1. 1 Algorithms for Port of Entry Inspection: Finding Optimal Binary Decision Trees Fred S. Roberts Rutgers University

2. 2 Happy 40 th Birthday Southeastern Conference

3. 3 Happy 38 th Southeastern Conference, Fred

4. 4 Happy 40 th Birthday Southeastern Conference In light of yesterday’s reminiscences, I wanted to see what would happen if I put “Southeastern Conference in Combinatorics, Graph Theory, and Computing” into Google.

5. 5 Happy 40 th Birthday Southeastern Conference

6. 6 In light of yesterday’s reminiscences, I wanted to see what would happen if I put “Fred Hoffman” into Google.

7. 7 Happy 40 th Birthday Southeastern Conference

8. 8 Port of Entry Inspection Algorithms Goal: Find ways to intercept illicit nuclear materials and weapons destined for the U.S. via the maritime transportation system Goal: “inspect all containers arriving at ports” Even carefully inspecting 8% of containers in Port of NY/NJ might bring international trade to a halt (Larrabbee 2002)

9. 9 Port of Entry Inspection Algorithms Aim: Develop decision support algorithms that will help us to “optimally” intercept illicit materials and weapons subject to limits on delays, manpower, and equipment Find inspection schemes that minimize total “cost” including “cost” of false alarms (“false positives”) and failed alarms (“false negatives”) Mobile Vacis: truck- mounted gamma ray imaging system

10. 10 Port of Entry Inspection Algorithms My work on port of entry inspection has gotten me and my students to some remarkable places. Me on a Coast Guard boat in a tour of the harbor in Philadelphia Thanks to Capt. David Scott, Captain of Port, for taking us on the tour

11. 11 Sequential Decision Making Problem Stream of containers arrives at a port The Decision Maker’s Problem: Which to inspect? Which inspections next based on previous results? Approach: –“decision logics” – Boolean methods –combinatorial optimization methods –Builds on ideas of Stroud and Saeger at Los Alamos National Laboratory –Need for new models and methods

12. 12 Sequential Diagnosis Problem Such sequential diagnosis problems arise in many areas: –Communication networks (testing connectivity, paging cellular customers, sequencing tasks, …) –Manufacturing (testing machines, fault diagnosis, routing customer service calls, …) –Medicine (diagnosing patients, sequencing treatments, …)

13. 13 Sequential Decision Making Problem Containers arriving to be classified into categories. Simple case: 0 = “ok”, 1 = “suspicious” Inspection scheme: specifies which inspections are to be made based on previous observations

14. 14 Sequential Decision Making Problem For Container Inspection Containers have attributes, each in a number of states Sample attributes: –Levels of certain kinds of chemicals or biological materials –Whether or not there are items of a certain kind in the cargo list –Whether cargo was picked up in a certain port

15. 15 Sequential Decision Making Problem Currently used attributes: –Does ship’s manifest set off an “alarm”? –What is the neutron or Gamma emission count? Is it above threshold? –Does a radiograph image come up positive? –Does an induced fission test come up positive? Gamma ray detector

16. 16 Sequential Decision Making Problem We can imagine many other attributes The project I have worked on is concerned with general algorithmic approaches. We seek a methodology not tied to today’s technology. Detectors are evolving quickly.

17. 17 Sequential Decision Making Problem Simplest Case: Attributes are in state 0 or 1 (absent or present) Then: Container is a bit string like 011001 So: Classification is a decision function F that assigns each bit string to a category. 011001F(011001) If attributes 2, 3, and 6 are present, assign container to category F(011001).

18. 18 Sequential Decision Making Problem If there are two categories, 0 and 1 (“safe” or “suspicious”), the decision function F is a Boolean function. Example: F(000) = F(111) = 1, F(abc) = 0 otherwise This classifies a container as positive iff it has none of the attributes or all of them. 1 =

19. 19 Sequential Decision Making Problem What if there are three categories, 0, ½, and 1?. Example: F(000) = 0, F(111) = 1, F(abc) = 1/2 otherwise This classifies a container as positive if it has all of the attributes, negative if it has none of the attributes, and uncertain if it has some but not all of the attributes. I won’t discuss this case.

20. 20 Sequential Decision Making Problem Given a container, test its attributes until know enough to calculate the value of F. An inspection scheme tells us in which order to test the attributes to minimize cost. Even this simplified problem is hard computationally.

