Signs of Hamiltonian Circles in Simple Plane Signed Graphs

SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS XIYONG Y AN Abstract. W e study whic h signs can o ccur among Hamiltonian circles in simple plane signed graphs. Using a face-based viewp oin t, we relate the sign of a Hamiltonian circle to the pro duct of the signs of the faces inside it, and w e introduce co-Hamiltonian sequences. This yields a criterion for the existence of opposite-sign Hamiltonian circles via tw o co- Hamiltonian sequences with opp osite face-pro ducts. Motiv ated by signed grid graphs, we dev elop local structural theorems that allow one to certify the existence of b oth signs with- out explicitly constructing the full sequences, including a ladder-type configuration where toggling along t wo 4-circles produces Hamiltonian circles of opp osite sign, as w ell as hexagon configurations that realize b oth signs. 1. Introduction W e in v estigate whic h signs can o ccur among Hamiltonian circles in simple plane signed graphs. Motiv ated b y a question of Zaslavsky [1], w e ask when a signed plane graph that con tains a Hamiltonian circle m ust con tain b oth a p ositive and a negativ e one. Our approach is based on a face-oriented viewp oin t. In a plane signed graph, the sign of a Hamiltonian circle can b e expressed as the pro duct of the signs of the b ounded faces it encloses. This leads to the notion of a c o-Hamiltonian se quenc e , whic h describes a systematic w a y to remo v e faces while preserving 2-connectedness and ultimately lea ving a unique Hamil- tonian circle. Using this framew ork, we obtain a criterion for the existence of opposite-sign Hamiltonian circles in terms of tw o co-Hamiltonian sequences whose face-pro ducts differ. Motiv ated b y signed grid graphs, we further dev elop lo cal structural results that allow one to certify the existence of b oth signs without constructing full sequences. In particular, ladder-t yp e configurations and certain hexagon structures force the realization of b oth p os- itiv e and negative Hamiltonian circles. These results sho w that the global sign b ehavior of Hamiltonian circles in plane signed graphs is gov erned by lo cal face configurations. 2. Hamil tonian Circle in Simple Plane Signed Graphs W e no w study Hamiltonian circles in simple plane signed graphs. In con trast to the complete graph case, the fixed embedding restricts how Hamiltonian circles can run through 2020 Mathematics Subje ct Classific ation. Primary: 05C45 (Hamiltonian graphs); Secondary: 05C22 (Signed and w eighted graphs). 1 2 XIYONG Y AN the graph, but it also pro vides additional structure through faces. Our approac h is to express the sign of a Hamiltonian circle in terms of the signs of the faces it encloses, and then to con trol these face pro ducts by removing faces from the outside in w ard. T o formalize this pro cess, we introduce co-Hamiltonian sequences and the asso ciated Hamiltonian sets, together with the weak dual (face graph) as a b o okk eeping to ol. This yields a criterion for the existence of Hamiltonian circles of opp osite sign in terms of t w o co-Hamiltonian sequences with opp osite face pro ducts, and it motiv ates later practical tests based on lo cal configurations in signed grid graphs and ladder-t yp e subgraphs. Definition 1 (Outer face, outer edges, exterior and in terior v ertices) . Let Σ b e a simple plane signed graph, and let F 0 denote its (unique) un b ounded face, called the outer fac e . An edge e of Σ is an outer e dge if it is inciden t with F 0 , equiv alen tly , if e lies on the b oundary ∂ F 0 . A v ertex v of Σ is an exterior vertex if it is inciden t with F 0 , and an interior vertex otherwise. Definition 2 (Co-Hamiltonian edge sequence, co-Hamiltonian face sequence, and Hamil- tonian set) . Let G b e a 2-connected plane graph with outer face F 0 , and let B b e the set of b ounded faces of G . (1) Co-Hamiltonian edge sequence. Let G b e a 2-connected plane graph with outer face F 0 ( G ). F or an ordered edge sequence E ′ = ( e 1 , e 2 , . . . , e k ) , set G 0 := G and G t := G − { e 1 , . . . , e t } ( t = 1 , . . . , k ) . W e call E ′ a c o-Hamiltonian e dge se quenc e if for eac h t = 1 , . . . , k , (1) e t lies on the b oundary of the outer face of G t − 1 , and (2) G t is 2-connected, and in the final graph G k ev ery vertex is exterior (i.e., incident with the outer face) and G k con tains exactly one Hamiltonian circle. (2) Co-Hamiltonian face sequence. During the deletion pro cess ab ov e, eac h deleted edge e t is required to lie on the b oundary of the outer face of G t − 1 . Hence deleting e t merges the outer face of G t − 1 with a unique b ounded face of G t − 1 ; denote that face by F t . The resulting ordered face sequence L = ( F 1 , F 2 , . . . , F k ) is called the c o-Hamiltonian (fac e) se quenc e induced b y E ′ . (Equiv alen tly , F t is the b ounded face that b ecomes part of the outer face at step t .) (3) Hamiltonian set. Let H b e th e set of b ounded faces of the final graph G k (equiv alently , the faces in B that are not merged into the outer face during the pro cess). Then H is called a Hamiltonian set of G , and its unique Hamiltonian circle is called the Hamiltonian cir cle determine d by H . SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 3 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 F 3 F 2 F 5 F 4 F 1 Figure 1. A plane graph with three interior vertices Example. In Figure 1 the graph has three in terior v ertices v 7 , v 8 , v 9 . Consider the face se- quence