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Graph-theoretical Models of the Spread and Control of Disease Fred Roberts, DIMACS

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### Graph-theoretical Models of the Spread and Control of DiseaseFred Roberts, DIMACS

### The Stages Networks

### Basic Majority Process Networks

### Irreversible Majority Process Networks

### Basic 2-Threshold Process Networks

### Irreversible 2-Threshold Process Networks

### Vaccination Strategy I: Worst Case (Adversary Infects Two) NetworksTwo Strategies for Adversary

### Vaccination Strategy I strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory. Adversary Strategy Ia

### Vaccination Strategy I strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory. Adversary Strategy Ib

### Vaccination Strategy II: Worst Case (Adversary Infects Two) strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.Two Strategies for Adversary

### Vaccination Strategy II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory. Adversary Strategy IIa

### Vaccination Strategy II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory. Adversary Strategy IIb

### Conclusions about Strategies I and II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

### Vaccination Strategies strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

### How do we Analyze this or More Complex Models for Graphs? strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

### What-if Experiments: strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

smallpox

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 bioterrorist attacks.

The Basic Model Networks

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)

More About States Networks

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.

The State Diagram for a Smallpox Model Networks

The following diagram is from a Kaplan-Craft-Wein (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.

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

Aside: Related Work Networks

- 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.

The Model: Moving From State to State Networks

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, …

First Try: Majority Processes Networks

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 neighbors are uninfected. Does this make sense?

Majority Processes II Networks

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.

Aside: Distributed Computing Networks

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, quorum systems, fault local mending.

Second Try: Threshold Processes Networks

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 sense?

Threshold Processes II Networks

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.

Complications to Add to Model Networks

- 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 re-enter state 1) once you enter and leave state 1.
- A public health authority has the ability to “vaccinate” a certain number of vertices, making them immune from infection.

Periodicity Networks

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.

Periodicity II Networks

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 period is either 1 or 2.

Periodicity III Networks

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?

Periodicity IV Networks

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 disease? (s(t) 0,0,…,0))

Conversion Sets Networks

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-conversionset if irreversible process.

1-Conversion Sets Networks

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?

Irreversible 1-Conversion Sets Networks

k = 1.

Every single vertex x is an irreversible 1-conversion set if the graph is connected. We make it 1 and eventually all vertices become 1 by following paths from x.

Conversion Sets for Odd Cycles Networks

C2p+1.

2-threshold process.

Place p+1 1’s in “alternating” positions.

Conversion Sets for Odd Cycles Networks

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.

Irreversible Conversion Sets for Odd Cycles Networks

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.

Vaccination Strategies Networks

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 bioterrorist.

Vaccination Strategies Networks

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 2-threshold process. Suppose your adversary has enough supply to infect two individuals.

Strategy 1: “Mass vaccination”: make everyone 0 and immune in initial state.

Vaccination Strategies Networks

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 use mass vaccination.

Vaccination Strategies Networks

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

Adversary Strategy Ia

Adversary Strategy Ib

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.

Adversary Strategy IIa

Adversary Strategy IIb

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.

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?

The Mathematics of k-Conversion Sets strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

k-conversion sets are complex.

Consider the graph K4 x K2.

k-Conversion Sets II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Exercise: (a). The vertices a, b, c, d, e form a 2-conversion 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.

k-Conversion Sets in Regular Graphs strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 independent set and D is not an independent set.

k-Conversion Sets in Regular Graphs II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Corollary (Dreyer 2000):

(a). The size of the smallest irreversible 2- conversion 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.

k-Conversion Sets in Regular Graphs III strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

k-Conversion Sets in Graphs of Maximum Degree r strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 SD and V-D is an independent set.

(b). D is an r-conversion set iff SD, V-D is an independent set, and if S = , then D is not an independent set.

NP-Completeness strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Problem k-CONVERSION SET: Given a positive integer d and a graph G, does G have a k-conversion 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.)

NP-Completeness II strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 k3, INDEPENDENT SET is NP-complete for the class of k-regular, non-bipartite graphs.

(The problem is in P for bipartite graphs.)

Given

NP-Completeness III strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 k-CONVERSION 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.

NP-Completeness IV strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 non-bipartite graph.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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?

All leaves have to be in it. strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

This will suffice.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 k-conversion set (an irreversible k-conversion set) iff S = CLT where C is a vertex cover of the edges in V-LT.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 k.

k-Conversion Sets in Trees strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

k-Conversion Sets in Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Let G(m,n) be the rectangular grid graph with m rows and n columns.

G(3,4)

Toroidal Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

T(3,4) strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

4-Conversion Sets in Toroidal Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 4-conversion set is as above when m or n is

odd, and it is mn/2 when m and n are even.

{

Part of the Proof strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.: 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])} vertices in an irreversible 4-conversion set.

4-Conversion Sets for Rectangular Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

More complicated methods give:

Theorem (Dreyer 2000): The size of the smallest 4-conversion set and smallest irreversible 4-conversion set in a grid graph G(m,n) is

2m + 2n - 4 + floor[(m-2)(n-2)/2]

4-Conversion Sets for Rectangular Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Consider G(3,3):

2m + 2n - 4 + floor[(m-2)(n-2)/2] = 8.

What is a smallest 4-conversion set and why 8?

4-Conversion Sets for Rectangular Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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

a conversion set.

2-Conversion Sets for Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

3-Conversion Sets for Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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)

2- and 3-Conversion Sets for Small Grids strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

For the special case of 2xn and 3xn grids, Dreyer has more precise results.

Bounds on the Size of the Smallest Conversion Sets strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

In general, it is difficult to get exact values for the size of the smallest k-conversion and irreversible k-conversion 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.

Bounds on the Size of the Smallest Conversion Sets strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

More Realistic Models strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

- 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 uninfected?

More Realistic Models strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 ...

More Realistic Models strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

- 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?

Alternative Model with Probabilities strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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?

The Case d = 2, p = 1/2 strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Consider the following initial state:

The Case d = 2, p = 1/2 strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

The Case d = 2, p = 1/2 strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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 state at time 2:

The Case d = 2, p = 1/2 strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

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.

The Case d = 2, p = 1/2 strategy and your adversary’s choice of an offensive strategy suggests we consider the problem from the point of view of game theory.

Is this realistic? What might we do to modify the model to make it more realistic?

We can use the model to do “what if” experiments.

What if we adopt a particular vaccination strategy?

What happens if we try different plans for quarantining infectious individuals?

There is much more analysis of a similar nature that can be done with graph-theoretical models. Let your imagination run free!

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