Graph-theoretical Models of the
Spread and Control of Disease
Fred Roberts, DIMACS
smallpox
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Models of the Spread and Control
of Disease through Social Networks
•Diseases are spread through social networks.
•“Contact tracing” is an important part of any
strategy to combat outbreaks of infectious diseases,
whether naturally occurring or resulting from
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bioterrorist attacks.
The Basic Model
Social Network = Graph
Vertices = People
Edges = contact
State of a Vertex:
simplest model: 1 if infected, 0 if not infected
(SI Model)
More complex models: SI, SEI, SEIR, etc.
S = susceptible, E = exposed, I = infected, R =
recovered (or removed)
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More About States
Once you are infected, can you be cured?
If you are cured, do you become immune or can
you re-enter the infected state?
We can build a digraph reflecting the possible ways
to move from state to state in the model.
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The State Diagram for a
Smallpox Model
The following diagram is from a Kaplan-CraftWein (2002) model for comparing alternative
responses to a smallpox attack. This has been
considered by the CDC, Dept. of Homeland
Security, Dept. of Health and Human Services, etc.
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The Stages
Row 1: “Untraced” and in various stages of
susceptibility or infectiousness.
Row 2: Traced and in various stages of the queue
for vaccination.
Row 3: Unsuccessfully vaccinated and in various
stages of infectiousness.
Row 4: Successfully vaccinated; dead
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Aside: Related Work
•The model to be presented arose from the study of
opinion formulation in groups.
•Similar models are used in distributed computing.
•The models we will study are sometimes called
threshold networks.
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The Model: Moving From State to State
Let si(t) give the state of vertex i at time t.
Simplified Model: Two states 0 and 1.
Times are discrete: t = 0, 1, 2, …
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First Try: Majority Processes
Basic Majority Process: You change your state at
time t+1 if a majority of your neighbors have the
opposite state at time t.
(No change in case of “ties”)
Useful in models of spread of opinion.
Disease interpretation? Cure if majority of your
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neighbors are uninfected. Does this make sense?
Majority Processes II
Irreversible Majority Process: You change your
state from 0 to 1 at time t+1 if a majority of
your neighbors have state 1 at time t. You never
leave state 1.
(No change in case of “ties”)
Disease interpretation? Infected if sufficiently
many of your neighbors are infected.
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Basic Majority Process
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Irreversible Majority Process
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Aside: Distributed Computing
Majority processes are studied in distributed
computing.
Goal: Eliminate damage caused by failed
processors (vertices) or at least to restrict their
influence.
Do this by maintaining replicated copies of crucial
data and, when a fault occurs, letting a processor
change “state” if a majority of its neighbors are in a
different state.
Other applications of similar ideas in distributed
computing: distributed database management,
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quorum systems, fault local mending.
Second Try: Threshold Processes
Basic k-Threshold Process: You change your state
at time t+1 if at least k of your neighbors have
the opposite state at time t.
Disease interpretation? Cure if sufficiently many of
your neighbors are uninfected. Does this make
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sense?
Threshold Processes II
Irreversible k-Threshold Process: You change
your state from 0 to 1 at time t+1 if at least k
of your neighbors have state 1 at time t. You
never leave state 1.
Disease interpretation? Infected if sufficiently
many of your neighbors are infected.
Special Case k = 1: Infected if any of your
neighbors is infected.
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Basic 2-Threshold Process
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Irreversible 2-Threshold Process
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Complications to Add to Model
•k = 1, but you only get infected with a certain
probability.
•You are automatically cured after you are in the
infected state for d time periods.
•You become immune from infection (can’t reenter state 1) once you enter and leave state 1.
•A public health authority has the ability to
“vaccinate” a certain number of vertices, making
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them immune from infection.
Periodicity
State vector: s(t) = (s1(t), s2(t), …, sn(t)).
First example, s(1) = s(3) = s(5) = …,
s(0) = s(2) = s(4) = s(6) = …
Second example: s(1) = s(2) = s(3) = ...
In all of these processes, because there is a finite
set of vertices, for any initial state vector s(0), the
state vector will eventually become periodic, i.e.,
for some P and T, s(t+P) = s(t) for all t > T.
The smallest such P is called the period.
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Periodicity II
First example: the period is 2.
