Expanding Search for one or more Hiders
Lorentz Centre, Leiden, May 2012
Tom Lidbetter (PhD candidate, London School of Economics)
Joint work with Prof Steve Alpern (London School of Economics)
Pathwise search
An pathwise search 𝑆 on a network with root 𝑂 is a continuous unit speed path
starting at the root.
𝐻
5
1
5
1
2
1
𝑂
For a Hider 𝐻 located at a node, the search time 𝑇(𝐻, 𝑆) is first time the path reaches 𝐻.
In this example, 𝑇 𝐻, 𝑆 = 1 + 1 + 1 + 1 + 1 + 1 + 2 + 5 = 13.
Expanding search
An expanding search on a rooted network is a nested family of sets 𝑆𝑡 increasing at
unit speed, starting with only the root and ending with the whole network.
𝐻
5
1
5
1
2
1
𝑂
For a Hider 𝐻 at a node, the search time 𝑇(𝐻, 𝑆) is first time 𝐻 is contained in some 𝑆𝑡.
In this example, 𝑇 𝐻, 𝑆 = 1 + 1 + 2 + 1 + 5 = 10.
Connection to variable speed networks
If an expanding search on a tree is depth first, it is equivalent to a pathwise search on
a variable speed network.
0
5
5
0
1
1
1
0
0
2
0
0
𝑂
If the Hider is located on 𝑄 according to a known distribution 𝜈, we can calculate the
expected search time for an expanding search 𝑆. We are interested in the Bayesian
Problem of how the Searcher should minimise the expected search time for a given Hider
distribution.
8/24
1/12
7/24
5
2/12
1
1
5
1/24
1/12
2
1
𝑂
𝑇 𝜈, 𝑆 ≔ Expected search time = 1
1
1
2
12 1 + 12 2 + 24 4 + 12 5
+ 8 24 10 + 7 24 15
= 8 23
24
Search density
Definition: The search density (or just density) 𝜌 𝐴 of a region 𝐴 is the probability
the Hider is located in 𝐴 divided by the time required to search 𝐴, which in
expanding search is just the total length of 𝐴.
Example: The density of the highlighted region 𝐴 in the network below is
1
2
1
1
+
+
/(1 + 1 + 1) =
12 12 12
9
𝜌 𝐴 =
8/24
1/12
7/24
5
2/12
1
𝐴
1
5
1/24
1/12
2
1
𝑂
Search Density Lemma
Lemma 1: Suppose 𝐴 and 𝐵 are disjoint regions of a rooted network 𝑄 with Hider
distribution 𝜈. Let 𝑆𝐴𝐵 be search of 𝑄 that searches 𝐴 immediately before 𝐵, and let
𝑆𝐵𝐴 be an identical search except that it searches 𝐵 before 𝐴. Then
if 𝜌 𝐴 ≥ 𝜌 𝐵 , 𝑇(𝑆𝐴𝐵, 𝜈) ≤ 𝑇(𝑆𝐵𝐴, 𝜈).
I.e. the region with the largest search density should be searched first.
𝐴
𝐵
Theorem: Let 𝑄 be a tree with root 𝑂 and suppose a Hider is located on the nodes of
𝑄 according to some distribution. Let 𝑀 be the subtree of maximum density*. Then
every optimal expanding search 𝑆 of 𝑄 begins by searching 𝑀.
Proof (sketch): By induction on number of arcs (obvious for 1 arc). Suppose 𝑆 is optimal.
Case 1: 𝑆 doesn’t begin in 𝑀
𝑄
𝑀
𝑄′
𝐴
𝑚
𝑂
𝑀
𝑚
𝑂
Then 𝑆 searches a subtree 𝐴 with smaller density than 𝑀 before searching the initial arc
𝑚 of 𝑀. By induction it then searches the maximum density subtree of 𝑄’, which must
be 𝑀. By the Search Density Lemma, it would have been better to search 𝑀 first.
*We are assuming that 𝑀 is unique, and there are no “density ties”.
Case 2: 𝑆 begins in 𝑀
𝑄′′
𝑄
𝐵
𝑀
𝑏
𝑂
𝑀′
𝑏
𝑂
If 𝑆 doesn’t begin by searching all of 𝑀 then it searches some of 𝑀 and then, by
induction searches the maximum density subtree 𝐵 of 𝑄’’, which must have higher
density than the remainder of 𝑀 (call this 𝑀’). Easy to show 𝑀’ has higher density
than 𝑀, and 𝐵 ∪ 𝑀 has higher density then 𝑀, a contradiction.
