as pathwidth, treewidth, linear width, bandwidth, and modified bandwidth [2,8,23,24], among others.For instance, for any graph G with vertices of degree bounded by an integer d ≥ 1, pw(G) ≤ c(G) ≤ d • pw(G), where c(G) and pw(G) are cutwidth value and pathwidth value, respectively.In this paper, we mainly study the critical unicyclic graph set T with a cutwidth of four that contains fifty elements.
For an integer n > 0, define S n = {1, 2, ..., n}.A labeling of a graph G = (V(G), E(G)) with |V(G)| = n is a bijection f : V(G) → S n , viewed as an embedding of G into a path P n with vertices in S n , where consecutive integers are the adjacent vertices.The cutwidth of G with respect to f is which is also the congestion of the labeling.The cutwidth of G is defined by where the minimum is taken over all labelings f .If k = c(G, f ), then f , and the embedding induced by f is called a k-cutwidth embedding of G.A labeling f attaining the minimum in (2) is an optimal labeling.For each i with i ∈ S n , let u i = f −1 (i) and S j = {u 1 , u 2 , ..., u j }.
Define ∇ f (S j ) = {u i u h ∈ E : i ≤ j < h} is called the (edge) cut at [j, j + 1] with respect to f .From (2), we then have c(G, f ) = max 1≤j<n |∇ f (S j )|. ( An f -max cut of G is a ∇ f (S j ), achieving the maximum in (3).For a graph G and integer i ≥ 0, let and the graph obtained from G − v by adding a new edge v 1 v 2 , is called a series reduction of G.A graph G is homeomorphically minimal if G does not have any series reductions.Two graphs G and G are homeomorphic if both of them can be obtained from the same graph G by inserting new vertices of degree two into its edges.A graph G is said to be k-cutwidth critical if G is homeomorphically minimal with c(G) = k such that every proper subgraph G of G satisfies c(G ) < k.The basic properties of cutwidth follow immediately from this definition.
¡ ¡ e e ¡ ¡ e e r r r r r r £ £ g g £ £ g g g g £ £ g g £ £ £ £ g g £ £ g g r r r r r r r r r r r r £ g g r r r r r r r r r r r r r r r r r r r r £ g g r r r r r r r r r r r r r r r r r r r £ £ g g g g £ £ g g £ £ r r r r r r r r r r The following observation is immediate from Lemma 1.1, Definition 2.1 (i) Let r ≥ 0 be an integer, and v be a vertex of graph .., v r ) to be the component of G − {vv 1 , vv 2 , ..., vv r } that contains v (see an illustration in Fig. 4(a)).
(ii) Let G, H be two disjoint graphs with u ∈ V (G) and v ∈ V (H).To identify u and v, denoted as G ⊕ u,v H, is to replace u, v by a single vertex z (i.e., u = v = z) incident to all the edges which were incident to u and v, where z is called the identified vertex.
as the graph obtained from the disjoint union G 1 , G 2 , G 3 and K 1,3 by identifying u j with v j (again denoted as v j ) for each j ∈ S 3 (see Fig. 4(b)).
} to be the family of all proper maximal subgraphs of G.
be an optimal labeling of H i , and let a labeling f : V (G) → S n of G be as follows: for v ∈ V (G), Then the labeling f is called a labeling by the order  Lemma 1.For graphs G and G , each of the following holds.
The purpose of this paper is to characterize critical unicyclic graphs with a cutwidth of four and to present a forbidden subgraph characterization for unicyclic graphs with a cutwidth of three.Let T = {τ i : 1 ≤ i ≤ 50} be the collection of the critical unicyclic graphs depicted in Figure 3.The main results of this paper are the following: Corollary 1.A unicyclic graph G has a cutwidth of three if and only if it does not contain any subgraph homeomorphic to any member in T .
The rest of this paper is as follows.Section 2 presents some preliminary results.The proof of Theorem 1 is given in Section 3 by a series of lemmas.We give a short remark in Section 4.

