Algorithms. Social Graphs. Algorithms


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1 Algorithms Social Graphs Algorithms
2 Social Graphs Definition I: A social graph contains all the friendship relations (edges) among a group of n people (vertices). The friendship relationship is symmetric. Two vertices with no edge between them are enemies. Definition ( II: In a social graph with n vertices, all the n ) 2 edges are colored either by blue (friends) or by red (enemies). Algorithms 1
3 Statement I Theorem: There is no group of 7 people such that each person in the group has exactly 3 friends in the group. Proof: The handshaking lemma: In any graph the sum of all the degrees is even. Because each edge contributes exactly 2 to this sum. The sum of all degrees in a graph with 7 vertices for which all degrees are 3 is 21 which is odd. Therefore, such a graph is impossible. Algorithms 2
4 Statement II Theorem: In any group of n 2 people, there are at least 2 people with the same number of friends in the group. Proof: In a graph with n vertices, 0,1,...,n 1 are the only possible values for a degree of a vertex. If each vertex has a distinct degree, then there must be a vertex v with degree d(v) = n 1 and a vertex u with degree d(u) = 0. But d(v) = n 1 implies that d(u) 1; a contradiction. Algorithms 3
5 Statement III Theorem: There exists a group of 5 people for which no 3 are mutual friends and no 3 are mutual enemies. Proof: The following coloring of the edges of K 5 with blue or red is the only (up to a symmetry) coloring with no unicolored triangle. Algorithms 4
6 Statement IV Theorem: Every group of 6 people contains either three mutual friends or three mutual enemies. Proof: Consider vertex A. A has either 3 friends or 3 enemies. Assume B,C,D are 3 friends of A. If any of the 3 pairs BC,CD,BD are friends, then this pair with A form a friendship triangle. Otherwise, BCD is a triangle of enemies. Algorithms 5
7 Statement IV Corollary I: If a vertex found a triangle that contains itself, then the following is a subcoloring of the graph: Corollary II: If a vertex found a triangle that does not contain itself, then the following is a subcoloring of the graph: Algorithms 6
8 Statement V Theorem: A social graph with 6 vertices contains at least 2 unicolored triangles (not necessarily of the same color). Remark: There are only 3 possible colorings (up to a symmetry) for a graph with 6 vertices and only 2 unicolored triangles. Algorithms 7
9 Proof I: Case Analysis Let ABC be a unicolored triangle found by Statement IV and let D, E, and F be the other 3 vertices. If DEF is also a unicolored triangle, then it is the second unicolored triangle. Otherwise, one of DE, DF, EF is not colored as the unicolored triangle. Assume ABC is a blue triangle and the DE is a red edge. Algorithms 8
10 Proof I: Case Analysis If either D or E is connected to ABC with 2 blue edges, then a second blue triangle is found. Algorithms 9
11 Proof I: Case Analysis Otherwise, D and E each is connected to ABC with at least 2 red edges. Therefore, at least 1 vertex from ABC is connected to both D and E with red edges. Let this vertex be A. ADE is a red triangle. Algorithms 10
12 Proof II: Based on Statement III Assume that there exists only one unicolored triangle. Omit one of the vertices of this triangle. By Statement III, the remaining 5 vertices are colored with a blue 5ring and a red 5ring. Let the omitted vertex be A. Algorithms 11
13 Proof II: Based on Statement III A has either at least 3 blue edges or at least 3 red edges. Assume B,C,D are 3 blue neighbors of A. If B,C,D are 3 consecutive vertices on the blue 5ring, then ABC and ACD are blue triangles. Algorithms 12
14 Proof II: Based on Statement III If A has at least 4 blue neighbors, then at least 3 of them are consecutive on the blue 5ring. In the remaining case, A has exactly 2 red neighbors, say D and E that must be adjacent on the red 5ring. ADE is a red triangle. Algorithms 13
15 Proof III: Based on Statement IV Assume towards contradiction that there exists only 1 unicolored triangle ABC. Assume ABC is a blue triangle. Case I: There exists a blue edge connected to the triangle. Assume that A has another blue neighbor D. Corollaries I and II of Statement IV imply that D must find another unicolored triangle. Algorithms 14
16 Proof III: Based on Statement IV Case II: Assume that the blue triangle ABC is connected to the rest of the vertices D,E,F with red edges. Algorithms 15
17 Proof III: Based on Statement IV If any of the pair DE,DF,EF is a red pair, then this pair form 3 red triangles with A,B,C. Otherwise, DEF is the second blue triangle. Algorithms 16
18 Proof IV: Counting There are exactly 20 = ( 6 3) distinct triangles in the complete graph K 6. Assume a coloring of the edges of K 6 with blue and red. Let T be the number of non unicolored triangles. Let S be the number of pairs of edges that intersect such that each edge is colored with a different color. Algorithms 17
19 Proof IV: Counting Each unicolored triangle contributes nothing to S wheras a non unicolored triangle contributes 2 to S. S = 2T. Each vertex contributes at most 2 3 = 6 to S in case it has 2 edges of one color and 3 edges of the other color. S 6 6 = 36. It follows that 2T 36 implying that T 18. There are at least 2 unicolored triangles. Algorithms 18