21. 21 Sequential Decision Making Problem This assumes F is known. Simplifying assumption: Attributes are independent. At any point we stop inspecting and output the value of F based on outcomes of inspections so far. Complications: May be precedence relations in the components (e.g., can’t test attribute a 4 before testing a 6. Or: cost may depend on attributes tested before. F may depend on variables that cannot be directly tested or for which tests are too costly.

22. 22 Sequential Decision Making Problem Such problems are hard computationally. There are many possible Boolean functions F. Even if F is fixed, problem of finding a good classification scheme (to be defined precisely below) is NP-complete. Several classes of Boolean functions F allow for efficient inspection schemes: - k-out-of-n systems - Certain series-parallel systems - Read-once systems - “regular” systems - Horn systems

23. 23 Sensors and Inspection Lanes n types of sensors measure presence or absence of the n attributes. Many copies of each sensor. Complication: different characteristics of sensors. Entities come for inspection. Which sensor of a given type to use? Think of inspection lanes and waiting on line for inspection Besides efficient inspection schemes, could decrease costs by: –Buying more sensors –Change allocation of containers to sensor lanes.

24. 24 Binary Decision Tree Approach Sensors measure presence/absence of attributes: so 0 or 1 Use two categories: 0, 1 (safe or suspicious) Binary Decision Tree: –Nodes are sensors or categories –Two arcs exit from each sensor node, labeled left and right. –Take the right arc when sensor says the attribute is present, left arc otherwise

25. 25 Binary Decision Tree Approach Reach category 1 from the root only through the path a 0 to a 1 to 1. Container is classified in category 1 iff it has both attributes a 0 and a 1. Corresponding Boolean function: F(11) = 1, F(10) = F(01) = F(00) = 0. Figure 1

26. 26 Binary Decision Tree Approach Reach category 1 from the root only through the path a 1 to a 0 to 1. Container is classified in category 1 iff it has both attributes a 0 and a 1. Corresponding Boolean function: F(11) = 1, F(10) = F(01) = F(00) = 0. Note: Different tree, same function Figure 1

27. 27 Binary Decision Tree Approach Reach category 1 from the root only through the path a 0 to 1 or a 0 to a 1 to 1. Container is classified in category 1 iff it has attribute a 0 or attribute a 1. Corresponding Boolean function: F(11) = 1, F(10) = F(01) = 1, F(00) = 0. Figure 1

28. 28 Binary Decision Tree Approach Reach category 1 from the root by: a 0 L to a 1 R a 2 R 1 or a 0 R a 2 R 1 Container classified in category 1 iff it has a 1 and a 2 and not a 0 or a 0 and a 2 and possibly a 1. Corresponding Boolean function: F(111) = F(101) = F(011) = 1, F(abc) = 0 otherwise. Figure 2

29. 29 Binary Decision Tree Approach This binary decision tree corresponds to the same Boolean function F(111) = F(101) = F(011) = 1, F(abc) = 0 otherwise. However, it has one less observation node a i. So, it is more efficient if all observations are equally costly and equally likely. Figure 3

30. 30 Binary Decision Tree Approach So we have seen that a given Boolean function may correspond to different binary decision trees. How do we find a low-cost or least-cost binary decision tree corresponding to a Boolean function?

31. 31 Binary Decision Tree Approach Even if the Boolean function F is fixed, the problem of finding the “least cost” binary decision tree for it is very hard (NP-complete). For small n = number of attributes, can try to solve it by trying all possible binary decision trees corresponding to the Boolean function F. Even for n = 4, not practical. (n = 4 at Port of Long Beach-Los Angeles) Port of Long Beach

32. 32 Binary Decision Tree Approach Promising Approaches: Heuristic algorithms, approximations to optimal. Special assumptions about the Boolean function F. For “monotone” Boolean functions, integer programming formulations give promising heuristics. Stroud and Saeger (Los Alamos National Lab) enumerate all “complete, monotone” Boolean functions and calculate the least expensive corresponding binary decision trees. Their method practical for n up to 4, not n = 5.

33. 33 Binary Decision Tree Approach Monotone Boolean Functions: Given two bit strings x 1 x 2 …x n, y 1 y 2 …y n Suppose that x i  y i for all i implies that F(x 1 x 2 …x n )  F(y 1 y 2 …y n ). Then we say that F is monotone. Then 11…1 has highest probability of being in category 1.

