Role Assignments and Social Networks Fred S Roberts

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

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. 2

• 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. 3

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

• 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 fundraiser. 5

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). 6

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 of x r(x) = role assigned to vertex x For simplicity: r(x) {1, 2, . . . , k}. 7

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. 8

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

Indifference Graphs 10

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. 11

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

k-Role Assignments 13

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. 14

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. 15

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: R 1: R 2 : R 3 : R 4 : R 5 : R 6 : 16

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 R 1 iff I = V(G). R 1 G has a 2 -role assignment with role graph R 2 iff I V(G) and I . R 2 17

G has a 2 -role assignment with role graph R 3 iff I = and G is disconnected. R 3 G has a 2 -role assignment with role graph R 4 iff I = and G is bipartite. R 4 18

What about R 5? R 5 : 19

Theorem (Roberts and Sheng): The problem of determining if G has a 2 -role assignment with role graph R 5 is NP-complete. (Proof is by reducing 3 -satisfiability to this problem. ) 20

Theorem (Roberts and Sheng): The problem of determining if G has a 2 -role assignment with role graph R 5 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 R 6 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. 22

Role Assignments and Indifference Graphs 23

Role Assignments and Indifference Graphs G = (V, E). Let x 1 , x 2 , . . . , xn be an ordering of V. We say the ordering is compatible if whenever i j < k l and {xi, x 1} E, then {xj, xk} E. x 5 x 1 x 2 x 3 x 4 x 1 , x 2 , x 3 , x 4 , x 5 , x 6 is a compatible ordering. x 6 24

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

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 26

Let G be an indifference graph and x 1 , x 2 , . . . , 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. 27

Example: x 5 2 1 2 x 1 x 2 1 x 3 2 x 4 1 x 6 x 1 1 since 1 r(N 1(x 1)), |N 1(x 1)| = 1, x 425 212 since xr(N 1(x 1 r(N (x|N ))3(x 3)| > 1 x 3 1 xsince ))r(N and since 1 (x )) 6 5 5 3 3 4 2 x 2 N 1(x 1), but 2 r(N 2(x 2)). 28

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 R 5. However, not every graph with a compatible vertex ordering has a 2 -role assignment: Example: 29

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 x 1 , x 2 , . . . , 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: 30

(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 31 always 2 -role assignable.

x 5 Consider previous example: x 1 x 2 x 3 x 4 x 6 x 7 x 8 x 9 A compatible order is given by x 1 , x 2 , . . . , x 9. 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. 32

Role Primitive Graphs 33

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 at least 3 vertices. Theorem (Everett and Borgatti (1991)): There is a role primitive graph: Question: What is the smallest role primitive graph? 34

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

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: a 0 Gp, q, w x 1 x 2 x 3 … xp+1 =a 1 a 2 a 3 … yq+1 =aw … y 3 y 2 y 1 36

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

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 2 k-1) or 2) one arm has length 0 and the other has length k-1 (mod 2 k-1) or 3) p = q = k-2. 38

Corollary: G 3, 12, 2 is a role primitive indifference graph. a 0 G 3, 12, 2 … x 1 x 2 x 3 a 1 a 2 y 11 y 3 y 2 y 1 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. 39

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 V 1 , V 2 , . . , 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, 40 r(N(x)) r(N(y)). Similarly, r(N(y)) r(N(x)). Q. E. D.

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) = Z 2 , contradicting the fact that G has 3 or more vertices. Q. E. D. 41

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

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. 43

Variants on the Role Assignment Model: Threshold Role Assignments 44

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. 45

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 kthreshold role assignment. Note that in contrast to role assignments, the proximity of numbers representing roles now means something. 46

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

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

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

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

r is a threshold close role assignment if r(x) = r(y) d. H(r(N(x)), r(N(y))) 1. k-threshold close role assignment if in addition r(V) = {1, 2, . . . , k} 51

Observe: Every graph of at least 2 vertices is 2 -threshold close role assignable. Why: Use role 1 on all isolated vertices, role 2 on all others. If no isolated vertices, use role 1 on one vertex and role 2 on all others. The result follows because d. H({1, 2}, {1}) = d. H({1, 2}, {2}) = d. H({1}, {2}) = 1. Theorem (Roberts): Every graph of at least 3 vertices is 3 threshold close role assignable. Theorem (Roberts and Sheng): Every graph of at least 4 vertices is 4 -threshold close role assignable. Every graph of at least 5 vertices is 5 -threshold close role assignable. What about every graph of at least 6 vertices? k vertices? 52

Fitting the Role Assignment Model Approximately 53

Fitting the Role Assignment Model Approximately Rarely does a mathematical model fit data perfectly. One is often satisfied if the number of inconsistencies is negligible. To make this precise, let r be a function from V onto {1, 2, . . . , k}. To express how close r is to a role assignment, we count the number of pairs of vertices for which the requirement (*) r(x) = r(y) r(N(x)) = r(N(y)) holds. 54

Let Vi = {x: r(x) = i}. Let Mi count the fraction of all pairs of vertices x and y of role i so that the condition for a role assignment holds for x and y: (&) r(N(x)) = r(N(y)). Let A(x, y) = 1 if (&) holds and A(x, y) = 0 otherwise. Let ni = | Vi |. Then We take Mi = 1 if |Vi | = 1. 55