Appendix
¢ s e es

Preliminary Results
Throughout this section, for any integer n > 1, we always use P n to denote the path with V(P n ) = S n such that for all 1 ≤ i < n, i and i + 1 are adjacent vertices in P n .In addition, because K 1,2k−1 is k-cutwidth critical, as demonstrated by [10], we can let The following observation is immediate from Lemma 1: Definition 1.
(ii) Let G, H be two disjoint graphs with u ∈ V(G) and v ∈ V(H).To identify u and v, denoted as G ⊕ u,v H, is to replace u, v with a single vertex z (i.e., u = v = z) incident to all the edges which are incident to u and v, where z is called the identified vertex. (iii as the graph obtained from the disjoint union G 1 , G 2 , G 3 and K 1,3 by identifying u j with v j (again denoted as v j ) for each j ∈ S 3 (see Figure 4b).
be an optimal labeling of H i , and let a labeling f : V(G) → S n of G be as follows: for Then the labeling f is called a labeling by the order Fig. 4. Illustrations of Definition 2.1(i) and (iii) Proof We first give a Claim.
Then there exists an optimal labeling f * such that the vertices in each of V 1 and V 2 are labeled consecutively.
In fact, if f is an optimal labeling of G with f (v 1 ) < f (v 2 ), then we can construct a labeling f * as follows.First label the vertices of V 1 in the same order as f , and then label the vertice of V 2 in the same order as f .Since the edges in G[V 1 ] and those in G[V 2 ] are not overlapped, i follows that c(G, f * ) ≤ c(G, f ).Thus f * is also an optimal labeling of G. Now, by using this observation, we proceed to prove Theorem 2.3.By the assumption tha Then there exists an optimal labeling f * such that the vertices o each of Proof.We first provide a claim.
Then, there exists an optimal labeling f * such that the vertices in each of V 1 and V 2 are labeled consecutively.
In fact, if f is an optimal labeling of G with f (v 1 ) < f (v 2 ), then we can construct a labeling f * as follows.First, label the vertices of V 1 in the same order as f , and then label the vertices of V 2 in the same order as f .Because the edges in ).Thus, f * is also an optimal labeling of G. Now, by using this observation, we proceed to prove Theorem 2. From the assumption that vv 1 , vv 2 are cut edges of G, let V 0 , V 1 , V 2 be the vertex sets of three components of Then, there exists an optimal labeling f * such that each of the vertices of V 0 , V 1 , V 2 are labeled consecutively.If c(G, f * ) ≤ k, then, because the edges vv 1 and vv 2 give a congestion of one to then there must be two cut edges v v 1 , v v 2 in G by assumption.In this way, V(G) can be further decomposed into a sequence Corollary 2. With the notation in Theorem 2, for graph G, if there exists a vertex v ∈ D ≥3 (G) such that c(G(v; v i , v j )) ≥ k − 1 holds for any v i , v j ∈ N G (v), then c(G) ≥k, where vv i , vv j are cut edges in G. Lemma 4. Let graph G be k-cutwidth critical with D 1 (G) = ∅, and P l = u 0 u 1 ...u l be a path with length l.Then, c(G ⊕ v 0 ,u 0 P l ) = k for v 0 ∈ V(G).
Proof.Let v 0 z be a pendant edge of G, where z is a pendant vertex.We subdivide the edge v 0 z into a path P with length l and denote the resulting graph with G .Then, by Lemma 1, (5), an optimal labeling f : V(G) → S n obtained by the order (2).On the other hand, it is routine to verify that c(G) ≥ k using Corollary 2 because c(G(u 0 ; v i , v j )) = k − 1 for any v i , v j ∈ N G (u 0 ).Thus, c(G) = k, and the proof is complete.

Corollary 3. With the notation of Definition
for each j ∈ S 3 , three series reductions are carried out first.Furthermore, we still let First, by assumption and Theorem 3, c(G) = k.Second, we prove that G is k-cutwidth critical.It remains to be shown that, for any G ∈M(G), c(G ) ≤ k − 1.Because any G is obtained by deleting a pendent edge xy or an nonpendent edge xy ∈ (2), and G is k-cutwidth critical.

