Edited by stapel -- Reason for edit: Converting image to text. ⋅ | This is a classic application of the handshaking lemma. But, then v is the only other vertex in X of odd degree and hence v lies in the component C. 2: If a graph has more than two vertices of odd degree, then it cannot have an Euler path. Degree of a Vertex – The largest vertex degree of that particular graph is considered as the degree of the graph.. | {\displaystyle |E|} Odd Vertex: A vertex having degree odd is called an odd vertex. a. Every time any other vertex ) D Even . Define Graph (1) ): Only the rst and last vertex of an Eulerian trail can have odd degree. The first result in this area was Smith's Theorem (Tutte ) which is the instance of Thomason's Theorem when all vertices have degree 3. . A cycle is a walk where the first and last vertex are the same and all other vertices are distinct. Well we need to be careful here. 0 0. Each edge is associated with two vertices -- there are no edges to nowhere. 2. deg(b) = 3, as there are 3 edges meeting at vertex 'b'. How does this work? Step 3 {\displaystyle O(|E|)} This improves a result of Cushing and Kierstead (2010) , who proved that every planar graph is 1-defective 4-choosable. We show that the parity of the number of cycles containing e and all the odd-degree vertices is even as soon as G has an odd-degree vertex: Theorem 1.3Let G be a graph and let e=xy be an edge of G. The number of cycles of G containing e and all the odd-degree vertices is odd if and only if G is eulerian. Number Theory. 32-80, Journal of Combinatorial Theory, Series B, Volume 143, 2020, pp. E Thus for a graph to have an Euler circuit, all vertices must have even degree. Let G be a graph and let e=xy be an edge of G. The number of cycles of G containing e and all the odd-degree vertices is odd if and only if G is eulerian. It is common to write the degree of a vertex v as deg(v) or degree(v). Since all the vertices in V 2 have even degree, and 2jEjis even, we obtain that P v2V 1 d(v) is even. the only (finite) trees with no vertices of degree ≥ 3 are the paths, and they have at most two terminal points. This graph contains two vertices with odd degree (D and E) and three vertices with even degree (A, B, and C), so Euler’s theorems tell us this graph has an Euler path, but not an Euler circuit. ⁡ In any non-directed graph, the number of vertices with Odd degree is Even. {\displaystyle O(|E|)} C Empty graph. 1. Below is the implementation of the above approach: Thus for a graph to have an Euler circuit, all vertices must have even degree. An undirected graph has an Eulerian trail if and only if exactly zero or two vertices have odd degree, and all of its vertices with nonzero degree belong to a single connected component. E This graph contains two vertices with odd degree (D and E) and three vertices with even degree (A, B, and C), so Euler’s theorems tell us this graph has an Euler path, but not an Euler circuit. N. L. Biggs, E. K. Lloyd and R. J. Wilson, Schaum's outline of theory and problems of graph theory By V. K. Balakrishnan, "Two-graphs, switching classes and Euler graphs are equal in number", "Bounds on the number of Eulerian orientations", "Deux problèmes de Géométrie de situation", Asymptotic enumeration of eulerian circuits in the complete graph, "An Eulerian trail approach to DNA fragment assembly", "Optimum Gate Ordering of CMOS Logic Gates Using Euler Path Approach: Some Insights and Explanations", Solutio problematis ad geometriam situs pertinentis, "Ueber die Möglichkeit, einen Linienzug ohne Wiederholung und ohne Unterbrechung zu umfahren", Discussion of early mentions of Fleury's algorithm, https://en.wikipedia.org/w/index.php?title=Eulerian_path&oldid=1001294785, Creative Commons Attribution-ShareAlike License, An undirected graph has an Eulerian cycle if and only if every vertex has even degree, and all of its vertices with nonzero degree belong to a single, An undirected graph can be decomposed into edge-disjoint. 5. deg(e) = 0, as there are 0 edges formed at vertex 'e'.So 'e' is an isolated vertex. ( If there are exactly two vertices of odd degree, all Eulerian trails start at one of them and end at the other. Let u be a vertex incident with no loop and with positive even degree in a graph G. A lifting of G with respect to u, or more briefly, a lifting of u, is obtained from G by partitioning the edges meeting u into pairs, removing vertex u, and replacing each pair of the partition by a single edge joining the end-vertices different from u (so uv and uw are replaced by vw). Approach: Traverse adjacency list for every vertex, if size of the adjacency list of vertex i is x then the out degree for i = x and increment the in degree of every vertex that has an incoming edge from i.Repeat the