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Role Assignments and Social Networks

Role Assignments and Social Networks

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Role Assignments and Social Networks

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  1. Role Assignments and Social Networks Fred S. Roberts Rutgers University Piscataway, NJ

  2. Role Assignments Role assignments arise from the effort to model the social roles that individuals play. The motivating idea: Individuals with the same role will relate in the same way to other individuals playing counterpart roles. Individuals occupying the same position do not necessarily have similar ties with the same other individuals, but they do have the same ties with the same types of others.

  3. Doctors have the same role-relations with patients, nurses, suppliers, and other doctors. • They do not necessarily have the same role-relations with the same patients, nurses, etc.

  4. Mothers do not have the same children. But they all have children.

  5. The leader of a terrorist group does not necessarily relate to the fund-raiser for all other terrorist groups, but the leader relates to some terrorist group fund-raiser.

  6. Role assignments were formalized using concepts of graph homomorphisms by Sailer (1978) and White and Reitz (1983). We will follow the definition of a role assignment (also called a role coloring) given by Everett and Borgatti (1991).

  7. Social Network is represented by a graph G = (V,E). V = individuals; edge {x,y} means x, y are related in some way. N(x) = {y: {x,y}  E} open neighborhood ofx r(x) = role assigned to vertex x For simplicity: r(x)  {1,2,...,k}. r(N(x)) = {r(y): y  N(x)}

  8. Role Assignment: r(x) = r(y)  r(N(x)) = r(N(y)) If two individuals have the same role, they are related to individuals with the same sets of roles.

  9. r(N(a)) = r(N(e)) = r(N(h)) = {2,3} r(N(b)) = r(N(d)) = r(N(f)) = r(N(i)) = {1,2,3} r(N(c)) = r(N(g)) = {1,2} 2 d g 2 f h 1 3 a e 1 1 c 2 b 3 i 2 Example

  10. Indifference Graphs

  11. Indifference Graphs In role assignment model, we place no significance on number defining role. We don't ask if x has smaller role than y or if x and y have roles that are close. A different kind of model: try to assign numbers to individuals so that individuals who are related are exactly the ones whose role-defining numbers are close. The model: Fix  > 0. {x,y}  E |r(x) - r(y)| < . Graphs for which we can find such an r are called indifference graphs.

  12. 8 1 3 5 7 9 11 13 15 Example: Indifference graphs have been widely studied. They are easy to recognize. We shall discuss the relation between the role assignment and indifference graph models. = 3

  13. k-Role Assignments

  14. k-Role Assignments Recall: role assignment means r(x) = r(y)  r(N(x)) = r(N(y)). If r(V) = {1,2,...,k}, the role assignment r is called a k-role assignment. G = (V,E) is k-role-assignable if it has a k-role assignment.

  15. 1-Role Assignable Graphs r(x)  1 is a role assignment iff G has no isolated vertices or all isolated vertices. |V(G)| -Role Assignable Graphs r(x) all different is always a role assignment, so every graph of n vertices is n-role assignable.

  16. R2 : R1: R3 : R4 : R5 : R6 : 2-Role Assignable Graphs Given a k-role assignment, build the corresponding role graph by letting the vertices be {1,2,...,k} and taking an edge between i and j iff some vertex of role i is adjacent to some vertex of role j. If k = 2, the possible role graphs (unlabeled) are:

  17. R2 R1 It is easy to check if G has a 2-role assignment with role graph Ri , i = 1, 2, 3, or 4. Let I = set of isolated vertices in G. G has a 2-role assignment with role graph R1 iff I = V(G). G has a 2-role assignment with role graph R2 iff I  V(G) and I .

  18. R3 R4 G has a 2-role assignment with role graph R3 iff I =  and G is disconnected. G has a 2-role assignment with role graph R4 iff I =  and G is bipartite.

  19. R5 : What about R5?

  20. Theorem (Roberts and Sheng): The problem of determining if G has a 2-role assignment with role graph R5 is NP-complete. (Proof is by reducing 3-satisfiability to this problem.)

  21. Theorem (Roberts and Sheng): The problem of determining if G has a 2-role assignment with role graph R5 is NP-complete. (Proof is by reducing 3-satisfiability to this problem.)