Proof of Theorem 1
We verify our main results by using a series of lemmas.Throughout this section, G denotes an unicyclic graph and C t = v 1 v 2 ...v t v 1 denotes the unique cycle of G with t ≥ 3. Furthermore, we have a convention that a graph H is designated to be homeomorphic to a subgraph of G if H can be obtained by deleting vertices or edges and some series reductions of G and c(H) = c(G).Because a cutwidth critical graph is homeomorphically minimal, if 5a), and F 2 = {K 1,5 ⊕ u 0 ,u 0 K 1,5 , K 1,5 ⊕ u 0 ,x 0 F 1 , F 1 ⊕ x 0 ,x 0 F 1 } (see Figure 5b), where K 1,5 , F 1 are copies of K 1,5 and F 1 , u 0 ∈ D 1 (K 1,5 ), x 0 ∈ D 3 (F 1 ) are copies of u 0 and x 0 , respectively.For an integer p > 1, we call a star K s s s s s s s s s s s s , then the series reduction can be implemented without effecting c(G) = k.Since u 0 v 2 is a pendent edge of the subgraph (2).On the other hand, it is routine to verify that c(G) ≥ k by Corollary 2.4, since c(G(u 0 ; v i , v j )) = k − 1 for any v i , v j ∈ N G (u 0 ).Thus c(G) = k, and the proof is complete.2 Corollary 2.7 With the notation of Definition 2.1(iii), for each for each j ∈ S 3 , three series reductions are carried out first.And we still let First, by assumption and Theorem 2.6, c(G) = k.
Second, we prove that G is k-cutwidth critical.It remains to show that, for any G ∈M(G), c(G ) ≤ k − 1.Since any G is obtained by deleting a pendent edge xy or an nonpendent edge 3 Proof of Theorem 1.4 We verify our main results by using a series of lemmas.Throughout this section, G denotes an unicyclic graph and C t = v 1 v 2 ...v t v 1 denotes the unique cycle of G with t ≥ 3.And we have a convention that a graph H is called to be homeomorphic to a subgraph of G if H can be obtained by deleting vertices or edges and some series reductions of G and c(H unless v i is a special vertex.Lemma 5.Each member of set T is 4-cutwidth critical in Figure 3.
Proof.For a unicyclic graph G, let is a forest of t subtrees T 1 , T 2 , ..., T t where T i is called the v i -branch leading from v i .We first consider the case of t = 3 in which G − E(C 3 ) has three subtrees T 1 , T 2 , T 3 .For an optimal labeling f of G, suppose that f (v 1 ) < f (v 2 ) < f (v 3 ); then, the number set S n is divided into three intervals are then embedded into I 1 , I 2 , I 3 in different manners.We have the following classifications of 4-cutwidth critical unicyclic graphs.
(1) Type 3A (including τ 1 to τ 4 ): T 1 is embedded in I 1 with a congestion of three, T 2 is embedded in I 2 with a congestion of four, and T 3 is embedded in I 3 with a congestion of three.Herein, T 1 and T 3 are the star K 1,3 with center v i or the two stars K 1,3 with an identifying leaf at v i (i=1,3) (see F 2 in Figure 5a).Let Ti denote T i combining with the two edges in C 3 incident with v i .Then, T1 and T3 are the 3-cutwidth critical tree H 1 = K 1,5 or the 3-cutwidth critical tree H 2 with a central edge (i.e., similar to w 0 x 0 in H 2 ) contracted.As to T 2 embedded in I 2 with a congestion of four, the cycle C 3 yields a congestion of two in this interval, and we have to choose T 2 as a 2-cutwidth critical tree, namely, a K 1,3 such that either d G (v 2 ) = 3 or d G (v 2 ) = 5.For this type of construction, the maximum congestion is four, that is, c(G) = 4. Furthermore, for any edge e ∈ E(G), if e ∈ E(C 3 ), then the deletion of e reduces