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A New Analysis of the LebMeasure Algorithm for Calculating Hypervolume Lyndon While Walking Fish Group School of Computer Science & Software Engineering.

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Presentation on theme: "A New Analysis of the LebMeasure Algorithm for Calculating Hypervolume Lyndon While Walking Fish Group School of Computer Science & Software Engineering."— Presentation transcript:

1 A New Analysis of the LebMeasure Algorithm for Calculating Hypervolume Lyndon While Walking Fish Group School of Computer Science & Software Engineering The University of Western Australia

2 A New Analysis of LebMeasure Page 2 of 25 Overview l Metrics for MOEAs l Hypervolume l LebMeasure and its behaviour l Empirical data on the performance of LebMeasure l A lower-bound on the complexity of LebMeasure l The general case l Conclusions and future work

3 A New Analysis of LebMeasure Page 3 of 25 Metrics for MOEAs l A MOEA produces a front of mutually non-dominating solutions to a given problem u m points in n objectives l To compare the performance of MOEAs, we need metrics to compare fronts l Many metrics have been proposed, of several types u cardinality-based metrics u convergence-based metrics u spread-based metrics u volume-based metrics

4 A New Analysis of LebMeasure Page 4 of 25 Hypervolume (S-metric, Lebesgue measure) l The hypervolume of a front is the size of the portion of objective space collectively dominated by the points on the front l Hypervolume captures in one scalar both the convergence and the spread of the front l Hypervolume has nicer mathematical properties than many other metrics l Hypervolume can be sensitive to scaling of objectives and to extremal values l Hypervolume is expensive to calculate u enter LebMeasure

5 A New Analysis of LebMeasure Page 5 of 25 LebMeasure (LM) l Given a mutually non-dominating front S, LM u calculates the hypervolume dominated exclusively by the first point p, then u discards p and processes the rest of S l If the hypervolume dominated exclusively by p is not “hyper-cuboid”, LM u lops off a hyper-cuboid that is dominated exclusively by p, and u replaces p with up to n “spawns” that collectively dominate the remainder of p’s exclusive hypervolume l A spawn is discarded immediately if it dominates no exclusive hypervolume, either because u it has a “zero” objective, or u it is dominated by an unprocessed point

6 A New Analysis of LebMeasure Page 6 of 25 LebMeasure in action A dominates exclusively the yellow shape A lops off the pink hyper-cuboid A has three potential spawns: A1 = (4,9,4) A2 = (6,7,4) A3 = (6,9,3) But A2 is dominated by B, so it is discarded immediately

7 A New Analysis of LebMeasure Page 7 of 25 A boost for LebMeasure Some “spawns of spawns” are guaranteed to be dominated, so LM doesn’t need to generate them at all l This limits the maximum depth of the stack to m + n – 1 (6, 9, 4) (9, 7, 5) (1,12, 3) (4, 2, 9) (4, 9, 4) (6, 9, 3) (9, 7, 5) (1,12, 3) (4, 2, 9) (1, 9, 4) (4, 7, 4) (4, 9, 3) (6, 9, 3) (9, 7, 5) (1,12, 3) (4, 2, 9) A13 A12      A11 guaranteed to be dominated } A3 B C DDD CC BB A A1

8 A New Analysis of LebMeasure Page 8 of 25 But… l This boost greatly reduces the space complexity of LM u the maximum depth of the stack is linear in both m and n l But it does far less for the time complexity of LM u note that the time complexity depends not only on the number of stack slots used, but also on how many times each slot is used l We shall measure the time complexity of LM in terms of the number of points (and spawns, and spawns of spawns, etc) that actually contribute to the hypervolume u i.e. the number of hyper-cuboids that must be summed

9 A New Analysis of LebMeasure Page 9 of 25 Running LebMeasure nm = 2m = 5m = 8m = , ,09610, ,12532,768100, ,625262,1441,000, ,1252,097,15210,000, ,62516,777,216100,000,000 No. of hyper-cuboids = m n−1 m points in n objectives

