APPENDIX A.8 Proof of Lemma 8 A ... - auchunkit.github.io · APPENDIX A. PROOFS A.1 Proof of Lemma...
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APPENDIX A. PROOFS A.1 Proof of Lemma 1 Proof. If x0 + x lies in the closed positive halfspace H+(n, x0), n T x must be greater than or equal to zero. Putting a reverse of x into the equation (1) inverts the sign and hence -n T x must be less than or equal to zero. Therefore, by def- inition, x0 - x must lie in the negative closed halfspace. A.2 Proof of Lemma 2 Proof. We prove on a geometrically based argumenta- tion. Assume a minimisation for the objective function f . Given a point x0 2 R n , by definition, the tangent hyperplane T (x0) at x0 separates the whole space into two halfspaces. The positive halfspace H+(n, x0) contains all the points in C-(x0) with objective function values smaller than or equal to f (x0) while the negative halfspace contains the points with larger or equal objective values. Hence the optimal halfspaces for the better solutions are delimited by the tan- gent hyperplane T (x0) at x0. A.3 Proof of Lemma 3 Proof. By Lemma 2, the tangent hyperplane T (x0) sep- arates the space into the optimal halfspaces. By definition of a tangent hyperplane, the normal of a tangent hyperplane at x0 is ⇣ @f (x 0 ) @x 1 ,..., @f (x 0 ) @xn ⌘ . Reversing the direction of this normal gives the normal of the supporting hyperplane that contains all the points in C-(x0). A.4 Proof of Lemma 4 Proof. By Lemma 3, the normal vector n ⇤ has the gradi- ent descent direction and is a form of ⇣ - @f (x 0 ) @x 1 ,..., - @f (x 0 ) @xn ⌘ . At the point e1, the gradient descent direction is -2e1. Af- ter normalisation, it gives -e1. A.5 Proof of Lemma 5 Proof. By definition of a positive halfspace, if x + e 1 2 H ⇤ + (n ⇤ , e1), then (n ⇤ ) T ( x + e 1 - e1 ) ≥ 0. By Lemma 4, sub- stituting -e1 for n ⇤ and e1 + σN for x + e 1 gives -e T 1 σN ≥ 0. Projecting the N onto the coordinate e1 and rearrang- ing the terms, the result σ [N ] 1 0 holds. The event x + e 1 2 H ⇤ - (n ⇤ , e1) can be derived similarly. A.6 Proof of Lemma 6 Proof. The proof follows similarly the proof of Lemma 5. A.7 Proof of Lemma 7 Proof. To prove equation (9), we consider the event U 0,W + 0 . Given σ 2 R + , the event W + 0 happens only if U 0. Hence the event U 0,W + 0 can be written as W + 0 . Similarly, we can write the event U ≥ 0,W - 0 as W - 0 . Replace the terms in equation (9) and the result holds. To prove equation (10), consider the event U 0,W + 0 . If U is negative, W + is the same as V . Similarly in the event U ≥ 0,W - 0 , if U is positive, then W - is the same as V . Putting these together, the equation (10) is simplified to E ⇥ ln(1 + σV 1 {U0,V 0} + σV 1 {U≥0,V 0} ). ⇤ Adding up the second and third terms in the logarithm gives σV 1 {V 0} and hence the result holds. A.8 Proof of Lemma 8 Proof. Because of the isotropy of the distribution of N and of sphere function, we can first replace Y + k , Y - k and Yk by e + 1 , e - 1 and e1 respectively, where e + 1 = e1 + σN and e - 1 = e1 - σN . The distribution Zk equals to Zk = 1 2 ln h ||e + 1 || 