Locating Sensors to Detect
Chem, Bio, or Nuclear Threats
Fred Roberts
DyDAn 1
Sensors are finding many uses in homeland
security.
We place sensors in buildings, on bridges, at
border crossings, and even on uniforms of
police.
They sense radiation, dangerous chemicals,
biological agents, etc.
2
The Bioterrorism Sensor
Location Problem
3
The Bioterrorism Sensor
Location Problem
Great concerns about deliberate release of
disease-causing pathogens has caused us to
Worry about “bioterrorism.”
anthrax
4
• Early warning is
critical in defense
against terrorism
• This is a crucial
factor underlying the
government’s plans
to place networks of
sensors/detectors to
warn of a bioterrorist
attack
5
The BASIS System – Salt Lake City
Types of Sensor Problems
• I first got involved in sensor placement for
biological attacks when I was approached by the
Defense Threat Reduction Agency.
• I ended up learning a lot about the problem from
the Institute for Defense Analyses.
• And from the NYC Department of Health
• Similar issues arise in placing sensors to protect
against or give early warning about attacks with
chemical or nuclear weapons.
• Or attacks on our networks: communication,
financial, etc.
6
Types of Sensor Problems
• Similar sensor placement problems arise in
applications in:
– Civil engineering: monitoring temperature,
humidity, structural stability in a building,
bridge, etc.
– Manufacturing
– Fault detection in distributed or
multiprocessor computer systems
7
Locating Sensors is not Easy
• Sensors are expensive
• How do we select them and where
do we place them to maximize
“coverage,” expedite an alarm, and
keep the cost down?
• Approaches that improve upon
existing, ad hoc location methods
could save countless lives in the
case of an attack and also money in
capital and operational costs.
8
Two Fundamental Problems
• Sensor Location
Problem
– Choose an
appropriate mix of
kinds of sensors
– decide where to
locate them for
best protection and
early warning
9
Two Fundamental Problems
• Pattern Interpretation
Problem: When sensors
set off an alarm, help public
health decision makers
decide
– Has an attack taken
place?
– What additional
monitoring is needed?
– What was its extent and
location?
– What is an appropriate
response?
10
The SLP: What is a Measure of
Success of a Solution?
•
•
•
•
A modeling problem.
Needs to be made precise.
Many possible formulations.
Good explorations for your students.
11
The SLP: What is a Measure of
Success of a Solution?
• Identify and ameliorate false alarms.
• Defending against a “worst case” attack or
an “average case” attack.
• Minimize time to first alarm? (Worst case?
(Average case?)
• Maximize “coverage” of the area.
– Minimize geographical area not covered
– Minimize size of population not covered
– Minimize probability of missing an attack
12
The SLP: What is a Measure of
Success of a Solution?
•Cost: Given a mix of available sensors and
a fixed budget, what mix will best
accomplish our other goals?
13
The SLP: What is a Measure of
Success of a Solution?
•It’s hard to separate the goals.
•Even a small number of sensors might
detect an attack if there is no constraint on
time to alarm.
•Without budgetary restrictions, a lot more
can be accomplished.
14
The Sensor Location Problem
•Approach is to develop new algorithmic
methods.
•Developing new algorithms involves
fundamental mathematical analysis.
•Analyzing how efficient algorithms are involves
fundamental mathematical methods.
•Implementing the algorithms on a computer is
often a separate problem – which needs to go
hand in hand with the basic mathematics of
algorithm development.
15
Algorithmic Approaches I :
Greedy Algorithms
16
Greedy Algorithms
• Find the “most important” location to place a
sensor first and locate a sensor there.
• What things could “most important” mean?
• Find second-most important location.
• Etc.
• Builds on earlier mathematical work at Institute
for Defense Analyses (Grotte, Platt)
• “Steepest ascent approach.’’
• No guarantee of “optimal” or best solution.
• In practice, gets pretty close to optimal solution.
17
Algorithmic Approaches II :
Variants of Classic Facility
Location Theory Methods
18
Location Theory
• Old problem in Operations research: Where
to locate facilities (fire houses, garbage
dumps, etc.) to best serve “users”
• Often deal with a network with vertices,
edges, and distances along edges
• Users u1, u2, …, un are located at vertices
• One approach: locate the facility at vertex x
chosen so that sum of distances to users is
minimized.
n
d ( x, ui )
• Minimize:

i 1
19
Location Theory: A Network
f
1
a
1
1
e
b
1
1
d
Vertices are places
for users or facilities
1
c
1’s represent
distances along
20
edges
Location Theory: A Network
f
1
a
1
1
e
b
1
1
d
1
d(x,y) = length of shortest route
from x to y
So, d(a,c) = 2.
c
1’s represent
distances along
21
edges
1
f
1
a
1
u1
e
b
u2
1
1
d
1
c
u3
Given users at u1, u2, u3, where do we place a facility
to minimize the sum of distances to the users?