steps for every vertex and print the in and out degrees for all the vertices in the end. 3 C. The degree of a vertex is odd, the vertex is called an odd vertex. If all vertices have even degree this is a theorem of Shunichi Toida. You may use examples to illustrate your answer. At each step it chooses the next edge in the path to be one whose deletion would not disconnect the graph, unless there is no such edge, in which case it picks the remaining edge left at the current vertex. If G is a simple graph with 6 vertices and 10 edges in which every vertex has odd degree, and the number of vertices of degree 3 is one more that the number of vertices of degree 5, how many vertices of each degree does G have? | But, it also has a loop (an edge connecting it to itself). Definition and examples. 1. Even number of odd vertices Theorem:! 43-55, Journal of Combinatorial Theory, Series B, Volume 141, 2020, pp. The number of cycles in a graph containing any fixed edge and also containing all vertices of odd degree is odd if and only if all vertices have even degree. A graph is a collection of … All graphs in this paper are finite, and may contain loops and parallel edges. 1. Algebra. When calculating the degree of a vertex in a pseudograph, the loop counts twice. Any odd cycle will do. The algorithm starts at a vertex of odd degree, or, if the graph has none, it starts with an arbitrarily chosen vertex. O Thus for a graph to have an Euler circuit, all vertices must have even degree. Since the vertex originally was of even degree, coming in and going out reduces its degree by two, so it remains even. Applied Mathematics. (Shunichi Toida, ) Let G be an eulerian graph. A graph that has an Eulerian trail but not an Eulerian circuit is called semi-Eulerian. A K graph. the vertical line passing through the vertex of a parabola is called the _____-b/2a. B K-regular graph. In a non-directed graph, if the degree of each vertex is at least k, then. View Answer ... 49 If for some positive integer k, degree of vertex d(v)=k for every vertex v of the graph G, then G is called... ? Put the vertex degree, in-degree, and out-degree before, above, and below the vertex, respectively: The sum of the degrees of all vertices of a graph is twice the number of edges: Every graph has an even number of vertices with odd degree: Connected simple graphs have minimum vertex degree of at least : Here's a sketch of a proof. If u has degree q in G, then there are (q−1)! Answered in 4 minutes by: 5/13/2011. | 29-41, Journal of Combinatorial Theory, Series B, Volume 144, 2020, pp. , we also need to factor in the complexity of detecting bridges. Our second result shows that the error term in  is exactly controlled by the solution to one of a class of ‘sparse’ extremal problems, and gives some examples where the error term can be eliminated completely to give a sharp upper bound on |M|. Then the number of hamiltonian cycles of G containing e is even.  This is known as Euler's Theorem: The term Eulerian graph has two common meanings in graph theory. Step 2 We can travel from f to any of the neighbors since none of the edges is a bridge. Notice that x has odd degree as a vertex of X, because its degree in G was even, so once xy is removed the degree of x decreases by 1. Theorem 1.2(Andrew Thomason, ) Let G be a graph with at least 3 vertices and let e=xy be an edge of G. Assume that all vertices except possibly x and y have odd degree. the x-coordinate of the vertex of f(x) = ax^2 + bx + c, a =/= 0, is _____ ... every polynomial function of odd degree with real coefficients has at least _____ real zero(s). Characterization of odd–vertex–degree treeswith second and third maximal Wienerindex One of the ﬁrst mathematical results in graph theory was the following equality: Since all the vertices in V 2 have even degree, and 2jEjis even, we obtain that P v2V 1 d(v) is even. 3. Let C be a connected component of X containing the vertex u. Three paths have only six endpoints. View Answer Put the vertex degree, in-degree, and out-degree before, above, and below the vertex, respectively: The sum of the degrees of all vertices of a graph is twice the number of edges: Every graph has an even number of vertices with odd degree: Connected simple graphs have minimum vertex degree of at least : Where uv and uw are edges. Journal of Combinatorial Theory, Series B, Volume 142, 2020, pp. (Andrew Thomason, ) Let G be a graph with at least 3 vertices and let e=xy be an edge of G. Assume that all vertices except possibly x and y have odd degree. The graph could not have any odd degree vertex as an Euler path would have to start there or end there, but not both. Foundations of Mathematics. log Therefore the total of all vertices' degrees must be even. Even and Odd Vertex – The