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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 Page 2 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
A New Analysis of LebMeasure Page 3 of 25
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
A New Analysis of LebMeasure Page 4 of 25
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
A New Analysis of LebMeasure Page 5 of 25
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
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
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
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
A New Analysis of LebMeasure Page 8 of 25
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
A New Analysis of LebMeasure Page 9 of 25
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
A New Analysis of LebMeasure Page 10 of 25
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
m points in n objectives
A New Analysis of LebMeasure Page 11 of 25
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)!
m points in n objectives
A New Analysis of LebMeasure Page 12 of 25
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)
m points in n objectives
A New Analysis of LebMeasure Page 13 of 25
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
A New Analysis of LebMeasure Page 14 of 25
The simple picture
12222
11222 12122 12212 12221
11112
11221 122111212111212 1211211122
11211 1211111121
A New Analysis of LebMeasure Page 15 of 25
The recursive picture
12222
11212 12112
11112
11122
11222 12122 12212
1211111211
12211
11121
11221 12121
12221
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
1
0),1(1),(
1),1(k
iinhknh
kh
1)1,1()( nnhnhcs
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)]
A New Analysis of LebMeasure Page 18 of 25
The recurrence solved
Simple expansion shows that
The paper gives a formal proof using mathematical induction
1
1
2)(
12),(
n
k
nhcs
kkh
A New Analysis of LebMeasure Page 19 of 25
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
A New Analysis of LebMeasure Page 20 of 25
m points in 2 objectives
xi denotes the ith best value in objective x
Each vertical list has length m Total size m2
u1v1
A New Analysis of LebMeasure Page 21 of 25
m points in 3 objectives
Each vertical list has length m Each 2-way sub-tree has size m2
Total size m3
u1v1w1
A New Analysis of LebMeasure Page 22 of 25
m points in 4 objectives
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 New Analysis of LebMeasure Page 23 of 25
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
),(
A New Analysis of LebMeasure Page 24 of 25
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
A New Analysis of LebMeasure Page 25 of 25
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
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