Coupled Bisection for Root-Ordering
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1 Peter I. Frazier Shane G. Henderson Cornell University ORIE Cornell University ORIE Cornell University ORIE July 17, 2015
2 One-Dimensional Root-Finding 0 1 Problem Statement Structural Results Bisection Policy 2/28
3 One-Dimensional Root-Finding f 0 1 x Problem Statement Structural Results Bisection Policy 2/28
4 One-Dimensional Root-Finding f 0 1 x x Problem Statement Structural Results Bisection Policy 2/28
5 Coupled Root-Ordering 0 1 Problem Statement Structural Results Bisection Policy 3/28
6 Coupled Root-Ordering f(s 1, ) f(s 2, ) f(s 3, ) 0 1 x (s 1 ) x (s 2 ) x (s 3 ) Problem Statement Structural Results Bisection Policy 3/28
7 Coupled Root-Ordering f(s 1, ) f(s 2, ) f(s 3, ) 0 1 x (s 1 ) x (s 2 ) x (s 3 ) x Problem Statement Structural Results Bisection Policy 3/28
8 Coupled Root-Ordering Let S denote a set of elements (large but finite). Let f : S [0, 1) R, monotonic in its second argument. For every element s S, denote x (s) as the unique root of f (s, ). Suppose when we evaluate f for a given x [0, 1), we find f (s, x) for every s S. Goal: find the complete ordering of elements s by their roots x (s) with the least number of function evaluations as possible. Problem Statement Structural Results Bisection Policy 4/28
9 Applications Multi-armed Bandit Problems Sorting states in state space according to their Gittens or Whittle indices. Ordering of states is all that is required to implement optimal policy. Remote Sensing E.g., finding the boundary of a forest from an airborne laser scanner. Problem Statement Structural Results Bisection Policy 5/28
10 The Naive Approach Problem Statement Structural Results Bisection Policy 6/28
11 The Naive Approach Problem Statement Structural Results Bisection Policy 6/28
12 The Naive Approach Problems with Discretizing the Interval Non-adaptive in nature; unclear how to determine subinterval length. May require more computational work than necessary. Problem Statement Structural Results Bisection Policy 6/28
13 The Clever Approach Problem Statement Structural Results Bisection Policy 7/28
14 The Clever Approach 4 Problem Statement Structural Results Bisection Policy 7/28
15 The Clever Approach 2 2 Problem Statement Structural Results Bisection Policy 7/28
16 The Clever Approach Problem Statement Structural Results Bisection Policy 7/28
17 The Clever Approach How can we adaptively choose where to query f? Problem Statement Structural Results Bisection Policy 7/28
18 Formal Definition State Variables 2 2 Problem Statement Structural Results Bisection Policy 8/28
19 Formal Definition State Variables n (0) n (1) 2 2 x (0) x (1) x (2) Problem Statement Structural Results Bisection Policy 8/28
20 Formal Definition State Variables n (0) n (1) n (2) x (0) x (1) x (2) x (3) Problem Statement Structural Results Bisection Policy 8/28
21 Formal Definition State Variables n (0) n (1) n (2) x (0) x (1) x (2) x (3) At time epoch j: { } Let X j = [x (0), x (1) ),..., [x (j), x (j+1) ) denote the current partition of the interval. Let N j = (n (0), n (1),..., n (j) ) denote the number of roots in each subinterval. The pair (X j, N j ) denotes the computational state. Problem Statement Structural Results Bisection Policy 8/28
22 Formal Definition Transition We assume i.i.d. priors on x (s) for every element s S. Without loss of generality, x (s) Uniform[0, 1). Problem Statement Structural Results Bisection Policy 9/28
23 Formal Definition Transition We assume i.i.d. priors on x (s) for every element s S. Without loss of generality, x (s) Uniform[0, 1). Problem Statement Structural Results Bisection Policy 9/28
24 Formal Definition Transition We assume i.i.d. priors on x (s) for every element s S. Without loss of generality, x (s) Uniform[0, 1). Problem Statement Structural Results Bisection Policy 9/28
25 Formal Definition Transition We assume i.i.d. priors on x (s) for every element s S. Without loss of generality, x (s) Uniform[0, 1). 2 2 Problem Statement Structural Results Bisection Policy 9/28
