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A survey on temporal network dynamics with incomplete data


  • Received: 29 May 2022 Revised: 21 July 2022 Accepted: 01 August 2022 Published: 23 August 2022
  • With the development of complex network theory, many phenomena on complex networks, such as infectious disease transmission, information spreading and transportation management, can be explained by temporal network dynamics, to reveal the evolution of the real world. Due to the failure of equipment for collecting data, human subjectivity, and false decisions made by machines when the high accuracy is required, data from temporal networks is usually incomplete, which makes the samples unrepresentative and the model analysis more challenging. This survey concentrates on the pre-processing strategies of incomplete data and overviews two categories of methods on data imputation and prediction, respectively. According to whether each layer in temporal networks has the coupling process, this survey overviews the dynamic modeling approaches in terms of both a single process and coupling processes on complex temporal networks. Moreover, for complex temporal networks with incomplete data, this survey summarizes various characteristic analysis methods, which concentrate on critical nodes identification, network reconstruction, network recoverity, and criticality. Finally, some future directions are discussed for temporal networks dynamics with incomplete data.

    Citation: Xing Wu, Shuai Mao, Luolin Xiong, Yang Tang. A survey on temporal network dynamics with incomplete data[J]. Electronic Research Archive, 2022, 30(10): 3786-3810. doi: 10.3934/era.2022193

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  • With the development of complex network theory, many phenomena on complex networks, such as infectious disease transmission, information spreading and transportation management, can be explained by temporal network dynamics, to reveal the evolution of the real world. Due to the failure of equipment for collecting data, human subjectivity, and false decisions made by machines when the high accuracy is required, data from temporal networks is usually incomplete, which makes the samples unrepresentative and the model analysis more challenging. This survey concentrates on the pre-processing strategies of incomplete data and overviews two categories of methods on data imputation and prediction, respectively. According to whether each layer in temporal networks has the coupling process, this survey overviews the dynamic modeling approaches in terms of both a single process and coupling processes on complex temporal networks. Moreover, for complex temporal networks with incomplete data, this survey summarizes various characteristic analysis methods, which concentrate on critical nodes identification, network reconstruction, network recoverity, and criticality. Finally, some future directions are discussed for temporal networks dynamics with incomplete data.



    All graphs considered in this paper are finite, simple, and undirected. Given a graph G, we use V(G), E(G), and F(G) to denote the vertex set of G, edge set of G and face set of G, respectively. We say that two faces are intersecting if they have at least one vertex in common. Let g(G) denote the girth of G, which is the length of a shortest cycle in G.

    Given a graph G, let Gi be a class of graphs for 1im. A (G1, G2, , Gm)-partition of a graph G is the partition of V(G) into m sets V1,V2,,Vm, such that for all 1im, the induced subgraph G[Vi] belongs to Gi. We use I, Ok, Pk, F and Fd to denote the class of edgeless graphs (independent sets), the class of graphs whose components have order at most k, the class of graphs whose components are paths of order at most k, the class of forests and the class of forests with maximum degree d. In particular, an (I, Ok)-partition of a graph G is the partition of V(G) into two non-empty subsets V1 and V2, such that G[V1] is an edgeless graph and G[V2] is a graph with components of order at most k. A planar graph G, equipped with a drawing in the plane so that two edges intersect only at their ends, ia called a plane graph.

    There are many results on partitions of graphs. The Four Color Theorem [1,2] implies that every planar graph has an (I, I, I, I)-partition. Alon et al. [5] showed that there is no finite k such that every planar graph has an (Ok, Ok, Ok)-partition. Poh [6] showed that every planar graphs admit an (F2, F2, F2)-partition. Borodin [8] proved that every planar graph admits an (I, F, F)-partition.

    We focus on partitions of planar graphs with girth restrictions. Borodin, Kostochka, and Yancey [4] proved that a planar graph with girth at least 7 has a (P2, P2)-partition. Borodin and Glebov [7] showed that every planar graph with girth 5 admits an (I, F)-partition. Dross, Montassier, Pinlou [9] proved that every triangle-free planar graph admits an (F5, F)-partition. Choi, Dross and Ochem [3] proved that every planar graph with girth at least 10 admits an (I, P3)-partition and every planar graph with girth at least 9 admits an (I, O9)-partition. Choi, Dross and Ochem [3] give an example that a planar graph with girth 7 and maximum degree 4 that has no (I, P3)-partition.

