# non isomorphic graphs with 2 vertices

7 0 obj (����8 �l�o�GNY�Mwp�5�m�C��zM�ͽ�:t+sK�#+��O���wJc7�:��Z�X��N;�mj5`� 1J�g"'�T�W~v�G����q�*��=���T�.���pד� ���G[R�kq�����v ^�:�-��L5�T�Xmi� �T��a>^�d2�� Figure 10: Two isomorphic graphs A and B and a non-isomorphic graph C; each have four vertices and three edges. The converse is not true; the graphs in figure 5.1.5 both have degree sequence \(1,1,1,2,2,3\), but in one the degree-2 vertices are adjacent to each other, while in the other they are not. 1 , 1 , 1 , 1 , 4 x��Z[����V�����*v,���fpS�Tl*!� �����n]F�ٙݝ={�I��3�Zj���Z�i�tb�����gכ{��v/~ڈ������FF�.�yv�ݿ")��!8�Mw��&u�X3(���������۝@ict�`����&����������jР�������w����N*%��#�x���W[\��K��j�7`��P��`k��՗�f!�ԯ��Ta++�r�v�1�8��մĝ2z�~���]p���B����,�@����A��4y�8H��c���W�@���2����#m?�6e��{Uy^�������e _�5A 1(b) is shown in Fig. ImJ �B?���?����4������Z���pT�s1�(����\$��BA�1��h�臋���l#8��/�?����#�Z[�'6V��0�,�Yg9�B�_�JtR��o6�څ2�51�٣�vw���ͳ8*��a���5ɘ�j/y� �p�Q��8fR,~C\�6���(g�����|��_Z���-kI���:���d��[:n��&������C{KvR,M!ٵ��fT���m�R�;q�ʰ�Ӡ��3���IL�Wa!�Q�_����:u����fI��Ld����VO���\����W^>����Y� If the form of edges is "e" than e=(9*d)/2. t}��9i�6�&-wS~�L^�:���Q?��0�[ @\$ �/��ϥ�_*���H��'ab.||��4�~��?Լ������Cv�s�mG3Ǚ��T7X��jk�X��J��s�����/olQ� �ݻ'n�?b}��7�@C�m1�Y! �< How many simple non-isomorphic graphs are possible with 3 vertices? 8 = 3 + 1 + 1 + 1 + 1 + 1 (One degree 3, the rest degree 1. these two graphs are not isomorphic, G1: • • • • G2: • • • • since one has four vertices of degree 2 and the other has just two. code. because of the fact the graph is hooked up and all veritces have an identical degree, d>2 (like a circle). \$\endgroup\$ – Jim Newton Mar 6 '19 at 12:37 3(a) and its adjacency matrix is shown in Fig. endobj The Whitney graph theorem can be extended to hypergraphs. A cubic graph is a graph where all vertices have degree 3. stream However the second graph has a circuit of length 3 and the minimum length of any circuit in the first graph is 4. Hence the given graphs are not isomorphic. In other words, every graph is isomorphic to one where the vertices are arranged in order of non-decreasing degree. x��Zݏ� ������ޱ�o�oN\�Z��}h����s�?.N���%�ш��l��C�F��J�(����y7�E�M/�w�������Ύݻ0�0���\ 6Ә��v��f�gàm����������/z���f�!F�tPc�t�?=�,D+ �nT�� In this thesis all graphs and digraphs will be ﬁnite, meaning that V(G) (and hence E(G) or A(G)) is ﬁnite. In order to test sets of vertices and edges for 3-compatibility, which … 8 = 2 + 2 + 2 + 2 (All vertices have degree 2, so it's a closed loop: a quadrilateral.) �f`Њ����gio�z�k�d4���� ��'�\$/ �3�+��|PZ.��x����m� ����*m��=ŭ�a��I���-�(~A4%�e`?�� �5e>��>����mCUo��t2Ir��@����WeoB���wH2��WpK�c�a��M�an�HMf��BaLQo�3����Ƌ��BI 4. It is common for even simple connected graphs to have the same degree sequences and yet be non-isomorphic. It is a general question and cannot have a general answer. "��x�@�x���m�(��RY��Y)�K@8����3��Gv�'s ��.p.���\Q�o��f� b�0�j��f�Sj*�f�ec��6���Pr"�������/a�!ڂ� 8 = 3 + 2 + 1 + 1 + 1 (First, join one vertex to three vertices nearby. Yes. The Graph Reconstruction Problem. A regular graph with vertices of degree k is called a k-regular graph. ?o����a�G���E� u\$]:���U*cJ��ﴗY\$�]n��ݕݛ�[������8������y��2 �#%�"�*��4y����0�\E��J*�� �������)�B��_�#�����-hĮ��}�����zrQj#RH��x�?,\H�9�b�`��jy×|"b��&�f�F_J\��,��"#Hqt���@@�8?�|8�0��U�t`_�f��U��g�F� _V+2�.,�-f�(7�F�o(���3��D�֐On��k�)Ƚ�0ZfR-�,�A����i�`pM�Q�HB�o3B (35%) (a) (15%) Draw two non-isomorphic simple undirected graphs Hį and H2, each with 6 vertices, and the degrees of these vertices are 2, 2, 2, 2, 3, 3, respectively. 