L 1 = ( F 1 , F 2 ). First delete the outer edge v 5 v 6 . This merges the bounded face F 1 with the outer face, so F 1 is no longer a b ounded face and the vertices v 7 and v 9 b ecome exterior. Next delete the edge v 7 v 9 ; after the first deletion this edge lies on the b oundary of the outer face, so its deletion merges the bounded face F 2 with the outer face and mak es v 8 exterior as well. After these tw o deletions the remaining graph is still 2-connected. Moreo v er, the remaining bounded faces form a Hamiltonian set and the associated subgraph con tains ex- actly one Hamiltonian circle. Th us L 1 is a co-Hamiltonian sequence (Definition 2). Similarly , L 2 = ( F 3 , F 4 , F 5 ) is another co-Hamiltonian sequence. Definition 3 (W eak dual or face graph) . Let G b e a plane graph with outer face F 0 . The we ak dual of G , also called the fac e gr aph , and denoted by D ( G ), is the graph whose vertices corresp ond to the b ounded faces of G . Tw o vertices of D ( G ) are adjacen t if the corresp onding b ounded faces of G share an edge. Definition 4 (Outerplane graph) . A simple outerplane gr aph is a simple plane graph in whic h ev ery v ertex lies on the b oundary of the outer face F 0 . Definition 5 (Outer b oundary circle) . Let G b e a 2-connected plane graph with outer face F 0 . The b oundary ∂ F 0 is the closed w alk formed b y the edges inciden t with F 0 . In a 2-connected plane graph, ∂ F 0 is a simple circle, called the outer b oundary cir cle of G . Lemma 2.1. L et G b e a 2 -c onne cte d simple outerplane gr aph with outer fac e F 0 , and assume that G has at le ast one b ounde d fac e. Then D ( G ) is a tr e e. Pr o of. W e sho w that D ( G ) is connected and has no circle. (1) D ( G ) is c onne cte d. Assume for a con tradiction that D ( G ) is disconnected, and let F 1 , F 2 , ..., F k b e nonempt y comp onents of D ( G ). If k = 1 , then D ( G ) is connected. Th us, 4 XIYONG Y AN w e ma y assume k > 1. F or i = 1 , 2 , ..., k set G i := [ f ∈F i ∂ f . If a b ounded face in F 1 shared an edge with a b ounded face in F 2 , then the corresp onding v ertices of D ( G ) would b e adjacen t, imp ossible. Hence G 1 and G 2 are edge disjoin t. Because G is 2-connected, it has no bridges, so ev ery edge of G lies on the b oundary of some b ounded face. If V ( G 1 ) ∩ V ( G 2 ) = ∅ , then G is disconnected, con tradiction. So V ( G 1 ) ∩ V ( G 2 )  = ∅ . W e claim that | V ( G 1 ) ∩ V ( G 2 ) | ≤ 1. Supp ose G 1 and G 2 share t w o distinct v ertices v 1  = v 2 . Let O i b e the b oundary circle of the outer face of G i in the inherited em b edding. Since G is outerplane, all v ertices lie on ∂ F 0 , hence O 1 and O 2 lie on ∂ F 0 as w ell. Cho ose v 1 , v 2 so that the v 1 – v 2 arcs P 1 ⊆ O 1 and P 2 ⊆ O 2 satisfy V ( P 1 ) ∩ V ( P 2 ) = { v 1 , v 2 } . Then P 1 ∪ P 2 is a simple circle in G . If b oth P 1 and P 2 are the single edge v 1 v 2 , then G has tw o parallel edges b etw een v 1 and v 2 , con tradicting that G is simple (Figure 2, left). Otherwise, at least one of P 1 , P 2 has an in terior v ertex; say P 2 con tains v 3 / ∈ { v 1 , v 2 } . Since P 1 ∪ P 2 is a circle, the v ertex v 3 lies inside it, hence v 3 is not inciden t with F 0 , con tradicting that G is outerplane (Figure 2, middle). This pro v es | V ( G 1 ) ∩ V ( G 2 ) | ≤ 1. Th us V ( G 1 ) ∩ V ( G 2 ) = { v } for some vertex v . Suppose k = 2. Since G 1 and G 2 are edge-disjoin t and meet only at v , the graph G − v is disconnected. Hence v is a cut vertex of G , con tradicting 2-connectedness. Supp ose k > 2. If v is a vertex cut, then w e get a con tradiction. If v is not a v ertex cut, then there exist G i 1 , . . . , G i r ∈ { G 1 , . . . , G k } such that the sequence of subgraphs G 1 , G 2 , G i 1 , . . . , G i r , G 1 has the prop ert y that each pair of consecutiv e subgraphs shares a common v ertex. Ho w ev er, these subgraphs enclose a b ounded face F (see Figure 2, right), and F connects all G i , a con tradiction. Therefore D ( G ) is connected. (2) D ( G ) is acyclic. Supp ose for a contradiction that D ( G ) contains a circle f 1 , f 2 , . . . , f k , f 1 ( k ≥ 3) , where consecutiv e faces share an edge. Let e i b e the edge common to f i and f i +1 (indices mo d k ). Then the dual edges e ∗ 1 , . . . , e ∗ k form a circle in the planar dual G ∗ , hence they trace a simple closed curve in the plane. By the Jordan curve theorem, this curv e b ounds a region that is disjoint from the outer face. In particular, that b ounded region con tains a vertex of G that is not inciden t with F 0 , contradicting that G is outerplane. Hence D ( G ) has no circle, so it is acyclic. Since D ( G ) is connected and acyclic, it is a tree. □ SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 5 v 1 v 2 O 1 O 2 v 2 v 1 v 3 O 1 O 2 F G i 2 G 1 G i 1 G 2 Figure 2. If O 1 and O 2 share t wo v ertices v 1 , v 2 , then either parallel edges o ccur (left) or an interior v ertex v 3 is forced (middle). The righ tmost sk etc h illustrates the decomp osition into G 1 , G 2 , G i 1 , G i 2 around a face F . Lemma 2.2. L et G b e a 2 -c onne cte d simple outerplane gr aph with outer fac e F 0 , and assume G has at le ast one b ounde d fac e. Then G c ontains a unique Hamiltonian cir cle, namely the outer b oundary cir cle ∂ F 0 . Pr o of. Since G is 2-connected and outerplane, ∂ F 0 is a simple circle con taining every vertex of G , so ∂ F 0 is a Hamiltonian circle. W e prov e uniqueness by induction on the n um b er b of b ounded faces. Base b = 1 . Then G has exactly one b ounded face f . Every edge of G lies on the boundary of a b ounded face (no bridges in a 2-connected plane graph), hence on ∂ f . Th us G = ∂ f = ∂ F 0 , so there is exactly one Hamiltonian circle. Inductive step. Assume b ≥ 2 and the statement holds for all such graphs with few er than b b ounded faces. By Lemma 2.1, D ( G ) is a tree, so it has a leaf face f . Let g = xy b e the unique edge shared b y f and another b ounded face; equiv alen tly , g is the unique edge of ∂ f not con tained in ∂ F 0 . Let P := ∂ f − g , the x – y path along ∂ f . Claim. Every Hamiltonian circle C of G con tains all edges of P and do es not con tain g . Since f is a leaf of D ( G ), the edge g is the only edge of ∂ f that is not on ∂ F 0 . Equiv alen tly , ev ery edge of P lies on the outer b oundary ∂ F 0 . Let u b e an in ternal vertex of P (so u  = x, y ). In an outerplane embedding, all edges inciden t with u lie in the outer face except those b elonging to faces that con tain u . Because f is the only b ounded face on the side of P , there is no edge of G that leav es u to the in terior of the disk b ounded b y ∂ f . Consequently , the only wa y for a Hamiltonian circle to pass through u is to use the tw o b oundary edges of P incident with u . Th us every Hamiltonian circle m ust con tain those t w o edges, and hence P ⊆ C . 6 XIYONG Y AN Finally , if g ∈ C as w ell, then C contains every edge of ∂ f = P ∪ { g } . Since ∂ f is a circle and f is b ounded, this w ould make ∂ f a prop er sub circle of the Hamiltonian circle C , which is imp ossible. Hence g / ∈ C . No w form G 1 b y con tracting the path P to a single edge xy em b edded on the outer b oundary . Then G 1 is still simple, outerplane, and 2-connected, and it has exactly b − 1 b ounded faces (the leaf face f is remov ed). By induction, G 1 has a unique Hamiltonian circle, namely ∂ F 0 ( G 1 ). By the Claim, every Hamiltonian circle of G con tains P and av oids g , so contracting P gives a Hamiltonian circle of G 1 . Con versely , expanding the edge xy in ∂ F 0 ( G 1 ) back to the path P yields a Hamiltonian circle of G . Hence G has exactly one Hamiltonian circle, and it is ∂ F 0 . □ Lemma 2.3. L et Σ b e a 2 -c onne cte d simple plane signe d gr aph, let H ′ b e the set of al l b ounde d fac es of Σ , and let C 0 b e the b oundary cir cle of the outer fac e. L et σ : E ( G ) → { + , −} b e an e dge-signing. F or e ach b ounde d fac e f ∈ H ′ , define its fac e-sign by σ ( f ) = Y g ∈ ∂ f σ ( g ) . Then σ ( C 0 ) = Y f ∈ H ′ σ ( f ) . Pr o of. By definition, σ ( C 0 ) = Y g ∈ E ( C 0 ) σ ( g ) . On the other hand, Y f ∈ H ′ σ ( f ) = Y f ∈ H ′ Y g ∈ ∂ f σ ( g ) . Reordering the pro duct, each edge g of G app ears once for eac h b ounded face incident with g . W e distinguish tw o cases. (i) Boundary edges. If an edge g lies on the boundary circle C 0 , then it is inciden t with exactly one b ounded face. Hence σ ( g ) app ears exactly once in the pro duct. (ii) Interior edges. If an edge g is shared b y tw o b ounded faces, then σ ( g ) appears twice in the pro duct. Since σ ( g ) ∈ { + , −} , we ha ve σ ( g ) 2 = + . Th us all in terior-edge con tributions cancel. SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 7 Therefore, only b oundary edges contribute nontrivially , and we obtain Y f ∈ H ′ σ ( f ) = Y g ∈ E ( C 0 ) σ ( g ) = σ ( C 0 ) . □ Lemma 2.4 (P eeling Lemma) . L et G b e a 2 -c onne cte d plane gr aph that c ontains a Hamilton- ian cir cle C . Assume that C is not the b oundary cir cle of the outer fac e of G (e quivalently, some e dge of G on the b oundary of the outer fac e is not in E ( C ) ). Then ther e exists an e dge e / ∈ E ( C ) on the b oundary of the outer fac e of G such that G − e is stil l 2 -c onne cte d. Pr o of. Since C is not the b oundary circle of the outer face, the boundary walk of the outer face uses at least one edge that is not in C . Choose suc h an edge and call it e . Then e lies on the b oundary of the outer face and e / ∈ E ( C ). No w consider G − e . The circle C is still presen t as a Hamiltonian circle in G − e (b ecause w e did not delete an y edge of C ). A graph that con tains a Hamiltonian circle is 2-connected: for any t w o distinct vertices x, y on C , the circle C provides t w o internally v ertex-disjoin t x – y paths (the tw o arcs of C b et w een x and y ). Hence G − e is 2-connected. □ Lemma 2.5. L et G b e a 2 -c onne cte d plane gr aph with outer fac e F 0 , and let B b e the set of b ounde d fac es of G . Then G c ontains a Hamiltonian cir cle if and only if G admits a c o-Hamiltonian (fac e) se quenc e in the sense of Definition 2. Pr o of. ( ⇒ ) Assume G con tains a Hamiltonian circle C , and let H b e the set of b ounded faces inside the closed disk b ounded by C . Let B out := B \ H denote the b ounded faces outside C . If B out = ∅ , then C = ∂ F 0 and the empty sequence is a co-Hamiltonian face sequence. Otherwise, we iterativ ely delete edges to merge the faces of B out in to the outer face, while preserving 2-connectedness and never deleting an edge of C . A t each step, the current graph G t − 1 is 2-connected and still contains the Hamiltonian circle C (as a subgraph), and G t − 1  = C b ecause there remains at least one edge outside C . Therefore, by the Peeling Lemma (Lemma 2.4), there exists an edge e t / ∈ E ( C ) that lies on the boundary of the outer face of G t − 1 suc h that deleting e t preserv es 2-connectedness. Define G t := G t − 1 − e t . Let F t b e the b ounded face of G t − 1 that is merged into the outer face when e t is deleted. Rep eating this pro cess pro duces a co-Hamiltonian edge sequence ( e 1 , . . . , e k ) and the corre- sp onding co-Hamiltonian face sequence ( F 1 , . . . , F k ). The pro cess terminates when all faces in B out ha v e b een merged into the outer face, i.e. when the b ounded faces of the final graph are exactly H . By construction, ev ery vertex of the final graph is exterior and it con tains exactly one Hamiltonian circle (namely C ), so H is the Hamiltonian set determined b y the sequence. Hence G admits a co-Hamiltonian (face) sequence. 