22
1
f
1
a
1
u1
e
b
u2
1
1
d
1
c
x=a: d(x,ui)=1+1+2=4
u3
x=b: d(x,ui)=2+0+1=3
x=c: d(x,ui)=3+1+0=4
x=d: d(x,ui)=2+2+1=5
x=e: d(x,ui)=1+3+2=6
x=f: d(x,ui)=0+2+3=5
x=b is optimal
23
Location Theory
• Users u1, u2, …, un are located at vertices
• Alternative approach: Locate the facility at
vertex x chosen so that maximum distance
to one of the users is minimized.
24
1
f
1
a
1
u1
e
b
u2
1
1
d
1
c
u3
Given users at u1, u2, u3, where do we place a facility
to minimize the maximum distance a user has to
travel to get to the facility?
25
1
f
1
a
1
u1
e
b
u2
1
1
d
x = a: maxid(ui,x) = 2
x = c: maxid(ui,c) = 3
x = e: maxid(ui,e) = 3
1
c
u3
x = b: maxid(ui,b) = 2
x = d: maxid(ui,d) = 2
x = f: maxid(ui,f) = 3
There are three optimal solutions: a, b, d
26
Variants of Classic Facility
Location Theory Methods:
Complications
• We don’t have a network with vertices and edges;
we have points in a city
• Sensors can only be at certain locations (size,
weight, power source, hiding place)
• We need to place more than one sensor
• Instead of “users,” we have places where potential
attacks take place.
• Potential attacks take place with certain
probabilities.
• Wind, buildings, mountains, etc. add
27
complications.
The Pattern Interpretation
Problem
28
The Pattern Interpretation
Problem (PIP)
• It will be up to the
Decision Maker to
decide how to
respond to an
alarm from the
sensor network.
29
Approaching the PIP: Minimizing
False Alarms
30
Approaching the PIP: Minimizing
False Alarms
One approach: Redundancy.
• Could require two or more
sensors to make a detection
before an alarm is
considered confirmed
• Could require same sensor
to register two alarms:
Biosensor “Portal Shield”
requires two positives for
the same agent during a
specific time period.
31
Approaching the PIP: Minimizing
False Alarms
• Could place two or more sensors at or
near the same location. Require two
proximate sensors to give off an alarm
before we consider it confirmed.
• Redundancy has drawbacks: cost, delay in
confirming an alarm.
• We need mathematical methods to
analyze the tradeoff between lowered
false alarm rate and extra cost/delay
32
Sensor Placement in Networks
• From now on, we will concentrate on
networks or graphs.
• G = (V,E) = graph; V = vertices, E = edges
• Assume all edges have distance 1
• Limit to “attacks” on vertices
• Assume detectors can only be placed at
vertices
• Two goals:
– Detection: Detect that an attack on a
vertex has taken place
– Identification: Identify where the attack
33
took place
Sensor Placement in Networks
(Joint work with David L. Roberts)
• Assume that a detector can detect an
attack up to distance r away.
• We often think of distance = time.
• Assume that there are no errors in
detection: If there is an attack up to
distance r away, the detector will detect it
with 100% certainty.
• Nr[x] = set of all vertices in V to which
there is a path in the graph of length at
most r from x. (r-neighborhood of x)
34
Sensor Placement in Networks
f
a
e
b
d
c
• N2(c) = ?
• N2(c) = {c, b, d, a, e}
35
Sensor Placement in Networks
• Let D be the set of vertices where we
place a detector.
• Let Dr(x) = Nr(x)  D
36
Sensor Placement in Networks
f
a
e
b
d
D = {a, b, f}
D2(d) = ?
D2(d) = {b, f}
c
37
Sensor Placement in Networks
• Detection: D is an r-dominating set in G
if for every vertex x of G, Dr(x)  .
• That is, for all x, there is a path of length at
most r from x to some y in D.
38
Sensor Placement in Networks
f
a
e
b
d
D = {a, b, f}
D is a 2-dominating set.
It is not a 1-dominating set.
c
39
Sensor Placement in Networks
• Unique Identification: D is an ridentifying code (r-IC) in G if it is an rdominating set and if, whenever x  y are
vertices, Dr(x)  Dr(y).
