A Two-Dimenstional Bisection Envelope Algorithm for Fixed Pointsehan/cs7940/presentation.pdf ·...
Transcript of A Two-Dimenstional Bisection Envelope Algorithm for Fixed Pointsehan/cs7940/presentation.pdf ·...
4/24/2007School of Computing
EunGyoung Han 1
A TwoA Two--Dimensional Bisection Dimensional Bisection
Envelope Algorithm Envelope Algorithm
for Fixed Pointsfor Fixed Points
Kris Sikorski and Spencer ShellmanFrom published Journal of Complexity 18, 641-659(2002)
EunGyoung Han
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IntroductionIntroduction
� How we solve for two-dimensional – domain: [0, 1]X[0, 1]
– f : Lipschitz continuous function (q = 1).
� Previous method– Time complexity was bad
� Paper introduce new algorithm– Computes approximate satisfying
– Tolerance
– Upper bound on the function evaluations
x~
xxfrr
=)(
ε≤−∞
xxf ~)~(
5.0<ε
.1)/1(log2 2 +ε
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HistoryHistory
� 1920s - Present
– Banach’s simple iteration algorithm
– Homotopy continuation
– Simplicial and Newton type methods
� Time complexity
– Lipschitz function (q>1)
• Exponential in the worst case
• Lower bound is also exponential (best case)
1,)()( >−≤−∞∞
qyxqyfxf
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Problem FormulationProblem Formulation
� Class of Lipschitz continuous functions
� By the Brouwer fixed point theorem
{ }∞∞
−≤−∈∀→= yxyfxfDyxDDf babababaF )()(,|: ,,,,
.)(such that , into maps **
,
*
,,, xxfDxDDFf babababa =∈∃∈∀
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Problem FormulationProblem Formulation
� We know a solution exists, we just need a
constructive algorithm…
� Two different criteria to satisfy
– Residual criterion
• Can always be satisfied
– Absolute criterion
• Can sometimes be satisfied
.~)~( ε≤−∞
xxf
.~ εα ≤−∞
x
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Problem FormulationProblem Formulation
� To find the fixed point using the Bisection
Envelope Algorithm, we are required nfunction evaluations of f, where
.11
log21 2 +
≤≤
εn
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Envelope TheoremEnvelope Theorem
� Define the fixed point sets forbaFf ,∈
)()(
},)(|{
},)(|{
21
22,2
11,1
ff
xxfDx
xxfDx
ba
ba
FFF
F
F
∩=
=∈=
=∈=rr
rr
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Theorem 3.5Theorem 3.5
� …
. satifies ~ then , of
point fixed a contains addition,in If,
. satifies )(~Then
.intersect )( and )(both and
2)()(such that Let
21
11,
reterianabsolute cyf
R
reterianresidual cRcy
Rff
RlRlDR ba
=
≤+⊂ −
FF
ε
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Theorem 3.5Theorem 3.5
ε≤
ε≤)(Rc
)(2 fF
)(1 fF
)(1 Rl
)(1 Rl−
ε.x)-xf(
c(R)
ε(R)l(R)l
R
≤
≤+
∞
−
~~
satifies so
2
satifies
11
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The The BEFixBEFix Algorithm: DefinitionAlgorithm: Definition
� ..
� ..
}.1)5.0,5.0(:{ and
domains theDefine
1
2
0
5.1,5.0
≤−ℜ∈=
= −
xxD
DD
points. fixed all contains -
.D within existspoint fixed oneleast At -
on )1( continuous Lipschitz is ,: If
D
DqfDDf =→
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The The BEFixBEFix Algorithm: Figure D, et.Algorithm: Figure D, et.
5.1
D D
DD
0D
0D
0D
0DD
5.1
5.0−
5.0−
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The The BEFixBEFix Algorithm: ProjectionAlgorithm: Projection
� Projection
– ..
– ..