8 XIYONG Y AN ( ⇐ ) Conv ersely , assume G admits a co-Hamiltonian edge sequence E ′ = ( e 1 , . . . , e k ), with induced face sequence L = ( F 1 , . . . , F k ) and final graph G k as in Definition 2. By definition, G k con tains exactly one Hamiltonian circle. In particular, G k con tains a Hamiltonian circle. Since G k is a spanning subgraph of G (obtained b y deleting edges only), that Hamiltonian circle is also a Hamiltonian circle of G . Therefore G con tains a Hamiltonian circle. □ 2.1. Grid graphs. Let Σ b e an m × n signed grid. W e lab el the v ertices b y V = { ( i, j ) | i ∈ { 1 , 2 , . . . , m } , j ∈ { 1 , 2 , . . . , n }} . The grid con tains ( m − 1)( n − 1) unit b o xes (or faces). Let B denote the set of all boxes. W e lab el the b o xes by B = { [ i, j ] | i ∈ { 1 , 2 , . . . , m − 1 } , j ∈ { 1 , 2 , . . . , n − 1 }} . W e lab el the edges of the grid as follows. F or each vertex ( i, j ) with 1 ≤ i ≤ m and 1 ≤ j ≤ n : (1) the horizontal e dge joining ( i, j ) and ( i, j + 1), for 1 ≤ i ≤ m and 1 ≤ j ≤ n − 1, is denoted b y e h i,j =  ( i, j ) , ( i, j + 1)  ; (2) the vertic al e dge joining ( i, j ) and ( i + 1 , j ), for 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n , is denoted b y e v i,j =  ( i, j ) , ( i + 1 , j )  . W e present t wo examples: one that contains a Hamiltonian circle and one that do es not. In Figure 3, there are exactly tw o interior vertices, namely (2 , 2) and (2 , 3). All other v ertices lie on the outer face. If w e remo v e the v ertical edge e v 1 , 2 , equiv alently remo ve the b o x [1 , 2], then it is straigh tforw ard to verify that the resulting graph contains exactly one Hamiltonian circle. Moreo ver, in the resulting graph the vertices (2 , 2) and (2 , 3) b ecome exterior v ertices. This illustrates a general phenomenon. Remo ving an edge incident with the outer face con v erts t wo in terior v ertices in to exterior v ertices in the resulting graph. (1 , 1) (1 , 2) (1 , 3) (1 , 4) (2 , 1) (2 , 2) (2 , 3) (2 , 4) (3 , 1) (3 , 2) (3 , 3) (3 , 4) [1 , 1] [2 , 1] [1 , 2] [2 , 2] [1 , 3] [2 , 3] Figure 3. 3 by 4 grid with face (b ox) lab els in blue and v ertex lab els in paren theses SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 9 (1 , 1) (1 , 2) (1 , 3) (2 , 1) (2 , 2) (2 , 3) (3 , 1) (3 , 2) (3 , 3) [1 , 1] [1 , 2] [2 , 1] [2 , 2] Figure 4. 3 b y 3 grid with lab els ( x, y ) replaced by ( y , x ) Note that not every grid graph con tains a Hamiltonian circle. F or e xample, consider the graph in Figure 4. If we remov e an y exterior edge from the graph, say the vertical edge e v 1 , 2 , then the v ertex (1 , 3) has degree 1. Consequen tly , the remaining graph cannot contain a Hamiltonian circle. Alternativ ely , observ e that remo ving an exterior edge from a grid graph turns tw o interior v ertices into exterior v ertices. Ho w ev er, the graph in Figure 4 has only one in terior vertex. Th us, after the remo v al of any exterior edge, the resulting graph has no in terior vertex structure compatible with a Hamiltonian circle. The com binatorial structure of rectangular grid graphs imp oses strong parit y constraints. In particular, the parity of the in terior v ertices yields an immediate obstruction to Hamil- tonicit y , as stated in the following lemma. Lemma 2.6. A r e ctangular m × n grid gr aph with an o dd numb er of interior vertic es c annot c ontain a Hamiltonian cir cle. Pr o of. In the m × n grid graph, the interior vertices are exactly those ( i, j ) with 2 ≤ i ≤ m − 1 and 2 ≤ j ≤ n − 1, hence the num b er of interior v ertices is ( m − 2)( n − 2) . If ( m − 2)( n − 2) is o dd, then b oth m − 2 and n − 2 are o dd, so m and n are o dd. Therefore the total n um b er of vertices is | V | = mn, whic h is o dd. No w the grid graph is bipartite under the coloring given by the parit y of i + j : ( i, j ) ∈ X ⇐ ⇒ i + j is even , ( i, j ) ∈ Y ⇐ ⇒ i + j is o dd . Ev ery edge joins a v ertex in X to a v ertex in Y , so this is a bipartition. In an y circle of a bipartite graph, vertices alternate b etw een the t w o parts, so every circle uses the same num ber of vertices from X and from Y . In particular, a Hamiltonian circle w ould use all v ertices, forcing | X | = | Y | . 