• In an r-IC, the set of detectors activated by
an attack provides a unique signature that
allows us to determine where the attack
took place.
• We shall look for the smallest d so that
there is an r-IC of d vertices
40
Sensor Placement in Networks
f
a
e
b
d
D = {a, b, f}
D is a 2-dominating set.
Is D a 2-IC?
c
41
Sensor Placement in Networks
f
a
e
b
d
c
D = {a, b, f}
D2(a) = {a, b, f}
D2(b) = {a, b, f}
So: D is not a 2-IC
42
Sensor Placement in Networks
f
a
e
b
d
c
Can you find a 2-IC?
43
Sensor Placement in Networks
f
a
e
b
d
c
Can you find a 2-IC?
D = {b, d, f} is not a 2-IC. Why?
44
Sensor Placement in Networks
f
a
e
b
d
c
Can you find a 2-IC?
D = {b, d, f} is not a 2-IC. Why?
D2(b) = D2(d)
45
Sensor Placement in Networks
f
a
e
d
D = {a, b, c, d, e} is a 2-IC.
D2(a) = {a, b, c, e}
D2(c) = {a, b, c, d, e}
D2(e) = {a, c, d, e}
b
c
D2(b) = {a, b, c, d}
D2(d) = {b, c, d, e}
D2(f) = {a, b, d, e}
46
Sensor Placement in Networks
f
a
e
b
d
c
D = {a, b, c, d, e} is a 2-IC.
Can you find a smaller one?
47
Sensor Placement in Networks
f
a
e
b
d
c
Can you find a smaller one?
Answer: No. But the proof is a bit
complicated.
48
Sensor Placement in Networks
In general, the problem of finding optimal rICs is difficult in a precise sense:
It is NP-complete in general and for many
special classes of graphs
49
Sensor Placement in Networks:
1-Identifying Codes
Let us concentrate first on 1-IC’s.
• Many network structures have been studied:
– Paths and cycles
– Binary cubes and nonbinary cubes and
hypercubes and meshes (of interest in CS
applications)
– Trees
– Square lattices
– Hexagonal and triangular grids
– Complete multipartite graphs
50
– Planar and outerplanar graphs
Sensor Placement in Networks:
1-Identifying Codes
• We will study paths and cycles.
P6
a
b
c
d
e
f
f
a
e
b
Path Pk has k vertices
Cycle Ck has k vertices
C6
c
51
Sensor Placement in Networks:
1-Identifying Codes
• We will study paths and cycles.
• Why are these important?
• Typical application: subway tunnels are
paths
• We use undirected graphs: Trains go in
one direction, but chemicals or biological
agents used in an attack go in both
directions.
• Another application: airport trams go in
cycles or paths.
52
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C4?
53
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C4?
a
b
D = {a, b, c} is a 1-IC
d
c
D1(a) = {a, b}
D1(b) = {a, b, c}
D1(c) = {b, c}
D1(d) = {a, c}
54
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C4?
a
b
No 2-element set D is a 1-IC
d
c
If D = {a,c}, then
D1(b) = D1(d)
55
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C4?
a
b
No 2-element set D is a 1-IC
d
c
If D = {a,b}, then
D1(a) = D1(b)
56
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C6?
f
a
e
b
d
c
57
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C6?
f
a
This is a 1-IC
e
b
d
c
58
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C6?
f
a
This is a 1-IC
e
b
d
Similarly, for C2k,
k > 2, there is a
1-IC of k vertices.
c
59
1-Identifying Codes in Cycles
• Exercise: What is the smallest 1-IC for C6?
f
a
e
b
d
Let us prove that
there can be no
1-IC of less than
k vertices.
c
60
1-Identifying Codes in Cycles
Lemma: Suppose no vertex of G has more
than 2 neighbors. Suppose y1,y2,y3,y4 is a
path in G and D is a 1-IC for G. Then it is
not possible to have y1 not in D and y4 not
in D.
Proof: If y1 and y4 are not in D, then D1(y2) =
D1(y3). This is a contradiction.
Q.E.D.
a
y1
y2
y3
y4
f
61
1-Identifying Codes in Cycles
Suppose that D is a 1-IC in C2k, k > 2.
We show that D has at least k vertices.
Label vertices around the cycle as x1, x2, …, x2k.
By the lemma:
x1  D or x4  D
x2  D or x5  D
x3  D or x6  D
…
x2k-2  D or x1  D
x2k-1  D or x2  D
x2k  D or x3  D
There are 2k constraints. Each xi is in only 2 of
them.