Second example: the period is 1.
Both basic and irreversible threshold processes are
special cases of symmetric synchronous neural
networks.
Theorem (Goles and Olivos, Poljak and Sura): For
symmetric, synchronous neural networks, the
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period is either 1 or 2.
Periodicity III
When period is 1, we call the ultimate state vector
a fixed point.
When the fixed point is the vector s(t) = (1,1,…,1)
or (0,0,…,0), we talk about a final common state.
One problem of interest: Given a graph, what
subsets S of the vertices can force one of our
processes to a final common state with entries
equal to the state shared by all the vertices in S in
the initial state?
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Periodicity IV
Interpretation: Given a graph, what subsets S of
the vertices should we plant a disease with so that
ultimately everyone will get it? (s(t)  (1,1,…,1))
Economic interpretation: What set of people do we
place a new product with to guarantee “saturation”
of the product in the population?
Interpretation: Given a graph, what subsets S of
the vertices should we vaccinate to guarantee that
ultimately everyone will end up without the
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disease? (s(t)  0,0,…,0))
Conversion Sets
Conversion set: Subset S of the vertices that can
force a k-threshold process to a final common state
with entries equal to the state shared by all the
vertices in S in the initial state. (In other words, if
all vertices of S start in same state x = 1 or 0, then
the process goes to a state where all vertices are in
state x.)
Irreversible k-conversion set if irreversible
process.
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1-Conversion Sets
k = 1.
What are the conversion sets in a basic 1-threshold
process?
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1-Conversion Sets
k = 1.
The only conversion set in a basic 1-threshold
process is the set of all vertices. For, if any two
adjacent vertices have 0 and 1 in the initial state,
then they keep switching between 0 and 1 forever.
What are the irreversible 1-conversion sets?
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Irreversible 1-Conversion Sets
k = 1.
Every single vertex x is an irreversible 1conversion set if the graph is connected. We make
it 1 and eventually all vertices become 1 by
following paths from x.
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Conversion Sets for Odd Cycles
C2p+1
2-threshold process.
What is a conversion set?
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Conversion Sets for Odd Cycles
C2p+1.
2-threshold process.
Place p+1 1’s in “alternating” positions.
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Conversion Sets for Odd Cycles
We have to be careful where we put the initial 1’s.
p+1 1’s do not suffice if they are next to each
other.
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Irreversible Conversion Sets for
Odd Cycles
What if we want an irreversible conversion set
under an irreversible 2-threshold process?
Same set of p+1 vertices is an irreversible
conversion set. Moreover, everyone gets infected in
one step.
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Vaccination Strategies
Mathematical models are very helpful in
comparing alternative vaccination strategies. The
problem is especially interesting if we think of
protecting against deliberate infection by a
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bioterrorist.
Vaccination Strategies
If you didn’t know whom a bioterrorist might
infect, what people would you vaccinate to be sure
that a disease doesn’t spread very much?
(Vaccinated vertices stay at state 0 regardless of
the state of their neighbors.)
Try odd cycles again. Consider an irreversible 2threshold process. Suppose your adversary has
enough supply to infect two individuals.
Strategy 1: “Mass vaccination”: make everyone 0
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and immune in initial state.
Vaccination Strategies
In C5, mass vaccination means vaccinate 5
vertices. This obviously works.
In practice, vaccination is only effective with a
certain probability, so results could be different.
Can we do better than mass vaccination?
What does better mean? If vaccine has no cost and
is unlimited and has no side effects, of course we
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use mass vaccination.
Vaccination Strategies
What if vaccine is in limited supply? Suppose we
only have enough vaccine to vaccinate 2 vertices.
Consider two different vaccination strategies:
Vaccination Strategy I
Vaccination Strategy II
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Vaccination Strategy I: Worst
Case (Adversary Infects Two)
Two Strategies for Adversary
Adversary Strategy Ia
Adversary Strategy Ib
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The “alternation” between your choice of a
defensive strategy and your adversary’s choice
of an offensive strategy suggests we consider
the problem from the point of view of game
theory.
The Food and Drug Administration is studying
the use of game-theoretic models in the
defense against bioterrorism.