2 ⇥ 1 {e + 1 2H ⇤ + (n ⇤ ,e 1 ),||e + 1 ||1} + ||e - 1 || 2 ⇥ 1 {e + 1 / 2H ⇤ + (n ⇤ ,e 1 ),||e - 1 ||1} +1 {e + 1 2H ⇤ + (n ⇤ ,e 1 ),||e + 1 ||>1} +1 {e + 1 / 2H ⇤ + (n ⇤ ,e 1 ),||e - 1 ||>1} i (29) We then substitute the notations U , W + and W - and ex- pand ||e + 1 || 2 as 1 + 2σ [N ] 1 + σ 2 ||N || 2 and ||e - 1 || 2 as 1 + 2σ [N ] 1 - σ 2 ||N || 2 . By Lemma 5 and 6, the term in the logarithm of equation (29) is simplified into 1 {U0,W + 0} + σW + 1 {U0,W + 0} +1 {U≥0,W - 0} +σW - 1 {U≥0,W - 0} +1 {U0,W + >0} +1 {U≥0,W - >0} (30) We substitute 1 {U0,W + 0} +1 {U≥0,W - 0} +1 {U0,W + >0} + 1 {U≥0,W - >0} = 1 and the equation (30) is simplified as 1+ σW + 1 {U0,W + 0} + σW - 1 {U≥0,W - 0} (31) By Lemma 7, we can simplify equation (31) into 1+σV 1 {V 0} . Injecting it into equation (29), we then obtain the result. A.9 Proof of Lemma 9 Proof. We first start from the expression of p s (1+1 hs ) (σ/d): p s (1+1 hs ) (σ/d) = Pr ⇣ -2| [N ] 1 | + σ n ||N || 2 0 ⌘ (32) =E h 1 {-2|[N] 1 |+ σ n ||N|| 2 0} i (33) By LLN, we can derive that -2| [N ] 1 | + σ n ||N || 2 ----! n!1 -2| [N ] 1 | + σ (34) Since 1 {-2|[N] 1 |+ σ n ||N|| 2 0} 1, we can apply Lebesgue dominated theorem and therefore E h 1 {-2|[N] 1 |+ σ n ||N|| 2 0} i ----! n!1 E h 1 {-2|[N] 1 |+σ0} i (35) RHS of equation (35) can be rewrite as E h 1 {-2|[N] 1 |+σ0} i = Pr ( -2| [N ] 1 | + σ 0 ) = Pr ( | [N ] 1 | ≥ σ/2 ) Lastly Pr ( | [N ] 1 | ≥ σ/2 ) =1 - erf ⇣ σ 2 p 2 ⌘ = ¯ ΦHN( σ 2 ) A.10 Proof of Lemma 10 Proof. First the LHS of equation (18) can be written as the density of a standard half-normal distribution as E h | [N ] 1 |1 {|[N] 1 |≥σ/2} i = r 2 ⇡ Z 1 σ/2 x exp(- x 2 2 )dx (36) Integrating the RHS of equation (36) gives the result. B. BBOB EXPERIMENTAL RESULTS
Transcript of APPENDIX A.8 Proof of Lemma 8 A ... - auchunkit.github.io · APPENDIX A. PROOFS A.1 Proof of Lemma...
APPENDIXA. PROOFSA.1 Proof of Lemma 1
Proof. If x0 + x lies in the closed positive halfspaceH+(n,x0), n
Txmust be greater than or equal to zero. Puttinga reverse of x into the equation (1) inverts the sign and hence�nTx must be less than or equal to zero. Therefore, by def-inition, x0 �x must lie in the negative closed halfspace.
A.2 Proof of Lemma 2Proof. We prove on a geometrically based argumenta-
tion. Assume a minimisation for the objective function f .Given a point x0 2 Rn, by definition, the tangent hyperplaneT (x0) at x0 separates the whole space into two halfspaces.The positive halfspace H+(n,x0) contains all the points inC�(x0) with objective function values smaller than or equalto f(x0) while the negative halfspace contains the pointswith larger or equal objective values. Hence the optimalhalfspaces for the better solutions are delimited by the tan-gent hyperplane T (x0) at x0.
A.3 Proof of Lemma 3Proof. By Lemma 2, the tangent hyperplane T (x0) sep-
arates the space into the optimal halfspaces. By definitionof a tangent hyperplane, the normal of a tangent hyperplane
at x0 is⇣
@f(x0)@x1
, . . . , @f(x0)@x
n
⌘
. Reversing the direction of this
normal gives the normal of the supporting hyperplane thatcontains all the points in C�(x0).