    In this paper, we establish the following result.

    Theorem 1. Every plane graph with girth at least 9 and without intersecting 9-face admits an (I, O6)-partition.

    Assume that G is the counterexample to Theorem 1 such that G is vertex-minimal. The graph G is connected, since otherwise at least one of its components would be a counterexample with fewer vertices than G. This further implies that every vertex of G has degree at least 1.

    For an element xV(G)F(G), the degree of x is denoted by d(x). A vertex v is called a k-vertex, k+-vertex, or k-vertex if d(v)=k,d(v)k, or d(v)k, respectively. We define a k-face, k+-face, or k-face analogously. Let N(v) denote the set of the neighbours of v. Let N[v] denote N(v){v}. A neighbour of the vertex v with degree k, at least k, or at most k is called a k-neighbour, k+-neighbour, or k-neighbour of v, respectively. We use dk(f), dk+(f) and dk(f) to denote the number of k-vertices incident with f, k+-vertices incident with f and k-vertices incident with f respectively. For fF(G), we use b(f) to denote the boundary walk of f, and f=[v1v2vm] if v1,v2,,vm are the boundary vertices of f in cyclic order. An (1, 2, , k)-face is a k-face [v1v2vk] with d(vi)=i for each i{1,2,,k}. An (1, 2, , k)-path is a k-path v1v2vk with d(vi)=i for each i{1,2,,k}, analogously.

    Given an (I, Ok)-partition of G, we will assume that V(G) is partitioned into two parts I and O where I is an independent set and O induces a graph whose components have order at most k; we also call the sets I and O colors, and a vertex in I and O is said to have color I and O, respectively.

    Claim 1. Every vertex in G has degree at least 2.

    Proof. Let v be a vertex of degree 1 in G. Since the graph Gv has fewer vertices than G, it admits an (I, O6)-partition, which can be extended to an (I, O6)-partition of G by giving to v the color distinct from that of its neighbour. This contradicts G as a counterexample.

    Claim 2. Every 6-vertex in G has at least one 3+-neighbour.

    Proof. Let v be a 6-vertex where every neighbour has degree 2 and let G=GN[v]. Because the girth of graph G is at least 9, every 2-neighbour of v can not have neighbours in N(v) and the neighbours of 2-neighbour in G are different. Since G has fewer vertices than G, it admits an (I, O6)-partition. For every neighbour u of v that has a neighbour u in G, color u with the color distinct from that of u. And color v with color O. Obviously, it does not give an (I, O6)-partition of G only when all uncoloured vertices with O. Therefore, we can recolor v with I to obtain an (I, O6)-partition of G, which is a contradiction.

    In G, a chain is a longest induced path whose internal vertices all have degree 2. A chain with k internal vertices is a k-chain. Every end-vertex of a chain is a 3+-vertex. By Claim 2, there are no 3-chains in G. A 3-vertex is weak if it has two 2-neighbours; a 3-vertex is bad if it is weak and incident with a 2-chain; and a 3-vertex is good otherwise.

    Remark 3. Let v1, v2, v3, v4 be four vertices of 2-chain, where v2 and v3 are 2-vertices. Whether v1 has been colored I or O, we choose one of the four coloring methods in Table 1 to color the other three uncolored vertices of the 2-chain in the following proofs.

    Table 1.  Four coloring methods of 2-chain.
    v1 v2 v3 v4
    I O I O
    I O O I
    O I O O
    O I O I

     | Show Table
    DownLoad: CSV

    Claim 4. Every d(v)-vertex v(3d(v)6)in G is incident with at most (d(v)2) 2-chain.

    Proof. By Claim 2, v has at least one 3+-neighbour v1. Assume to the contrary that v is incident with (d(v)1) 2-chains. Let graph G be a graph obtained from G by deleting v and all 2-vertices of 2-chains incident with v. By the minimality of G, G has an (I, O6)-partition. For all the 3+-vertices other than v of 2-chains that have been colored, we let them correspond to v1 in the Table 1. Now we color the uncolored vertices. Firstly, we color v with the color distinct from that of v1. Then according to Remark 3, no matter what color v and all the 3+-vertices other than v of 2-chains are colored, we can always choose appropriate methods from Table 1 to color all the uncolored 2-vertices such that G admits an (I, O6)-partition. This is a contradiction.