Example
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1/12
7/24
2/12
5
1
1
5
Max density =
1
2
+
12 12
/2 = 1/8
1/24
1/12
2
1
𝑂
8/24
7/24
5
Max density =
1
12
/1 = 1/12
1/12
5
1/24
2
1
𝑂
This doesn’t work for non-trees
c
12/20
7/20
1
1
b
Max density subtree = abc
Optimal search begins with d.
d
2
1/20
1
a
𝑂
Continuous Hider distribution
The ‘max tree first’ theorem is true for continuous Hider distributions on trees.
However, it cannot always be applied, as there may not be a maximum density
subtree:
Same density on each arc,
increasing towards the root
𝑂
To get around this problem, we can add an arc of length 𝐿 to the root and find the
maximum density subtree of the new network. This will indicate how to begin
searching the original network
𝑂
𝐿
𝑂′
As 𝐿 increases the max density
subtree gets bigger.
8/24
1/12
7/24
5
2/12
1
1
5
1/24
1/12
2
1
𝑂
31
𝑂′
𝑂′
𝐿 = 0: max density is
1
2
+
12 12
𝐿 = 1: max density is
1
2
1
+
+
12 12 12
/2 = 1/8
/(4 + 1) = 1/12
𝐿 = 3: max density is 1/(15+3) = 1/18
Search games
For both pathwise and expanding search, we define zero sum search games in which
a Hider picks a point on a rooted network and a Searcher picks a unit speed path
(respectively expanding search) starting at the root. The payoff is the search time,
which the Hider seeks to maximise, and the Searcher to minimise.
Pathwise search
Expanding search
Bayesian problem
Linear Search Problem (Beck) Alpern & Lidbetter
Otherwise not much known
Search game
Gal theory
Alpern & Lidbetter
Pathwise search game: Weakly Eulerian networks
Q is Weakly Eulerian if it
contains a number of
disjoint Eulerian networks
which, when each is
shrunk to a point, leave a
tree.
𝑂
Optimal Search strategy is a Random Chinese Postman Tour , i.e. equiprobable choice of
a minimal tour and its reverse.
Value of the game = 𝜇 2 = half the length of a Chinese Postman Tour (Gal 2000)
Pathwise search game: Trees
A tree is a special case of a Weakly Eulerian network.
The length of a Chinese Postman
Tour is 2𝜇, twice the measure of
the whole network.
Value of the game = 𝜇 2 = 𝜇.
𝑂
The Hider hides at the nodes according to the Equal Branch Density (EBD)
distribution. This is the unique distribution on the leaf nodes such that at every
branch node, the density of each branch is the same.
Expanding search game: Trees
EBD is the best strategy for the Hider in Expanding Search, as it makes the Searcher
indifferent as to which subtree to choose.
Minimax theorem ⇒ EBD strategy is optimal for the Hider.
2/5
1/10
2/5
1/10
1
5
5
1
2
1
𝑂
Value = expected search time when the Hider uses EBD = ½(𝜇 + 𝐷)
(D = mean distance to the leaf nodes with respect to the EBD.)
Searcher’s optimal strategy in Expanding Search Game on trees
If the Searcher performs a depth-first search according to the following branching
strategy, he ensures an expected search time of not more than 𝑉 = ½(𝜇 + 𝐷)
𝑏
𝑎
𝛽
𝑥
1−𝛽
𝛽=
1
𝐷 − 𝐷𝑏
𝜇𝑥 𝑎
General networks: the three-arc network
For pathwise search, the three-arc network search game is notoriously difficult
to solve (see Pavlovic 1995). The solution involves a complicated Hider
distribution, and a Searcher strategy involving backtracking.
1
1
𝑂
1
However, in expanding search the problem is easy!
∗
There is an expanding search, 𝑆 on the three-arc network whose reverse, 𝑆 is
also an expanding search. Hence for a Hider found by 𝑆 at time 𝑡, the expected
search time is
1
1 ∗
𝑇 𝑆 + 𝑆 , 𝐻 = 1/2(𝑡) + 1/2(𝜇 − 𝑡) = 𝜇/2
2
2
By hiding uniformly on the network, the Hider can ensure a search time of no
more than 𝜇/2.
1
1
𝑂
1
Not all expanding searches are reversible...