A.4 Proof of Lemma 4Proof. By Lemma 3, the normal vector n⇤ has the gradi-
ent descent direction and is a form of⇣
� @f(x0)@x1
, . . . ,� @f(x0)@x
n
⌘
.
At the point e1, the gradient descent direction is �2e1. Af-ter normalisation, it gives �e1.
A.5 Proof of Lemma 5Proof. By definition of a positive halfspace, if x+
e12
H⇤+(n
⇤, e1), then (n⇤)T�
x+e1
� e1
�
� 0. By Lemma 4, sub-
stituting �e1 for n⇤ and e1 + �N for x+e1
gives �eT1 �N �
0. Projecting the N onto the coordinate e1 and rearrang-ing the terms, the result � [N ]1 0 holds. The event�
x+e1
2 H⇤�(n
⇤, e1)
can be derived similarly.
A.6 Proof of Lemma 6Proof. The proof follows similarly the proof of Lemma
5.
A.7 Proof of Lemma 7Proof. To prove equation (9), we consider the event
�
U 0,W+ 0
. Given � 2 R+, the event�
W+ 0
happens only if U 0. Hence the event�
U 0,W+ 0
can be written as�
W+ 0
. Similarly, we can write theevent
�
U � 0,W� 0
as�
W� 0
. Replace the termsin equation (9) and the result holds.To prove equation (10), consider the event
�
U 0,W+ 0
.If U is negative, W+ is the same as V . Similarly in the event�
U � 0,W� 0
, if U is positive, then W� is the same asV . Putting these together, the equation (10) is simplified to
E⇥
ln(1 + �V 1{U0,V 0} + �V 1{U�0,V 0}).⇤
Adding up the second and third terms in the logarithm gives�V 1{V 0} and hence the result holds.
A.8 Proof of Lemma 8Proof. Because of the isotropy of the distribution of N
and of sphere function, we can first replace Y+k , Y�
k and Yk
by e+1 , e
�1 and e1 respectively, where e+
1 = e1 + �N ande�1 = e1 � �N . The distribution Zk equals to
Zk =12lnh
||e+1 ||
2 ⇥ 1{e+1 2H⇤+(n⇤,e1),||e+1 ||1}
+ ||e�1 ||
2 ⇥ 1{e+1 /2H⇤+(n⇤,e1),||e�1 ||1}
+ 1{e+1 2H⇤+(n⇤,e1),||e+1 ||>1}
+1{e+1 /2H⇤+(n⇤,e1),||e�1 ||>1}
i
(29)
We then substitute the notations U , W+ and W� and ex-pand ||e+
1 ||2 as 1 + 2� [N ]1 + �2||N ||2 and ||e�1 ||2 as 1 +
2� [N ]1 � �2||N ||2. By Lemma 5 and 6, the term in thelogarithm of equation (29) is simplified into
1{U0,W+0} + �W+1{U0,W+0} + 1{U�0,W�0}+�W�1{U�0,W�0} + 1{U0,W+>0} + 1{U�0,W�>0}
(30)
We substitute 1{U0,W+0}+1{U�0,W�0}+1{U0,W+>0}+1{U�0,W�>0} = 1 and the equation (30) is simplified as
1 + �W+1{U0,W+0} + �W�1{U�0,W�0} (31)
By Lemma 7, we can simplify equation (31) into 1+�V 1{V 0}.Injecting it into equation (29), we then obtain the result.