    Claim 5. Every 3-vertex v is adjacent to at most one weak 3-vertex.

    Proof. Let v1, v2 and v3 be the neighbours of v. Assume to the contrary that v is adjacent to 2 weak 3-vertices. That is, d(v1)=d(v2)=3 and d(v3)2. Let u1 and u2 be two 2-neighbours of v1. Let w1 and w2 be two 2-neighbours of v2. Let z1 and z2 be the neighbours other than v1 of u1 and u2, respectively. By the minimality of G, G=G{v,v1,v2,u1,u2,w1,w2} has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v with the color distinct from that of v3. Secondly, we consider the coloring methods of v1, u1 and u2. We give the following three coloring methods. If z1 and z2 are colored O, then we assign I to u1, u2 and assign O to v1. If z1 and z2 are colored I, then we assign O to u1, u2 and assign O to v1. If z1 and z2 are colored I and O respectively, then we assign O to v1, u1 and assign I to u2. In all of the above cases, we can assign O to v1. The coloring methods of v2, w1 and w2 are similar to those of v1, u1 and u2. We can color the remaining uncolored vertices according to the given three coloring methods. It does not give an (I, O6)-partition of G only when every vertex in {v,v1,v2,u1,u2,w1,w2} with O. Therefore, we can recolor v1 and v2 with I to obtain an (I, O6)-partition of G, which is a contradiction.

    Claim 6. Every 4-vertex v incident with two 2-chains can not be adjacent to a weak 3-vertex.

    Proof. Let v1, v2, v3 and v4 be the neighbours of v. Assume to the contrary that v is adjacent to at least weak 3-vertex. That is, d(v1)=d(v2)=2, d(v3)=3 and d(v4)2. Let ui be the 2-vertex adjacent to vi for i=1,2. Let w1 and w2 be two 2-neighbours of v3. By the minimality of G, G=G{v,v1,v2,v3,u1,u2,w1,w2} has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v with the color distinct from that of v4. Then according to Remark 3 and the given three coloring methods of v1, u1 and u2 in the proof of Claim 5, we can always choose appropriate methods to color the remaining uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    Claim 7. Let v1 and v2 be two adjacent 3-vertices.

    (1)These two vertices can not both be weak 3-vertices.

    (2)If v1 is a weak 3-vertex, then v2 can not be incident with 2-chain.

    Proof. (1)Assume to the contrary that v1 and v2 be two weak 3-vertices. Let u1 and u2 be two 2-neighbours of v1. Let w1 and w2 be two 2-neighbours of v2. By the minimality of G, G=G{v1,v2,u1,u2,w1,w2} has an (I, O6)-partition. Now we color the uncolored vertices. According to the given three coloring methods of v1, u1 and u2 in the proof of Claim 5, we can always choose appropriate methods to color all uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    (2) By (1), we know v2 has a 3+-neighbour z1. Assume to the contrary that v2 is incident with a 2-chain. Let u1 and u2 be two 2-neighbours of v1. Let w1 and w2 be two 2-vertices of 2-chain. Here w1 is a neighbour of v2. By the minimality of G, G=G{v1,v2,u1,u2,w1,w2} has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v2 with the color distinct from that of z1. Then according to Remark 3 and the given three coloring methods of v1, u1 and u2 in the proof of Claim 5, we can always choose appropriate methods to color the remaining uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    Claim 8. Let v1, v2 and v3 be three 3-vertices such that vivi+1 E(G), where i=1,2.

    (1)If v1 is a weak 3-vertex and v2 is incident with one 1-chain, then v3 can not be incident with 2-chain.

    (2)If v2 is adjacent to a 2-vertex, then v1 and v3 can not both be incident with 2-chain.