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Vaccination Strategy I
Adversary Strategy Ia
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Vaccination Strategy I
Adversary Strategy Ib
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Vaccination Strategy II: Worst
Case (Adversary Infects Two)
Two Strategies for Adversary
Adversary Strategy IIa
Adversary Strategy IIb
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Vaccination Strategy II
Adversary Strategy IIa
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Vaccination Strategy II
Adversary Strategy IIb
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Conclusions about Strategies I
and II
If you can only vaccinate two individuals:
Vaccination Strategy II never leads to more than
two infected individuals, while Vaccination
Strategy I sometimes leads to three infected
individuals (depending upon strategy used by
adversary).
Thus, Vaccination Strategy II is better.
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Vaccination Strategies
Stephen Hartke (2004) worked on a different
problem:
You can only vaccinate one person per time period
and you can only infect one person per time period.
The vaccinator and infector alternate turns.
What is a good strategy for the vaccinator?
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The Mathematics of k-Conversion Sets
k-conversion sets are complex.
Consider the graph K4 x K2.
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k-Conversion Sets II
Exercise: (a). The vertices a, b, c, d, e form a 2conversion set.
(b). However, the vertices a,b,c,d,e,f do not.
Interpretation: Immunizing one more person can be
worse! (Planting a disease with one more person
can be worse if you want to infect everyone.)
Note: the same does not hold true for irreversible
k-conversion sets.
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k-Conversion Sets in Regular Graphs
G is r-regular if every vertex has degree r.
Set of vertices is independent if there are no edges.
Theorem (Dreyer 2000): Let G = (V,E) be a
connected r-regular graph and D be a set of
vertices.
(a). D is an irreversible r-conversion set iff V-D
is an independent set.
(b). D is an r-conversion set iff V-D is an
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independent set and D is not an independent set.
k-Conversion Sets in Regular
Graphs II
Corollary (Dreyer 2000):
(a). The size of the smallest irreversible 2conversion set in Cn is ceiling[n/2].
(b). The size of the smallest 2-conversion set in Cn
is ceiling[(n+1)/2].
This result agrees with our observation.
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k-Conversion Sets in Regular
Graphs III
Proof:
(a). Cn is 2-regular. The largest independent set
has size floor[n/2]. Thus, the smallest D so that
V-D is independent has size ceiling[n/2].
(b). If n is odd, taking the first, third, …, nth
vertices around the cycle gives a set that is not
independent and whose complement is
independent. If n is even, every vertex set of size
n/2 with an independent complement is itself
independent, so an additional vertex is needed. 61
k-Conversion Sets in Graphs of
Maximum Degree r
Theorem (Dreyer 2000): Let G = (V,E) be a
connected graph with maximum degree r and S be
the set of all vertices of degree < r. If D is a set of
vertices, then:
(a). D is an irreversible r-conversion set iff SD
and V-D is an independent set.
(b). D is an r-conversion set iff SD, V-D is an
independent set, and if S = , then D is not an
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independent set.
NP-Completeness
Problem k-CONVERSION SET: Given a positive
integer d and a graph G, does G have a kconversion set of size at most d?
IRREVERSIBLE k-CONVERSION SET:
Similar.
Theorem (Dreyer 2000): k-CONVERSION SET
and IRREVERSIBLE k-CONVERSION SET are
NP-complete for fixed k > 2.
(Whether or not they are NP-complete for k = 2
remains open.)
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NP-Completeness II
Proof: Reduction from a case of the independent
set problem.
Problem INDEPENDENT SET: Given a positive
integer m and a graph G, does G have an
independent set of size  m?
Theorem (Fricke, Hedetniemi, Jacobs): For fixed
k3, INDEPENDENT SET is NP-complete for
the class of k-regular, non-bipartite graphs.
(The problem is in P for bipartite graphs.)
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NP-Completeness III
Given an instance of INDEPENDENT SET for a k-regular,
non-bipartite graph G:
Does G have an independent set of size  m?
Construct an instance of IRREVERSIBLE kCONVERSION SET:
Does G have an irreversible k-conversion set of size
 |V(G)| - m?
D is an irreversible k-conversion set iff V-D is
independent.
So: There is an irreversible k-conversion set of size 
|V(G)| - m iff there is an independent set of size  m.
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NP-Completeness IV
Non-bipartiteness is not used here.