34. 34 Binary Decision Tree Approach Monotone Boolean Functions: Given two bit strings x 1 x 2 …x n, y 1 y 2 …y n Suppose that x i  y i for all i implies that F(x 1 x 2 …x n )  F(y 1 y 2 …y n ). Then we say that F is monotone. Example: n = 4, F(x) = 1 iff x has at least two 1’s. F(1100) = F(0101) = F(1011) = 1, F(1000) = 0, etc.

35. 35 Binary Decision Tree Approach Incomplete Boolean Functions: Boolean function F is incomplete if F can be calculated by finding at most n-1 attributes and knowing the value of the input string on those attributes Example: F(111) = F(110) = F(101) = F(100) = 1, F(000) = F(001) = F(010) = F(011) = 0. F(abc) is determined without knowing b (or c). F is incomplete.

36. 36 Binary Decision Tree Approach Complete, Monotone Boolean Functions: Stroud and Saeger: algorithm for enumerating binary decision trees implementing complete, monotone Boolean functions. Feasible to implement up to n = 4. Then you can find least cost tree by enumerating all binary decision trees corresponding to a given complete, monotone Boolean function and repeating this for all complete, monotone Boolean functions.

37. 37 Binary Decision Tree Approach Complete, Monotone Boolean Functions: Stroud and Saeger: algorithm for enumerating binary decision trees implementing complete, monotone Boolean functions. n = 2: –There are 6 monotone Boolean functions. –Only 2 of them are complete, monotone –There are 4 binary decision trees for calculating these 2 complete, monotone Boolean functions.

38. 38 Binary Decision Tree Approach Complete, Monotone Boolean Functions: n = 3: –9 complete, monotone Boolean functions. –60 distinct binary trees for calculating them

39. 39 Binary Decision Tree Approach Complete, Monotone Boolean Functions: n = 4: –114 complete, monotone Boolean functions. –11,808 distinct binary decision trees for calculating them. –(Compare 1,079,779,602 BDTs for all Boolean functions)

40. 40 Binary Decision Tree Approach Complete, Monotone Boolean Functions: n = 5: –6894 complete, monotone Boolean functions –263,515,920 corresponding binary decision trees. Combinatorial explosion! Need alternative approaches; enumeration not feasible! (Even worse: compare 5 x 10 18 BDTs corresponding to all Boolean functions)

41. 41 Cost Functions So far, we have figured one binary decision tree is cheaper than another if it has fewer nodes. This is oversimplified. There are more complex costs involved than number of sensors in a tree.

42. 42 Cost Functions Stroud-Saeger method applies to more sophisticated cost models, not just cost = number of sensors in the BDT. Using a sensor has a cost: –Unit cost of inspecting one item with it –Fixed cost of purchasing and deploying it –Delay cost from queuing up at the sensor station Preliminary problem: disregard fixed and delay costs. Minimize unit costs.

43. 43 Cost Functions: Delay Costs Tradeoff between fixed costs and delay costs: Add more sensors cuts down on delays. More sophisticated models describe the process of containers arriving There are differing delay times for inspections Use “queuing theory” to find average delay times under different models

44. 44 Cost Functions Unit Cost Complication: How many nodes of the decision tree are actually visited during average container’s inspection? Depends on “distribution” of containers. Answer can also depend on probability of sensor errors and probability of “bomb” in a container.

45. 45 Cost Functions: Unit Costs Tree Utilization In our early models, we assume we are given probability of sensor errors and probability of bomb in a container. This allows us to calculate “expected” cost of utilization of the tree C util.

46. 46 Cost Functions OTHER COSTS: Cost of false positive: Cost of additional tests. –If it means opening the container, it’s expensive. Cost of false negative: –Complex issue. –What is cost of a bomb going off in Manhattan?

47. 47 Cost Functions: Sensor Errors One Approach to False Positives/Negatives: Assume there can be Sensor Errors Simplest model: assume that all sensors checking for attribute a i have same fixed probability of saying a i is 0 if in fact it is 1, and similarly saying it is 1 if in fact it is 0. More sophisticated analysis later describes a model for determining probabilities of sensor errors. Notation: X = state of nature (bomb or no bomb) Y = outcome (of sensor or entire inspection process).