))),1min(,0max()),,1min(,0(max()( 21 xxxP =
. where, onto project Let DDDDP ⊇
.~)~( and ~ where
,for solution residual a is)~(~then
,for solution residual a is ~ If
0 ε≤−∈
=
yyfDy
fyPx
fy
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The The BEFixBEFix Algorithm: Description Algorithm: Description
�
�
.~criterion absolute
satifies ~ ifonly trueiswhich
variablelogical returns Algorithm
εα ≤−y
y
abs
.~)~(
osolution t a as )~(~ takesAlgorithm
ε≤−
=
xxf
yPx
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The The BEFixBEFix Algorithm: ConstructionAlgorithm: Construction
� Constructs a algorithm
–
ε2)()( ifor -
at satified iscriterion residual a If-
: when stepat s terminateAlgorithm -
1)-or 1 (slope rectangle closed a is Each -
. through 1-or 1 slope with lines along bisecting
by rectangle a Constructs -
stepon at Evaluates -
11
1
1
≤+
=
⊂
−
−
−
kk
k
k
k
k
kk
k
DlDl
x
k
D
xD
DD
kxf
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BarycentricBarycentric Coordinate SystemCoordinate System
� Find the next centroid by using Barycentric coordinate system at .
�
)(DC k
r
bax
xba
DCx
lbla
k
k
k
k
rrr
MM
rM
rM
rMM
r
rr
βα
β
α
++=
==
−=
=
−
−
−
1
1
11
1
)(
1
1
2
2,
1
1
2
2
br
ar
12
−= lbr
12la =
r
)(DC 1-k
r
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BarycentricBarycentric coordinate systemcoordinate system
� Define the basis vectors of the
Barycentric coordinate system
relative to the origin defined by x.
� The vectors and point in the
directions of the and edges of the rectangle.
1l 1−l
br
ar
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Algorithm AnalysisAlgorithm Analysis
3x
4x
5x
1x2
x
)()(
0)(
01
11
1
DCDC
xxfV
>⇒
>−=
05 =V
)( 0DC
)()(0 122 DCDCV >⇒<
xz =
)(xfz =
)(1 xf
)( 1DC
1=x
axis- x
1=z
=
0
0
z
x
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Fixed PointFixed Point
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3D intersection of Pyramid function3D intersection of Pyramid function
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Visualize intersectionVisualize intersection
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Algorithm Analysis: ConvergenceAlgorithm Analysis: Convergence
br
ar
12
−= lbr
12la =
r
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Algorithm Analysis: ConvergenceAlgorithm Analysis: Convergence
� Exponential decay of infinity norm residuals.
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ComplexityComplexity
.1)/1(log21)/2(log2 22 +=−≤ εεk
where,2)()( since
,2)()(
and ,2
2)(satisfy
0101
11
max
==
≤+
≤
−
−
DlDl
DlDl
DlD
kk
kk
ε
ε
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Numerical Tests: Numerical Tests: Pyramid basis functionPyramid basis function
� Tests Pyramid Function defined as
]1,0[ and for ]1,0[:
function basis Pyramid where
)),0,max(,1min()(
∈∈→
−−=∞
hDbDP
bxhxP
h
b
h
b
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Numerical Tests: Numerical Tests: Pyramid basis functionPyramid basis function
� Plots of for several values of b and hh
bP
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Numerical Tests: Numerical Tests: Pyramid basis functionPyramid basis function
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Numerical Tests: 3DPyramid TestsNumerical Tests: 3DPyramid Tests
� Tests 3-Dimensional Pyramid function
{ } .41,13,,1 of and subsetsempty non
of pairs allfor ))(),(()(
functions on the algorithm theTested
.131,131,,,
integersdistinct given the where
)),(,),(max()(
21
21
1
1
1,,1
−=
=
−
∀≤≤≤≤
=
eSS
xPxPxf
jikii
xPxPxP
ss
jk
h
b
h
biiki
ki
i
ik
εL
L
LL
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Numerical Tests: 4DPyramid TestsNumerical Tests: 4DPyramid Tests
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Complex Complex abs(Cabs(C))
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Complex angleComplex angle
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Numerical TestsNumerical Tests
� Tests algorithm on the functions
– Average ratio of a test’s function evaluations to
– Total number of tests satisfying the absolute error criterion: 21,776.
– Average ratio of a test’s function evaluations to
, for satisfying the absolute error criterion:0.522.
– Minimum number of function evaluations achieved by a test: 1.
.759.0:291)/1(log2 2 =+ε
291)/1(log2 2 =+ε
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Future workFuture work
� Plan algorithm works for any dimension
– Complexity in lower bound
� Investigate the restricted function class may have finite complexity in the absolute
criterion.
.2≥d
.in polynomial is )( where
)),/1log()((
ddc
dcO ε
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� Thank You
� Questions ?