    Proof. (1)By Claim 5, we know v3 has at least a 3+-neighbour z1. Assume to the contrary that v3 is incident with a 2-chain. Let u1 and u2 be two 2-neighbours of v1. Let w1 and w2 be two 2-vertices of 2-chain. Let y1 and y2 be the neighbours other than v1 of u1 and u2, respectively. Let x1 be one 2-neighbour of v2. Let x2 be an another 3+-vertex of 1-chain incident with v2. By the minimality of G, G=G{v1,v2,v3,u1,u2,w1,w2,x1} has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v3 with the color distinct from that of z1. Secondly, we consider the coloring methods of x1 and v2. We give the following two coloring methods. If x2 is colored I, then we assign O to x1 and assign O to v2. If x2 is colored O, then we assign I to x1 and assign O to v2. So, we can assign O to v2 whatever x2 has been colored I or O. Then we consider the coloring methods of v1, u1 and u2. We give the following three coloring methods. If y1 and y2 are colored O, then we assign I to u1, u2 and assign O to v1. If y1 and y2 are colored I, then we assign O to u1, u2 and assign I to v1. If y1 and y2 are colored I and O respectively, then we assign O to v1, u1 and assign I to u2. Then according to Remark 3 and the given these coloring methods, we can always choose appropriate methods to color the remaining uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    (2) Assume to the contrary that v1 and v3 are both incident with a 2-chain. Let u1 and u2 be two 2-vertices of 2-chain incident with v1. Let w1 and w2 be two 2-vertices of 2-chain incident with v3. Let x1 be one 2-neighbour of v2. Let z1 and z2 be other neighbours of v1 and v3 respectively. By the minimality of G, G=G{v1,v2,v3,u1,u2,w1,w2,x1} has an (I, O6)-partition. Now we color the uncolored vertices. According to Remark 3 and the given two coloring methods of x1 and v2 in the proof of Claim 8(1), we can always choose appropriate methods to color all uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    Claim 9. Let v1 and v3 be two 3-vertices and v2 is the common 2-neighbor of v1 and v3. Then v1 and v3 can not both be incident with 2-chain.

    Proof. Assume to the contrary that v1 and v3 are both incident with a 2-chain. Let u1 and u2 be two 2-vertices of 2-chain incident with v1. Let w1 and w2 be two 2-vertices of 2-chain incident with v3. By the minimality of G, G=G{v1,v2,v3,u1,u2,w1,w2} has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we assign O to v2. Then by Remark 3, we can color all uncolored vertices such that G admits an (I, O6)-partition. This is a contradiction.

    Claim 10. Let v1, v2, v3 and v4 be four 3-vertices such that vivi+1 E(G), where i=1,2,3. If v2 and v3 are both incident with a 1-chain, then v1 and v4 can not both be weak 3-vertices.

    Proof. Assume to the contrary that v1 and v4 be two weak 3-vertices. Let u1 and u2 be two 2-neighbours of v1. Let w1 and w2 be two 2-neighbours of v4. Let z1 and z2 be 2-neighbours of v2 and v3 respectively. By the minimality of G, G=G{v1,v2,v3,v4,u1,u2,w1,w2,z1,z2} has an (I, O6)-partition. Now we color the uncolored vertices. We can color all uncolored vertices according to the given two coloring methods of x1, v2 and three coloring methods of v1, u1 and u2 in the proof of Claim 8(1). Obviously, it does not give an (I, O6)-partition of G only when v1, v2, v3 and v4 are colored with O and at least one of z1 and z2 is colored with O, say z1. Then we recolor 3-neighbour v2 of z1 with I. Therefore, we can know G admits an (I, O6)-partition. This is a contradiction.

    Claim 11. If f is a 9-face with d3(f)=9, then these 3-vertices on f can not all be incident with 2-chain.

    Proof. Assume to the contrary that these 3-vertices all be incident with 2-chain. According to Claim 8(2), this situation is impossible.

    Claim 12. There are no (3,2,2,3,2,3,2,3,2)-faces in G.

    Proof. Suppose to the contrary that G contains such a (3,2,2,3,2,3,2,3,2)-face f. By Claim 2, we know the neighbours of these 3-vertices that are not on f are 3+-vertices. Let graph G be a graph obtained from G by deleting all vertices on f. By the minimality of G, G has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color these 3-vertices on f with the color distinct from that of their 3+-neighbours. We know these 3-vertices on f color either I or O. Then we consider the coloring of 2-vertices on f. If two 3-vertices of 1-chain are colored O(I), then we assign I(O) to 2-vertex. If two 3-vertices of 1-chain are colored O and I respectively, then we assign O to 2-vertex. And we assign O to two 2-vertices of 2-chain. In this way, we can get an (I, O6)-partition of graph G, a contradiction.

    Claim 13. If f is a (3,2,2,3,3,2,3,2,3)-face, then the neighbours of v1 and v4 that are not on f can not both be 2-vertices.

