a new analysis of the lebmeasure algorithm for calculating hypervolume lyndon while walking fish...

A New Analysis ofA New Analysis ofthe LebMeasure Algorithmthe LebMeasure Algorithm

for Calculating Hypervolumefor Calculating Hypervolume

Lyndon While

Walking Fish GroupSchool of Computer Science & Software Engineering

The University of Western Australia

A New Analysis of LebMeasure of 25

Overview

Metrics for MOEAs Hypervolume LebMeasure and its behaviour Empirical data on the performance of LebMeasure A lower-bound on the complexity of LebMeasure The general case Conclusions and future work


Metrics for MOEAs

A MOEA produces a front of mutually non-dominating solutions to a given problem

m points in n objectives To compare the performance of MOEAs,

we need metrics to compare fronts Many metrics have been proposed, of several types

cardinality-based metrics convergence-based metrics spread-based metrics volume-based metrics


Hypervolume (S-metric, Lebesgue measure)

The hypervolume of a front is the size of the portion of objective space collectively dominated by the points on the front

Hypervolume captures in one scalar both the convergence and the spread of the front

Hypervolume has nicer mathematical properties than many other metrics

Hypervolume can be sensitive to scaling of objectivesand to extremal values

Hypervolume is expensive to calculate enter LebMeasure


LebMeasure (LM)

Given a mutually non-dominating front S, LM calculates the hypervolume dominated exclusively

by the first point p, then discards p and processes the rest of S

If the hypervolume dominated exclusively by p is not “hyper-cuboid”, LM

lops off a hyper-cuboid that isdominated exclusively by p, and

replaces p with up to n “spawns” that collectively dominate the remainder of p’s exclusive hypervolume

A spawn is discarded immediately if it dominates no exclusive hypervolume, either because

it has a “zero” objective, or it is dominated by an unprocessed point


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


A boost for LebMeasure

Some “spawns of spawns” are guaranteed to be dominated, so LM doesn’t need to generate them at all

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 tobe dominated}

A3

B

C

DD D

C C

BB

A A3

A1


But…

This boost greatly reduces the space complexity of LM the maximum depth of the stack is linear in both m and n

But it does far less for the time complexity of LM 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

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

i.e. the number of hyper-cuboids that must be summed


Running LebMeasure

1 5 ••• 5

2 4 ••• 4

3 3 ••• 3

4 2 ••• 2

5 1 ••• 1

n m = 2 m = 5 m = 8 m = 10

2 2 5 8 10

3 4 25 64 100

4 8 125 512 1,000

5 16 625 4,096 10,000

6 32 3,125 32,768 100,000

7 64 15,625 262,144 1,000,000

8 128 78,125 2,097,152 10,000,000

9 256 390,62516,777,21

6100,000,000

No. of hyper-cuboids = mn−1

m points in n objectives


Running LebMeasure (in reverse order)

1 5 ••• 5

2 4 ••• 4

3 3 ••• 3

4 2 ••• 2

5 1 ••• 1

No. of hyper-cuboids = m

n m = 2 m = 5 m = 8 m = 10

2 2 5 8 10

3 2 5 8 10

4 2 5 8 10

5 2 5 8 10

6 2 5 8 10

7 2 5 8 10

8 2 5 8 10

9 2 5 8 10



Running LebMeasure (in optimal order)

1 1 2 3 4 5 1 ••• 5

2 2 3 4 5 1 2 ••• 4

3 3 4 5 1 2 3 ••• 3

4 4 5 1 2 3 4 ••• 2

5 5 1 2 3 4 5 ••• 1

n m = 2 m = 3 m = 4 m = 5

2 2 3 4 5

3 2 3 4 5

4 4 6 8 10

5 4 17 23 29

6 8 17 88 112

7 8 35 88 549

8 16 105 180 549

9 16 105 558 1,115

10 32 213 2,248 3,421

11 32 641 2,248 14,083

12 64 641 4,528 70,899

13 64 1,289 13,708 70,889

14 128 3,873 54,976 142,309

15 128 3,873 54,976 428,449

16 256 7,761 110,160 1,721,605

17 256 23,297 331,128 8,618,577

No. of hyper-cuboids m(m!)((n−2)div m)((n – 2)mod m)!



Running LebMeasure (first point only)

1 5 ••• 5

2 4 ••• 4

3 3 ••• 3

4 2 ••• 2

5 1 ••• 1

n m = 2 m = 5 m = 8 m = 10

2 1 1 1 1

3 3 9 15 19

4 7 61 169 271

5 15 369 1,695 3,439

6 31 2,101 15,961 40,951

7 63 11,529 144,495 468,559

8 127 61,741 1,273,609 5,217,031

9 255 325,089 11,012,415 56,953,279

No. of hyper-cuboids = mn−1 – (m – 1)n−1, i.e. O(mn−2)



A lower-bound on the complexity of LebMeasure

We can determine a lower-bound on the worst-case complexity of LM by considering a single example

We will derive a recurrence for the number of hyper-cuboids summed for this example, then prove that the recurrence equals 2n−1

1 2 2 2 ••• 2

2 1 1 1 ••• 1


The simple picture

12222

11222 12122 12212 12221

11112

11221 122111212111212 1211211122

11211 1211111121


The recursive picture

12222

11212 12112

11112

11122

11222 12122 12212

1211111211

12211

11121

11221 12121

12221


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

1

0),1(1),(

1),1(k

iinhknh

kh

1)1,1()( nnhnhcs


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)]


The recurrence solved

Simple expansion shows that

The paper gives a formal proof using mathematical induction

1

1

2)(

12),(

n

k

nhcs

kkh


The general case

It is difficult to be certain what patterns of points will perform worst for LM

We will describe the behaviour of an illegal “beyond worst case” pattern

Illegal because some points dominate others

m m ••• m

m−1 m−1 ••• m−1

•••

•••

•••

1 1 ••• 1


m points in 2 objectives

xi denotes the ith best value in objective x

Each vertical list has length m Total size m2

u1v1



Each vertical list has length m Each 2-way sub-tree has size m2

Total size m3

u1v1w1



denotes a k-way sub-tree

Each k-way sub-tree has size mk

Total size m4

k

12

3

12

3

12

3

12

3

u1v1w1x1


A recurrence and its solution

Again, we can capture this behaviour as a recurrence

By simple expansion (and proved formally in the paper)

1

0

1

1

),,(

),,(11

10

m

i

y

i

iniqp(m,n)

zizq,y,z)q(xq(x,y,z)

,y,z)q(

m

i

ninmp1

),(


Conclusions

LM is exponential in the number of objectives, in the worst case

Re-ordering the points often makes LM go faster,but the worst case is still exponential

the proof technique used for the “simple” case will also work for the “unreorderable” case


Future work

Try to make LM faster re-order the points re-order the objectives

Develop and refine other algorithms (e.g. HSO) possibly develop a hybrid algorithm

Prove that no polynomial-time algorithm existsfor calculating hypervolume

a new analysis of the lebmeasure algorithm for calculating hypervolume lyndon while walking fish...

Documents

lebmeasure slide

lebmeasure lm

complexity of lebmeasure

running lebmeasure

hypervolume u

performance of lebmeasure

lebmeasure algorithm

time complexity of lm