1
1
𝑂
1
𝜇
Theorem: If a rooted network 𝑄 has a reversible expanding search then 𝑉 𝑄 = 2 .
Theorem: A rooted network 𝑄 is 2-edge-connected if and only if it has a reversible
expanding search.
Proof (⇒): We successively construct reversible expanding searches on subnetworks,
starting with 𝑆0 = a cycle containing the root. This is clearly reversible.
𝑂
Next, since the network is 2-arc-connected, we can find a path between two nodes 𝑥0
and 𝑦0 in the cycle. Let S1 be the expanding search that follows S0 up to x, then
follows the path from x to y, then follows the rest of S0. This is reversible.
𝑥0
𝑦0
𝑂
Find another pair of nodes, 𝑥1 and 𝑦1 on 𝑆1 with a path between them, and add this
path in to form 𝑆2. Continuing in this fashion produces a reversible expanding search
of the whole network.
𝑥1
𝑦1
𝑂
Decomposition of networks
The arcs of every rooted network 𝑄 can be decomposed into its ‘tree parts’ 𝑄1 (whose
removal disconnects 𝑄) and its ‘2-arc-connected parts’ 𝑄2 (whose removal doesn’t
disconnect 𝑄).
𝑄1
𝑄2
𝑂
A lower bound
We give the Searcher an advantage by
changing the network so the 2-arcconnected parts are connected to the
tree at a single point.
𝑂
𝑂
0
1
1
1
0
0
1
0
0
1
1
3/2
0
3/2
0
1
1
𝑂
1
This network is equivalent
to the variable speed
network shown here, with
the Hider measure on the
blue arcs moved to the leaf
nodes.
Theorem: For a weakly 2-arc-connected network (2 connected components are
connected to the tree part at a single point), the bound is tight.
𝑂
Expanding search for multiple objects on a rooted network
The same as the expanding search game already considered except:
• the Hider must choose 𝑘 points (objects) on the network,
• the payoff is the total time for the Searcher to find all the objects.
𝐻1
𝐻2
5
1
1
2
1
𝑂
In this example, 𝑇 𝐻, 𝑆 = 1 + 1 + 2 + 1 + 5 = 10.
5
For the star network with 𝑛 arcs, expanding search for 𝑘 multiple objects is equivalent
to a game where 𝑘 objects are hidden in 𝑛 boxes with designated search costs.
Eg. 𝑛 = 5, 𝑘 = 3
𝐻1
𝐻3
𝐻2
4
5
2
3
1
𝑂
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€1
€2
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A strategy for the Hider is a 𝑘-subset of boxes. A strategy for the Searcher is an ordering of
the boxes.
It turns out to be optimal for the Hider to pick a 𝑘-set of boxes with probability proportional
to the product of their search costs.
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€4
€1
€2
€5
Eg. the 3-subset above is picked with probability proportional to 4 × 2 × 5.
Under this Hider distribution, all Searcher strategies have the same total search cost.
Suppose we restrict the Searcher to strategies where he picks his first 𝑘 boxes (which he
searches in any order) and then he searches the rest of the boxes in a random order.
The Searcher’s strategy set is now the same as the Hider’s, and it can be shown that the
payoff matrix for this restricted game is symmetric.
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€4
€1
€2
€5
€2
€5
Searcher’s strategy
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€4
€1
Search cost =
1
4 + 2 + 5 + 1 + .3
2
Searcher’s strategy
It follows that if the Searcher uses the Hider’s optimal strategy, this is also optimal for him.
That is, it is optimal for the Searcher to pick his first k boxes with probability proportional to
the product of their search costs, then search the remaining boxes randomly.
For a general tree network, it may be possible to simplify the game, as in this
example with 𝑘 = 2
𝑎
𝑎
O
O
The Searcher must always traverse the arc 𝑎, so the problem is reduced to
the game on a star network.
For a general tree, the problem is hard. In this simple example where all the arcs
have unit length and k=2, the Search and Hider have only two strategies each, up to
symmetry: same or different.
O
Solving the 2×2 game, it is optimal for both players to choose same with probability
1/5 and different with probability 4/5. The value is 5.3.
However, a smart Searcher, who is permitted to change his strategy as he goes
along, can do better. He can ensure expected search time 5.2 against the Hider’s
strategy above.
Interesting fact: it’s no advantage for a Searcher to be smart on a star network
(proof by induction…)