the congestion two of cycle-edge in I 2 by one.Hence, T 2 embedded in I 2 has a congestion of three, and so c(G − e) < 4. If e / ∈ E(C 3 ), for τ 1 with T j = K 1,3 and d G (v j ) = 5 for each 1 ≤ j ≤ 3, we can let e ∈ E(T 2 ).Because T 2 − e = K 1,3 − e has a congestion of one, we can embed For τ 2 , τ 3 , and τ 4 , because d G (v 2 ) = 3, we can let e ∈ E(T 1 ); then, T 1 − e in I 1 has a congestion of two.Thus, we embed T 2 − v 2 in I 1 such that I 1 has a congestion of three (for example, T 1 is an F 2 at v 1 for τ 4 ), and the same is true for the case of e ∈ E(T 3 ).Hence, c(G − e) < 4. Thus, G is 4-cutwidth critical.
(2) Type 3B (including τ 5 to τ 13 ): f (v 1 ), f (v 2 ), f (v 3 ) are consecutive and I 2 = { f (v 2 )}, T 1 is embedded in I 1 with a congestion of three, T 2 is embedded in I 2 ∪ I 3 with a congestion of four, T 3 is embedded in I 3 with a congestion of three.Herein, we denote the subtree of H 2 obtained by deleting two leaves in the same branch (say y 0 , z 0 in Figure 1) with F 1 , and denote the subtree of H 2 obtained by deleting three vertices in the same branch (say x 0 , y 0 , z 0 in Figure 1) with F 2 (see F 1 , F 2 in Figure 5a).Then, T 1 is a star K 1,3 , K 1,5 , F 1 , F 2 , or T 2 and T 3 is a star K 1,5 or F 1 .Note that if where H 1 and H 2 are 3-cutwidth critical.Because T 2 and T 3 are embedded in I 3 consecutively and an edge of T 2 incident with v 2 strides over all edges of T 3 , we see that the overlapped edges of T 2 and T 3 give rise to a congestion of four in the embedding.Hence, c(G) = 4. Furthermore, for any edge e ∈ E(G), if e ∈ E(C 3 ), then G − e is a tree made up with H 1 and H 2 , which has a cutwidth of three.Thus, c(G − e) < 4. Otherwise, we may assume e ∈ E(T 2 ) (we Proof.(i) From ( 6), d G (v i ) ≥ 3. First, from the assumption that G is 4-cutwidth critical, it follows that c(G − xy) = 3 for any xy ∈ E(G) and c(T i ) ≤ 3 for each v i ∈ V(C t ).For the edge xy, there are three cases to consider.
Case 1 x ∈ V(C t ), y / ∈ V(C t ).In this case, C t ⊂ G − xy.So, by the minimality of H 3 and H 4 (see H 3 , H 4 in Figure 1), either H 3 or H 4 is a subgraph of G − xy resulting in that either F 2 or K 2 is contained in some T i , say T 1 .
Case 2 x / ∈ V(C t ), y / ∈ V(C t ).Similar to that of Case 1, we can conclude that F 2 or K 2 is also contained in some T i , say T 1 . Case Clearly, G − xy is a 3-cutwidth tree.So, by the minimality of H 1 and H 2 in Figure 1, either H 1 or H 2 is a subgraph of G − xy leading to the conclusion that either K 1,5 or F 1 is contained in some T i with v i = v 1 .
In addition, because K 1,3 is a proper subgraph of any of {K 1,5 , F 1 , F 2 } and t ≥ 4, we can conclude that there is at least a vertex In this case, we can verify that either c(G) = 3 (contradicting c(G) = 4) or one of {τ 44 , ..., τ 50 } is homeomorphic to a subgraph of G contradicting the minimality of G. Thus, T i ∈F 1 for i ∈ S t .
(ii) Assume that one member of F 2 is a subgraph of some T i 0 and v i 0 = u 0 (or x 0 ) by homeomorphism.Because t ≥ 4 and d G (v i ) ≥ 3 for each i = i 0 by (6), one of {τ 41 , τ 42 , τ 43 } must be either a subgraph of G or homeomorphic to a subgraph of G, contrary to the minimality of G. Hence, T i 0 ∈F 2 .Thus, by Lemma 5, G is 4-cutwidth critical with T i 0 ∈F 2 for v i 0 ∈ V(C t ) if and only if G ∈ {τ 41 , τ 42 , τ 43 }.This completes the proof.