Thus, D has at least k vertices.
62
1-Identifying Codes in Cycles
Thus:
Theorem: In C2k, k > 2, the smallest 1-IC has
k vertices. In C4, the smallest 1-IC has 3
vertices.
(First proved by Bertrand, Charon, Hudry,
and Lobstein 2004)
63
1-Identifying Codes in Cycles
For odd length cycles, we have the following
theorem:
Theorem: In C2k+1, k > 2, the smallest 1-IC
has k+2 vertices. In C5, the smallest 1-IC
has 3 vertices.
First proved by Daniel (2003) and Gravier,
Moncel, and Semri (2007).
64
1-Identifying Codes in Cycles
Exercise: For C5, find a 1-IC of 3 vertices.
65
1-Identifying Codes in Cycles
Exercise: For C5, find a 1-IC of 3 vertices.
66
1-Identifying Codes in Cycles
For C2k+1, k > 2, the following gives an
optimal 1-IC:
{x1, x2, x3, x5, x7, x9, …., x2k+1}
67
1-Identifying Codes in Cycles
A key idea in the proof that the minimum 1IC is at least this big is the following
lemma.
Lemma: If n > 4, then D is a 1-IC for a cycle
Cn if and only if:
(1) There are no four consecutive vertices
with the first and last not in D
(2) There are no three consecutive vertices
none of which is in D.
68
1-Identifying Codes in Cycles
Lemma: If n > 4, then D is a 1-IC for a cycle
Cn if and only if:
(1) There are no four consecutive vertices
with the first and last not in D
(2) There are no three consecutive vertices
none of which is in D.
Partial Proof: If condition (2) holds, then
D1(x)   for every x. If (2) fails, then
D1(x) =  for some x.
69
1-Identifying Codes in Paths
Turning next to paths:
What is the smallest 1-IC for P4?
70
1-Identifying Codes in Paths
Turning next to paths:
What is the smallest 1-IC for P4?
a
b
c
d
This 1-IC has 3 vertices. By considering
cases, it is easy to show that no 1-IC can
have 2 vertices.
71
1-Identifying Codes in Paths
A more formal proof uses the following
lemma.
Lemma: Let Pn be the path of n vertices,
labeled in order x1, x2, …, xn. If D is a 1IC for Pn, then x3 is in D and xn-2 is in D.
Proof: Do you see why?
72
1-Identifying Codes in Paths
A more formal proof uses the following
lemma.
Lemma: Let Pn be the path of n vertices,
labeled in order x1, x2, …, xn. If D is a 1IC for Pn, then x3 is in D and xn-2 is in D.
Proof: If x3 is not in D, then D1(x1) = D1(x2).
Similarly for xn-2.
Q.E.D.
…
x1
x2
x3
x4
xn-2
xn-1
xn73
1-Identifying Codes in Paths
What is the smallest 1-IC in P4?
Solution using the lemma:
Lemma: Let Pn be the path of n vertices,
labeled in order x1, x2, …, xn. If D is a 1IC for Pn, then x3 is in D and xn-2 is in D.
By the lemma, c and b are in D.
a
b
c
d
74
1-Identifying Codes in Paths
Now use another lemma we already saw:
Lemma: Suppose no vertex of G has more
than 2 neighbors. Suppose y1,y2,y3,y4 is
a path in G and D has is a 1-IC for G.
Then it is not possible to have y1 not in D
and y4 not in D.
Thus, a or d is in D. By symmetry, say d.
a
b
c
d
75
1-Identifying Codes in Paths
Lemma: Suppose no vertex of G has more
than 2 neighbors. Suppose y1,y2,y3,y4 is
a path in G and D has is a 1-IC for G.
Then it is not possible to have y1 not in D
and y4 not in D.
Thus, a or d is in D. By symmetry, say d.
Thus, D has at least 3 vertices. These 3
suffice.
a
b
c
d
76
1-Identifying Codes in Paths
This result generalizes:
Theorem: For P2k, k > 1, a smallest 1-IC is
obtained by using
x2, x3, x4, x6, x8, x10, …, x2k
Thus, it has k+1 vertices.
77
1-Identifying Codes in Paths
Exercise: For P2k+1, find a 1-IC.
78
1-Identifying Codes in Paths
Exercise: For P2k+1, find a 1-IC.
Solution: Use x1, x3, …, x2k+1
This turns out to be optimal.
So, k+1 is the size of the smallest 1-IC, as it
is for P2k.