10 A New Analysis of LebMeasure Page 10 of 25 Running LebMeasure (in reverse order) No. of hyper-cuboids = m nm = 2m = 5m = 8m = m points in n objectives

11 A New Analysis of LebMeasure Page 11 of 25 Running LebMeasure (in optimal order) nm = 2m = 3m = 4m = , ,2483, ,24814, ,52870, ,28913,70870, ,87354,976142, ,87354,976428, ,761110,1601,721, ,297331,1288,618,577 No. of hyper-cuboids  m(m!) ((n−2)div m) ((n – 2)mod m)! m points in n objectives

12 A New Analysis of LebMeasure Page 12 of 25 Running LebMeasure (first point only) nm = 2m = 5m = 8m = ,6953, ,10115,96140, ,529144,495468, ,7411,273,6095,217, ,08911,012,41556,953,279 No. of hyper-cuboids = m n−1 – (m – 1) n−1, i.e. O(m n−2 ) m points in n objectives

13 A New Analysis of LebMeasure Page 13 of 25 A lower-bound on the complexity of LebMeasure l We can determine a lower-bound on the worst-case complexity of LM by considering a single example l We will derive a recurrence for the number of hyper-cuboids summed for this example, then prove that the recurrence equals 2 n−

14 A New Analysis of LebMeasure Page 14 of 25 The simple picture

15 A New Analysis of LebMeasure Page 15 of 25 The recursive picture

16 A New Analysis of LebMeasure Page 16 of 25 A recurrence h(n,k) gives the number of hyper-cuboids summed for a point (or spawn) with n 2s, of which we can reduce k and still generate points that aren’t dominated by their relatives hcs(n) gives the total number of hyper-cuboids summed for the example, with n objectives

17 A New Analysis of LebMeasure Page 17 of 25 The recurrence in action [ h(4,4) ] (1,2,2,2,2) (1,2,2,1,2) [ h(3,2) ] (1,2,2,2,1) [ h(3,3) ] (1,2,1,2,2) [ h(3,1) ] (1,1,2,2,2) [ h(3,0) ]

18 A New Analysis of LebMeasure Page 18 of 25 The recurrence solved l Simple expansion shows that l The paper gives a formal proof using mathematical induction

19 A New Analysis of LebMeasure Page 19 of 25 The general case l It is difficult to be certain what patterns of points will perform worst for LM l We will describe the behaviour of an illegal “beyond worst case” pattern l Illegal because some points dominate others mmm m−1m−1m−1m−1m−1m−1 11 1

20 A New Analysis of LebMeasure Page 20 of 25 m points in 2 objectives x i denotes the i th best value in objective x l Each vertical list has length  m l Total size  m 2 u1v1u1v1

21 A New Analysis of LebMeasure Page 21 of 25 m points in 3 objectives l Each vertical list has length  m l Each 2-way sub-tree has size  m 2 l Total size  m 3 u1v1w1u1v1w1

22 A New Analysis of LebMeasure Page 22 of 25 m points in 4 objectives l denotes a k-way sub-tree l Each k-way sub-tree has size  m k l Total size  m 4 k u1v1w1x1u1v1w1x1

23 A New Analysis of LebMeasure Page 23 of 25 A recurrence and its solution l Again, we can capture this behaviour as a recurrence By simple expansion (and proved formally in the paper)

24 A New Analysis of LebMeasure Page 24 of 25 Conclusions l LM is exponential in the number of objectives, in the worst case l Re-ordering the points often makes LM go faster, but the worst case is still exponential u the proof technique used for the “simple” case will also work for the “unreorderable” case

25 A New Analysis of LebMeasure Page 25 of 25 Future work l Try to make LM faster u re-order the points u re-order the objectives l Develop and refine other algorithms (e.g. HSO) u possibly develop a hybrid algorithm l Prove that no polynomial-time algorithm exists for calculating hypervolume


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