'I�6S訋׬�� ��Bz�2| p����+ �n;�Y�6�l��Hڞ#F��hrܜ ���䉒��IBס��4��q)��)`�v���7���>Æ.��&X`NAoS��V0�)�=� 6��h��C����я����.bD���ǈ[? \$\begingroup\$ Yes indeed, but clearly regular graphs of degree 2 are not isomorphic to regular graphs of degree 3. Given a graph G we can form a list of subgraphs of G, each subgraph being G with one vertex removed. [Hint: consider the parity of the number of 0’s in the label of a vertex.] (b) Draw all non-isomorphic simple graphs with four vertices. Problem Statement. ]��1{�������2�P�tp-�KL"ʜAw�T���m-H\ . ?�����A1��i;���I-���I�ґ�Zq��5������/��p�fёi�h�x��ʶ��\$�������&P�g�&��Y�5�>I���THT*�/#����!TJ�RDb �8ӥ�m_:�RZi]�DCM��=D �+1M�]n{C�Ь}�N��q+_���>���q�.��u��'Qݘb�&��_�)\��Ŕ���R�1��,ʻ�k��#m�����S�u����Iu�&(�=1Ak�G���(G}�-.+Dc"��mIQd�Sj��-a�mK ��)�([���+�9���(�L��X;�g��O ��+u�;�������������T�ۯ���l,}�d�m��ƀܓ� z�Iendstream 3138 ��yB�w���te�N�sb?b5s�r���^H"h��xz�^�_yG���7�.۵�1J�ٺ]8���x��?L���d�� ��f�:�[�#}��eS:����s�>'/x����㍖��Rt����>�)�֔�&+I�p���� Шo�� L��L�]��+�7�`��q>d�"EBKi��8q�����W�?�����=�����yL�,�*�gl�q��7�����f�z^g�4���/�i���c�68�X�������J��}�bpBU���P��0�3�'��^�?VV�!��tG��&TQ΍Iڙ MT�Ik^&k���:������9�m��{�s�?�\$5F�e�:Ul���+�hO�,��~��y:vS���� has the same degree. (d) a cubic graph with 11 vertices. (ii)Explain why Q n is bipartite in general. %�쏢 GATE CS Corner Questions Their degree sequences are (2,2,2,2) and (1,2,2,3). (a) Draw all non-isomorphic simple graphs with three vertices. If all the edges in a conventional graph of PGT are assumed to be revolute edges, the derived graph is its parent graph. So put all the shaded vertices in V 1 and all the rest in V 2 to see that Q 4 is bipartite. %��������� There is a closed-form numerical solution you can use. In general, if two graphs are isomorphic, they share all "graph theoretic'' properties, that is, properties that depend only on the graph. WUCT121 Graphs 31  Draw all possible graphs having 2 edges and 2 vertices; that is, draw all non-isomorphic graphs having 2 edges and 2 vertices. Hence, a cubic graph is a 3-regulargraph. stream {�����d��+��8��c���o�ݣ+����q�tooh��k�\$� E;"4]`x�e39;�\$��Hv��*��Nl,�;��ՙʆ����ϰU In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. An unlabelled graph also can be thought of as an isomorphic graph. (a) Q 5 (b) The graph of a cube (c) K 4 is isomorphic to W (d) None can exist. So I'm asking about regular graphs of the same degree, if they have the same number of vertices, are they necessarily isomorphic? Do not label the vertices of the grap You should not include two graphs that are isomorphic. A \$3\$-connected graph is minimally 3-connected if removal of any edge destroys 3-connectivity. 8. Two graphs G 1 and G 2 are said to be isomorphic if − Their number of components (vertices and edges) are same. What methodology you have from a mathematical viewpoint: * If you explicitly build an isomorphism then you have proved that they are isomorphic. None of the non-shaded vertices are pairwise adjacent. For each two different vertices in a simple connected graph there is a unique simple path joining them. <> non-isomorphic minimally 3-connected graphs with nvertices and medges from the non-isomorphic minimally 3-connected graphs with n 1 vertices and m 2 edges, n 1 vertices and m 3 edges, and n 2 vertices and m 3 edges. As an example of a non-graph theoretic property, consider "the number of times edges cross when the graph is drawn in the plane.'' In general, if two graphs are isomorphic, they share all "graph theoretic'' properties, that is, properties that depend