A.9 Proof of Lemma 9Proof. We first start from the expression of ps(1+1hs)
(�/d):
ps(1+1hs)(�/d) = Pr
⇣
�2| [N ]1 |+�
n||N ||2 0
⌘
(32)
= Eh
1{�2|[N ]1|+�
n
||N ||20}i
(33)
By LLN, we can derive that
�2| [N ]1 |+�
n||N ||2 ����!
n!1�2| [N ]1 |+ � (34)
Since 1{�2|[N ]1|+�
n
||N ||20} 1, we can apply Lebesgue
dominated theorem and therefore
Eh
1{�2|[N ]1|+�
n
||N ||20}i
����!n!1
Eh
1{�2|[N ]1|+�0}i
(35)
RHS of equation (35) can be rewrite as
Eh
1{�2|[N ]1|+�0}i
= Pr�
�2| [N ]1 |+ � 0�
= Pr�
| [N ]1 | � �/2�
Lastly Pr�
| [N ]1 | � �/2�
= 1� erf⇣
�
2p2
⌘
= �̄HN(�2 )
A.10 Proof of Lemma 10Proof. First the LHS of equation (18) can be written as
the density of a standard half-normal distribution as
Eh
| [N ]1 |1{|[N ]1|��/2}i
=
r
2⇡
Z 1
�/2
x exp(�x2
2) dx (36)
Integrating the RHS of equation (36) gives the result.
B. BBOB EXPERIMENTAL RESULTS
Figure 4: Expected running time (ERT in number of f-evaluations) divided by dimension for target functionvalue 10�8 as log10 values versus dimension. Di↵erent symbols correspond to di↵erent algorithms given inthe legend of f1 and f24. Light symbols give the maximum number of function evaluations from the longesttrial divided by dimension. Horizontal lines give linear scaling, slanted dotted lines give quadratic scaling.Black stars indicate statistically better result compared to all other algorithms with p < 0.01 and Bonferronicorrection number of dimensions (six). Legend: �:(1+1)-CMA-ES with Halfspace Sampling, O:(1+1)-CMA-ES.
1Sphere
2Ellipsoidseparable
3Rastriginseparable
4Skew
Rastrigin-Bueche
5Linearslope
6Attractivesector
7Step-ellipsoid
8Rosenbrockoriginal
9Rosenbrockrotated
10Ellipsoid
11Discus
12Bentcigar
13Sharpridge
14Sum
ofdi↵.powers
15Rastrigin
16Weierstrass
17Scha↵erF7,cond.10
18Scha↵erF7,cond.1000
19Griewank-Rosenbrock
20Schwefelx*sin(x)
21Gallagher101peaks
22Gallagher21peaks
23Katsuuras
24Lunacekbi-Rastrigin
Figure 5: Expected running time (ERT in log10 of number of function evaluations) of (1+1)-CMA-ESwith Halfspace Sampling (x-axis) versus (1+1)-CMA-ES (y-axis) for 46 target values �f 2 [10�8, 10] in eachdimension on functions f1–f24. Markers on the upper or right edge indicate that the target value was neverreached. Markers represent dimension: 2:+, 3:O, 5:?, 10:�, 20:2, 40:3.
5-D 20-D
separab
lefcts
moderatefcts
ill-conditioned
fcts
multi-modal
fcts
weakstructure
fcts
allfunctions
Figure 6: Empirical cumulative distributions (ECDF) of run lengths and speed-up ratios in 5-D (left)and 20-D (right). Left sub-columns: ECDF of the number of function evaluations divided by dimension D(FEvals/D) to reach a target value fopt +�f with �f = 10k, where k 2 {1,�1,�4,�8} is given by the first valuein the legend, for (1+1)-CMA-ES with Halfspace Sampling (�) and (1+1)-CMA-ES (O). Light beige linesshow the ECDF of FEvals for target value �f = 10�8 of all algorithms benchmarked during BBOB-2009. Rightsub-columns: ECDF of FEval ratios of (1+1)-CMA-ES with Halfspace Sampling divided by (1+1)-CMA-ES,all trial pairs for each function. Pairs where both trials failed are disregarded, pairs where one trial failedare visible in the limits being > 0 or < 1. The legends indicate the number of functions that were solved inat least one trial ((1+1)-CMA-ES with Halfspace Sampling first).