79
2-Identifying Codes in Cycles
So far, 1-ICs.
Try 2-ICs.
Note that C4 cannot have a 2-IC.
Why??
80
2-Identifying Codes in Cycles
So far, 1-ICs.
Try 2-ICs.
Note that C4 cannot have a 2-IC.
Why??
Note that N2(a) = N2(c), so
D2(a) = D2(c).
a
b
d
c
81
2-Identifying Codes in Cycles
C6 is an interesting case.
We will see that the smallest 2-IC has 5
vertices.
Assume that it has a 2-IC of at most 4
vertices.
82
2-Identifying Codes in Cycles
C6 is an interesting case.
Lemma: Suppose no vertex in G has more
than 2 neighbors. Suppose y1, y2, …, y6
is a path in G and D is a 2-IC for G. Then
it is not possible to have y1 not in D and
y6 not in D.
Proof: If y1 is not in D and y6 is not in D, then
D2(y3) = D2(y4).
Q.E.D.
a
y1
y2
y3
y4
y5
y6
83
h
2-Identifying Codes in Cycles
f
e
•
•
•
•
a
b
c
d
Suppose D is a 2-IC of at most 4 vertices.
Without loss of generality a is not in D.
By the Lemma, f is in D.
Also by the lemma, b is in D.
84
2-Identifying Codes in Cycles
f
a
e
b
d
•
•
c
If d is not in D, then D2(a) = D2(d).
So, d is in D.
85
2-Identifying Codes in Cycles
f
a
e
b
d
•
•
c
Since D has at most 4 vertices, either c
or e is not in D.
By symmetry, assume c is not in D.
86
2-Identifying Codes in Cycles
f
a
e
b
d
•
c
It follows that D2(d) = D2(f).
87
2-Identifying Codes in Cycles
f
a
e
b
d
•
•
c
Hence the smallest 2-IC in C6 has at
least 5 vertices.
Can you find one with 5 vertices?
88
2-Identifying Codes in Cycles
f
a
e
b
d
•
c
Here is a 2-IC of 5 vertices.
89
2-Identifying Codes in Cycles
However, for C2k, k > 3, things are much
easier.
The optimal 2-IC is obtained using alternate
vertices.
Thus, the optimal 2-IC has k vertices.
90
2-Identifying Codes in Cycles
For C2k+1, we find that the result is more complicated.
Let k = 5p+q, where 0  q  4. Then: the smallest
2-IC in C2k+1 is given by:
k+2 if q = 0, p > 0
k+1 if q = 1, p > 0
k+3 if q = 2, p > 0
k+1 if q = 3, p  0
k+2 if q = 4, p > 0.
91
2-Identifying Codes in Cycles
Let k = 5p+q, where 0  q  4. Then: the smallest
2-IC in C2k+1 is given by:
k+2 if q = 0, p > 0
k+1 if q = 1, p > 0
k+3 if q = 2, p > 0
k+1 if q = 3, p  0
k+2 if q = 4, p > 0.
This covers all cases of C2k+1 except C9.
9 = 2(4) + 1, so k = 4.
k = 5(0) + 4, so p = 0, q = 4.
For C9 the smallest 2-IC has 5 elements.
92
2-Identifying Codes in Paths
Consider the path Pn with n vertices x1, x2, …,
xn in order.
Lemma: Suppose D is a 2-IC for Pn. Then x4,
x5, xn-3, xn-4 are in D.
Proof: Do you see why??
93
2-Identifying Codes in Paths
Consider the path Pn with n vertices x1, x2, …,
xn in order.
Lemma: Suppose D is a 2-IC for Pn. Then x4,
x5, xn-3, xn-4 are in D.
Proof: If x4 is not in D, then D2(x1) = D2(x2).
Similarly for xn-3.
If x5 is not in D, then D2(x2) = D2(x3).
Similarly for xn-4.
Q.E.D.
…
x1
x2
x3
x4
x5
xn-2
xn-1
xn
94
2-Identifying Codes in Paths
Let us apply what we know to calculate the
smallest 2-IC for P6.
First, we have
Lemma: Suppose no vertex in G has more
than 2 neighbors. Suppose y1, y2, …, y6 is
a path in G and D is a 2-IC for G. Then it is
not possible to have y1 not in D and y6 not
in D.
95
2-Identifying Codes in Paths
For P6.
First, we have
Lemma: Suppose no vertex in G has more
than 2 neighbors. Suppose y1, y2, …, y6 is
a path in G and D is a 2-IC for G. Then it is
not possible to have y1 not in D and y6 not
in D.