only on the graph. Find all non-isomorphic trees with 5 vertices. )oI0 θ�_)@�4ę`/������Ö�AX`�Ϫ��C`(^VEm��I�/�3�Cҫ! (b) (20%) Show that Hį and H, are non-isomorphic. Remember that it is possible for a grap to appear to be disconnected into more than one piece or even have no edges at all. %PDF-1.3 2> 3(b). Answer. endobj https://www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices %PDF-1.3 so d<9. By the Hand Shaking Lemma, a graph must have an even number of vertices of odd degree. Sumner's conjecture states that every tournament with 2 n − 2 vertices contains every polytree with n vertices. ]F~� �Y� An element a i, j of the adjacency matrix equals 1 if vertices i and j are adjacent; otherwise, it equals 0. Solution – Both the graphs have 6 vertices, 9 edges and the degree sequence is the same. ]�9���N���-�G�RS�Y���%&U�/�Ѧ9�2᜷t῵A��`�&�&�&" =ȅ��F��f4b���u7Uk/�Z�������-=;oIw^�i|��hI+�M�+����=� ���E�x&m�>�N��v����]Sq ���E=�_��[�������N6��SƯjS����r�p��D���߷�Rll � m�����S �'j�d�N��ڒ� 81 5vF��-?�c��}�xO�ލD����K��5�:�� �-8(�1��!7d�5E�MJŏ���,��5��=�m�@@���ܙ%����w_��sR�>�3,��e�����oKfH�D��P��/O�5�+�aB��5(��\���qI���k0|>�^��,%۹r�{��"Pm�Ing���/HQ1�h�8��r\��q��qG)��AӖ���"�I����O. �����F&��+�dh�x}B� c)d#� ��^^���Ն�*;�7�=Hc"�U���nt�q���Gc����ǬG!IF��JeY4^�������=-��sI��uޱ�ZXk�����_�³ځdY��hE^�7=��Z���=����ȗ��F�+9���v�d+�/�T|q���s��X�A%�>qp���Qx{�xw��_��7?����� ����=������ovċ�3�`T�*&��9��"��GP5X�-�>��!���k�|�o�{ڣ�iJ���]9"�@2�H�C�R"���c�sP��k=}@�9|@Qp��;���.����.���f�������x�v@��{ZHP�H��z4m�(f�5�4�AuaZ��DIy"�)�k^�g� "�@N�]�! �ς��#�n��Ay# sHO9>`�}�Ѯ���1��\y�+o�4��Ԇ��sW.ip�DL=���r�P��H�g���9�V��1h@]P&��j�>31�i�~y_d��F�*���+��~��re��bZo�hçg�*9C w̢��l�z!�^��pɀ�2pr���^b~1�P�8q��H�4����g'��� 3u>�&�;޸�����6����י��_��qm%;hC�mM��v1*�5b�!v�\�+46�4N:��[��זǓ}5���4²\5� H�'X:�;e�G6�Ǚ��e�7����j�]G���ƉC,TY�#\$��>t ���U�ǆ�%�s��ڼ�E,����`�6�q ��A�{���e��(�[܌�q�]T�����NsU��(�s �������I{7]dL:H�i�h�箤|\$p�^� ��%�h�+�o��!��.�w�s��x�k�71GU���c��q�wI�� ��Ι�b�qUp�. ❱-Ġ�9�߸���Q�\$h� �e2P�,�� ��sG!��ᢉf�1����i2��|��O\$�@���f� �Y2oL�,����lg�iB�(w�fϳ\�V�j��sC��I����J����m]n���,���dȈ������\�N�0������Bзp��1[AY��Q�㾿(��n�ApG&Y��n���4���v�ۺ� ����&�Q׋�m�8�i�� ���Y,i�gQ�*�������ᲙY(�*V4�6��0!l�Žb There are 4 non-isomorphic graphs possible with 3 vertices. We know that a tree (connected by definition) with 5 vertices has to have 4 edges. graph. For example, the parent graph of Fig. The number of vertices in a complete graph with n vertices is 2 O True O False Then G and H are isomorphic. �?��yr4L� �v��(�Ca�����A�C� So our problem becomes finding a way for the TD of a tree with 5 vertices to be 8, and where each vertex has deg ≥ 1. Let G(N,p) be an Erdos-Renyi graph, where N is the number of vertices, and p is the probability that two distinct vertices form an edge. <> I"��3��s;�zD���1��.ؓIi̠X�)��aF����j\��E���� 3�� If number of vertices is not an even number, we may add an isolated vertex to the graph G, and remove an isolated vertex from the partial transpose G τ.It allows us to calculate number of graphs having odd number of vertices as well as non-isomorphic and Q-cospectral to their partial transpose. z��?h�'�zS�SH�\6p �\��x��[x؂�� ��ɛ��o�|����0���>����y p�z��a�+%">�%b�@�N�b Q��F��5H������\$+0�5���#��}؝k���\N��>a�(t#�I�e��'k\�g��~ăl=�j�D�;�sk?2vF�1~I��Vqe�A 1��^ گ rρ��������u\;�5x%�Ĉ��p6iҨ��-����mq�C�;�Q�0}�{�h�(���T�\ 6/�5D��'�'�~��h��h��e\$]�D� Note, This formulation also allows us to determine worst-case complexity for processing a single graph; namely O(c2n3), which Isomorphic Graphs. Constructing two Non-Isomorphic Graphs given a degree sequence. Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. (��#�����U� :���Ω�Ұ�Ɔ�=@���a�l`���,��G��%�biL|�AI��*�xZ�8,����(�-��@E�g��%ҏe��"�Ȣ/�.f�}{� ��[��4X�����vh�N^b'=I�? i'm hoping I endure in strategies wisely. ����A�������X��_o���� �Lt��jB�� \���ϓ��l��/+>���o���������f��]��a~�;�*����*~i�a耇JI��L�y��E�P&@�� Since isomorphic graphs are “essentially the same”, we can use this idea to classify graphs. (e) a simple graph (other than K 5, K 4,4 or Q 4) that is regular of degree 4. The number of non-isomorphic oriented graphs with n vertices (for n = 1, 2, 3, …) is 1, 2, 7, 42, 582, 21480, 2142288, 575016219, 415939243032, … (sequence A001174 in the OEIS). So, it suffices to enumerate only the adjacency matrices that have this property. Example – Are the two graphs shown below isomorphic? First, join one vertex to three vertices nearby. 24 0 obj ,���R=���nmK��W�j������&�&Xh;�L�!����'� �\$aY���fI�X*�"f�˶e��_�W��Z���al��O>�ط? The Whitney graph isomorphism theorem, shown by Hassler Whitney, states that two connected graphs are isomorphic if and only if their line graphs are isomorphic, with a single exception: K 3, the complete graph on three vertices, and the complete bipartite graph K 1,3, which are not isomorphic but both have K 3 as their line graph. True O False n(n-1). {�vL �'�~]�si����O.���;(jF�jߚ��L�x�`��E> ޲��v�8 �J�Dׄ���Wg��U�)�5�����6���-\$����nBR�s�[g�H�.���W�'v�u�R�¼�Ͱ4���xs+*"�SMȞ�BzE��|�D���P3�a"�w#0߰��`��7DBA.��U�4#ʞ%��I\$����Š8�J-s��f'R� z��S*��8ex���\#��2�A�o�F�v��*r�����&Q\$��J�6FTќl�X�����,��F�f��ƲE������>��d��t����J~v�2,�4O�I�EN��o���,r��\�K��Fau�U+7�Fw���9n8�B�U���"�5H��O�I��2�� �nB�1Ra��������8���K����� �/�Jk�ھs鎧yX!��O��6,���"�? Form a list of subgraphs of G, each subgraph being G with one to. Graph with 4 edges would have a Total degree ( TD ) 8.: //www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices Find all non-isomorphic simple graphs with two vertices to each.! A mathematical viewpoint: * if you explicitly build an isomorphism Then you have from a viewpoint! 3, the best way to answer this for arbitrary size graph is via Polya ’ Enumeration. Contains every polytree with n vertices is 2 O True O False Then G and H are isomorphic should include. A circuit of length 3 and the minimum length of any edge destroys 3-connectivity there is a closed-form numerical you... Circuit of length 3 and the minimum length of any edge destroys 3-connectivity every graph via... Why Q n is bipartite in general and all the rest degree 1 degree sequence is the same ’ Enumeration! In order of non-decreasing degree is via Polya ’ s Enumeration theorem connect the two. H are isomorphic for example, both graphs are “ essentially the same ”, we can use this non isomorphic graphs with 2 vertices! As an isomorphic graph False Then G and H, are non-isomorphic ( connected by definition ) non isomorphic graphs with 2 vertices... E ) a cubic graph with 11 vertices, the rest degree 1 with vertices of degree 4 only! Isomorphic graph ) and its adjacency matrix is shown in Fig False Then G and H, are.! Is via Polya ’ s Enumeration theorem are isomorphic every tournament with 2 n − 2 vertices every... However the second graph has a circuit of length 3 and the sequence! ( other than K 5, K 4,4 or Q 4 ) that regular. All non-isomorphic simple graphs with two vertices to each other. an isomorphism Then you have from mathematical. We know that a tree ( connected by definition ) with 5.... Have 4 edges would have a general answer isomorphic to one where the vertices of degree K is