10 XIYONG Y AN But if | V | = | X | + | Y | is o dd, then | X |  = | Y | , a contradiction. Hence the grid graph cannot con tain a Hamiltonian circle. □ W e illustrate the next lemma (Lemma 2.7) with a 4 × 6 grid graph: after deleting the sequence L , the remaining graph is shown in Figure 5. (4 , 1) (4 , 6) (1 , 1) (1 , 6) [1 , 1] [3 , 5] Figure 5. The resulting graph obtained from a 4 × 6 grid after deleting a co-Hamiltonian sequence. Lemma 2.7. L et Σ b e an m × n r e ctangular grid with n even. Ther e exists a c o-Hamiltonian se quenc e L c onsisting of  n 2 − 1  ( m − 2) b oxes r emove d fr om the outside of Σ such that the r emaining gr aph c ontains exactly one Hamiltonian cir cle. Pr o of. W e giv e an explicit c hoice of L and pro v e that it forces a unique Hamiltonian circle. Construction of L . View Σ as an ( m − 1) × ( n − 1) array of unit b o xes lab eled [ i, j ] (1 ≤ i ≤ m − 1 , 1 ≤ j ≤ n − 1) . Since n is even, the set { 2 , 4 , 6 , . . . , n − 2 } has size n 2 − 1. F or each ro w i ∈ { 1 , 2 , . . . , m − 2 } , remov e the b oxes [ i, 2] , [ i, 4] , [ i, 6] , . . . , [ i, n − 2] in this order . Th us w e remo v e exactly n 2 − 1 b oxes in each of the m − 2 rows, so | L | =  n 2 − 1  ( m − 2) . Observ e that the resulting graph G L is 2-connected and that every vertex of G L is an exterior v ertex. Hence G L satisfies the h yp otheses of Lemma 2.2. Therefore G L con tains a unique Hamiltonian circle, namely the b oundary circle of its outer face. In particular, the remaining graph contains exactly one Hamiltonian circle, as desired. □ SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 11 Notation. F or tw o sets of b o xes H ′ 1 and H ′ 2 , the symmetric difference H ′ 1 △ H ′ 2 denotes the set of b o xes that are contained in H ′ 1 or H ′ 2 , but not in b oth. Lemma 2.8. L et Σ b e an m × n grid with m, n > 3 and n even. L et H 1 and H 2 b e two distinct Hamiltonian cir cles with c orr esp onding Hamiltonian sets H ′ 1 and H ′ 2 , r esp e ctively. Supp ose H ′ 1 △ H ′ 2 = { a, b } , wher e a ∈ H ′ 1 \ H ′ 2 and b ∈ H ′ 2 \ H ′ 1 ar e b oxes. Then σ ( H 1 ) = σ ( H 2 ) ⇐ ⇒ σ ( a ) = σ ( b ) . Pr o of. Let S = H ′ 1 ∩ H ′ 2 . Since H ′ 1 △ H ′ 2 = { a, b } , w e ha ve H ′ 1 = S ∪ { a } , H ′ 2 = S ∪ { b } . W rite S = { c 1 , c 2 , . . . , c k } . Since σ is m ultiplicativ e, b y Lemma 2.3 w e obtain σ ( H 1 ) = σ ( H ′ 1 ) =  k Y i =1 σ ( c i )  σ ( a ) , σ ( H 2 ) = σ ( H ′ 2 ) =  k Y i =1 σ ( c i )  σ ( b ) . Therefore, σ ( H 1 ) = σ ( H 2 ) ⇐ ⇒  k Y i =1 σ ( c i )  σ ( a ) =  k Y i =1 σ ( c i )  σ ( b ) ⇐ ⇒ σ ( a ) = σ ( b ) , as claimed. □ Theorem 2.9. L et Σ b e an m × n grid with m even and m, n > 3 . Al l Hamiltonian cir cles in Σ have the same sign if and only if al l b oxes of Σ exc ept the four c orner b oxes have the same sign. Pr o of. Supp ose all Hamiltonian circles in Σ ha ve the same sign. W e will sho w that all b oxes of Σ, except the four corner b o xes, hav e the same sign. W e pro v e this direction in t w o cases. Case 1: n is even. Let B denote the set of all b ounded faces of the graph Σ. Let L := { [ i, j ] | i ∈ { 1 , 2 , . . . , m − 2 } , j ∈ { 2 , 4 , . . . , n − 2 }} b e a co-Hamiltonian set of B , and let H = B \ L . Fix a column index j ∈ { 2 , 4 , . . . , n − 2 } , and each i ∈ { 1 , 2 , . . . , m − 2 } , we ma y replace the b o x [ m − 1 , j ] in H b y the b ox [ i, j ] (that is, delete [ m − 1 , j ] and add [ i, j ]). Denote the resulting Hamiltonian set b y H i,j . Since the Hamiltonian circle in H i,j and the Hamiltonian circle in H hav e the same signs and H i,j △ H = { [ i, j ] , [ m − 1 , j ] } , b y Lemma 2.8, σ ([ i, j ]) = σ ([ m − 1 , j ]). Since i is arbitrary , all the b o xes in the column j ha v e the same sign. Consider H again. Delete the b o x [2 , 1] and add the b o x [1 , 2] (equiv alently , mo v e the b o x [2 , 1] to p osition [1 , 2]). Denote the resulting Hamiltonian set b y H 2 . Then, in H 2 , for an y 12 XIYONG Y AN k ∈ { 3 , 4 , . . . , m − 2 } , we may replace the b o x [ k , 1] by the b o x [2 , 1] (that is, mo v e [ k , 1] to p osition [2 , 1]). Denote the resulting Hamiltonian set by H k, 1 . Since all the Hamiltonian circles hav e the same sign and H k, 1 △ H 2 = { [2 , 1] , [ k , 1] } , b y Lemma 2.8, σ ([ k , 1]) = σ ([2 , 1]). Hence, all non-corner b o xes in the first column m ust hav e the same sign. Similarly , all non-corner b o xes in the last column must hav e the same sign. Next, consider Σ again. Let L 2 := { [ i, j ] | i ∈ { 1 , 2 , . . . , m − 2 } , j ∈ { 3 , 5 , . . . , n − 3 }} ∪ { [ i, 1] , [ i, n − 1] | i ∈ { 2 , 4 , . . . , m − 2 }} . Let H ′ = B \ L 2 . Then H ′ con tains exactly one Hamiltonian circle. R ow movements in H ′ . Fix an o dd column index j ∈ { 3 , 5 , . . . , n − 3 } . F or an y i ∈ { 1 , 2 , . . . , m − 2 } , replace the b ox [ m − 1 , j ] in H ′ b y the b o x [ i, j ], and denote the resulting Hamiltonian set b y H i,j . Since we assume that all Hamiltonian circles in Σ hav e the same sign, b y Lemma 2.8 w e hav e σ ([ i, j ]) = σ ([ m − 1 , j ]). Since i is arbitrary , w e conclude that all b o xes in column j ha v e the same sign. R elations b etwe en differ ent c olumns. Finally , w e compare boxes lying in different columns. In H , fix a n um b er j ∈ { 1 , 3 , . . . , n − 3 } , w e ma y mo v e the b ox [2 , j ] to p osition [1 , j + 1]. Denote the resulting Hamiltonian set by H 1 ,j +1 . Since all the Hamiltonian circles ha v e the same sign, by Lemma 2.8, σ ([2 , j ]) = σ ([1 , j + 1]). This implies the non-corner b o x sign in column l is the same as the non-corner b o x sign in column l + 1, for l is o dd. Similarly , in H again, for eac h j ∈ { 3 , 5 , . . . , n − 1 } , w e ma y mo v e the b o x [2 , j ] to p osition [1 , j − 1]. Denote the resulting Hamiltonian set b y H 1 ,j − 1 . With the same reasoning, σ ([2 , j ]) = σ ([1 , j − 1]). This implies the non-corner b ox sign in column l is the same as the non-corner b o x sign in column l + 1, for l is even. Hence, any t w o non-corner b ox in differen t column will ha v e the same sign. Case 2: n is o dd. Let L := { [ i, j ] | i ∈ { 1 , 2 , . . . , m − 2 } , j ∈ { 2 , 4 , . . . , n − 3 }} ∪ { [ i, n − 1] | i ∈ { 2 , 4 , . . . , m − 2 }} b e a co-Hamiltonian set of Σ, and let H = B \ L . Then H con tains exactly one Hamiltonian circle. Using the same row and column mo vemen t arguments as in Case 