48. 48 Probability of Error for The Entire Tree State of nature is zero (X = 0), absence of a bomb State of nature is one (X = 1), presence of a bomb Probability of false positive (P(Y=1|X=0)) for this tree is given by Probability of false negative (P(Y=0|X=1)) for this tree is given by A B C 0 1 0 1 A B C 0 1 0 1 P(Y=1|X=0) = P(Y A =1|X=0) * P(Y B =1|X=0) + P(Y A =1|X=0) *P(Y B =0|X=0)* P(Y C =1|X=0) P falsepositive P(Y=0|X=1) = P(Y A =0|X=1) + P(Y A =1|X=1) *P(Y B =0|X=1)*P(Y C =0|X=1) P falsenegative

49. 49 Cost Function used for Evaluating the Decision Trees. C Tot = C FalsePositive *P FalsePositive + C FalseNegative *P FalseNegative + C util C FalsePositive is the cost of false positive (Type I error) C FalseNegative is the cost of false negative (Type II error) P FalsePositive is the probability of a false positive occurring P FalseNegative is the probability of a false negative occurring C util is the expected cost of utilization of the tree.

50. 50 Cost Function used for Evaluating the Decision Trees. C FalsePositive is the cost of false positive (Type I error) C FalseNegative is the cost of false negative (Type II error) P FalsePositive is the probability of a false positive occurring P FalseNegative is the probability of a false negative occurring C util is the expected cost of utilization of the tree. P FalsePositive and P FalseNegative are calculated from the tree. C util is calculated from tree and probabilities of bomb in container and probability of sensor errors. C FalsePositive, C FalseNegative are input – given information.

51. 51 Stroud Saeger Experiments Stroud-Saeger ranked all trees formed from 3 or 4 sensors A, B, C and D according to increasing tree costs. Used cost function defined above. Values used in their experiments: –C A =.25; P(Y A =1|X=1) =.90; P(Y A =1|X=0) =.10; –C B = 10; P(Y C =1|X=1) =.99; P(Y B =1|X=0) =.01; –C C = 30; P(Y D =1|X=1) =.999; P(Y C =1|X=0) =.001; –C D = 1; P(Y D =1|X=1) =.95; P(Y D =1|X=0) =.05; –Here, C i = unit cost of utilization of sensor i. Also fixed were: C FalseNegative, C FalsePositive, P(X=1)

52. 52 Sensitivity Analysis When parameters in a model are not known exactly, the results of a mathematical analysis can change depending on the values of the parameters. It is important to do a sensitivity analysis: let the parameter values vary and see if the results change. So, do the least cost trees change if we change values like probability of a bomb, cost of a false positive, etc?

53. 53 Stroud Saeger Experiments: Our Sensitivity Analysis We have explored sensitivity of the Stroud- Saeger conclusions to variations in values of the three parameters: C FalseNegative, C FalsePositive, P(X=1) Extensive computer experimentation. Fascinating results. To start, we estimated high and low values for the parameters.

54. 54 Stroud Saeger Experiments: Our Sensitivity Analysis –C FalseNegative was varied between 25 million and 10 billion dollars Low and high estimates of direct and indirect costs incurred due to a false negative. –C FalsePositive was varied between $180 and $720 Cost incurred due to false positive (4 men * (3 -6 hrs) * (15 – 30 $/hr) –P(X=1) was varied between 1/10,000,000 and 1/100,000

55. 55 Stroud Saeger Experiments: Our Sensitivity Analysis n = 3 (use sensors A, B, C) Varied the parameters C FalseNegative, C FalsePositive, P(X=1) We chose the value of one of these parameters from the interval of values Then explored the highest ranked tree as the other two parameters were chosen at random in the interval of values. 10,000 experiments for each fixed value. We looked for the variation in the top-ranked tree and how the top-rank related to choice of parameter values. Very surprising results.

56. 56 Frequency of Top-ranked Trees when C FalseNegative and C FalsePositive are Varied 10,000 randomized experiments (randomly selected values of C FalseNegative and C FalsePositive from the specified range of values) for the median value of P(X=1). The above graph has frequency counts of the number of experiments when a particular tree was ranked first or second or third and so on. Only three trees (7, 55 and 1) ever came first. 6 trees came second, 10 came third, 13 came fourth.

