Positive and Negative Relationship

Chapter 5
Positive & Negative Relationships
Saeid Ghasemshirazi

Android Malware Adventures
Relationships
2
Networks, Crowds, Markets: Reasoning about a Highly Connected World
 Positive Relationships
 Negative Relationships

Positive Relationships
3

Negative Relationships
4

Structural Balance
5

Structural Balance: Complete Graphs
6

Structural Balance Property
7
For every set of three nodes, if we consider the
three edges connecting them, either all three of
these edges are labeled +, or else exactly one of
them is labeled +

Balance theorem
8
If a complete graph is balanced then either all pairs of nodes are friends or else the
nodes can be divided into two groups, X and Y, such that:
 The people in X all like one another, likewise in Y
 Everyone in X is the enemy of everyone in Y

Proof of the Balance Theorem, arbitrary A
9
Recall:
 Every two nodes in X are friends.
 Every two nodes in Y are friends.
 Every node in X is an enemy of every node in Y .

Applications of Structural Balance
10
Use for Friendships
--------------------
Use for Antagonist
the nations GB, Fr, Ru, It, Ge,and AH are Great Britain, France, Russia, Italy, Germany, and Austria-Hungary respectively

Applications of Structural Balance
11

Weak Structured Balanced Networks
12
There is no set of three nodes
such that the edges amongst
them consist of two positive
edges and one negative edge
We allow a triangle that is all
negative, unlike structured
balanced networks
Characterization theorem:
If a labeled complete graph is
weakly balanced then it’s nodes
can be divided into groups that
that every two nodes in the same
group are friends and every two
nodes in different groups are
enemies

Allowing negative triangles : Weakly Balanced Networks
13

Essentially same proof as before
14

Structural Balance in Non-Complete(Arbitrary) Networks
15

Structural Balance in Non-Complete Networks (Disallowing negative cycles)
16
Two equivalent definitions:
Local view :
a (non-complete) graph is balanced if it can be “completed” by adding edges to form a signed complete graph that is
balanced .
Global view :
we could define a signed graph to be balanced if it is possible to divide the nodes into two sets X and Y ,such that any
edge with both ends inside X or both ends inside Y is positive, and any edge with one end in X and the other in Y is
negative

Structural Balance in Non-Complete Networks (Disallowing negative cycles)
17
Two equivalent definitions

Characterization
18
Theorem:
A signed graph is balanced if and only if it contains no cycle with an odd number(2k+1) of negative edges

Proof(Create Positive Supernodes)
19
Convert the graph to a reduced one in which there are only negative edges
Reduced graph

Negative cycle in supernode graph gives negative cycle in original graph
20

Negative cycle in supernode graph gives negative cycle in original graph
21

BFS of the Reduces Graph
22
The BFS has a cross edge if the graph is not bipartite
If the graph is bipartite then we can add even(2k) layers to X,
odd(2k+1) to Y, and all negative edges go between X and Y
If the graph is not bipartite, there is a negative odd cycle (2k+1)
A more standard drawing of the reduced graph
node G as the starting root node
The D-A,D-B path have the same length k

What if we only know that most triangles are balanced?
23
Claim:
If at least 99.9% of all triangles in a labeled complete graph are balanced, then either
• there is a set consisting of at least 90% of the nodes in which at least 90% of all pairs are friends, or
else.
• the nodes can be divided into two groups, X and Y , such that:
(i)at least 90% of the pairs in X like each other.
(ii) at least 90% of the pairs in Y like each other.
(iii) at least 90% of the pairs with one end in X and the other end in Y are enemies.
.

Approximately Balanced Networks(General Statement)
24
Theorem:
Let 𝜖 be any real 0 ≤ 𝜖 < 1/8, and let 𝛿 = 𝜖
1
3.
If at least 1 − 𝜖 of the triangles in a complete labeled graph are balanced then either
 There is a set consisting of 1 − 𝛿 of the nodes where at least 1 − 𝛿 of the pairs are
friends, or else
 The nodes can be divided into two sets X,Y, such that
At least 1 − 𝛿 of the pairs in X like each other
At least 1 − 𝛿 of the pairs in Y like each other
At least 1 − 𝛿 of the pairs with one edge in X and the other in Y are enemies.
In Balance Theorem: 𝜖=0 , 𝛿=0
In Approximately Balance Network:𝜖=0.001 , 𝛿=0.1

Approximately Balanced Networks Proof
25
Assume graph is balanced, show it has one of the two forms
• As before, define two sets X and Y to be friends and enemies of some node A
• A must be a “good” node: a node that isn’t involved in too many unbalanced triangles
• Given a complete graph with N nodes
How many edges?
N(N-1)/2
How many triangles?
N(N-1)(N-2)/6
Triangles with A,B,C : ABC; ACB; BAC; BCA; CAB; and CBA

26
There are at most ε N(N-1)(N-2)/6 unbalanced triangles
• Find each node’s weight: the number of unbalanced triangles it participates in
• Total weight of all nodes : 3N(N − 1)(N − 2)/6 =N(N − 1)(N − 2)/2
• Average weight of all nodes : N(N − 1)(N − 2)/2 *1/N=ε (N-1)(N-2)/2
• Pick some node A with weight less than the average They actually use at most ε N^2/2
• Since A is involved in less than ε N^2/2 unbalanced triangles, properties
within and between X and Y fail < ε N^2/2 times .
• Determine if one of X or Y is “small”
If so, most nodes are in a single mutually friendly group
If not, the nodes are divided into two mutually friendly groups who are mutual enemies

27
Let x =|X| be the number of nodes in X and y = |Y| be the number of nodes in Y
• Suppose first that X is big, x≥(1-δ)N
•
X has x(x-1)/2 internal edges
•
X contains at least (1-δ)N of the nodes
• At least (1-δ)N of the pairs in X are friends
• Same arguments hold if Y is big

28
Suppose neither X nor Y is big:
 x<(1-δ)N
 y<(1-δ)N
• There are XY edges between, at least xy≤δ N^2/2
• There are at most ε N2/2 positive edges with one end in X and the other
in Y , so as a fraction of the total this is at most:
• There are x(x-1)/2 internal edges in X (same for Y)
• Since x>δN, this total is at most δ^2 N^2/2 (internal edges)
• There are at most ε N2/2 negative edges with one end in X and the other
in Y , so as a fraction of the total this is at most :

Positive and Negative Relationship

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