20 Coloring Edges with 3Colors Theorem: 17 people discuss 3 topics among themselves where each pair discusses only 1 topic. Then, there are at least 3 people who discuss among themselves the same topic. Proof: Consider vertex A. There is a topic that A discusses with at least 6 people. If 2 of these 6 people, B and C, discuss this topic, then A,B,C are 3 people that discuss the same topic. Otherwise, these 6 people discuss among themselves only 2 topics. By Statement IV, there are 3 people among these 6 people who discuss the same topic. Algorithms 19
21 Ramsey Numbers R(h,l) is the minimum integer n such that any graph with n vertices contains either a clique with h vertices or an independent set with l vertices. R(h,l) is the minimum integer n such that any coloring of the edges of the complete graph K n with the colors blue and red creates either a red clique of size h or a blue clique of size l. Statements III and IV imply that R(3,3) = 6. Algorithms 20
22 Ramsey Numbers: Observations and Known Bounds R(h,l) = R(l,h). R(2,l) = l R(h,h) 4R(h 2,h) + 2. h2h/2 e 2 R(h,h) c 1 l 2 ln l R(3,l) c 2 l2 ln l for some constants c 1 < c 2. Algorithms 21
23 Generalized Ramsey Numbers Let G be a family of graphs. Examples: K n, N n, C n, trees, binary trees,... R(G,k) is the minimum integer n such that any coloring of the edges of the complete graph K n with k colors has a monochromatic subgraph G from the family G. R(G,2) is the minimum integer n such that for any graph with n vertices either the graph or its complement has a subgraph isomorphic to a graph G from the family G. Algorithms 22
24 Examples R({K 3 },2) = 6. Statement III implies that R({K 3 },2) > 5. Statement IV implies that R({K 3 },2) 6. R({K 3,C 5 },2) = 5. Statement III implies that R({K 3,C 5 },2) 5. Trivially R({K 3 },2) > 4. R({K 3 },3) 17. From the 17 people who discuss 3 topics statement. Algorithms 23
25 Statement VI Theorem: In any group of n 1 people, there exists a committee of 1 k n members such that no 2 committee members are friends and every person not included in the committee is a friend of at least 1 committee member. Proof: Construct the following committee. As long as there exists a vertex v in the graph: Add v to the committee. Omit v and all of its neighbors from the graph. Algorithms 24
26 Statement VI The algorithm is correct: Let u and v be 2 members in the committee. Assume v joined the committee first. Then u cannot be a neighbor of v. The committee is an independent set. Let w be a noncommittee vertex. Then w has a neighbor v that is a committee member that caused w not to be a committee member. The committee is a dominating set. Algorithms 25
27 Statement VII Theorem: Any group of n 2 people can be partitioned into 2 subgroups such that at least half the friends of each person belong to the subgroup of which that person is not a member. Notations: Let P be a partition of the set of all vertices V into 2 disjoint sets: Denote by d s (v) the degree of v in its own set. Èv V Let d s(v) 2 be the number of contained edges in P. Denote by d o (v) the degree of v in the other set. Èv V Let d o(v) 2 be the number of crossed edges in P. Algorithms 26
28 Statement VII: A Constructive Proof Algorithm: Start with an arbitrary partition P. As long as there exists a vertex v such that d s (v) > d o (v), transfer v to the other set. Correctness: If the algorithm terminates, then in the final partition d s (v) d o (v) for any vertex v. Algorithms 27
29 Statement VII: Termination Proof Define D(P) to be the difference between the number of cross edges and contained edges for the partition P. D(P) = 1 2 v V (d o(v) d s (v)). Let Q be the partition after transferring v to the other set in teh partition P. D(Q) = D(P) + 2(d s (v) d o (v)) > D(P). Since D(Q) terminates. n2 4 for any partition, the algorithm Algorithms 28
30 Statement VIII Theorem: Suppose everyone in a group of n 3 people is a friend of at least half the people in the group. It is possible to seat the group around a table in such a way that everyone is seated between 2 of their friends. Equivalent Theorem: Let G be a simple undirected graph with n 3 vertices. Assume that d(v) n 2 for any vertex v. Then there exists a Hamilton Circuit in G. Proof: A greedy algorithm finds the Hamilton circuit. Algorithms 29
31 Statement IX Theorem: Suppose in a group of n 2 people, any pair has precisely one common friend. Then there is always a person (the politician ) who is everybody s friend. Equivalent Theorem: Let G be a simple undirected graph with n vertices. Assume that any 2 vertices have precisely one common neighbor. Then there exists a vertex v that is adjacent to all other vertices. Algorithms 30
32 Statement IX: The Windmill Graph A stronger statement: The only possible graphs are the windmill graphs for odd n 3. Algorithms 31
33 Statement IX: Proof Outline Assume that there exists a graph G obeying the theorem s conditions that is not a windmill graph. G must be a dregular graph: d(v) = d for each vertex v. d 2 d + 1 = n. This is impossible. Algorithms 32
34 Statement IX: Infinite Graphs There exists an infinite graph without a vertex that is a neighbor to all other vertices such that every pair of vertices has exactly one common neighbor. Start with a 5ring and add a new vertex when needed to satisfy the condition for any pair of vertices. Algorithms 33
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