  22. Theorem (Roberts and Sheng): The problem of determining if G has a 2-role assignment with role graph R6 is NP-complete. Corollary: The problem of determining if G has a 2-role assignment is NP-complete. Thus, there are probably no good algorithms for determining whether or not the 2-role model fits data.

  23. Role Assignments and Indifference Graphs

  24. x5 x4 x1 x6 x3 x2 Role Assignments and Indifference Graphs G = (V,E). Let x1 , x2 , ..., xn be an ordering of V. We say the ordering is compatible if whenever i  j < k  l and {xi,x1}  E, then {xj,xk}  E. x1 , x2 , x3 , x4 , x5 , x6 is a compatible ordering.

  25. Theorem (Roberts 1968): A graph G is an indifference graph iff there is a compatible order of vertices.

  26. Theorem (Roberts 1968): A graph G is an indifference graph iff there is a compatible order of vertices. Role Coloring an Indifference Graph: A Greedy Algorithm

  27. Let G be an indifference graph and x1 , x2 , ..., xn be a compatible order. Let Nk(xk ) denote N(xk)  {xk ,xk+1 , ..., xn}. Greedy Algorithm with 2 Roles: r(xn )  1 For i  n-1 to 1 step -1, do if 1  r(Nk(xk)) or |Nk (xk)| = 1 and xm Nk(xk ) and 2  r(Nm(xm)), r(xk)  2. otherwise r(xk)  1.

  28. x5 x4 x1 x6 x3 x2 Example: 2 1 1 1 2 2 x1 1 since 1  r(N1(x1)), |N1(x1)| = 1, x2 N1(x1), but 2  r(N2(x2)). x5 2 since 1  r(N5(x5)) x6 1 x3 1 since 1  r(N3(x3)) and |N3(x3)| > 1 x4 2 since 1  r(N4(x4)) x2 2 since 1  r(N2(x2))

  29. Theorem (Sheng): If G is an indifference graph with at most one vertex with only one neighbor, then G has a 2-role assignment, obtainable by using the greedy algorithm based on the compatible ordering. If there are no isolated vertices, the role graph is R5 . However, not every graph with a compatible vertex ordering has a 2-role assignment: Example:

  30. 2-Role Assignable Indifference Graphs Simple paths are 2-role assignable since they are bipartite. Theorem (Sheng): Let G = (V,E) be a connected indifference graph with n > 2 vertices and two or more pendant vertices and assume that G is not a simple path. Let x1 ,x2 ,...,xn be a compatible order and let s,t  [1,n] be the first and last i  [1,n] s.t. {xi ,xi+2 }  E. Then:

  31. (1). If s-1  0 (mod 3) and n-t  0 (mod 3), G is 2-role assignable iff {xs,xt}  E. (2). If s-1  2 (mod 3) and n-t  0 (mod 3), G is 2-role assignable iff {xs,xt}  E or {xs+1 ,xt}  E. (3). If s-1  0 (mod 3) and n-t  2 (mod 3), G is 2-role assignable iff {xs,xt}  E or {xs,xt-1}  E. (4). If s-1  2 (mod 3) and n-t  2 (mod 3), and m  [s,t] is the last i s.t. {xs ,xi }  E, and u  [s,t] is the first i s.t. {xi,xt}  E, then G is 2-role assignable iff u  m or {xs+1,xt+1}  E. (5). If s-1  1 (mod 3) or n-t  1 (mod 3), then G is always 2-role assignable.

  32. x5 x1 x2 x3 x4 x6 x7 x8 x9 Consider previous example: A compatible order is given by x1 ,x2 ,..., x9. s = 4, t = 4 s-1 = 3  0 (mod 3), n-t = 5  2 (mod 3) {xs,xt}  E, {xs,xt-1}  E. Thus G is not 2-role assignable by part (3). Comment: Sheng has related results for triangulated graphs.

  33. Role Primitive Graphs

  34. Role Primitive Graphs G is role primitive if the only role assignments are the trivial ones where all vertices get the same role or all vertices get different roles, and there are at least 3 vertices. Theorem (Everett and Borgatti (1991)): There is a role primitive graph: Question: What is the smallest role primitive graph?

  35. This graph is not an indifference graph; it does not have a compatible vertex ordering. Question: Are there role primitive indifference graphs?