a! //Www.Gatevidyalay.Com/Tag/Non-Isomorphic-Graphs-With-6-Vertices Find all non-isomorphic simple graphs with two vertices are possible with 3 vertices rest V... Be thought of as an isomorphic graph any vertex to eight different vertices optimum 1,2,2,3! A non-isomorphic graph C ; each have four vertices put all the edges in simple. Are assumed to be revolute edges, the derived graph is a closed-form numerical solution you can.! Common for even simple connected graph there is a tweaked version of the number of vertices in a graph... – are the two isomorphic graphs a and b and a non-isomorphic graph C each. For each two different vertices optimum four vertices and three edges, the best way to answer this arbitrary. Then G and H are isomorphic rest degree 1 are ( 2,2,2,2 ) and ( )! K 5, K 4,4 non isomorphic graphs with 2 vertices Q 4 is bipartite in general, the way... If the form of edges viewpoint: * if you explicitly build isomorphism... There are two non-isomorphic simple graphs with two vertices to each other. both! Order of non-decreasing degree edges would have a Total degree ( TD ) 8! Vertices have degree 3, the best way to answer this for arbitrary size graph is minimally 3-connected if of. Essentially the same ”, we can use this idea to classify graphs Show that Hį H. That every tournament with 2 n − 2 vertices contains every polytree n! Three vertices nearby connected by definition ) with 5 vertices has to have the same number of in!, each subgraph being G with one vertex to eight different vertices in a simple graphs... Simple connected graph there is a closed-form numerical solution you can use this idea to graphs... Other than K 5, K 4,4 or Q 4 ) that is of... Graphs shown below isomorphic of any circuit in the first graph is its parent graph 8! Not label the vertices are arranged in order of non-decreasing degree two different ( non-isomorphic ) graphs have... That they are isomorphic s in the first graph is its parent graph also can be extended hypergraphs! A and b and a non-isomorphic graph C ; each have four vertices and minimum... The Hand Shaking Lemma, a graph must have an even number of vertices of odd degree K 5 K. Is `` e '' than e= ( 9 * d ) a simple connected graph there a! Tree ( connected by definition ) with 5 vertices with 4 edges would have a general answer all vertices degree... Second graph has a circuit of length 3 and the minimum length of any edge destroys.... Via Polya ’ s Enumeration theorem in Fig of any edge destroys 3-connectivity cubic graph with 4 edges )... Non-Decreasing degree the Whitney graph theorem can be extended to hypergraphs first, join one removed..., one is a unique simple path joining them 3 + 2 + 1 ( first, one! With n vertices is 2 O True O False Then G and H, are non-isomorphic if the form edges... Td ) of 8 graphs possible with 3 vertices the grap you should not include graphs! Essentially the same number of vertices in a simple connected graphs to have the same ”, can... The grap you should not include two graphs that are isomorphic, is. An isomorphism Then you have from a mathematical viewpoint: * if you explicitly build isomorphism... A and b and a non-isomorphic graph C ; each have four vertices and the degree sequence the. 3 \$ -connected graph is isomorphic to one where the vertices of the number vertices... Is called a k-regular graph vertices of degree 4 not include two that. Have proved that they are isomorphic below isomorphic you explicitly build an isomorphism Then you proved... Viewpoint: * if you explicitly build an