57. 57 10,000 randomized experiments for the median value of C FalsePositive. Only 2 trees (7 and 55) ever came first. 4 trees came second. 7 trees came third. 10 and 13 trees came 4 th and 5 th respectively. Frequency of Top-ranked Trees when C FalseNegative and P(X=1) are Varied

58. 58 10,000 randomized experiments for the median value of C FalseNegative. Only 3 trees (7, 55 and 1) ever came first. 6 trees came second. 10 trees came third. 13 and 16 trees came 4 th and 5 th respectively. Frequency of Top-ranked Trees when P(X=1) and C FalsePositive are Varied

59. 59 Most Frequent Tree Groups Attaining the Top Three Ranks. Trees 7, 9 and 10 A B C 0 1 0 1 B A C A 1 01 0 0 B A A C 1 01 0 0 All the three decision trees have been generated from the same Boolean function: 00000111 representing F(000)F(001)…F(111) Both Tree 9 and Tree 10 are ranked second and third more than 99% of the times when Tree 7 is ranked first.

60. 60 Most Frequent Tree Groups Attaining the Top Three Ranks Trees 55, 57 and 58 A 1 C B 1 0 1 B 1 C A 1 0 1 B 1 A C 1 0 1 All three trees correspond to the same Boolean function: 01111111 Tree ranked 57 is second 96% of the times and tree 58 is third 79 % of the times when tree 55 is ranked first.

61. 61 Most Frequent Tree Groups Attaining the Top Three Ranks Trees 1, 3, and 2 A B C 0 0 0 1 A C B 0 0 0 1 B A C 0 0 0 1 All three trees correspond to the same Boolean function: 00000001 Tree 3 is ranked second 98% of times and tree 2 is ranked third 80 % of the times when tree 1 is ranked first.

62. 62 Most Frequent Tree Groups Attaining the Top Three Ranks Challenge: Why so few trees? Why these trees? Why so few Boolean functions? Why these Boolean functions?

63. 63 Stroud Saeger Experiments: Sensitivity Analysis: 4 Sensors Second set of computer experiments: n = 4 (use sensors, A, B, C, D). Same values as before. Experiment 1: Fix values of two of C FalseNegative, C FalsePositive, P(X=1) and vary the third through their interval of possible values. Experiment 2: Fix a value of one of C FalseNegative, C FalsePositive, P(X=1) and vary the other two. Do 10,000 experiments each time. Look for the variation in the highest ranked tree.

64. 64 Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors Experiment 1: Fix values of two of C FalseNegative, C FalsePositive, P(X=1) and vary the third.

65. 65 C Tot vs C FalseNegative for Ranked 1 Trees (Trees 11485 (9651) and 10129 (349) ) Only two trees ever were ranked first, and one, tree 11485, was ranked first in 9651 out of 10,000 runs.

66. 66 C Tot vs C FalsePositive for Ranked 1 Trees (Tree no. 11485 (10000) ) One tree, number 11485, was ranked first every time.

67. 67 C Tot vs P(X=1) for Ranked 1 Trees (Tree no. 11485 (8372), 10129 (488), 11521 (1056) ) Three trees dominated first place. Trees 10201 (60), 10225 (17) and 10153 (7) also achieved first rank but with relatively low frequency.

68. 68 Tree Structure For Top Trees 1 a b b dc d1 c d 1 1 0 01 0 1 Tree number 11485 Boolean Expr: 0101011101111111 1 a b b c d1 c d 1 1 0 0 1 c 0 d 1 0 Tree number 10129 Boolean Expr: 0001011101111111 Note how close the Boolean expressions are

69. 69 Most Frequent Tree Groups Attaining the Top Three Ranks Same challenge as before: Why so few trees? Why these trees? Why so few Boolean functions? Why these Boolean functions?

70. 70 Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors Experiment 2: Fix the values of one of C FalseNegative, C FalsePositive, P(X=1) and vary the others.

71. 71 Stroud Saeger Experiments: Our Sensitivity Analysis: 4 Sensors Experiment 2: Fix the values of one of C FalseNegative, C FalsePositive, P(X=1) and vary the others. Similar results

72. 72 Conclusions from Sensitivity Analysis Considerable lack of sensitivity to modification in parameters for trees using 3 or 4 sensors. Very few optimal trees. Very few Boolean functions arise among optimal and near-optimal trees. Surprising results.