1, we obtain that all non-corner boxes in the first n − 2 columns must hav e the same sign since all Hamiltonian circles in Σ to hav e the same sign. Next, consider a different co-Hamiltonian set L ′ := { [ i, j ] | i ∈ { 1 , 2 , . . . , m − 2 } , j ∈ { 3 , 5 , . . . , n − 2 }} ∪ { [ i, 1] | i ∈ { 2 , 4 , . . . , m − 2 }} . Using the same ro w and column mov emen t argumen ts as in Case 1, we conclude that all non-corner b oxes in columns 2 , 3 , . . . , n − 1 must ha v e the same sign, since all Hamiltonian circles in Σ are assumed to hav e the same sign. SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 13 Hence, from b oth cases w e conclude that all b o xes of Σ, except the four corner b o xes, m ust ha v e the same sign. Con v ersely , assume that all b o xes of Σ, except the four corner b oxes, ha v e the same sign. W e show that all Hamiltonian circles in Σ ha v e the same sign. F rom the forward direction, w e mak e the follo wing tw o observ ations: (1) every Hamiltonian set contains all four corner b o xes; (2) all Hamiltonian sets contain the same num ber of b o xes. Let H ′ b e an y Hamiltonian set, and let C denote its corresp onding Hamiltonian circle. By Lemma 2.3, σ ( C ) = Y [ i,j ] ∈ H ′ σ ([ i, j ]) . W rite H ′ = { four corner b o xes } ∪ { non-corner b oxes in H ′ } . Since all non-corner b o xes hav e the same sign, say α , and since every Hamiltonian set con tains the same num b er of non-corner b oxes, the contribution of the non-corner b oxes to σ ( C ) is the same for ev ery Hamiltonian circle. Moreov er, the contr ibution of the four corner b o xes is fixed, since every Hamiltonian set contains all four of them. Therefore, for an y t w o Hamiltonian circles C 1 and C 2 , w e ha v e σ ( C 1 ) = σ ( C 2 ) . Hence, all Hamiltonian circles in Σ hav e the same sign. □ (1 , 1) (1 , 2) (1 , 3) (2 , 1) (2 , 2) (2 , 3) (3 , 1) (3 , 2) (3 , 3) Figure 6. A 3 × 3 grid with diagonals added to each square 2.2. T riangulated grid graphs. The 3 × 3 grid graph do es not con tain a Hamiltonian circle (Figure 4). If w e add a diagonal to eac h unit squa re, w e obtain the triangulated grid sho wn in Figure 6, whose b ounded faces are triangles. Let G △ denote this triangulated grid. Deleting a b oundary edge of G △ merges one b ound- ary triangle in to the outer face. F or example, if we delete the horizon tal edge e h 3 , 2 (as in Figure 6), then the resulting graph contains a unique Hamiltonian circle C . 14 XIYONG Y AN In Figure 7 below, deleting the v ertical edge e v 2 , 3 , equiv alen tly deleting the negative face on the righ t hand side, leav es a graph that is still 2-connected, and it contains a unique Hamiltonian circle. In the corresp onding face graph, the v ertex lab eled − 2 on the righ t is remo v ed. Consequently , the resulting face graph is a tree. (1 , 1) (1 , 2) (1 , 3) (2 , 1) (2 , 2) (2 , 3) (3 , 1) (3 , 2) (3 , 3) (4 , 1) (4 , 2) (4 , 3) − − − 2 2 2 − 2 2 2 Figure 7. A 4 × 3 grid with t w o negative and four p ositiv e faces, and its face graph. Definition 6 (T riangulation and face map) . Let Σ b e a 2-connected plane signed graph with outer face F 0 , and let B denote the set of b ounded faces of Σ. F or a b ounded face f ∈ B , a triangulation of f is a sub division of f in to triangles b y adding noncrossing diagonals b et w een vertices on the boundary of f . Let τ ( f ) denote the n um b er of triangles in any such triangulation of f . Since f is a polygon, the v alue τ ( f ) is indep enden t of the chosen triangulation. Define the fac e map ϕ : B → Z b y ϕ ( f ) = σ ( f ) τ ( f ) , where σ ( f ) ∈ {± 1 } is the sign of the face f . Definition 7 (Remo v able face-vertex) . Let D (Σ) be the face graph, and let v f b e the v ertex corresp onding to a b ounded face f . W e call v f r emovable if (1) deg D (Σ) ( v f ) = | ϕ ( f ) | + 1 . T o help a computer searc h for a Hamiltonian set, w e can w ork in the face graph D (Σ). Lab el each vertex v f of D (Σ) by the pair  ϕ ( f ) , deg D (Σ) ( v f )  , where f is the corresponding b ounded face of Σ. Using (1), w e ma y delete remo v able v ertices one at a time. The goal is to con tin ue this elimination process until the remaining face graph SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 15 is a tree (or reduces to a single vertex), at which p oin t the surviving faces form a candidate Hamiltonian set. Remark 1. After deleting a vertex from the face graph, we obtain a new face graph (the induced subgraph on the remaining face-v ertices). In particular, the degrees of the remaining v ertices ma y c hange after eac h deletion. 2.3. Criteria and lo cal constructions for opp osite-sign Hamiltonian circles. In The- orem 2.10, our goal is to find t w o co-Hamiltonian sequences with distinct signs, whic h in turn yield tw o Hamiltonian circles of opp osite sign. In practice, ho w ev er, it may not b e necessary to explicitly construct such sequences. Motiv ated by the b ehavior of signed complete graphs and signed grid graphs, w e obtain the follo wing theorems. Under certain lo cal face configurations, one can quic kly determine the existence of b oth a p ositiv e Hamiltonian circle and a negativ e Hamiltonian circle. Theorem 2.10. L et Σ = ( G, σ ) b e a simple plane signe d gr aph that c ontains a Hamiltonian cir cle C . Then Σ c ontains b oth a p ositive Hamiltonian cir cle and a ne gative Hamiltonian cir cle if and only if ther e exist two c o-Hamiltonian se quenc es of fac es F 1 = ( F 1 , . . . , F k ) and F 2 = ( F ′ 1 , . . . , F ′ ℓ ) , such that k Y i =1 σ ( F i ) = − ℓ Y j =1 σ ( F ′ j ) . Pr o of. Let B denote the set of all b ounded faces of G . F or an y (finite) set S of b ounded faces, write σ ( S ) := Y f ∈ S σ ( f ) . ( ⇐ ). Assume there exist tw o co-Hamiltonian sequences F 1 = ( F 1 , . . . , F k ) and F 2 = ( F ′ 1 , . . . , F ′ ℓ ) suc h that k Y i =1 σ ( F i ) = − ℓ Y j =1 σ ( F ′ j ) . Let L 1 := { F 1 , . . . , F k } and L 2 := { F ′ 1 , . . . , F ′ ℓ } , and define H 1 := B \ L 1 , H 2 := B \ L 2 . By the definition of a co-Hamiltonian sequence, H 1 determines a Hamiltonian circle C 1 and H 2 determines a Hamiltonian circle C 2 . By Lemma 2.3, σ ( C 1 ) = σ ( H 1 ) , σ ( C 2 ) = σ ( H 2 ) . 