  36. a0 Gp,q,w … … … x1 x2 x3 xp+1 a2 a3 yq+1 y3 y2 y1 =aw =a1 This graph is not an indifference graph; it does not have a compatible vertex ordering. Question: Are there role primitive indifference graphs? Answer (Roberts and Sheng): Yes:

  37. Theorem (Roberts and Sheng): Gp,q,w is not role primitive if w  2. Gp,q,2 is an indifference graph.

  38. Theorem (Roberts and Sheng): Gp,q,2 is not 2-role assignable iff p > 0, q > 0, p  0 (mod 3), and q  0 (mod 3). Theorem (Roberts and Sheng): Let k  [1,n-1] where n = |V(G)|. Then Gp,q,2 is k-role assignable iff 1) p > 0, q > 0, and {p,q}  {k-2,k-1} or {k-2,k} (mod 2k-1) or 2) one arm has length 0 and the other has length k-1 (mod 2k-1) or 3) p = q = k-2.

  39. a0 G3,12,2 … x1 x2 x3 a1 a2 y12 y11 y3 y2 y1 Corollary: G3,12,2 is a role primitive indifference graph. It is not 2-role assignable by the first theorem. It is not k-role assignable for any k  [1,n-1] by the second theorem.

  40. Automorphisms of Role Primitive Graphs Theorem (Everett and Borgatti 1991): If G is role primitive, the only automorphism of G is the identity. Sketch of Proof: Lemma: Let H be a subgroup of Aut(G). Then the orbits of H form a partition of V corresponding to the sets of vertices of a given role in some role assignment. Proof: Suppose V1 , V2 , .., Vk are the orbits and r(x) = i if x  Vi . Suppose r(x) = r(y). Therefore ∃  H s.t. (x) = y. If u  N(x), then (u)  N((x)), so (u)  N(y). But r(u) = r((u)) by definition. Hence, r(N(x))  r(N(y)). Similarly, r(N(y))  r(N(x)). Q.E.D.

  41. Proof of the Theorem: By the Lemma, if G is role primitive, either Aut(G) is the identity or Aut(G) acts transitively. Suppose the latter. By the lemma, the stabilizers must be trivial so Aut(G) acts regularly. Since no subgroup of a regular group can be transitive, Aut(G) cannot contain subgroups. Thus, Aut(G) is of prime order and therefore Abelian. But the only Abelian automorphism groups which can act regularly on the vertices of a graph are the elementary Abelian 2-groups. Hence, Aut(G) = Z2 , contradicting the fact that G has 3 or more vertices. Q.E.D.

  42. 1 1 2 2 2 1 The converse is false. Example: a 2-role assignment

  43. Question of Everett and Borgatti: How common are role-primitive graphs? Everett conjectured that, asymptotically, almost all graphs are role primitive. Pekec and Roberts showed that this is in fact quite wrong: Asymptotically, almost all graphs are not role primitive. So: While it had been believed that for most social networks, only trivial role assignments were feasible, this shows the opposite.

  44. Variants on the Role Assignment Model: Threshold Role Assignments

  45. Variants on the Role Assignment Model: Threshold Role Assignments If S and T are two sets of numbers, let distance d(S,T) be defined by d(S,T) = min{|s-t| : s  S, t  T}. Convention: d(,) = 0, d(S,) =  if S . Note: not necessarily a metric: d(S,T) can be 0 if S  T; also triangle inequality can fail.

  46. r is a threshold role assignment if r(x) = r(y)  d(r(N(x)),r(N(y)))  1. If in addition r(V) = {1,2,...,k}, we say it is a k-threshold role assignment. Note that in contrast to role assignments, the proximity of numbers representing roles now means something.

  47. f a 3 3 h 2 b c 2 1 1 g d e 2 1 Example: r(e) = r(g) r(N(e)) = {1}, r(N(g)) = {2,3} d({1},{2,3}) = 1.

  48. Theorem (Roberts): Every graph is k-threshold role assignable for all k s.t. 2  k |V(G)|.

  49. Theorem (Roberts): Every graph is k-threshold role assignable for all k s.t. 2  k |V(G)|. Boring

  50. An Alternative Notion of Distance dH(S,T) = smallest p s.t. s  S ∃ t  T s.t. |s-t|  p and  t  T ∃ s  S s.t. |s-t| p. Convention: dH(,) = 0, dH (S,) =  if S . dH is called the Hausdorff distance. d({1,2,3},{1}) = 0, dH({1,2,3},{1}) = 2.