73. 73 New Idea: Searching through a Generalized Tree Space Sometimes adding more possibilities results in being able to do more efficient searches. We expand the space of trees from those corresponding to Stroud and Saeger’s “Complete and Monotonic” Boolean Functions to “Complete and Monotonic” BDTs. Advantages: –Unlike Boolean functions, BDTs may not have to consider all sensor inputs to give a final decision. –Allow more potentially useful trees to participate in the analysis –Help define an irreducible tree space for search operations

74. 74 Revisiting Monotonicity Monotonic Decision Trees –A binary decision tree will be called monotonic if all the left leaves are class “0” and all the right leaves are class “1”. Example: a b c F(abc) 0 0 0 0 1 0 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 a b b b c a c c a 0 1 0 1 0 c a 1 0 a 1 c 0 1 1 0 0 1 0 1 b c c c b b 0 a a 1 0 a a 1 0 1 1 0 1 0 0 1 c c a b b a b 0 a 1 0 a b 1 0 1 0 1 1 0 0 1 All these trees correspond to same monotonic Boolean function Only one is a monotonic BDT.

75. 75 Revisiting Completeness Complete Decision Trees –A binary decision tree will be called complete if every sensor occurs at least once in the tree and, at any non- leaf node in the tree, its left and right sub-trees are not identical. Example: a b c F(abc) 0 0 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 a b c c 1 b 1 0 1 0 1 a c b b 1 c 1 0 1 0 1 a c c b 1 b 1 0 1 0 1 a b b c 1 c 1 0 1 0 1

76. 76 The CM Tree Space No. of attributes Distinct BDTs Trees From CM Boolean Functions Complete, Monotonic BDTs 27444 316,43060114 41,079,779,60211,80866,936 complete, monotonic BDTs

77. 77 Tree Neighborhood and Tree Space Define tree neighborhood by giving operations for moving from one tree in CM Tree Space to another. We have developed an algorithm for finding low- cost BDTs by searching through CM Tree Space from a tree to one of its neighbors.

78. 78 Search Operations in Tree Space Split Pick a leaf node and replace it with a sensor that is not already present in that branch, and then insert arcs from that sensor to 0 and to 1. a b c 0 c d 1 d 1 0 1 0 1 SPLIT a b c 0 c d 1 d 1 b 1 0 1 0 1

79. 79 Search Operations Swap Pick a non-leaf node in the tree and swap it with its parent node such that the new tree is still monotonic and complete and no sensor occurs more than once in any branch. a b c 0 c d 1 d 1 0 1 0 1 SWAP a b c 0 d d 1 c 1 0 1 0 1

80. 80 Search Operations Merge Pick a parent node of two leaf nodes and make it a leaf node by collapsing the two leaf nodes below it, or pick a parent node with one leaf node, collapse both the parent node and its one leaf node, and shift the sub-tree up in the tree by one level. a b c 0 c d 1 d 1 0 1 0 1 MERGE a b c 0 d d 1 0 1 0 1 a b c 0 c d 1 0 1 0 1

81. 81 Search Operations Replace Pick a node with a sensor occurring more than once in the tree and replace it with any other sensor such that no sensor occurs more than once in any branch. a b c 0 c d 1 d 1 0 1 0 1 REPLACE a b c 0 c b 1 d 1 0 1 0 1

82. 82

83. 83 Tree Neighborhood and Tree Space Define tree neighborhood by using these four operations for moving from one tree in CM Tree Space to another. Irreducibility –Theorem: Any tree in the CM tree space can be reached from any other tree by using these neighborhood operations repetitively –An irreducible CM tree space helps “search” for the cheapest trees using neighborhood operations

84. 84 Tree Neighborhood and Tree Space Sketch of Proof of the Theorem: Simple Tree: – A simple tree is defined as a CM tree in which every sensor occurs exactly once in such a way that there is exactly one path in the tree with all sensors in it.

85. 85 Tree Neighborhood and Tree Space Sketch of Proof of the Theorem: To Prove: Given any two trees τ 1, τ 2 in CM tree space, τ 2 can be reached from τ 1 by a sequence of neighborhood operations We prove this in three different steps: –1. Any tree τ 1 can be converted to a simple tree τs 1 –2. Any simple tree τs 1 can be converted to any other simple tree τs 2 –3. Any simple tree τs 2 can be converted to any tree τ 2