5-D 20-D�f 1e+1 1e-1 1e-3 1e-5 1e-7 #succf1 11 12 12 12 12 15/15
1: hs 1.5(1) 10(3) 17(3) 24(3) 32(5) 15/152: 1p1 2.3(2) 10(3) 17(3) 25(3) 32(3) 15/15
f2 83 88 90 92 94 15/151: hs 8.0(3) 11(1) 12(1)? 13(2)? 13(1)? 15/152: 1p1 9.4(4) 12(3) 14(1) 14(2) 15(2) 15/15
f3 716 1637 1646 1650 1654 15/151: hs 8.4(7) 1 1 1 15.0e4 0/152: 1p1 9.1(9) 1 1 1 15.0e4 0/15
f4 809 1688 1817 1886 1903 15/151: hs 8.2(8) 1 1 1 15.0e4 0/152: 1p115(22) 1 1 1 15.0e4 0/15
f5 10 10 10 10 10 15/151: hs 1.7(1) 2.3(0.9) 2.3(0.9) 2.3(0.9) 2.3(0.9) 15/152: 1p1 2.3(1) 3.2(1) 3.2(1) 3.2(1) 3.2(1) 15/15
f6 114 281 580 1038 1332 15/151: hs 1.3(0.6) 1.3(0.5) 1.0(0.3) 0.84(0.2) 0.89(0.3) 15/152: 1p1 1.4(0.9) 1.6(1) 1.6(1) 1.6(1) 1.8(1) 15/15
f7 24 1171 1572 1572 1597 15/151: hs 4.6(2) 2.3(1) 6.2(7) 6.2(7) 6.1(7) 14/152: 1p1 3.5(3) 2.0(2) 3.1(4) 3.1(4) 3.1(4) 15/15
f8 73 336 391 410 422 15/151: hs 2.3(1) 2.9(2) 3.0(2) 3.2(2) 3.3(2) 15/152: 1p1 2.1(2) 5.0(5) 4.9(4) 4.9(4) 5.0(4) 15/15
f9 35 214 300 335 369 15/151: hs 4.2(1) 5.8(4) 4.8(3) 4.6(2) 4.5(2) 15/152: 1p1 4.2(1) 5.9(3) 5.0(2) 4.9(2) 4.7(2) 15/15f10 349 574 626 829 880 15/15
1: hs 2.2(0.6) 1.7(0.1)? 1.7(0.1)?2 1.4(0.1)?2 1.4(0.1)?2 15/152: 1p1 2.5(0.9) 2.0(0.3) 2.1(0.3) 1.7(0.2) 1.7(0.2) 15/15f11 143 763 1177 1467 1673 15/15
1: hs 5.8(3) 1.7(0.3)?2 1.3(0.1)?3 1.1(0.1)?3 0.99(0.1)?315/152: 1p1 6.5(2) 2.1(0.3) 1.5(0.2) 1.3(0.1) 1.2(0.1) 15/15f12 108 371 461 1303 1494 15/151: hs 8.4(10) 7.0(6) 7.0(6) 2.9(2) 2.9(2) 15/152: 1p1 4.0(4) 3.5(2) 4.1(3) 1.9(1) 1.9(1) 15/15f13 132 250 1310 1752 2255 15/151: hs 3.2(3) 5.0(2) 1.5(1) 1.8(0.9) 2.0(1) 15/152: 1p1 4.4(4) 6.1(3) 1.9(1) 3.2(4) 3.1(3) 15/15f14 10 58 139 251 476 15/15
1: hs 1.2(2) 2.2(0.3) 2.8(0.5) 3.6(0.8) 2.8(0.2)?2 15/152: 1p1 1.8(3) 2.3(0.5) 2.6(0.8) 4.0(1.0) 3.2(0.3) 15/15f15 511 19369 20073 20769 21359 14/151: hs 12(12) 1 1 1 15.0e4 0/152: 1p110(13) 1 1 1 15.0e4 0/15f16 120 2662 10449 11644 12095 15/151: hs 2.7(4) 22(25) 1 1 15.0e4 0/152: 1p1 2.5(5) 17(18) 31(38) 60(67) 15.0e4 0/15f17 5.2 899 3669 6351 7934 15/151: hs 3.0(4) 24(25) 1 1 15.0e4 0/152: 1p1 4.5(5) 108(114) 1 1 15.0e4 0/15f18 103 3968 