It is used in the similar proof that k-CONVERSION SET
is NP-complete.
D is a k-conversion set iff D is independent and V(G) – D
is independent.
The construction is the same.
We need the fact that if D is not a k-conversion set in G,
then either V(G) – D is not independent or both V(G) – D
and D are independent. The latter is impossible in a nonbipartite graph.
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k-Conversion Sets in Trees
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k-Conversion Sets in Trees
The simplest case is when every internal vertex of
the tree has degree > k.
Leaf = vertex of degree 1; internal vertex = not a
leaf.
What is a 2-conversion set here?
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All leaves have to be in it.
This will suffice.
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k-Conversion Sets in Trees
So k = 2 is easy. What about k > 2? Also easy.
Proposition (Dreyer 2000): Let T be a tree and
every internal vertex have degree > k, where k > 1.
Then the smallest k-conversion set and the
smallest irreversible k-conversion set have size
equal to the number of leaves of the tree.
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k-Conversion Sets in Trees
A vertex cover of graph G is a set S of vertices so
that for every edge {u,v}, either u or v is in S.
If T is a tree, let LT denote the set of leaves.
Proposition (Dreyer 2000): Let T be a tree with
every internal vertex of degree > k. Then S is a kconversion set (an irreversible k-conversion set) iff
S = CLT where C is a vertex cover of the edges
in V-LT.
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k-Conversion Sets in Trees
What if not every internal vertex has degree > k?
If there is an internal vertex of degree < k, it will have
to be in any k-conversion set and will never change
sign.
So, to every neighbor, this vertex v acts like a leaf, and
we can break T into deg(v) subtrees with v a leaf in
each.
If every internal vertex has degree  k, one can obtain
analogous results to those for the > k case by looking
at maximal connected subsets of vertices of degree 73k.
k-Conversion Sets in Trees
Dreyer presents an O(n) algorithm for finding the
size of the smallest irreversible 2-conversion set in
a tree of n vertices.
The method builds on the generalization of the
theorem characterizing irreversible k-conversion
sets in the case where every internal vertex has
degree  k.
The algorithm can be readily extended to find the
size of the smallest irreversible k-conversion set.74
k-Conversion Sets in Grids
Let G(m,n) be the rectangular grid graph with m
rows and n columns.
G(3,4)
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Toroidal Grids
The toroidal grid T(m,n) is obtained from the
rectangular grid G(m,n) by adding edges from the
first vertex in each row to the last and from the first
vertex in each column to the last.
Toroidal grids are easier to deal with than
rectangular grids because they form regular graphs:
Every vertex has degree 4. Thus, we can make use
of the results about regular graphs.
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T(3,4)
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4-Conversion Sets in Toroidal Grids
Theorem (Dreyer 2000): In a toroidal grid T(m,n)
(a). The size of the smallest 4-conversion set is
max{n(ceiling[m/2]), m(ceiling[n/2])} m or n odd
{
mn/2 + 1
m, n even
(b). The size of the smallest irreversible 4conversion set is as above when m or n is
odd, and it is mn/2 when m and n are even.78
Part of the Proof: Recall that D is an irreversible
4-conversion set in a 4-regular graph iff V-D is
independent.
V-D independent means that every edge {u,v} in
G has u or v in D. In particular, the ith row
must contain at least ceiling[n/2] vertices in D and
the ith column at least ceiling[m/2] vertices in D
(alternating starting with the end vertex of the row
or column).
We must cover all rows and all columns, and so
need at least max{n(ceiling[m/2]), m(ceiling[n/2])}
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vertices in an irreversible 4-conversion set.
4-Conversion Sets for Rectangular
Grids
More complicated methods give:
Theorem (Dreyer 2000): The size of the smallest 4conversion set and smallest irreversible 4conversion set in a grid graph G(m,n) is
2m + 2n - 4 + floor[(m-2)(n-2)/2]
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4-Conversion Sets for Rectangular
Grids
Consider G(3,3):
2m + 2n - 4 + floor[(m-2)(n-2)/2] = 8.
What is a smallest 4-conversion set and why 8? 81
4-Conversion Sets for Rectangular Grids
Consider G(3,3):
2m + 2n - 4 + floor[(m-2)(n-2)/2] = 8.