16 XIYONG Y AN Since σ ( B ) = σ ( H 1 ) σ ( L 1 ) = σ ( H 2 ) σ ( L 2 ) , w e ha v e σ ( H 1 ) = σ ( B ) σ ( L 1 ) , σ ( H 2 ) = σ ( B ) σ ( L 2 ) . Therefore, σ ( C 1 ) σ ( C 2 ) = σ ( H 1 ) σ ( H 2 ) =  σ ( B ) σ ( L 1 )  σ ( B ) σ ( L 2 )  = σ ( L 1 ) σ ( L 2 ) , b ecause σ ( B ) 2 = 1. The hypothesis σ ( L 1 ) = − σ ( L 2 ) implies σ ( L 1 ) σ ( L 2 ) = − 1, hence σ ( C 1 ) σ ( C 2 ) = − 1 , so C 1 and C 2 ha v e opp osite signs. ( ⇒ ). Conv ersely , assume G contains a p ositiv e Hamiltonian circle C + and a negative Hamil- tonian circle C − . Let H + and H − b e the sets of b ounded faces inside C + and C − , respectively , and set L + := B \ H + , L − := B \ H − . Since C + is a Hamiltonian circle, there exists a co-Hamiltonian sequence whose set of remov ed faces is L + . Lik ewise, there exists a co-Hamiltonian sequence whose set of remov ed faces is L − . Cho ose suc h sequences and write them as F 1 = ( F 1 , . . . , F k ) and F 2 = ( F ′ 1 , . . . , F ′ ℓ ) , so that L 1 := { F 1 , . . . , F k } = L + and L 2 := { F ′ 1 , . . . , F ′ ℓ } = L − . By Lemma 2.3, σ ( C + ) = σ ( H + ) = σ ( B ) σ ( L + ) , σ ( C − ) = σ ( H − ) = σ ( B ) σ ( L − ) . Multiplying giv es σ ( C + ) σ ( C − ) = σ ( L + ) σ ( L − ) . Since σ ( C + ) = +1 and σ ( C − ) = − 1, we ha ve σ ( C + ) σ ( C − ) = − 1, hence σ ( L + ) σ ( L − ) = − 1 ⇐ ⇒ σ ( L + ) = − σ ( L − ) . Equiv alently , k Y i =1 σ ( F i ) = − ℓ Y j =1 σ ( F ′ j ) , as desired. □ SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 17 p 2 p ′ l p ′ 3 p 3 = p ′ 2 p 1 v 1 v 2 q 1 = q ′ 1 q ′ 2 v s v 3 v i +2 v i +1 v i − 1 v i p k q r = q ′ r Figure 8. The graph G with the central ladder r egion used in the setup of Theorem 2.11. Let Σ = ( G, σ ) b e a simple plane signed graph (See Figure 8) whose outer face b oundary is the circle C out := v 1 v 2 q ′ 1 q ′ 2 · · · q ′ r v i v i +1 p k p ′ ℓ p ′ ℓ − 1 · · · p ′ 3 p ′ 2 p ′ 1 v 1 . Assume that Σ contains a circle C := v 1 v 2 v 3 · · · v i − 1 v i v i +1 v i +2 · · · v s v 1 suc h that the closed disk b ounded by C contains no v ertices of Σ in its in terior. Define the circle L := v 1 v s v s − 1 · · · v i +2 v i +1 p k p ′ ℓ p ′ ℓ − 1 · · · p ′ 3 p ′ 2 p ′ 1 v 1 . Assume that the edges of the path P L := p 1 v 1 v s v s − 1 · · · v i +2 v i +1 p k are fixed, i.e., none of the edges of P L is allo wed to b e deleted. Supp ose that there exists an edge set E L ⊆ E (Σ) disjoin t from E ( P L ) suc h that, after deleting E L , (1) every vertex inside L b ecomes inciden t with the outer face, and (2) the resulting graph Σ − E L is still 2-connected. In particular, in Σ − E L there is a new outerface path from p 1 to p k . Assume symmetrically that the analogous statemen t holds on the righ t side: after deleting some edge set E R , while keeping the b oundary path q 1 v 2 v 3 · · · v i − 1 v i q r fixed, every vertex in the corresp onding right region b ecomes incident with the outer face, the resulting graph remains 2-connected, and in the end the circle R := v 2 q 1 q 2 · · · q r v i v i − 1 · · · v 3 v 2 b ounds a closed disk containing no vertic es of G in its in terior. 18 XIYONG Y AN Theorem 2.11. Under the assumptions ab ove, let C 1 := v 1 v 2 v 3 v s v 1 and C 2 := v i − 1 v i v i +1 v i +2 v i − 1 . If σ ( C 1 )  = σ ( C 2 ) , then Σ c ontains two Hamiltonian cir cles of opp osite sign. Pr o of. Let Σ = ( G, σ ) satisfy all h yp otheses, and k eep the notation E L , E R , L, R from the setup. Step 1: Reduce to the released graph. Let Σ ′ := Σ − ( E L ∪ E R ) . By assumption, Σ ′ is still 2-connected. Moreo v er, the left and righ t release assumptions imply that every vertex that w as inside L (resp ectively , inside R ) b ecomes inciden t with the outer face of Σ ′ . In particular, all vertices on the p -c hain (:= p 1 p 2 ...p k ) and q -chain lie on the outer b oundary of Σ ′ . Only vertices in the ladder region (Figure 8) may remain non-outer. Step 2: Define H 1 and H 2 . Consider the Hamiltonian circle H 1 giv en b y H 1 := v 1 v s · · · v i +2 v i − 1 · · · v 3 v 2 q 1 q 2 · · · q r v i v i +1 p k · · · p 2 p 1 v 1 . Th us H 1 tra v erses ev ery v ertex exactly once. No w define H 2 to b e the circle obtained from H 1 b y toggling along C 1 and C 2 , that is, E ( H 2 ) := E ( H 1 ) ⊕ E ( C 1 ) ⊕ E ( C 2 ) , where ⊕ denotes symmetric difference. Observ e that H 2 is spanning and 2-regular, that is, it con tains all v ertices and every v ertex has degree 2 in H 2 . Moreo ver, the toggling op eration only reroutes the trav ersal lo cally and do es not disconnect the circle. Hence H 2 is a Hamiltonian circle. Step 3: Compare the signs. Since E ( H 2 ) = E ( H 1 ) ⊕ E ( C 1 ) ⊕ E ( C 2 ) , w e ha v e σ ( H 2 ) = σ ( H 1 ) σ ( C 1 ) σ ( C 2 ) . Because eac h circle sign is ± 1, the hypothesis σ ( C 1 )  = σ ( C 2 ) implies σ ( C 1 ) σ ( C 2 ) = − 1, and hence σ ( H 2 ) = − σ ( H 1 ). Therefore Σ con tains t w o Hamiltonian circles