    Proof. Let v1 and v4 be neighbours of v1 and v4 that are not on f, respectively. Assume to the contrary that v1 and v4 are both 2-vertices. Let z1 be another neighbour other than v1 of v1. By Claim 7(2), we know the neighbours of v5 and v9 that are not on f are 3+-vertices. By Claim 2, we know the neighbour of v7 that is not on f is a 3+-vertex. Let graph G be a graph obtained from G by deleting v1, v4 and all vertices on f. By the minimality of G, G has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v5, v7 and v9 to make their colors different from their 3+-neighbours that are not on f. Then we consider the coloring of v1 and v1. If z1 is colored I, then we assign O to v1 and assign O to v1. If z1 is colored O, then we assign I to v1 and assign O to v1. So, we can assign O to v1 whatever z1 has been colored I or O. The coloring methods of v4 and v4 are similar to those of v1 and v1. Finally, we consider the coloring of 2-vertices on f. We assign O and I to v2 and v3, respectively. If two 3-vertices of 1-chain are colored O(I), then we assign I(O) to 2-vertex. If two 3-vertices of 1-chain are colored O and I respectively, then we assign O to 2-vertex. In this way, we can get an (I, O6)-partition of graph G, a contradiction.

    Claim 14. If f is a (3,3,2,3,2,3,2,3,2)-face, then the neighbors of v1 and v2 that are not on f are both 3+-vertices.

    Proof. Let v1 and v2 be neighbours of v1 and v2 that are not on f, respectively. By Claim 7(1), we know that one of v1 and v2 is a 3+-vertex. Without loss of generality, let v1 be a 3+-vertex. Assume to the contrary that v2 is a 2-vertex. By Claim 2, we know the neighbours of v4, v6 and v8 that are not on f are 3+-vertices. Let graph G be a graph obtained from G by deleting v2 and all vertices on f. By the minimality of G, G has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color v1, v4, v6 and v8 to make their colors different from their 3+-neighbours that are not on f. Then we consider the coloring of v2 and v2. According to the coloring methods of v1 and v1 in the proof of Claim 13, we can color v2 and v2. Finally, we consider the coloring of 2-vertices on f. If two 3-vertices of 1-chain are colored O(I), then we assign I(O) to 2-vertex. If two 3-vertices of 1-chain are colored O and I respectively, then we assign O to 2-vertex. In this way, we can get an (I, O6)-partition of graph G, a contradiction.

    Claim 15. If f is a (3,3,2,3,3,2,3,3,2)-face, then at least a pair of adjacent 3-vertices on f have two 3+-neighbours that are not on f.

    Proof. By Claim 7(1), we know one neighbour of each pair of adjacent 3-vertices that is not on f is a 3+-vertex. Assume to the contrary that the other neighbour of each pair of adjacent 3-vertices that is not on f is a 2-vertex. Let graph G be a graph obtained from G by deleting all vertices on f and 2-vertices which are not on f and are incident with 3-vertices on f. By the minimality of G, G has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color non-weak 3-vertices on f to make their colors different from their 3+-neighbours that are not on f. Then, we consider the coloring of weak 3-vertices and their 2-neighbours that are not f. According to the coloring methods of v1 and v1 in the proof of Claim 13, we can color weak 3-vertices and their 2-neighbours that are not f. Finally, we consider the coloring of 2-vertices on f. If two 3-vertices of 1-chain are colored O(I), then we assign I(O) to 2-vertex. If two 3-vertices of 1-chain are colored O and I respectively, then we assign O to 2-vertex. In this way, we can get an (I, O6)-partition of graph G, a contradiction.

    Claim 16. If f is a (3,3,2,3,3,2,3,2,2)-face, then at least a pair of adjacent 3-vertices on f have two 3+-neighbours that are not on f.

