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
fund-raiser.
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
2d
Example
2f
1a
2 b
3c
3g
1h
1e
r(N(a)) = r(N(e)) = r(N(h)) = {2,3}
i2
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:
R1:
R2 :
R3 :
R4 :
R5 :
R6 :
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 R1 iff I =
V(G).
R1
G has a 2-role assignment with role graph R2 iff I 
V(G) and I  .
R2
17
G has a 2-role assignment with role graph R3 iff I =
 and G is disconnected.
R3
G has a 2-role assignment with role graph R4 iff I =
 and G is bipartite.
R4
18
What about R5?
R5 :
19
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.)
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 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.
22
Role Assignments and Indifference
Graphs
23
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.
x5
x1
x2
x3
x4
x1 , x2 , x3 , x4 , x5 , x6 is a compatible ordering.
x6
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 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.
27
Example:
x5
2
1
2
x1
x2
1
x3
2
x4
1
x6
x1  1 since 1  r(N1(x1)), |N1(x1)| = 1,
x425
122
since
xr(N

1(x
1
r(N
(x|N
))3(x3)| > 1
x3  1 xsince
))r(N
and
since
1

(x
))
6
5
5
3
3
4
2
4
2
x2  N1(x1), but 2  r(N2(x2)).
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 R5 .
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 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:
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 2role 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 2role 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.
x5
Consider previous
example:
x1
x2
x3
x4
x6
x7
x8
x9
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.
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 are 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:
a0
Gp,q,w
…
x1
x2
x3
xp+1
=a1
…
a2
a3
yq+1
=aw
…
y3
y2
y1
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 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.
38
Corollary: G3,12,2 is a role primitive indifference graph.
a0
G3,12,2
…
x1
x2
x3
a1
a2 y12 y11
y3
y2
y1
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 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,
40
r(N(x))  r(N(y)). Similarly, r(N(y))  r(N(x)).
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.
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
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.
50
r is a threshold close role assignment if
r(x) = r(y)  dH(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 dH({1,2},{1}) =
dH({1,2},{2}) = dH({1},{2}) = 1.
Theorem (Roberts): Every graph of at least 3 vertices is 3threshold 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
M i   x, yV
i
We take Mi = 1 if |Vi | = 1.
A(x, y)
 ni 
 
2 
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 the 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: C5
x5
x1
x2
x4
x3
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 are at most two 1's.
58
Case 1: One 1:
V1 = {x1}, V2 = {x2 ,x3 ,x4 ,x5 }.
M1 = 1 since |V1| = 1.
x1
The requirement (&) r(N(x)) = r(N(y))
is satisfied in V2 only for
x5 2
x2, x5 and x3, x4 so
1
2
M2 
1
3
 4
 
 2
Thus, M(r) = 1/3.
2 x2
2
x4
2
x3
59
Case 2: Two 1's on adjacent vertices.
M1 = 1 since the requirement
(&) r(N(x)) = r(N(y)) holds
for the one pair of
vertices x1, x2 in V1 .
x5 2
M2 = 1/3: the requirement
holds for x3, x5 but fails
for x3, x4 and x4, x5 .
1
1 x2
2
x4
Thus, M(r) = 1/3.
x1
2
x3
60
Case 3: Two 1's not on adjacent vertices.
x1
Again, M1 =1 since the
requirement holds on
V1 = {x1, x3}.
On V2, the requirement
holds for x4, x5, but fails
for x2, x4 and x2,x5.
Thus, M2 = 1/3.
M(r) = 1/3.
1
x5
2
x4
x2
2
2
1
x3
Conclusion: 2(C5) = 1/3.
61
In fact: C5 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/(2n-2)
n  1 (mod 4)  2(G)  1/2 - 1/(2n-4)
n  3 (mod 4)  2(G)  1/2 – (n-5)/(2n2 6n+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 the 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 log2n /(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
 2n /(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 the 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 are 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!