9280 10905 12469 15/151: hs 5.1(13) 43(44) 1 1 15.0e4 0/152: 1p1 4.9(5) 84(101) 1 1 15.0e4 0/15f19 1 242 1.2e5 1.2e5 1.2e5 15/151: hs 22(20) 497(558) 1 1 15.0e4 0/152: 1p120(17) 971(1085) 1 1 15.0e4 0/15f20 16 38111 54470 54861 55313 14/151: hs 3.3(2) 18(21) 13(14) 13(14) 13(15) 1/152: 1p1 2.4(2) 5.9(6) 4.1(5) 4.1(5) 4.1(5) 3/15f21 41 1674 1705 1729 1757 14/151: hs 1.3(0.9) 5.8(8) 5.7(8) 5.7(8) 5.6(8) 15/152: 1p1 4.2(10) 6.6(9) 6.5(9) 6.5(9) 6.4(9) 15/15f22 71 938 1008 1040 1068 14/151: hs 1.6(0.7) 6.5(8) 6.2(7) 6.1(7) 6.0(7) 15/152: 1p1 2.8(7) 4.7(6) 4.5(6) 4.4(5) 4.4(5) 15/15f23 3.0 14249 31654 33030 34256 15/151: hs 5.3(7) 2.5(2) 1 1 15.0e4 0/152: 1p1 4.2(5) 7.5(7) 1 1 15.0e4 0/15f24 1622 6.4e6 9.6e6 1.3e7 1.3e7 3/151: hs 4.1(5) 1 1 1 15.0e4 0/152: 1p1 6.7(8) 1 1 1 15.0e4 0/15
�f 1e+1 1e-1 1e-3 1e-5 1e-7 #succf1 43 43 43 43 43 15/15
1: hs 5.6(0.7) 13(1) 20(1) 28(2) 36(3) 15/152: 1p1 5.4(0.8) 13(1) 21(2) 29(2) 37(2) 15/15
f2 385 387 390 391 393 15/15
1: hs 28(6) 33(3)?3 35(1)?3 36(1)?3 37(2)?3 15/152: 1p1 30(5) 39(3) 41(3) 43(2) 44(2) 15/15
f3 5066 7635 7643 7646 7651 15/151: hs 1 1 1 1 12.0e5 0/152: 1p1 1 1 1 1 12.0e5 0/15
f4 4722 7666 7700 7758 1.4e5 9/151: hs 1 1 1 1 12.0e5 0/152: 1p1 1 1 1 1 12.0e5 0/15
f5 41 41 41 41 41 15/151: hs 2.5(0.6) 3.1(0.8) 3.1(0.8) 3.1(0.8) 3.1(0.8) 15/152: 1p1 3.1(0.7) 3.7(0.7) 3.7(0.7) 3.7(0.7) 3.7(0.7) 15/15
f6 1296 3413 5220 6728 8409 15/151: hs 3.5(2) 179(193) 1 1 12.0e5 0/152: 1p1 13(15) 835(982) 1 1 12.0e5 0/15
f7 1351 9503 16524 16524 16969 15/151: hs 24(13) 96(104) 1 1 12.0e5 0/152: 1p1 30(17) 304(316) 1 1 12.0e5 0/15
f8 2039 4040 4219 4371 4484 15/151: hs 2.6(1) 4.8(4) 4.9(4) 4.9(3) 4.9(3) 15/152: 1p1 3.7(1) 6.6(5) 6.6(4) 6.6(4) 6.6(4) 15/15
f9 1716 3277 3455 3594 3727 15/151: hs 3.2(1) 5.4(3) 5.5(3) 5.5(3) 5.5(3) 15/152: 1p1 4.5(2) 7.0(4) 7.0(4) 7.0(4) 6.9(4) 15/15f10 7413 10735 14920 17073 17476 15/15
1: hs 1.4(0.2)? 1.2(0.1)?2 0.93(0.1)?3 0.84(0.1)?3#4 0.84(0.1)?3#415/152: 1p1 1.7(0.3) 1.4(0.2) 1.1(0.1) 1(0.1) 1(0.1) 15/15f11 1002 6278 9762 12285 14831 15/15