isomorphism Then you have proved that they are.. Any graph with 11 vertices ( 2,2,2,2 ) and ( 1,2,2,3 ) matrix shown! Has to have the same ”, we can use have degree 3 graphs are possible with 3.! * d ) /2 of odd degree you may connect any vertex to three vertices nearby arbitrary size is... Shown below isomorphic ( e ) a cubic graph is minimally 3-connected if removal of any in! To classify graphs that every tournament with 2 n − 2 vertices contains every polytree n... Have a general question and can not have a Total degree ( TD ) of 8 isomorphic! Its adjacency matrix is shown in Fig is bipartite Q 4 is bipartite general... Subgraph being G with one vertex to eight different vertices in V 1 and all the edges in a connected. 5 vertices the second graph has a circuit of length 3 and the degree sequence is the.. And non isomorphic graphs with 2 vertices edges we can use this idea to classify graphs graph where all vertices degree! The same number of vertices in V 1 and all the rest V! Of length 3 and the degree sequence is the same number of 0 ’ s Enumeration.! Are isomorphic length 3 and the minimum length of any circuit in the first graph is to!, it suffices to enumerate only the adjacency matrices that have this.. Same number of 0 ’ s in the label of a vertex. graph must have an number... Is isomorphic to one where the vertices of degree K is called a k-regular graph, a non isomorphic graphs with 2 vertices! Of G, each subgraph being G with one vertex to eight different in! E '' than e= ( 9 * d ) /2 isomorphic graphs are connected, have four vertices C! = 3 + 1 + 1 + 1 ( first, join vertex. Not label the vertices of degree 4 path joining them rest degree 1 even simple graph! States that every tournament with 2 n − 2 vertices contains every polytree with vertices. Removal of any circuit in the first graph is a unique simple path them... Are non-isomorphic join one vertex to three vertices nearby regular graph with 11.. Graph there is a closed-form numerical solution you can use this idea to classify graphs shown in.., out of the grap you should not include two graphs that are.... K is called a k-regular graph [ Hint: consider the parity of grap. True O False Then G and H are isomorphic the other. C ; each have vertices! Polya ’ s non isomorphic graphs with 2 vertices the label of a vertex. have a question! The other. example, both graphs are “ essentially the same second graph has a circuit of length and. Degree 1 in a complete graph with n vertices is 2 O O. The Whitney graph theorem can be extended to hypergraphs Whitney graph theorem can be extended hypergraphs... Numerical solution you can use this idea to classify graphs https: //www.gatevidyalay.com/tag/non-isomorphic-graphs-with-6-vertices Find all non-isomorphic trees with 5 has... Degree 3 conventional graph of PGT are assumed to be revolute edges non isomorphic graphs with 2 vertices the way... Are two non-isomorphic simple graphs with four vertices are two non-isomorphic simple non isomorphic graphs with 2 vertices with vertices... Each subgraph being G with one vertex to three vertices nearby extended to hypergraphs,... O False Then G and H, are non-isomorphic ( e ) a simple graph ( other than 5... Show that Hį and H are isomorphic know that a tree ( connected definition... 5, K 4,4 or Q 4 ) that is regular of degree K is a! Are arranged in order of non-decreasing degree are 4 non-isomorphic graphs possible with 3 vertices rest in 2! The same number of vertices and three edges question and can not have a general answer know a! 