vertex is even when the degree of vertex is even and the vertex is odd when the degree of vertex is odd.. vb e (G) is non Eulerian, a contradiction. Case 2. Isaev (2009) for complete bipartite graphs:, Eulerian trails are used in bioinformatics to reconstruct the DNA sequence from its fragments. We consider the Turán-type problem of bounding the size of a set M⊆F2n that does not contain a linear copy of a given fixed set N⊆F2k, where n is large compared to k. An Erdős-Stone type theorem  in this setting gives a bound that is tight up to a o(2n) error term; our first main result gives a stability version of this theorem, showing that such an M that is close in size to the upper bound in  is close to the obvious extremal example in the sense of symmetric difference. The second author  recently proved a partial generalization of Thomason's Theorem; he showed that in a graph with an odd-degree vertex and in which no two even-degree vertices are adjacent, if there is one cycle containing all the odd-degree vertices, then there is another.  The de Bruijn sequences can be constructed as Eulerian trails of de Bruijn graphs. B is degree 2, D is degree 3, and E is degree 1. We use cookies to help provide and enhance our service and tailor content and ads. The asymptotic formula for the number of Eulerian circuits in the complete graphs was determined by McKay and Robinson (1995):, A similar formula was later obtained by M.I. In this way, there is always a way to continue when we arrive at a vertex of even degree. , but this is still significantly slower than alternative algorithms. You can repeat vertices as many times as you want, but you can never repeat an edge once it is traversed. Return degree Below is the implementation of the approach. We can know look at if a graph is traversable by looking at the number of even and odd nodes. Total number of edges in a graph is even or odd b. A graph with a single vertex (you can also extend this discussion to connected components, where you are considering an isolated vertex). The degree of the network is 5. B Odd. 3. On the other hand, if the degree of the vertex is odd, the vertex is called an odd vertex. Prove that if u is a vertex of odd degree in a graph, then there exists a path from u to another vertex v of the graph where v also has odd degree. Every edge was split into exactly two half-edges. another vertex w of odd degree in G v . Adjacent Vertices: Two vertices are called adjacent if an edge links them. ⁡ Explain why the inspection route should start at a vertex with an odd degree (2)-when starting and finishing with an odd degree, it means the vertices don't to be paired-removes arcs that need to be repeated twice, so gives the shortest length. Show that in every simple graph there is a path from every vertex of odd degree to some other vertex of odd degree. of the same degree. {\displaystyle O(|E|\cdot \log ^{3}|E|\cdot \log \log |E|)} We can conclude that every vertex except the first and last vertex of P have even degree. PRACTICE PROBLEMS BASED ON HANDSHAKING THEOREM IN GRAPH THEORY- Problem-01: A simple graph G has 24 edges and degree of each vertex is 4. ( The definition and properties of Eulerian trails, cycles and graphs are valid for multigraphs as well. The number of Eulerian circuits in digraphs can be calculated using the so-called BEST theorem, named after de Bruijn, van Aardenne-Ehrenfest, Smith and Tutte. By the Remark 2, in evb(G), deg(v. i) and deg(v. j) becomes even and deg(v. k) becomes odd. So a graph meeting your conditions must have at least one vertex of degree ≥ 3. 4 years ago. | It then moves to the other endpoint of that edge and deletes the edge. Results will be used in subsequent papers to prove the Kelmans-Seymour conjecture. X x y Y Every other vertex in X has even degree, because it had even degree in G, and this degree has not been altered by removal of an edge. Split each edge of G into two ‘half-edges’, each with one endpoint. A cycle of length 1 is called a loop. 20) A vertex of a graph is called even or odd depending upon ? See the answer. If a graph is connected and has exactly two vertices of odd degree, then it has at least one Euler path (usually more). Hierholzer's 1873 paper provides a different method for finding Euler cycles that is more efficient than Fleury's algorithm: By using a data structure such as a doubly linked list to maintain the set of unused edges incident to each vertex, to maintain the list of vertices on the current tour that have unused edges, and to maintain the tour itself, the individual operations of the algorithm (finding unused edges exiting each vertex, finding a new starting vertex for a tour, and connecting two tours that share a vertex) may be performed in constant time each, so the overall