86. 86 Tree Space Traversal Naïve Idea: Greedy Search 1.Randomly start at any tree in the CM tree space 2.Find its neighboring trees using the above operations 3.Move to the neighbor with the lowest cost 4.Iterate until we find a minimum –Problem: The CM Tree space is highly multi- modal (more than one local minimum)! –Therefore, we implement a stochastic search algorithm with simulated annealing to find the best tree

87. 87 Tree Space Traversal Stochastic Search –Randomly start at any tree in CM space –Find its neighboring trees, and evaluate each one for its total cost –Select next move according to a probability distribution over the neighboring trees To deal with the multimodality of the tree space, we introduce Simulated Annealing: –Make more random jumps initially, gradually decrease the randomness and finally converge at the overall minimum

88. 88 Results: Searching CM Tree Space We were able to perform experiments for 3, 4 and 5 sensors, successfully Results show improvement compared to the extensive search method. E.g., for 4 sensors (66,936 trees) –100 different experiments were performed –Each experiment was started 10 times randomly at some tree and chains were formed by making stochastic moves in the neighborhood, until we find a local minimum –Only 4890 trees were examined on average for every experiment –Global minimum was found 82 out of 100 times while the second best tree was found 10 times –The method found trees that were less costly than those found by earlier searches of BDTs corresponding to complete, monotonic Boolean functions.

89. 89 Genetic Algorithms-based Approach Structure-based neighborhood moves allow very short moves only. Therefore,… Techniques like Genetic Algorithms and Evolutionary Techniques may suggest ways for getting more efficiently to better trees, given a population of good trees

90. 90 Genetic Algorithms-based Approach Started implementing genetic algorithms-based techniques for tree space traversal Basically, we try to get “better” trees from the current population of “good” trees using the basic genetic operations on them: –Selection –Crossover –Mutation Here, “better” decision trees correspond to lower cost decision trees than the ones in the current population (“good”).

91. 91 Genetic Algorithms-based Approach Selection: – Select a random, initial population of N trees from CM tree space Crossover: – Performed k times between every pair of trees in the current best population, bestPop

92. 92 Genetic Algorithms-based Approach For each crossover operation between two trees, we randomly select a node in each tree and exchange their subtrees – However, we impose certain restrictions on the selection of nodes, so that the resultant trees still lie in CM tree space

93. 93 Genetic Algorithms-based Approach Mutation: – Performed after every m generations of the algorithm – We do two types of mutations: 1. Generate all neighbors of the current best population and put them into the gene pool 2. Replace a fraction of the trees of bestPop with random trees from the CM tree space

94. 94 Genetic Algorithms-based Approach Only ~1600 trees had to be examined to obtain the 10 best trees for 4 sensors!

95. 95 Modeling Sensor Errors One Approach to Sensor Errors: Modeling Sensor Operation Threshold Model: –Sensors have different discriminating power –Many use counts (e.g., Gamma radiation counts) –See if count exceeds threshold –If so, say attribute is present.

96. 96 Modeling Sensor Errors Threshold Model: Sensor i has discriminating power K i, threshold T i Attribute present if counts exceed T i Seek threshold values that minimize the overall cost function, including costs of inspection, false positive/negative Assume readings of category 0 containers follow a Gaussian distribution and similarly category 1 containers Simulation approach

97. 97 Probability of Error for Individual Sensors For i th sensor, the type 1 (P(Y i =1|X=0)) and type 2 (P(Y i =0|X=1)) errors are modeled using Gaussian distributions. –State of nature X=0 represents absence of a bomb. –State of nature X=1 represents presence of a bomb. –  i represents the outcome (count) of sensor i. –Σ i is variance of the distributions –P D = prob. of detection, P F = prob. of false pos. KiKi P(  i |X=1) P(  i |X=0) TiTi Characteristics of a typical sensor ii

98. 98 Modeling Sensor Errors The probability of false positive for the i th sensor is computed as: P(Y i =1|X=0) = 0.5 erfc[T i /√2] The probability of detection for the i th sensor is computed as: P(Y i =1|X=1) = 0.5 erfc[(T i -K i )/(Σ√2)] erfc = complementary error function erfc(x) =  (1/2,x 2 )/sqrt(  ) The following experiments have been done using sensors A, B, C and using: K A = 4.37; Σ A = 1 K B = 2.9; Σ B = 1 K C = 4.6; Σ C = 1 We then varied the individual sensor thresholds T A, T B and T C from -4.0 to +4.0 in steps of 0.4. These values were chosen since they gave us an “ROC curve” for the individual sensors over a complete range P(Y i =1|X=0) and P(Y i =1|X=1)

99. 99 Frequency of First Ranked Trees for Variations in Sensor Thresholds 68,921 experiments were conducted, as each T i was varied through its entire range. (n = 3) The above graph has frequency counts of the number of experiments when a particular tree was ranked first. There are 15 such trees. Tree 37 had the highest frequency of attaining rank one.