1: hs 6.9(2) 2.3(0.4) 2.1(0.3) 1.9(0.2)?2 1.7(0.1)?3 15/152: 1p1 7.0(1) 2.4(0.3) 2.4(0.4) 2.3(0.2) 2.0(0.1) 15/15f12 1042 2740 4140 12407 13827 15/151: hs 3.1(4) 5.1(5) 4.7(4) 2.1(1) 2.3(1) 15/152: 1p1 7.7(12) 10(9) 8.3(6) 3.4(2) 3.6(2) 15/15f13 652 2751 18749 24455 30201 15/151: hs 3.5(3) 8.2(6) 2.5(2) 3.9(4) 10(12) 4/152: 1p1 4.9(6) 10(14) 4.2(4) 6.3(6) 14(15) 4/15f14 75 304 932 1648 15661 15/151: hs 3.1(1) 2.2(0.3) 2.0(0.4) 4.6(0.8) 1.0(0.2) 15/152: 1p1 3.1(1) 2.3(0.5) 2.3(0.5) 5.6(0.9) 1.2(0.2) 15/15f15 30378 3.1e5 3.2e5 4.5e5 4.6e5 15/151: hs 1 1 1 1 12.0e5 0/152: 1p1 1 1 1 1 12.0e5 0/15f16 1384 77015 1.9e5 2.0e5 2.2e5 15/151: hs 36(44) 1 1 1 12.0e5 0/152: 1p1 34(37) 1 1 1 12.0e5 0/15f17 63 4005 30677 56288 80472 15/151: hs 19(33) 1 1 1 12.0e5 0/152: 1p1 29(10) 1 1 1 12.0e5 0/15f18 621 19561 67569 1.3e5 1.5e5 15/151: hs 969(1177) 1 1 1 12.0e5 0/152: 1p14824(4995) 1 1 1 12.0e5 0/15f19 1 3.4e5 6.2e6 6.7e6 6.7e6 15/151: hs 558(522) 1 1 1 12.0e5 0/152: 1p11365(1312) 1 1 1 12.0e5 0/15f20 82 3.1e6 5.5e6 5.6e6 5.6e6 14/151: hs 3.2(0.6) 1 1 1 12.0e5 0/152: 1p1 3.4(0.9) 1 1 1 12.0e5 0/15f21 561 14103 14643 15567 17589 15/151: hs 3.6(6) 2.9(2) 2.8(2) 2.7(2) 2.4(2) 15/152: 1p1 3.6(7) 5.8(5) 5.6(5) 5.3(5) 4.7(4) 14/15f22 467 23491 24948 26847 1.3e5 12/151: hs 12(16) 5.5(6) 5.2(6) 4.8(6) 0.97(1) 11/152: 1p1 3.5(4) 5.6(6) 5.3(6) 5.0(5) 1(1) 12/15f23 3.2 67457 4.9e5 8.1e5 8.4e5 15/151: hs 9.2(15) 1 1 1 12.0e5 0/152: 1p1 5.8(9) 1 1 1 12.0e5 0/15f24 1.3e6 5.2e7 5.2e7 5.2e7 5.2e7 3/151: hs 1 1 1 1 12.0e5 0/152: 1p1 1 1 1 1 12.0e5 0/15
Table 2: ERT in number of function evaluations divided by the best ERT measured during BBOB-2009given in the respective first row with the central 80% range divided by two in brackets for di↵erent �fvalues. #succ is the number of trials that reached the final target fopt + 10�8. 1:hs is (1+1)-CMA-ES withHalfspace Sampling and 2:1p1 is (1+1)-CMA-ES. Bold entries are statistically significantly better comparedto the other algorithm, with p = 0.05 or p = 10�k where k 2 {2, 3, 4, . . . } is the number following the ? symbol,with Bonferroni correction of 48. A # indicates the same tested against the best BBOB-2009.