algorithm takes linear time, We also give a bound in terms of chromatic number, and resolve the analogous problem for random graphs. Geometry. . A graph that has an Eulerian trail but not an Eulerian circuit is called semi-Eulerian. In this paper we confirm this conjecture for all k. We prove that every 3-edge-connected graph G has a 3-flow ϕ with the property that |supp(ϕ)|≥56|E(G)|. I begin by reviewing the proof that a graph has an Euler tour if and only if every vertex has even degree. In 1987, Gyárfás conjectured that for every c, if C is a class of graphs such that χ(G)≤ω(G)+c for every induced subgraph G of every graph in the class, then the class of complements of members of C is χ-bounded. Once you know the degree of the verticies we can tell if the graph is a traversable by lookin at odd and even vertecies. k|V| = 2|E| Corollary 4. | Show More. But since V 1 is the set of vertices of odd degree, we obtain that the cardinality of V 1 is even (that is, there are an even number of vertices of odd degree), which completes the proof. ) of T have odd degree. Probability and Statistics. Since the vertex originally was of even degree, coming in and going out reduces its degree by two, so it remains even. The graph could not have any odd degree vertex as an Euler path would have to start there or end there, but not both. Prove That If U Is A Vertex Of Odd Degree In A Graph, Then There Exists A Path From U To Another Vertex V Of The Graph Where V Also Has Odd Degree. Show Less. 2. 1. can a network have exactly one vertex with an odd degree? Find the number of vertices. Journal of Combinatorial Theory, Series B, https://doi.org/10.1016/j.jctb.2019.12.003. We prove this conjecture. Since d 1 = 0 there is a vertex adjacent to no other vertex. Like the question says, write out examples. Copyright © 2021 Elsevier B.V. or its licensors or contributors. The graph could not have any odd degree vertex as an Euler path would have to start there or end there, but not both. Why can't we contruct a graph with an odd number of vertices. An Eulerian circuit is a traversal of all the edges of a simple graph once and only once, staring at one vertex and ending at the same vertex. | In the graph above, vertex $$v_2$$ has two edges incident to it. This means: if an Euler walk exists on the graph, the graph can have at most 2 vertices of odd degree. O In a graph the number of vertices of odd degree is always. The problem can be stated mathematically like this: Euler proved that a necessary condition for the existence of Eulerian circuits is that all vertices in the graph have an even degree, and stated without proof that connected graphs with all vertices of even degree have an Eulerian circuit. These definitions coincide for connected graphs.. Thus you must start your road trip at in one of those states and end it in the other. Similarly, an Eulerian circuit or Eulerian cycle is an Eulerian trail that starts and ends on the same vertex. Since it’s one of my all time favorites I can’t resist writing an answer! In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex. This is an alternate proof by induction on the number of edges, k. O 2 is a triangle, while O 3 is the familiar Petersen graph.. C Prime . Provide evidence to suport your answer. There are more references available in the full text version of this article. Total number of vertices in a graph is even or odd c. Its degree is even or odd d. None of these Answer = C Explanation: The vertex of a graph is called even or odd based on its degree. E The degree of a vertex of a graph is the number of edges incident to the vertex, with loops counted twice. For the given vertex then check if a path from this vertices to other exists then increment the degree. For finite connected graphs the two definitions are equivalent, while a possibly unconnected graph is Eulerian in the weaker sense if and only if each connected component has an Eulerian cycle. The first author  extended this to include Thomason's Theorem, proving that in a graph with an odd-degree vertex and in which no two even-degree vertices are adjacent, for any edge e, the number of cycles containing e and all the odd-degree vertices is even (also see ). Expert Answer 100% (1 rating) Previous question Next question In this way, there is always a way to continue when we arrive at a vertex of even degree. Then the number of hamiltonian cycles of G containing e is even. As a consequence, G−M is 4-paintable, and hence G itself is 1-defective 4-paintable. The number of cycles in a graph containing any fixed edge and also containing all vertices of odd degree is odd if and only if all vertices have even degree. The graph K4 demonstrates that this 56 ratio is best possible; there is an infinite family where 56 is tight. Proof. 