Lemma 7.
Let G be a 4-cutwidth critical graph with cycle C t .Then, t ≤ 6.
Proof.This is a proof by contradiction.Assume that t ≥ 7; then, T i / ∈F 2 for each i ∈ S t .This is because, otherwise, one of {τ 41 , τ 42 , τ 43 } is homeomorphic to a subgraph G of G in which the cycle C 4 is subdivided into C t .So, c(G ) = 4, contradicting the conclusion that G is 4-cutwidth critical.Thus, T i ∈F 1 by Lemma 6.
For each i ∈ S t , if T i = K 2 , then direct computation yields that c(G) = 3.This implies that at least a T i ∈F 1 \{K 2 }.In addition, because c(τ 1 ) = 4, there are at most two vertices Thus, the congestions of T 1 and T h are at most three under f .Because G is 4-cutwidth critical and each cycle edge v i v i+1 of C t has a congestion of two, the subtree T i for i = 1, h must be 1-or 2-cutwidth critical, namely, then c(G) = 3 by direct computation.So, there are at least a vertex v i 0 (i 0 = 1, h) such that T i 0 = K 1,3 in G, which results in the conclusion that one of {τ 2 , τ 3 , τ 4 } is homeomorphic to a subgraph of G; this is a contradiction.Hence, this case is not possible. Case From the minimality of G, in this case, T 1 is either a K 1,3 centered at v 1 or F 2 , T h = K 1,3 not centered at v h .For each i = 1, h, if T i = K 2 then c(G) = 3 by the direct computation, contrary to c(G) = 4. So, there is at least a T i except T 1 and T h such that T i ∈ {K 1,3 , K 1,5 , F 1 }.This results in the conclusion that one of {τ 44 , τ 45 } must be homeomorphic to a subgraph of G, contrary to the minimality of G.For example, if t = 7 and h = 4, τ 44 must be homeomorphic to Figure 6a (or Figure 6b), while τ 45 is homeomorphic to Figure 6c.So, this case is impossible. Case By Lemma 6 and the minimality of G, T i ∈ F 1 \{F 2 } for each 1 ≤ i ≤ t.Because τ 11 , τ 12 , and τ 13 are 4-cutwidth critical, at most two subtrees of G − E(C t ), say T 1 and T h , are in {K 1,5 , F 1 }.So, similar to Cases 1 and 2, either one of {τ 46 , ..., τ 50 } is homeomorphic to a subgraph of G or c(G) < 4 (see seven typical cases in Figure 7 each of whose cutwidth is three by homemorphism), contradicting the conclusion that G is 4-cutwidth critical.So, this case is also impossible.
To sum up, we have t ≤ 6.This completes the proof.or K 1,3 .If each T i = K 2 then c(G) = 3 by direct computation.So there is at least a vertex v i 0 (i 0 = 1, h) such that T i 0 = K 1,3 in G, which results in that one of {τ 2 , τ 3 , τ 4 } is homeomorphic to a subgraph of G, a contradiction.Hence this case is not possible.
By the minimality of G, in this case, T 1 is either a K 1,3 centered at v 1 or F 2 , T h = K 1,3 not centered at v h .For each i = 1, h, if T i = K 2 then c(G) = 3 by the direct computation, contrary to c(G) = 4.So there is at least a T i except T 1 and T h such that T i ∈ {K 1,3 , K 1,5 , F 1 }.This results in that one of {τ 44 , τ 45 } must be homeomorphic to a subgraph of G, contrary to the minimality of G.For example, if t = 7 and h = 4, τ 44 must be homeomorphic to Fig. 6(a) (or Fig. 6(b)), while τ 45 is homeomorphic to Fig. 6(c).So this case is impossible.
By Lemma 3.2 and the minimality of G, T i ∈ F 1 \{F 2 } for each 1 ≤ i ≤ t.Since τ 11 , τ 12 and τ 13 are 4-cutwidth critical, at most two subtrees of G − E(C t ), say T 1 and T h , are in {K 1,5 , F 1 }.So, similar to that of Cases 1 and 2, either one of {τ 46 , ..., τ 50 } is homeomorphic to a subgraph of G or c(G) < 4 (see 7 typical cases in Fig. 7 each of whose cutwidth is 3 by homemorphism), contradicting that G is 4-cutwidth critical.So this case is also impossible.or K 1,3 .If each T i = K 2 then c(G) = 3 by direct computation.So there is at least a vertex v i 0 (i 0 = 1, h) such that T i 0 = K 1,3 in G, which results in that one of {τ 2 , τ 3 , τ 4 } is homeomorphic to a subgraph of G, a contradiction.Hence this case is not possible.
By the minimality of G, in this case, T 1 is either a K 1,3 centered at v 1 or F 2 , T h = K 1,3 not centered at v h .For each i = 1, h, if T i = K 2 then c(G) = 3 by the direct computation, contrary to c(G) = 4.So there is at least a T i except T 1 and T h such that T i ∈ {K 1,3 , K 1,5 , F 1 }.This results in that one of {τ 44 , τ 45 } must be homeomorphic to a subgraph of G, contrary to the minimality of G.For example, if t = 7 and h = 4, τ 44 must be homeomorphic to Fig. 6(a) (or Fig. 6(b)), while τ 45 is homeomorphic to Fig. 6(c).So this case is impossible.
By Lemma 3.2 and the minimality of G, T i ∈ F 1 \{F 2 } for each 1 ≤ i ≤ t.Since τ 11 , τ 12 and τ 13 are 4-cutwidth critical, at most two subtrees of G − E(C t ), say T 1 and T h , are in {K 1,5 , F 1 }.So, similar to that of Cases 1 and 2, either one of {τ 46 , ..., τ 50 } is homeomorphic to a subgraph of G or c(G) < 4 (see 7 typical cases in Fig. 7 each of whose cutwidth is 3 by homemorphism), contradicting that G is 4-cutwidth critical.So this case is also impossible.