26 Formal Definition Transition We assume i.i.d. priors on x (s) for every element s S. Without loss of generality, x (s) Uniform[0, 1). 2 2 N j+1 = ( n (0),..., n (l 1), N (l), n (l) N (l), n (l+1),..., n (j)), where N (l) Binomial ( n (l), x x (l) ) x (l+1) x (l). Problem Statement Structural Results Bisection Policy 9/28
27 Formal Definition Objective Function Problem Statement Structural Results Bisection Policy 10/28
28 Formal Definition Objective Function Problem Statement Structural Results Bisection Policy 10/28
29 Formal Definition Objective Function Problem Statement Structural Results Bisection Policy 10/28
30 Formal Definition Objective Function Let τ denote the number of function evaluations to sort all elements, i.e., τ = inf{j N : n (i) j 1 i}. Problem Statement Structural Results Bisection Policy 10/28
31 Formal Definition Defining the Optimal Policy Problem Statement Structural Results Bisection Policy 11/28
32 Formal Definition Defining the Optimal Policy A policy π is a mapping from computational states (X, N ) to a refined partition X that adds one additional subinterval. Problem Statement Structural Results Bisection Policy 11/28
33 Formal Definition Defining the Optimal Policy A policy π is a mapping from computational states (X, N ) to a refined partition X that adds one additional subinterval. Let W π (X, N ) denote the expected number of function evaluations required to sort elements, where we start at state (X, N ), i.e., W π (X, N ) = E π [τ (X 0, N 0 ) = (X, N )]. Problem Statement Structural Results Bisection Policy 11/28
34 Formal Definition Defining the Optimal Policy A policy π is a mapping from computational states (X, N ) to a refined partition X that adds one additional subinterval. Let W π (X, N ) denote the expected number of function evaluations required to sort elements, where we start at state (X, N ), i.e., W π (X, N ) = E π [τ (X 0, N 0 ) = (X, N )]. Let W (X, N ) = inf π W π (X, N ) denote the expected number of evaluations required under the optimal policy. Problem Statement Structural Results Bisection Policy 11/28
35 Dynamic Programming By the principle of optimality, we can iteratively take the best partition X, giving us a recursion. Problem Statement Structural Results Bisection Policy 12/28
36 Dynamic Programming By the principle of optimality, we can iteratively take the best partition X, giving us a recursion. Theorem (Basic Recursion) For a given computational state (X, N ), we have W (X, N ) = 1 + inf X [ E W ( X, N ] ) (X 0, N 0 ) = (X, N ), with stopping conditions W (X, N ) = 0 if n (i) 1 for all i. Problem Statement Structural Results Bisection Policy 12/28
37 Divide and Conqueor Theorem (Decomposition and Interval Invariance) Under the optimal policy, the value function W has the following two properties. Decomposition: W (X, N ) = N 1 i=0 W Interval Invariance: ({ } W [x (i), x (i+1) ), n (i)) = W ({ } [x (i), x (i+1) ), n (i)) ( {[0, 1)}, n (i)). Problem Statement Structural Results Bisection Policy 13/28
38 Decomposition Ex. Expected remaining number of iterations required to finish sorting 7 elements: ({ } ) W [x (0), x (1) ), [x (1), x (2) ), [x (2), x (3) ), (2, 2, 3) Problem Statement Structural Results Bisection Policy 14/28
39 Decomposition Ex. Expected remaining number of iterations required to finish sorting 7 elements: ({ } ) W [x (0), x (1) ), [x (1), x (2) ), [x (2), x (3) ), (2, 2, 3) }{{} sort 2 elements }{{} sort 2 elements } {{ } sort 3 elements Problem Statement Structural Results Bisection Policy 14/28
40 Decomposition Ex. Expected remaining number of iterations required to finish sorting 7 elements: ({ } ) W [x (0), x (1) ), [x (1), x (2) ), [x (2), x (3) ), (2, 2, 3) }{{} sort 2 elements }{{} sort 2 elements } {{ } sort 3 elements = W ( {[x (0), x (1) )}, 2 ) + W ( {[x (1), x (2) )}, 2 ) + W ( {[x (2), x (3) )}, 3 ) Problem Statement Structural Results Bisection Policy 14/28
41 Interval Invariance 3 a b Problem Statement Structural Results Bisection Policy 15/28
42 Interval Invariance 3 a 1 2 b a b Problem Statement Structural Results Bisection Policy 15/28
43 Interval Invariance 3 a 1 2 b a b a b Problem Statement Structural Results Bisection Policy 15/28
44 Interval Invariance 3 a 1 2 b a b a b a b Problem Statement Structural Results Bisection Policy 15/28
