Expanding Search for one or more Hiders Lorentz Centre, Leiden, May 2012

1. 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)

2. Pathwise search An pathwise search on a network with root is a continuous unit speed path starting at the root. 5 5 1 1 2 1 For a Hider located at a node, the search time is first time the path reaches . In this example,

3. Expanding search An expanding search on a rooted network is a nested family of sets increasing at unit speed, starting with only the root and ending with the whole network. 5 5 1 1 2 1 For a Hider at a node, the search time is first time is contained in some . In this example,

4. 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 0 1 0 1 2 0 1 0

5. If the Hider is located on according to a known distribution , we can calculate the expected search time for an expanding search . We are interested in the Bayesian Problem of how the Searcher should minimise the expected search time for a given Hider distribution. 8/24 7/24 5 5 1/12 2/12 1 1 1/24 1/12 2 1 Expected search time

6. Search density Definition: The search density (or just density) of a region is the probability the Hider is located in divided by the time required to search , which in expanding searchis just the total length of . Example: The density of the highlighted region in the network below is 8/24 7/24 5 5 1/12 2/12 1 1 1/24 1/12 2 1

7. Search Density Lemma Lemma 1: Suppose and are disjoint regions of a rooted network with Hider distribution . Let be search of that searches immediately before , and let be an identical search except that it searches before . Then if )). I.e. the region with the largest search density should be searched first.

8. Theorem: Let be a tree with root and suppose a Hider is located on the nodes of according to some distribution. Let be the subtree of maximum density*. Then every optimal expanding search of begins by searching. Proof (sketch): By induction on number of arcs (obvious for 1 arc). Suppose is optimal. Case 1: doesn’t begin in Then searches a subtree with smaller density than before searching the initial arc of . By induction it then searches the maximum density subtree of , which must be . By the Search Density Lemma, it would have been better to search first. *We are assuming that is unique, and there are no “density ties”.

9. Case 2: begins in If doesn’t begin by searching all of then it searches some of and then, by induction searches the maximum density subtree of , which must have higher density than the remainder of (call this ’). Easy to show ’ has higher density than , and has higher density then , a contradiction.

10. Example 8/24 7/24 5 5 1/12 2/12 Max density = /2 = 1/8 1 1 1 1/24 1/12 2 1 1 8/24 7/24 5 5 Max density = /1 = 1/12 1/12 1/24 2 1 1

11. This doesn’t work for non-trees c 12/20 7/20 Max density subtree = abc Optimal search begins with d. b d 1/20 a

12. 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

13. To get around this problem, we can add an arc of length to the root and find the maximum density subtree of the new network. This will indicate how to begin searching the original network As increases the max density subtree gets bigger.

14. 8/24 7/24 5 5 5 5 1/12 2/12 1 1 1 1/24 1/12 2 2 1 : max density is /2 1/8 1 3 : max density is /1/12 : max density is 1/(15+3

15. 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.

16. Pathwise search game: Weakly Eulerian networks Q is Weakly Eulerianif it contains a number of disjoint Eulerian networks which, when each is shrunk to a point, leave a tree. Optimal Search strategy is a Random Chinese Postman Tour , i.e. equiprobable choice of a minimal tour and its reverse. Value of the game half the length of a Chinese Postman Tour (Gal 2000)

17. Pathwise search game: Trees A tree is a special case of a Weakly Eulerian network. The length of a Chinese Postman Tour is 2, twice the measure of the whole network. Value of the game 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.

18. 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 2/5 5 5 1/10 1/10 1 1 2 1 Value = expected search time when the Hider uses EBD = ½( (D = mean distance to the leaf nodes with respect to the EBD.)

19. 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

20. 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 1

21. However, in expanding search the problem is easy! There is an expanding search, on the three-arc network whose reverse, is also an expanding search. Hence for a Hider found by at time , the expected search time is By hiding uniformly on the network, the Hider can ensure a search time of no more than . 1 1 1

22. Not all expanding searches are reversible... 1 1 1 Theorem: If a rooted network has a reversible expanding search then .

23. Theorem: A rooted network 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 = a cycle containing the root. This is clearly reversible.

24. Next, since the network is 2-arc-connected, we can find a path between two nodes and 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.

25. Find another pair of nodes, and on with a path between them, and add this path in to form . Continuing in this fashion produces a reversible expanding search of the whole network.

26. Decomposition of networks The arcs of every rooted network can be decomposed into its ‘tree parts’ (whose removal disconnects ) and its ‘2-arc-connected parts’ (whose removal doesn’t disconnect ).

27. A lower bound We give the Searcher an advantage by changing the network so the 2-arc-connected parts are connected to the tree at a single point. 0 1 1 0 This network is equivalent to the variable speed network shown here, with the Hider measure on the blue arcs moved to the leaf nodes. 0 1 1 0 0 1 1 0 0 3/2 1 3/2 1 1

28. 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.

29. Expanding search for multiple objects on a rooted network • The same as the expanding search game already considered except: • the Hider must choose points (objects) on the network, • the payoff is the total time for the Searcher to find all the objects. 1 5 5 1 1 2 1 In this example,

30. For the star network with arcs, expanding search for multiple objects is equivalent to a game where objects are hidden in boxes with designated search costs. Eg. 1 3 2 4 5 2 3 1 €3 €5 €4 €1 €2

31. A strategy for the Hider is a -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 -set of boxes with probability proportional to the product of their search costs. €3 €5 €4 €1 €2 Eg. the 3-subset above is picked with probability proportional to . Under this Hider distribution, all Searcher strategies have the same total search cost.

32. Suppose we restrict the Searcher to strategies where he picks his first 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. €3 €3 €5 €5 €4 €4 €1 €1 €2 €2 Searcher’s strategy Search cost 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.

33. For a general tree network, it may be possible to simplify the game, as in this example with O O The Searcher must always traverse the arc so the problem is reduced to the game on a star network.

34. 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 22game, it is optimal for both players to choose same with probability 1/5and 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…)