    Proof. By Claim 2, we know the neighbour of v7 that is not on f is a 3+-vertex. By Claim 7(2), we know the neighbour of v2 that is not on f is a 3+-vertex. By Claim 7(1), we know one neighbour of each pair of adjacent 3-vertices that is not on f is a 3+-vertex. Assume to the contrary that the other neighbour of each pair of adjacent 3-vertices that is not on f is a 2-vertex. Let graph G be a graph obtained from G by deleting all vertices on f and 2-vertices which are not on f and are incident with 3-vertices on f. By the minimality of G, G has an (I, O6)-partition. Now we color the uncolored vertices. Firstly, we color non-weak 3-vertices on f to make their colors different from their 3+-neighbours that are not on f. Then, we consider the coloring of weak 3-vertices and their 2-neighbours that are not f. According to the coloring methods of v1 and v1 in the proof of Claim 13, we can color weak 3-vertices and their 2-neighbours that are not f. Finally, we consider the coloring of 2-vertices on f. If two 3-vertices of 1-chain are colored O(I), then we assign I(O) to 2-vertex. If two 3-vertices of 1-chain are colored O and I respectively, then we assign O to 2-vertex. And we assign O and I to v8 and v9, respectively. In this way, we can get an (I, O6)-partition of graph G, a contradiction.

    To prove Theorem 1, we will get a contradiction by a discharging procedure. For all xV(G)F(G), we define an initial weight function ω: if vV(G), let ω(v)=2d(v)5; if fF(G), let ω(f)=12d(f)5. The total initial charge is negative, since Euler's formula implies

    vV(G)(2d(v)5)+fF(G)(12d(f)5)=10. (2.1)

    We then redistribute the charge at the vertices and faces according to carefully designed discharging rules, which preserve the total charge sum. Once the discharging is finished, a new charge function ω is produced. Finally, we can show that the final charge ω on V(G)F(G) satisfies xV(G)F(G)ω(x)0. Then it leads to a contradiction in the inequality:

    0xV(G)F(G)ω(x)=xV(G)F(G)ω(x)=10. (2.2)

    and thus this completes the proof of Theorem 1. The discharging rules are as follows.

    (R1) Every 2-vertex that belongs to a 1-chain gets charge 12 from its each ends, while each 2-vertex that belongs to a 2-chain gets charge 1 from its neighbour of degree greater than 3.

    (R2) Every weak 3-vertex sends charge 12 to its adjacent 2-vertex on 2-chain.

    (R3) Every good 3-vertex sends charge 1 to its adjacent 2-vertex on 2-chain.

    (R4) Each 3+-vertex along its adjacent bad 3-vertex v sends charge 12 to 2-vertex on 2-chain adjacent v

    For each 3-vertex v, let α(v) be the remaining charge of v after rules R1R4.

    (R5) Each 3-vertex v sends charge α(v) to each incident 9-face.

    (R6) Each 4+-vertex sends charge 12 to each incident 9-face.

    In the following, we will prove that ω(x)0 for all xV(G)F(G).

    Claim 17. Every vertex v has non-negative final charge.

    Proof. Let v be a 2-vertex, which has initial charge 1. If v belongs to a 1-chain, then ω(v)=1+12×2=0 by R1. If v belongs to a 2-chain, then ω(v)=1+12+12=0 or ω(v)=1+1=0 by R1, R2, R3, and R4.

    Let v be a 3-vertex, which has initial charge 1. By the discharging rules, we only need to show that α(v)0. By Claim 2, we know v has at least a 3+-neighbour v1. By Claim 4, we know v is incident with at most one 2-chain. Suppose v is a weak 3-vertex. By Claim 7(1), we know v1 can not be a weak 3-vertex. Then α(v)=11212=0 by R1 and R2. Suppose v is a good 3-vertex. By Claim 5, we know every 3-vertex v is adjacent to at most one weak 3-vertex. If v is not incident with chains, then α(v)112=12 by R4. If v is incident with a 1-chain, then α(v)11212=0 by R1 and R4. If v is incident with a 2-chain, then we know v can not be adjacent to weak 3-vertices by Claim 7(2). Thus, α(v)=11=0 by R3.

    Let v be a 4-vertex, which has initial charge 3. By Claim 2, we know v has at least a 3+-neighbour v1. By Claim 4, we know v is incident with at most two 2-chains. We also know v incident with two 2-chains can not be adjacent to weak 3-vertex by Claim 6. Then ω(v)3max{1×2+12+12,1+12×2+12+12}=0 by R1, R4 and R6.

    Let v be a 5-vertex, which has initial charge 5. By Claim 2, we know v has at least a 3+-neighbour v1. By Claim 4, we know v is incident with at most three 2-chains. Then ω(v)51×3121212=12 by R1, R4 and R6.