45 Interval Invariance 3 a 1 2 b a b Re-scaling a b a b Problem Statement Structural Results Bisection Policy 15/28
46 Interval Invariance 3 3 a 1 2 b a b Re-scaling a b a b 0 1 Problem Statement Structural Results Bisection Policy 15/28
47 A Simpler Recursion n 0 1 Problem Statement Structural Results Bisection Policy 16/28
48 A Simpler Recursion n 0 1 N x Binomial(n, x) N x n N x 0 x x 1 Problem Statement Structural Results Bisection Policy 16/28
49 A Simpler Recursion n 0 W ({[0, 1)}, n) 1 N x Binomial(n, x) N x n N x 0 x x 1 W ({[0, x)}, N x ) + W ({[x, 1)}, n N x ) Problem Statement Structural Results Bisection Policy 16/28
50 A Simpler Recursion n 0 W ({[0, 1)}, n) 1 N x Binomial(n, x) N x n N x W ({[0, 1)}, N x ) + W ({[0, 1)}, n N x ) Problem Statement Structural Results Bisection Policy 16/28
51 A Simpler Recursion n 0 W (n) 1 N x Binomial(n, x) N x n N x W (N x ) + W (n N x ) Problem Statement Structural Results Bisection Policy 16/28
52 A Simpler Recursion For notation purposes, let W (n) := W ({[0, 1)}, n). Problem Statement Structural Results Bisection Policy 16/28
53 A Simpler Recursion For notation purposes, let W (n) := W ({[0, 1)}, n). Corollary (Simple Recursion) W (n) = 1 + min x (0,1) E [W (N x) + W (n N x )], where N x Binomial (n, x), and W (0) = W (1) = 0. Problem Statement Structural Results Bisection Policy 16/28
54 Computing the Optimal Policy W (n) = 1 + min x (0,1) E [W (N x) + W (n N x )] Problem Statement Structural Results Bisection Policy 17/28
55 Computing the Optimal Policy W (n) = 1 + min x (0,1) E [W (N x) + W (n N x )] Using the stopping conditions W (0) = W (1) = 0, we can iteratively compute W ( ) for arbitrary n. Problem Statement Structural Results Bisection Policy 17/28
56 Computing the Optimal Policy W (n) = 1 + min x (0,1) E [W (N x) + W (n N x )] Using the stopping conditions W (0) = W (1) = 0, we can iteratively compute W ( ) for arbitrary n. Problem: N x {0, n} with positive probability. Problem Statement Structural Results Bisection Policy 17/28
57 Computing the Optimal Policy W (n) = 1 + min x (0,1) E [W (N x) + W (n N x )] Using the stopping conditions W (0) = W (1) = 0, we can iteratively compute W ( ) for arbitrary n. Problem: N x {0, n} with positive probability. This is an implicit definition of W. Problem Statement Structural Results Bisection Policy 17/28
58 Computing the Optimal Policy Theorem (Computational Recursion) For all n 2, we have { 1 W (n) = min x (0,1) 1 x n (1 x) n [ ]} + E W (N x ) + W (n N x ) 1 N x n 1, where N x Binomial (n, x), and W (0) = W (1) = 0. Problem Statement Structural Results Bisection Policy 17/28
59 Bisection Policy Suppose we queried the midpoint of the interval, regardless of the number of elements to sort, i.e., choose x = 1/2 for all n. We call this the Bisection Policy. Denote W β (n) the expected number of function evaluations to sort n elements by their roots, under the bisection policy. Problem Statement Structural Results Bisection Policy 18/28
60 Bisection Policy Theorem (Bisection Policy Recursion) For all n 2, we have [ ] W β (n) = E W β (N 1/2 ), where N 1/2 Binomial (n, 1/2), and W β (0) = W β (1) = 0. Problem Statement Structural Results Bisection Policy 18/28
61 Bisection vs. Optimal Policy 3000 W β (n) (Bisection Policy) W(n) (Optimal Policy) 2500 Expected Function Evaluations n Problem Statement Structural Results Bisection Policy 19/28
62 Is Bisection Optimal? Problem Statement Structural Results Bisection Policy 20/28
63 Is Bisection Optimal? No. Consider the case n = 6. Problem Statement Structural Results Bisection Policy 20/28
64 Is Bisection Optimal? The case n = Expected Function Evaluations x Problem Statement Structural Results Bisection Policy 20/28
65 Is Bisection Optimal? Linear Growth Rate W β (n) (Bisection Policy) W(n) (Optimal Policy) Linear Rate n Problem Statement Structural Results Bisection Policy 20/28
66 Theorem (Lower Bound) For all n 0, we have W (n) n 1. Problem Statement Structural Results Bisection Policy 21/28
67 Theorem (Lower Bound) For all n 0, we have W (n) n 1. Proof : If we want to sort n elements, we must perform at least n 1 function evaluations of f to seperate them. Problem Statement Structural Results Bisection Policy 21/28