Lemma 8. Suppose that C
Proof.By Lemma 5, it suffices to show its necessity.Because G is 4-cutwidth critical, T 1 , T 3 ∈ {K 1,3 , F 2 } by Lemmas 6 and 7.For an optimal labeling f of G with f (v 1 ) < f (v 2 ) < f (v 3 ), we can embed T 1 into the interval [1, f (v 1 )] with a congestion of three and T 3 into the interval [ f (v 3 ), n] with a congestion of three.So, T 2 must be K 1,3 , because the congestion of the cycle edge of C 3 is two, which is embedded into the interval ( f (v 1 ), f (v 3 )).Thus, G is one of {τ 1 , τ 2 , τ 3 , τ 4 }.
With an argument similar to that of Lemma 8, we can verify that the following two Lemmas 9 and 10 hold also.
Similar to Lemma 12, a class of critical unicyclic graphs with a cutwidth of four has an interesting structure (see Definition 3 below).This structure together with that of Lemma 12 is called the decomposability of the critical unicyclic graphs with a cutwidth of four.From Corollary 3, K 1,3 • (K 1,5 , K 1,5 , K 1,5 ) with v j ∈ D 1 (K 1,5 ) (1 ≤ j ≤ 3) is 4-cutwidth critical after that the series reductions are carried out, so we may assume that G − v has at most two K 1,4 's for any v ∈ V(G) in the sequel.Definition 3. Let C 3 be a unique cycle with a length of three in graph G In Definition 3, if d Ḡj (v) = 2 for some vertex v ∈ V( Ḡj ), and Ḡj − v + x j 1 x j 2 is 3cutwidth critical, then we also say that Ḡj is 3-cutwidth critical below, where Likewise, for Cases (ii) and (iii), we can let G = τ 30 and τ 29 , respectively.Lemma 13.With the notation in Definition 3, if Ḡj is 3-cutwidth critical for each j ∈ S 3 , then G is 4-cutwidth critical.
Proof.Without loss of generality, let G 1 , G 2 , G 3 satisfy (i) and C 3 ⊂ Ḡ3 by assumption.Then Ḡ1 , Ḡ2 are subtrees in G. Due to the fact that Ḡj is 3-cutwidth critical for each j ∈ S 3 , Ḡ3 ∈ {H  In fact, if there is some j 0 ∈ S 3 , say j 0 = 3, such that Ḡ3 is not 3-cutwidth critical, then two cases need to be considered: (1) there are at least an edge e ∈ E( Ḡ3 ) such that c( Ḡ3 − e) ≥ 3; (2) c( Ḡ3 ) ≤ 2. By assumption that G is 4-cutwidth critical, we can see that Case (1) is impossible by Lemma 13.Hence, it suffices to verify that Case (2) is also impossible.As Ḡ1 and Ḡ2 are 3-cutwidth critical, c(G 1 ) ≤ 2 and c(G 2 ) ≤ 2. Let f 1 , f 2 , f 3 be the optimal labelings of G 1 , Ḡ2 , and G 3 , respectively.Then, similar to that of (5) in Definition 2, we can obtain a 3-cutwidth labeling f : V(G) → S |V(G)| by the order Proof It suffices to show its necessity by Lemma 3.1.As the arguments are similar, we only consider the case that G 1 , G 2 , G 3 satisfy (i) of Definition 3.9.And without loss of generality, let cycle C 3 ⊂ Ḡ3 by assumption, then Ḡ1 , Ḡ2 are subtrees in G.
Claim 5. d G (v i ) = 3 for each v i ∈ V(C 5 ), and at most two subtrees T i 1 and T i 2 are in {K 1,5 , F 1 }.
In fact, if there exists at least one vertex, say v 1 , such that d G (v 1 ) ≥ 4, then, with an argument similar to that of Lemma 15, we can verify that one of {τ 1 , τ 2 , ..., τ 10 , τ 44 , τ 45 } is homeomorphic to a subgraph of G, which is a contradiction.If there is another T i 3 ∈ {K 1,5 , F 1 }, then one of {τ 11 , τ 12 , τ 13 } is homeomorphic to a subgraph of G, which is another contradiction.Claim 5 holds.