What is a smallest 4-conversion set and why 8?
All boundary vertices have degree < 4 and so must
be included in any 4-conversion set. They give 82
a conversion set.
2-Conversion Sets for Grids
Let Ci,k(G) be the smallest irreversible
k-conversion set in graph G.
Theorem (Flocchini, Lodi, Luccio, Pagli, and
Santoro):
Ci,2[G(m,n)] = ceiling([m+n]/2)
Ci,2[T(m,n)] = ceiling([m+n]/2) – 1.
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3-Conversion Sets for Grids
For 3-conversion sets, the best we have are bounds:
Theorem (Flocchini, Lodi, Luccio, Pagli, and
Santoro):
[(m-1)(n-1)+1]/3  Ci,3[G(m,n)] 
[(m-1)(n-1)+1]/3 +[3m+2n-3]/4 + 5
ceiling([mn+1]/2)  Ci,3[T(m,n)] 
ceiling(mn/2 + m/6 + 2/3)
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2- and 3-Conversion Sets for Small
Grids
For the special case of 2xn and 3xn grids, Dreyer has
more precise results.
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Bounds on the Size of the Smallest
Conversion Sets
In general, it is difficult to get exact values for the
size of the smallest k-conversion and irreversible kconversion set in a graph.
So, what about bounds?
Sample result:
Theorem (Dreyer, 2000): If G is an r-regular graph
with n vertices, then Ci,k(G)  (1 – r/2k)n for k  r
 2k.
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Bounds on the Size of the Smallest
Conversion Sets
Conjecture (Dreyer): For a graph G on n vertices,
the size of the smallest k-conversion set is  n/k,
and this bound is asymptotically sharp.
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More Realistic Models
Many oversimplifications in our models. For
instance:
•What if you stay infected only a certain number of
days?
•What if you are not necessarily infective for the
first few days you are sick?
•What if your threshold k for changes from 0 to 1
changes depending upon how long you have been
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uninfected?
More Realistic Models
Consider an irreversible process in which you stay
in the infected state (state 1) for d time periods
after entering it and then go back to the uninfected
state (state 0).
Consider a k-threshold process in which we
vaccinate a person in state 0 once k-1 neighbors are
infected (in state 1).
Etc. -- let your imagination roam free ...
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More Realistic Models
Our models are deterministic. How do probabilities
enter?
•What if you only get infected with a certain
probability if you meet an infected person?
•What if vaccines only work with a certain
probability?
•What if the amount of time you remain infective
exhibits a probability distribution?
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Alternative Model with Probabilities
Consider an irreversible 1-threshold process in
which you stay infected for d time periods and
then enter the uninfected state.
Assume that you get infected with probability p if
at least one of your neighbors is infected.
What is the probability that an epidemic will end
with no one infected?
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The Case d = 2, p = 1/2
Consider the following initial state:
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The Case d = 2, p = 1/2
With probability 1/2, vertex a does not get infected
at time 1.
Similarly for vertex b.
Thus, with probability 1/4, we stay in the same
states at time 1.
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The Case d = 2, p = 1/2
Suppose vertices are still in same states at time 1 as
they were at time 0. With probability 1/2, vertex a
does not get infected at time 2.
Similarly for vertex b.
Also after time 1, vertices c and d have been
infected for two time periods and thus enter the
uninfected state.
Thus, with probability 1/4, we get to the following
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state at time 2:
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The Case d = 2, p = 1/2
Thus, with probability 1/4 x 1/4 = 1/16, we enter
this state with no one infected at time 2.
However, we might enter this state at a later time.
It is not hard to show (using the theory of finite
Markov chains) that we will end in state (0,0,0,0).
(This is the only absorbing state in an absorbing
Markov chain.). Thus: with probability 1 we will
eventually kill the disease off entirely.
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The Case d = 2, p = 1/2
Is this realistic? What might we do to modify the
model to make it more realistic?
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How do we Analyze this or More
Complex Models for Graphs?
We can use the model to do “what if” experiments.
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What-if Experiments:
•What if we adopt a particular vaccination
strategy?
•What happens if we try different plans for
quarantining infectious individuals?
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There is much more analysis of a similar nature
that can be done with graph-theoretical models. Let
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your imagination run free!