of opp osite sign, namely H 1 and H 2 . □ W e now describ e a second lo cal configuration that forces the existence of Hamiltonian circles of b oth signs. SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 19 p 2 p 1 = p ′ 1 v 1 v 2 q 1 = q ′ 1 q 2 p k = p ′ k v 4 v 3 q l = q ′ l r 1 r t Figure 9. The resulting graph Σ 0 after remo ving the edge sets E L and E R from Σ. Let Σ = ( G, σ ) b e a simple plane signed graph with outer circle C out := v 1 v 2 q ′ 1 q ′ 2 · · · q ′ ℓ v 3 v 4 p ′ k p ′ k − 1 · · · p ′ 2 p ′ 1 v 1 . Assume that Σ contains an hexagon (it is p ossible that r 1 = r t ) H := v 1 v 2 r t v 3 v 4 r 1 v 1 . Assume moreo v er that ev ery v ertex in the in terior of H lies on a path R := r 1 r 2 · · · r t em b edded inside H ; in particular, r 1 and r t lie on H , while all other vertices r 2 , . . . , r t − 1 lie in the in terior of H . Consider the circle C ′ L := v 1 r 1 v 4 p ′ k p ′ k − 1 · · · p ′ 2 p ′ 1 v 1 . Assume that the four edges p 1 v 1 , v 1 r 1 , r 1 v 4 , v 4 p k , (here , p 1 = p ′ 1 , p k = p ′ k ) are fixed, that is, they are not allo wed to b e deleted. Supp ose that there exists a set (or sequence) of edges E L ⊆ E ( G ) suc h that deleting E L releases all v ertices strictly inside C ′ L to the outer face, and the resulting graph Σ − E L is still 2-connected. After deleting the edge set E L , the resulting graph Σ − E L con tains the circle C L := v 1 r 1 v 4 p k p k − 1 · · · p 2 p 1 v 1 . Similarly , consider the circle C ′ R := v 2 q ′ 1 q ′ 2 · · · q ′ ℓ v 3 r t v 2 . Assume that there exists a set (or sequence) of edges E R ⊆ E ( G ) such that deleting E R releases all v ertices strictly inside C ′ R to the outer face, and the resulting graph Σ − E R is still 2-connected. At the end of the righ t-side deletion pro cess, the resulting graph contains the circle C R := v 2 q 1 q 2 · · · q ℓ v 3 r t v 2 . 20 XIYONG Y AN Theorem 2.12. L et Σ = ( G, σ ) b e a simple plane signe d gr aph satisfying al l assump- tions state d ab ove. In p articular, Σ has outer cir cle C out that c ontains the hexagon H := v 1 v 2 r 4 v 3 v 4 r 1 v 1 , and every vertex in the interior of H lies on a p ath R = r 1 r 2 · · · r t emb e dde d inside H . Assume mor e over that the left and right r ele ase c onditions hold for the cir cles C ′ L and C ′ R , and after r ele asing e dge we obtain C L and C R . If two cir cles C 1 := v 1 v 2 r t r t − 1 · · · r 1 v 1 and C 2 := v 3 v 4 r 1 r 2 · · · r t v 3 have differ ent signs, then Σ c ontains b oth a p ositive Hamiltonian cir cle and a ne gative Hamiltonian cir cle. Pr o of. Let Σ = ( G, σ ) satisfy the assumptions stated ab ov e. By hypothesis, there exist edge sets E L , E R ⊆ E ( G ) such that deleting E L releases all v ertices inside C ′ L to the outer face and Σ − E L remains 2-connected, and similarly deleting E R releases all vertices inside C ′ R to the outer face and Σ − E R remains 2-connected. Let Σ 0 := Σ − ( E L ∪ E R ) . Then Σ 0 is still 2-connected (See Figure 9). Moreov er, by the tw o release conditions, ev ery v ertex of G outside the “central ” region b ounded b y the hexagon is no w inciden t with the outer face of Σ 0 ; hence the only v ertices that can remain in the in terior of the cen tral region are exactly the vertices on the path R = r 1 r 2 · · · r t . In Σ 0 consider the follo wing t w o circles: H 1 := v 1 v 2 q 1 q 2 · · · q ℓ v 3 r t r t − 1 · · · r 1 v 4 p k p k − 1 · · · p 2 p 1 v 1 , H 2 := v 1 r 1 r 2 · · · r t v 2 q 1 q 2 · · · q ℓ v 3 v 4 p k p k − 1 · · · p 2 p 1 v 1 . Both H 1 and H 2 tra v erse every vertex of G exactly once. Therefore, H 1 and H 2 are Hamil- tonian circles of Σ 0 . Since Σ 0 is obtained from Σ by deleting edges only , these circles are also Hamiltonian circles of Σ. It remains to compare their signs. Let Q := v 2 q 1 q 2 · · · q ℓ v 3 and P := v 4 p k p k − 1 · · · p 2 p 1 v 1 , and let R also denote the path r 1 r 2 · · · r t . W rite σ ( Q ), σ ( P ), and σ ( R ) for the products of the edge-signs along these paths. Then, b y insp ection, σ ( H 1 ) = σ ( P ) σ ( Q ) σ ( R ) σ ( v 1 v 2 ) σ ( v 3 r t ) σ ( v 4 r 1 ) , σ ( H 2 ) = σ ( P ) σ ( Q ) σ ( R ) σ ( v 1 r 1 ) σ ( v 2 r t ) σ ( v 3 v 4 ) . On the other hand, for the tw o circles in the statement, C 1 = v 1 v 2 r t r t − 1 · · · r 1 v 1 and C 2 = v 3 v 4 r 1 r 2 · · · r t v 3 , w e ha v e σ ( C 1 ) = σ ( R ) σ ( v 1 v 2 ) σ ( v 2 r t ) σ ( v 1 r 1 ) , σ ( C 2 ) = σ ( R ) σ ( v 3 v 4 ) σ ( v 4 r 1 ) σ ( v 3 r t ) . SIGNS OF HAMIL TONIAN CIR CLES IN SIMPLE PLANE SIGNED GRAPHS 21 Therefore σ ( H 1 ) σ ( H 2 ) = σ ( v 1 v 2 ) σ ( v 3 r t ) σ ( v 4 r 1 ) σ ( v 1 r 1 ) σ ( v 2 r t ) σ ( v 3 v 4 ) = σ ( C 2 ) σ ( C 1 ) . In particular, σ ( H 1 ) = σ ( H 2 ) if and only if σ ( C 1 ) = σ ( C 2 ). By h ypothesis, σ ( C 1 )  = σ ( C 2 ), so σ ( H 1 )  = σ ( H 2 ). Since each circle-sign is ± 1, this implies σ ( H 1 ) = − σ ( H 2 ). Hence Σ con tains one positive Hamil tonian circle and one negative Hamiltonian circle. □ 22 XIYONG Y AN 3. A cknowledgment The author thanks Professor Thomas Zasla vsky for his in v aluable suggestions and recom- mendations during the preparation of this draft. References [1] Thomas Zaslavsky . Negative (and p ositive) circles in signed graphs: A problem collection. AKCE In t. J. Graphs Com binatorics, 15 (2018), no. 1, 31–48. MR 3803228. Zbl 1390.05085. [2] Thomas Zaslavsky . Signed graphs. Discrete Appl. Math., 4 (1982), 47–74. MR 84e:05095a. Zbl 476.05080. Erratum, ibid., 5 (1983), 248. MR 84e:05095b. Zbl 503.05060. 89 P ark A ve Apt 29, Binghamton, NY, USA, 13903.