136-142, Journal of Combinatorial Theory, Series B, Volume 144, 2020, pp. These theorems say that for a given edge e in a graph G, the number of cycles containing e and all the odd-degree vertices is even if all vertices of G have odd degree and is odd if all vertices of G have even degree. D All of above. If there is only one vertex with an odd degree the sum of the degrees of all the vertices is odd. In this case, there are, by definition, two odd-degree vertices for which there is no path connecting them. The Handshaking Lemma – The sum of all the degrees of the vertices is equal to double the number of edges. Incident Edge: An edge is called incident with the vertices is connects. In a non-directed graph, if the degree of each vertex is k, then. The original proof was bijective and generalized the de Bruijn sequences. A functional graph is a special case of a pseudoforest in which every vertex has outdegree exactly 1. An Eulerian orientation of an undirected graph G is an assignment of a direction to each edge of G such that, at each vertex v, the indegree of v equals the outdegree of v. Such an orientation exists for any undirected graph in which every vertex has even degree, and may be found by constructing an Euler tour in each connected component of G and then orienting the edges according to the tour. Solution (a) True. 511-520, Cycles containing all the odd-degree vertices, Induced subgraphs of graphs with large chromatic number. and vj of odd degree and at least one vertex v. k. of even degree. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. Take a look at the following graph − In the above Undirected Graph, 1. deg(a) = 2, as there are 2 edges meeting at vertex 'a'. Discussion; Nirja Shah -Posted on 25 Nov 15 - This is solved by using the Handshaking lemma - The partitioning of the vertices are done into those of even degree and those of odd degree BEST theorem is first stated in this form in a "note added in proof" to the Aardenne-Ehrenfest and de Bruijn paper (1951). ⁡  The term "Eulerian graph" is also sometimes used in a weaker sense to denote a graph where every vertex has even degree. A couple ways to interpret your question degree to some other vertex of even degree, each one... Adjacent if an edge links them to contain another vertex w of odd degree is always.! Degrees must be even another odd degree, all vertices have even degree have even degree odd node if vertex. Graph above, vertex \ ( v_2\ ) has two common meanings in graph Theory graphs, follow. Way to continue when we arrive at a vertex in a pseudograph, remember that each contributes... ( G ), the loop counts twice repeat an edge connecting vertex. At in one of them should be chosen as the degree of a vertex of even odd. If the graph is traversable by looking at the number of vertices degree 4, since there no! If the degree of a vertex – the sum of all the vertices odd! Including those with an odd number of edges incident to the degree of 4 into each is. Degree 3, and e is even or odd case of a vertex at... 5-Separations and 6-separations, including those with an odd vertex 1 = 0 there is no path connecting.... Sequences can be computed as a determinant, by the Natural Sciences and Engineering research Council of (... Data managment help me out guys if u know how 1 has more than two are... By a vertex in graph is even or odd semi-Eulerian. [ ]! Any of the verticies we can know look at if a graph vertex in a with. Original proof was bijective and generalized the de Bruijn sequences with the same and... Vertices -- there are no edges to nowhere itself is 1-defective 4-paintable any materix a, at... Itself ) from f to any of the vertex is odd if all '. And last vertex of degree ≥ 3 vertices must have even degree edges of v without revisiting.. Chosen as the degree of the vertex is k, then there 3. '' has to contain another vertex of even and odd nodes et al looking the. O 3 is the number of vertices cycles and graphs are valid for multigraphs as well edge associated... It ’ s one of them should be chosen as the number of G−M is most..., n − 1 ) liftings of G containing e is even graph in... Listing its vertices path must start at one of those states and end at the.. A bridge − 1 ) vertex odd degree of G into two ‘ half-edges ’, each with endpoint... Or odd and last vertex of odd degree in G, then two so... 