Thus, one way to measure how close r is to a krole assignment is to use M(r) = min Mi. Taking the maximum over all possible r from V onto {1, 2, . . . , k} gives a measure of how close G is to being k-role assignable. We make this precise by using k(G) = max r M(r). Maximum is over all assignments onto {1, 2, . . . , k}. 56

A measure similar to k(G) arises in theory of “blockmodeling” in social network theory. Here, we try to map a social network "almost homomorphically" into a smaller network. General goal: replace a "large" network by a smaller one – fewer vertices – that reflects its structural relations. Implication: Use as few roles as possible. 57

Example: C 5 x 1 x 2 x 4 x 3 To calculate 2(G), we consider all possible assignments of 1's and 2's to the vertices with at least one of each. Without loss of generality, we consider only the cases where there at most two 1's. 58

Case 1: One 1: V 1 = {x 1}, V 2 = {x 2 , x 3 , x 4 , x 5 }. M 1 = 1 since |V 1| = 1. x 1 The requirement (&) r(N(x)) = r(N(y)) is satisfied in V 2 only for x 5 2 x 2, x 5 and x 3, x 4 so 1 2 x 2 2 x 4 Thus, M(r) = 1/3. 2 x 3 59

Case 2: Two 1's on adjacent vertices. M 1 = 1 since the requirement (&) r(N(x)) = r(N(y)) holds for the one pair of vertices x 1, x 2 in V 1. x 5 2 M 2 = 1/3: the requirement holds for x 3, x 5 but fails for x 3, x 4 and x 4, x 5. Thus, M(r) = 1/3. x 1 1 1 x 2 2 x 4 2 x 3 60

Case 3: Two 1's not on adjacent vertices. x 1 Again, M 1 =1 since the requirement holds on V 1 = {x 1, x 3}. On V 2, the requirement holds for x 4, x 5, but fails for x 2, x 4 and x 2, x 5. Thus, M 2 = 1/3. M(r) = 1/3. 1 x 5 2 x 4 x 2 2 2 1 x 3 Conclusion: 2(C 5) = 1/3. 61

In fact: C 5 is the only graph with 2 1/3. The proof uses: Theorem (Pekec and Roberts): For every graph G of n 3 vertices: n 0 (mod 2) 2(G) 1/2 - 1/(2 n-2) n 1 (mod 4) 2(G) 1/2 - 1/(2 n-4) n 3 (mod 4) 2(G) 1/2 – (n-5)/(2 n 2 6 n+4) 62

When is k(G) = 1? Let (G) = minimum degree of a vertex in G (minimum number of neighbors). Theorem (Pekec and Roberts): For k > 1, if G is a graph with n vertices and (G) > log(kn)/log(k(k-1)), then k(G) = 1. Corollary: For all k > 1, there is a constant ck such that if (G) > cklog n, then k(G) = 1. Interpretation: All social networks in which each individual is involved in a significant number of relationships can be captured by the role assignment model with k roles. 63

The proof is by the probabilistic method. It only shows that such a k-role assignment exists. There is no explicit construction that goes with this. The proof of theorem shows that, asymptotically, almost all graphs are not role primitive. This disproves a conjecture of Everett. 64

Approximate Role Assignments with k=2 Theorem (Pekec and Roberts). Let G have n 3 vertices. Then 2(G) > 1 – 2 log 2 n /(n-2). Corollary: For every > 0, all but finitely many graphs have 2(G) > 1 - . Proof: For every > 0, there is n such that for all n > n , 2 log 2 n /(n-2) < . Therefore, all graphs G of more than n vertices have 2(G) > 1 - . 65

Approximate k-Role Assignments Theorem (Pekec and Roberts). For all k > 0, there is a positive constant Ck such that for every graph G on n k vertices, k(G) 1 - (Ck log n)/n. Corollary. For every > 0, there is n such that n > n implies that k(G) > 1 - . In other words: For any positive integer k, all but finitely many graphs are "almost" k-role assignable. That is, for all but finitely many graphs, there is a k-role assignment such that the fraction of pairs x, y satisfying the condition for a role assignment is close to 1. 66

Connections to Ecology 67

Connections to Ecology Concepts of social network theory, in particular the role assignment model, have potentially useful applications in ecology. Relevant areas of ecology: Study of community organization, food webs, and biogeochemical cycles, with emphasis on network structure. In the study of food webs, graph-theoretic approaches similar to those used to study social networks have been used to study "trophic interactions. " Borgatti, Everett, Johnson, Luczkovich (2001) worked on defining and measuring "trophic similarity" in food webs. They found that theory of role assignments is relevant 68 to the definition of "structural role" in ecology.

Open Problems 69

Open Problems The theory of social networks is an old one and it has given rise to many fascinating graph-theoretical problems. Models of social role lead to such problems. Here a few of the open questions that remain. 1. Characterize or recognize 2 -role assignable graphs, at least under certain assumptions about the graphs. 2. Investigate k-role assignable graphs for 2 < k < |V(G)|. Very little is known about these. 3. Characterize role primitive graphs. 70

4. What is the smallest role primitive graph? 5. Develop methods for determining if a graph has a kthreshold close role assignment. 6. Is every graph of at least k vertices k-threshold role assignable? 7. Find constructions of graphs G for which k(G) is large. 71

Thank you!