Thus, a or f is in D, without loss of generality a.
a
b
c
d
e
f
96
2-Identifying Codes in Paths
For P6.
Now, use:
Lemma: Suppose D is a 2-IC for Pn. Then x4,
x5, xn-3, xn-4 are in D.
x4 = d, x5 = e, xn-3 = c, xn-4 = b.
a
b
c
d
e
f
97
2-Identifying Codes in Paths
a
b
c
d
e
f
If this is D, then D2(c) = D2(d).
So, D has at least one more element, either a
or f. By symmetry, we may assume a is in
D. (So, we don’t really need the earlier
argument that a or f is in D. Sometimes
good to do arguments in two ways.)
a
b
c
d
e
f
98
2-Identifying Codes in Paths
This is a 2-IC.
Easy to check.
It follows that this is the smallest 2-IC and it
has 5 elements.
a
b
c
d
e
f
99
2-Identifying Codes in Paths
Generalization of this result:
Suppose n = 5p+q, 0  q  4. If q = 1 and p is
odd, then the smallest 2-IC for Pn has
5(p-1)/2 + 5
vertices.
If n = 6, then p = 1, q = 1, and we get
5(p-1)/2 + 5 = 5.
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2-Identifying Codes in Paths
The previous result gives the size of the
smallest 2-IC for Pn for one case of values
for p and q if n = 5p + q.
Similar results hold for other values of p and q.
101
Identifying Codes in Other Graphs
Let’s find a smallest 2-IC in the 3x3 grid:
102
Identifying Codes in Other Graphs
Let’s find a smallest 2-IC in the 3x3 grid.
Suppose D has 3 elements. The number of
subsets of a 3-element set is 23. There are
23-1 = 7 nonempty subsets.
103
Identifying Codes in Other Graphs
Let’s find a smallest 2-IC in the 3x3 grid.
Suppose D has 3 elements. The number of
subsets of a 3-element set is 23. There are
23-1 = 7 nonempty subsets.
Set = {1,2,3}
Nonempty subsets: {1,2,3}, {1,2}, {1,3}, {2,3},
{1}, {2}, {3}.
104
Identifying Codes in Other Graphs
Let’s find a smallest 2-IC in the 3x3 grid.
Suppose D has 3 elements. There are 23-1 = 7
nonempty subsets.
Since D2(x)  D2(y) for all x  y, each vertex
of the grid gets a different nonempty subset
of D.
But there are 9 vertices and only 7 subsets.
105
So, D has to have at least 4 vertices.
Identifying Codes in Other Graphs
Let’s find a smallest 2-IC in the 3x3 grid.
So, D has to have at least 4 vertices.
A 4-vertex 2-IC is shown.
106
Some Paradoxes
Our intuition about sensor location is not
very good.
Even for paths and cycles, some rather
strange things can happen.
107
Some Paradoxes
Paradox 1: For k > 2, the smallest 1-IC for
C2k+1 is larger than the smallest 1-IC for
C2k+2.
For instance, for C7, the smallest 1-IC has 5
vertices, while for C8, it has 4 vertices.
Thus, we take the same “topology” of a
network, make it longer, but require
fewer detectors!
108
Some Paradoxes
Paradox 2: For n = 6, 11, 10p+5, 10p+9,
10p+11, p  1, the smallest 2-IC for Cn is
larger than the smallest 2-IC for Cn+1.
For instance, for C11, the smallest 2-IC has 7
vertices, while for C12, it has 6 vertices.
Thus, we take the same “topology” of a
network, make it longer, but require
fewer detectors!
109
Some Paradoxes
Paradox 3: For p > 0, the smallest 2-IC for
C10p+5 is larger than the smallest 1-IC for
C10p+5.
For instance, for C15, the smallest 2-IC has
10 vertices, while the smallest 1-IC has 9
vertices.
Thus, using detectors of greater range can
actually require more detectors!
110
Some Paradoxes
Paradox 4: For n = 5, 6, 7, 10p+3, 10p+5,
10p+6, 10p+7, p > 0, the smallest 2-IC
for Pn is larger than the smallest 1-IC for
Pn.
For instance, for P5, the smallest 2-IC has 4
vertices while the smallest 1-IC has 3
vertices.
As with last example, using stronger
detectors may require more detectors.
111
Mathematics and Homeland
Security
Our intuition about sensor location is
not very good.
Even for paths and cycles, some rather
strange things can happen.
This shows why you really need
precise mathematical analysis.
112