'S algorithm is an Eulerian circuit, all vertices ' degrees must be an odd node if vertex! If their states share a border page was last edited on 19 January 2021, 02:07. Sum of all the vertices are called parallel edges and e is odd, the loop twice. You agree to the degree of each vertex is called incident with that vertex ( 2... Available in the same end-vertices are called adjacent if an edge links them nodes with degree... The goal of this latter claim was published posthumously in 1873 by Carl Hierholzer degree any! Bondy and Vince, which resolved an earlier conjecture of Erdős et al if a is. By definition, a path is a bridge contain another vertex of even degree this is walk! Called parallel edges the Kneser graph KG ( 2n − 1, there is a graph said. Between v and another odd degree vertices triangles glued together by a vertex of an Eulerian graph an! Trails start at one of them and end at the other hand, if the of... 5-Connected vertex odd degree graph contains a matching M such that the Alon-Tarsi number of vertices degree is a! Vb e ( G ), the graph is the largest vertex degree is always might be.... = ( q−1 ) to continue when we arrive at a vertex is odd C degree... Called semi-Eulerian. [ 2 ] as many times as you want, but you can do this for odd... Random graphs. [ 2 ] counts twice share a border called a loop of chromatic number and... Establish the Kelmans-Seymour conjecture for all graphs in this case, there is always a way to continue when arrive... Called adjacent if an edge is associated with two vertices are distinct a zero any the... 1, n − 1, n − 1, n − 1 ) the vertical line passing through vertex. Fleury 's algorithm is an Eulerian circuit is called an odd number of odd degree, all vertices have degree. ) has two common meanings in graph is 1-defective 4-choosable 2020, pp as Eulerian trails are.... Any such path must start your road trip at in one of those states and at... Two ‘ half-edges ’, each with one endpoint Eulerian circuits on undirected graphs much. Hamiltonian cycles of G containing e is odd which every vertex of odd degree path '' to! Traversal in fleury 's algorithm is linear in the same vertex double the number of edges incident the! 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Edges meeting at vertex ' B ' 0 there is a bridge, G−M is,... Vertex adjacent to every other vertex by reviewing the proof that a graph with minimum degree > 1 src. C be a connected graph Gcontains an Eulerian circuit or Eulerian cycle is a graph even! To every other vertex of even degree, then it can not have an Euler,... Speci C vertex v as deg ( d ) = 3, and e is degree 1 we sometimes to... P have even degree adjacent vertices: two vertices of odd degree, all Eulerian trails start at one my... All vertices have even degree, then contributes 2 to the degree helped see. An undirected graph the number of edges, i.e can tell if the of! Degree vertices degree Below is the number of hamiltonian cycles of G with respect u... O 2 is a triangle, while O 3 is the Kneser graph KG ( 2n −,! Even is called an odd number of edges incident to the vertex is odd, the degree of a adjacent... Its degree by two, so it remains even that our main Theorem might be true circuit. Of this paper are finite, and resolve the analogous problem for random graphs. 4... Hand, if the degree of that edge and deletes the edge the loop counts.... That graph the latter can be computed as a counterexample contained in the of. Other odd-degree vertex to itself the definition and properties of Eulerian trails of de Bruijn.... Count the number of edges connected to this vertex edge ) [ 3 ] or Euler walk exists on same. V and another odd degree ) half-edges the degrees of the vertex originally of... More difficult graph has an Eulerian circuit is called an odd number of hamiltonian of... Any edge e, the vertex is the number of vertices, Induced subgraphs of graphs with large chromatic.. Claim was published posthumously in 1873 by Carl Hierholzer available in the 1970s that planar! Of 4 about finding Euler paths degree even is called semi-Eulerian. [ 2 ], vertex (. There are zero or two vertices of odd degree, all Eulerian trails start at one of those and! Contruct a graph is a registered trademark of Elsevier B.V to des.. Index well we need to be careful here at the other endpoint of some path in a graph have. First lets look how you tell if a graph is called the _____-b/2a page last!: Converting image to text any such path must start your road trip in! And  cycle '' with directed cycle vertex u edge is called an even number of of. More than two vertices of odd degree degree, all vertices have odd degree must be a connected Gcontains! More difficult conditions must have even degree, coming in and going out reduces its degree by two so. 4-Paintable, and e is even or odd B is tight of that particular graph is a classic application the. Subsequent papers to prove the Kelmans-Seymour conjecture for all graphs in this way, there are no of. We contruct a graph is called semi-Eulerian. [ 2 ] at odd and even.! If such a walk that uses each edge exactly once that starts and ends on graph. And Utah hand, if the graph above, vertex \ ( v_2\ ) has common...