68 Theorem (Lower Bound) For all n 0, we have W (n) n 1. Proof : If we want to sort n elements, we must perform at least n 1 function evaluations of f to seperate them. Can we find upper bounds on W β? Problem Statement Structural Results Bisection Policy 21/28
69 What s the difference? For notation purposes, define h(n) = h(n + 1) h(n). Theorem (Recursion for W β ) For all n 2, we have [ ] W β (n) = E W β (N 1/2 ), where N 1/2 Binomial (n, 1/2), and stopping conditions W β (0) = 0 and W β (1) = 2. Problem Statement Structural Results Bisection Policy 22/28
70 Bounding the Bisection Policy Goal: show for some m, γ > 0 that W β (n) γ for all n m. Problem Statement Structural Results Bisection Policy 23/28
71 Bounding the Bisection Policy Goal: show for some m, γ > 0 that W β (n) γ for all n m. Each subsequent term of W β is a weighted average of the previous terms in the sequence. Idea: Induction! Problem Statement Structural Results Bisection Policy 23/28
72 Bounding the Bisection Policy Goal: show for some m, γ > 0 that W β (n) γ for all n m. E [ W β (N 1/2 ) ] E [ W β (N 1/2 ) N 1/2 m 1 ] E [ W β (N 1/2 ) N 1/2 m ] Bound above using? Bound above using Induction Hypthosis Problem Statement Structural Results Bisection Policy 23/28
73 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 24/28
74 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 24/28
75 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 24/28
76 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 24/28
77 A Useful Property of Binomial RVs Problem Statement Structural Results Bisection Policy 25/28
78 A Useful Property of Binomial RVs Problem Statement Structural Results Bisection Policy 25/28
79 A Useful Property of Binomial RVs Problem Statement Structural Results Bisection Policy 25/28
80 A Useful Property of Binomial RVs Problem Statement Structural Results Bisection Policy 25/28
81 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 26/28
82 Computational Upper Bounds Growth Rate of Expected Evaluations Required to Sort W β (n) (Linear Growth Rate) Monotone Decreasing Upper Bound PMF of Truncated Binonmial Distribution Problem Statement Structural Results Bisection Policy 26/28
83 Computational Upper Bounds Theorem (Upper Bound on Bisection Policy) For some non-negative integer m, we define g m (l) = max k [l,m 1] W β (k), and define h as ( ) ] h m (n) = E [g m N 1/2 N1/2 m 1, where N 1/2 Binomial (n, 1/2). Suppose we have γ m > 0 so that the following condition holds: { } max W β (m), h m (m) γ m. Then for all n m, it must be that W β (n) γ m. Problem Statement Structural Results Bisection Policy 26/28
84 Computing Upper Bounds Use the computational recursion to compute W β (k) for k = {1, 2,..., m} for some positive integer m. Compute γ m = max{ W β (m), h m (m)}. Problem Statement Structural Results Bisection Policy 27/28
85 Computing Upper Bounds Use the computational recursion to compute W β (k) for k = {1, 2,..., m} for some positive integer m. Compute γ m = max{ W β (m), h m (m)}. Corollary (Bounds for Bisection Policy) For n 15, we have that n 1 W (n) W β (n) γ 15 (n 15) + W β (15), where γ 15 = and W β (15) = Problem Statement Structural Results Bisection Policy 27/28
86 Further Questions Problem Statement Structural Results Bisection Policy 28/28
87 Further Questions If we were allowed to evaluate the function at multiple points in parallel, where would we query the function? Which subintervals do we refine at each time epoch? How much computing power do we allocate to each subinterval? Problem Statement Structural Results Bisection Policy 28/28
88 Further Questions If we were allowed to evaluate the function at multiple points in parallel, where would we query the function? Which subintervals do we refine at each time epoch? How much computing power do we allocate to each subinterval? What if we were only interested in sorting the rankings of the top k zeros? What would a bisection policy look like? Can we effectively compute the optimal policy if the i.i.d. assumption does not hold? Problem Statement Structural Results Bisection Policy 28/28
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