    Let v be a 6-vertex, which has initial charge 7. By Claim 2, we know v has at least a 3+-neighbour v1. By Claim 4, we know v is incident with at most four 2-chains. Then ω(v)71×4121212=32 by R1, R4 and R6.

    Each 7+-vertex with degree d(v) has initial charge 2d(v)5. Then ω(v)2d(v)5d(v)12=d(v)11232 by R1 and R6.

    Claim 18. Every 10+-face f has non-negative final charge.

    Proof. Let f be a 10+-face. We know that a 10+-face is not involved in discharging rules, so ω(f)=ω(f)=12d(f)512×105=0.

    Before discussing 9-faces, we give the following two useful Lemmas.

    Lamma 19. If there is a (3,2,2,3,3,3)-path on f, then ω(f)0.

    Proof. Let vi be a neighbour of vi that is not on f(i=4,5). By Claim 5, we know every 3-vertex is adjacent to at most one weak 3-vertex.

    Suppose v4 is a weak 3-vertex. If v5 is a 3+-vertex, then α(v5)=112=12 by R4. So ω(f)12+12=0 by R5. If v5 is a 2-vertex, then we can know v5 belongs to 1-chain by Claim 7(2). Then we consider the case of v6. According to Claim 5 and Claim 8(2), we know v6 is not a weak 3-vertex and v6 can not be incident with 2-chain. If v6 is not adjacent 2-vertex, then α(v6)112=12 by R4. So ω(f)12+12=0 by R5. If v6 is incident with a 1-chain, then we can know another 3+-vertex other than v5 of v6 is not a weak 3-vertex by Claim 10. Then α(v6)=112=12 by R1. So ω(f)12+12=0 by R5.

    Suppose v4 is a good 3-vertex. If v5 is a 3+-vertex, then α(v5)112=12 by R4. Thus, ω(f)12+12=0. So we can assume v5 is a 2-vertex. If v5 belongs to 1-chain, then we can know v6 is not a weak 3-vertex by Claim 8(1). Then α(v5)=112=12 by R1. So ω(f)12+12=0 by R5. If v5 belongs to 2-chain, then we consider the case of v6. According to Claim 7(2) and Claim 8(2), we know v6 is not a weak 3-vertex and v6 can not be incident with 2-chain. If v6 is not adjacent 2-vertex, Then α(v6)112=12 by R4. So ω(f)12+12=0 by R5. If v6 is incident with a 1-chain, then we can know another 3+-vertex other than v5 of v6 is not a weak 3-vertex by Claim 8(1). Then α(v6)=112=12 by R1. So ω(f)12+12=0 by R5.

    Lamma 20. If there is a (3,2,3,3,3)-path on f, then ω(f)0.

    Proof. Let vi be a neighbour of vi that is not on f(i=3,4,). By Claim 5, we know every 3-vertex is adjacent to at most one weak 3-vertex.

    Suppose v3 is a weak 3-vertex. If v4 is a 3+-vertex, then α(v4)112=12 by R4. So ω(f)12+12=0 by R5. If v4 is a 2-vertex, then we can know v4 belongs to 1-chain by Claim 7(2). Then we consider the case of v5. According to Claim 5 and Claim 8(1), we know v5 is not a weak 3-vertex and v5 can not be incident with 2-chain. If v5 is not adjacent to 2-vertex, then α(v5)112=12 by R4. So ω(f)12+12=0 by R5. If v5 is incident with a 1-chain, then we can know another 3+-vertex other than v4 of v5 is not a weak 3-vertex by Claim 10. Then α(v5)=112=12 by R1. So ω(f)12+12=0 by R5.

    Suppose v3 is a good 3-vertex. If v3 is a 3+-vertex but not a bad 3-vertex, then α(v3)=112=12 by R1. So ω(f)12+12=0 by R5. If v3 is a bad 3-vertex, then we consider the case of v4. According to Claim 5 and Claim 8(2), we know v4 is not a weak 3-vertex and v4 can not be incident with 2-chain. If v4 is not an adjacent 2-vertex, then α(v4)112=12 by R4. So ω(f)12+12=0 by R5. If v4 is incident with a 1-chain, then we can know v5 is not a weak 3-vertex by Claim 10. Then α(v4)=112=12 by R1. So ω(f)12+12=0 by R5.

    Claim 21. Every 9-face f has non-negative final charge.