Remarks
In this paper, fifty critical unicyclic graphs with a cutwidth of four were obtained, during which a decomposable property of some 4-cutwidth critical unicyclic graphs was also obtained (see Lemma 12 and Definition 3).For an integer k ≥ 4, although it seems to be difficult to find all k-cutwidth critical graphs, some structural properties of some of them can be found definitively.In fact, as the decomposability of k-cutwidth critical trees [25] and some special non-tree graphs with uncomplicated structure [26], a similar decomposable property of 4-cutwidth critical unicyclic graphs, which is contained in Lemma 12 and Definition 3, can be generalized to k-cutwidth critical graphs even if these graphs are multicyclic graphs.For instance, in Lemma 12, if any element of {G 1 , G 2 , G 3 } is a critical unicyclic graph with a cutwidth of k − 1, then we can verify that K 1,3 • (G 1 , G 2 , G 3 ) is a critical unicyclic graph with a cutwidth of k.Clearly, if v j / ∈ V (j) p with 1 ≤ j ≤ 3, then {G j + u 0 v j : 1 ≤ j ≤ 3} is a decomposition of K 1,3 • (G 1 , G 2 , G 3 ).Regarding the critical multicyclic graphs with a cutwidth of at least four, their general structural properties have yet to be known.Additionally, the application of critical unicyclic graphs with a cutwidth of four to some realistic fields, such as social and biological networks, multivariate , |T 3 (tw(G))| = |T 3 (bw(G))| = 1, |T 4 (tw(G))| = |T 4 (bw(G))| = 4, |T 3 (pw(G))| = 110.As shown in

Figure 2 .
Figure 2. All elements of the 4-cutwidth critical tree set T 4 .

Figure 5 .
Figure 5. Two elements of F 1 and set F 2 .

Figure 6 .
Figure 6.(a-c) Three examples of Case 2 with the proof of Lemma 7.

Figure 7 .
Figure 7. (a-h) Seven typical examples on G of Case 3 with the proof of Lemma 7.

Proof.Claim 3 .
It suffices to show its necessity by Lemma 5.As the arguments are similar, we only consider the case that G 1 , G 2 , G 3 satisfy (i) of Definition 3. Furthermore, without loss of generality, let cycle C 3 ⊂ Ḡ3 by assumption, then Ḡ1 , Ḡ2 are subtrees in G.For each j ∈ S 3 , Ḡj is 3-cutwidth critical.

Fig. 8 .
Fig.8.Six possible graphs on G with the proof of Lemma 3.12

Figure 10 .
Figure 10.(a-c) Three possible graphs with the proof of Lemma 17.
The set of the critical unicyclic graphs with cutwidth 4 1,p centered at vertex x if d K 1,p (x) = p.