100. 100 Modeling Sensor Errors A number of trees ranking first in other experiments also ranked first here. Similar results in case of n = 4. 4,194,481 experiments. 244 different trees were ranked first in at least one experiment. Trees ranked first in other experiments also frequently appeared first here. Conclusion: considerable insensitivity to change of threshold.

101. 101 New Approaches to Optimum Threshold Computation Extensive search over a range of thresholds has some practical drawbacks: –Large number of threshold values for every sensor –Large step size –Grows exponentially with the number of sensors (computationally infeasible for n > 4) A non-linear optimization approach proves more satisfactory: –A combination of Gradient Descent and modified Newton’s methods

102. 102 Problems with Standard Approaches Gradient Descent Method: –Too small step size results in large number of iterations to reach the minimum –Too big step size results in skipping the minimum Newton’s Method: –The convergence depends largely on the starting point. This method occasionally drifts in the wrong direction and hence fails to converge. Solution: combination of gradient descent and Newton’s methods This works well.

103. 103 Results: Threshold Optimization Costs of false positive C FalsePositive and false negative C FalseNegative and prior probability of occurrence of a bad container, P(X=1), were fixed as medians of the min and max values given by Stroud and Saeger (same as we used in earlier experiments) We were able to converge to a (hopefully-close- to-minimum) cost every time with a modest number of iterations changing thresholds.

104. 104 Results: Threshold Optimization We were able to converge to a (hopefully-close-to- minimum) cost every time with a modest number of iterations changing thresholds. For example: –For 3 sensors, it took an average of 0.081 seconds (as opposed to 0.387 seconds using extensive search) to converge to a cost for all 114 trees studied –For 4 sensors, it took an average of 0.196 seconds (as opposed to more than 2 seconds using extensive search) to converge to a cost for all 66,936 trees studied In each case, min cost attained with new algorithm was lower, and often much lower, than that attained with extensive search.

105. 105 Results: Threshold Optimization Many times the minimum obtained using the optimization method was considerably less than the one from the extensive search technique.

106. 106 Closing Comments Very few optimal trees; optimality insensitive to changes in parameters. Extensive search techniques become practically infeasible beyond a very small number of sensors Studying an irreducible tree space helps us to “search” for the best trees rather than evaluating all the trees for their cost A new stochastic search algorithm allows us to search for optimum inspection schemes beyond 4 sensors successfully Our new threshold optimization algorithms provide faster ways to arrive at a low tree cost; cost is lower and often much lower than in extensive search

107. 107 Discussion and Future Work Future Work: Explain why conclusions are so insensitive to variation in parameter values. Future Work: Explore the structure of the optimal trees and compare the different optimal trees. Future Work: Develop methods for approximating the optimal tree. Pallet vacis

108. 108 Discussion and Future Work Future work: More than two values of an attribute -(present, absent, present with probability > 75%, absent with probability at least 75%) -(ok, not ok, ok with probability > 99%, ok with probability between 95% and 99%) Future work: In the Boolean function model: inferring the Boolean function from observations (partially defined Boolean functions)

109. 109 Discussion and Future Work Future work: Need for more complicated cost models; bringing in costs of delays

110. 110 Discussion and Future Work Future work: Because of the rapid growth in number of trees in CM Tree Space when the number of sensors grows, it is necessary to try to reduce the number of trees we need to search through. A notion of tree equivalence could be incorporated when the number of sensors go beyond 5 or 6 We hope that incorporating this into our model will enable us to extend our model to a large number of sensors

111. 111 Collaborators on this Work: Saket Anand David Madigan Richard Mammone Sushil Mittal Saumitr Pathak Research Support: Dept. of Homeland Security University Programs Domestic Nuclear Detection Office Office of Naval Research National Science Foundation Los Alamos National Laboratory: Rick Picard Kevin Saeger Phil Stroud

112. 112 This work has gotten me places I never thought I’d go.

113. 113 This work has also taken me to places to which I very much enjoy going.