    Proof. If f is incident with at least a 4+-vertex, then ω(f)12×95+12=0 by R6. Now we only need to consider the situation that f is only incident with 2-vertices and 3-vertices.

    Case 1. d2(f)=0.

    By Claim 11, we know these 3-vertices on f can not all be incident with 2-chain. So there is at least one vertex v not incident with 2-chain. If v is incident with a 1-chain, then α(v)=112=12 by R1. If v is adjacent to a 3+-vertex that is not on f, then α(v)112=12 by R4. So ω(f)12+12=0 by R5.

    Case 2. d2(f)=1.

    By Lemma 20, we know ω(f)0.

    Case 3. d2(f)=2.

    For (3,2,2,3,3,3,3,3,3)-face, we have ω(f)0 by Lemma 19.

    For (3,2,3,2,3,3,3,3,3)-face, (3,2,3,3,2,3,3,3,3)-face and (3,2,3,3,3,2,3,3,3)-face, we also have ω(f)0 by Lemma 20.

    Case 4. d2(f)=3.

    For (3,2,2,3,2,3,3,3,3)-face, (3,2,2,3,3,2,3,3,3)-face and (3,2,2,3,3,3,2,3,3)-face, we have ω(f)0 by Lemma 19.

    For (3,2,3,2,3,2,3,3,3)-face and (3,2,3,2,3,3,2,3,3)-face, we also have ω(f)0 by Lemma 20.

    For (3,3,2,3,3,2,3,3,2)-face, we can know at least a pair of adjacent 3-vertices on f have two 3+-neighbours that are not on f by Claim 15. By Claim 10, we know these two 3+-neighbours can not both be weak 3-vertices. So there is a 3-vertex v on f such that α(v)=112=12 by R1. So ω(f)12+12=0 by R5.

    Case 5. d2(f)=4.

    For (3,2,2,3,2,3,2,3,3)-face, we have ω(f)0 by Lemma 19.

    For (3,2,2,3,3,2,3,2,3)-face, we can know the neighbours of v1 and v4 that are not on f can not both be 2-vertices by Claim 13. Without loss of generality, let the neighbour of v4 that is not on f is a 3+-vertex. By Claim 7(2), v5 is not a weak 3-vertex. By Claim 8(1), we know 3+-neighbour of v5 that is not on f can not be a weak 3-vertex. Then α(v5)=112=12 by R1. So ω(f)12+12=0 by R5.

    For (3,3,2,3,2,3,2,3,2)-face, we can know the neighbors of v1 and v2 that are not on f are both 3+-vertices by Claim 14. By Claim 10, we know these two 3+-neighbours can not both be weak 3-vertices. So there is a 3-vertex v on f such that α(v)=112=12 by R1. So ω(f)12+12=0 by R5.

    For (3,3,2,3,3,2,3,2,2)-face, we can know at least a pair of adjacent 3-vertices on f, say v1 and v2, have two 3+-neighbours that are not on f by Claim 16. By Claim 8(1), we know 3+-neighbour of v2 that is not on f can not be a weak 3-vertex. Then α(v2)=112=12 by R1. So ω(f)12+12=0 by R5.

    For (3,3,3,2,2,3,3,2,2)-face, we know v2 can not have a 2-neighbour that is not on f by Claim 8(2). By Claim 5, we know every 3-vertex is adjacent to at most one weak 3-vertex. then α(v2)112=12 by R1. So ω(f)12+12=0 by R5.

    For (3,2,3,3,3,2,3,2,2)-face, there is a (3,2,3,3,3)-path. For (3,2,3,3,3)-path, we can conclude that ω(f)0 by using the same analysis method as Lemma 20.

    By Claim 12, there are no (3,2,2,3,2,3,2,3,2)-faces in G. We know there are no (3,2,2,3,3,2,2,3,2)-faces in G by Claim 9. So there is no case of d2(f)=5.

    According to Claim 2, we know that there are only 1-chains and 2-chains in G. According to Claim 4, every 3-vertex v in G is incident with at most one 2-chain. So there is no case of d2(f)6.

    Every planar graph with girth 9 and without intersecting 9-face admits an (I, O6)-partition.

    The research was partially supported by the National Natural Science Foundation of China (Grant No.12071265) and the Natural Science Foundation of Shandong Province (Grant No.ZR2019MA032).

    The authors declare there is no conflict of interests.



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