A Two-Dimensional Bisection Envelope Algorithm for Fixed Points
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1 A Two-imensional Bisection Envelope Algoithm fo Fied Points Kis Siosi and Spence Shellman Fom pulished Jounal of Compleity 8, (00
2 Intoduction How we solve fo two-dimensional f ( domain: [0, ]X[0, ] f : Lipschitz continuous function (q =. Pevious method Time compleity was ad Pape intoduce new algoithm Computes appoimate ~ satisfying Toleance ε < 0.5 Uppe ound on the function evaluations log (/ ε +. = f ( ~ ~ ε
3 Histoy 90s - Pesent Banach s simple iteation algoithm Homotopy continuation Simplicial and Newton type methods Time compleity Lipschitz function (q> f ( f ( y q y, q > Eponential in the wost case Lowe ound is also eponential (est case 3
4 4 Polem Fomulation Polem Fomulation Class of Lipschitz continuous functions By the Bouwe fied point theoem { } = y y f f y f a a a F a ( (, :,,,,. ( such that, into maps * *, *,,, f F f a a a a =
5 Polem Fomulation We now a solution eists, we just need a constuctive algoithm Two diffeent citeia to satisfy Residual citeion Can always e satisfied f ( ~ ~ ε. Asolute citeion Can sometimes e satisfied ~ α ε. 5
6 Polem Fomulation To find the fied point using the Bisection Envelope Algoithm, we ae equied n function evaluations of f, whee n log +. ε 6
7 7 Envelope Theoem Envelope Theoem efine the fied point sets fo F a f, ( ( }, ( { }, ( {,, f f f f a a F F F F F = = = = =
8 Theoem 3.5 Let R a, such that ( R and oth F ( f and F ( f intesect R. Then ~ y = c( R satifies esidual ceteian. If, in addition, R contains a fied point of f, then ~ y satifies asolute ceteian. l ( R + l ε 8
9 Theoem 3.5 F ( f l ( R R satifies l ( R l (R + l (R ε c(r ε so c(r satifies f(- ~ ~ ε. F ( f ε 9
10 The BEFi Algoithm: efinition.. efine the domains = 0.5,.5 and 0 = { R : (0.5,0.5 }... If f :, f is Lipschitz continuous ( q = on - At least one fied point eists within. - contains all fied points. 0
11 The BEFi Algoithm: Figue, et
12 The BEFi Algoithm: Pojection Pojection.. Let P poject onto, whee. P ( = (ma(0, min(,,ma(0,min(,.. If ~ y is a esidual solution fo f, then ~ = P( ~ yis a esidual solution fo whee ~ y and f ( ~ y ~ y. 0 ε f,
13 The BEFi Algoithm: esciption Algoithm taes f ( ~ ~ ε. ~ = P( ~ y as a solution to Algoithm etuns logical vaiale tue only if ~ y satifies asolute citeion ~ y α ε. which is as 3
14 The BEFi Algoithm: Constuction Constucts a algoithm - Evaluates f at on step - Constucts a ectangle isecting along lines with - Each y is a closed ectangle (slope = o - - Algoithm teminate s at step when : - If a esidual citeion is satified at - o if l ( + l ( ε slope o - though. 4
15 5 Baycentic Baycentic Coodinate System Coodinate System Find the net centoid y using Baycentic coodinate system at. ( C a a C l l a M M M M M M β α β α + + = = = = = (, a = l l a = ( C -
16 Baycentic coodinate system efine the asis vectos of the Baycentic coodinate system elative to the oigin defined y. The vectos and point in the a diections of the l and l edges of the ectangle. 6
17 V 5 = 0 z = z Algoithm Analysis = f ( V f ( = f ( C( > C( > 0 0 V < 0 C( > C( z = C( 0 C( - ais 0 = z = 7
18 Fied Point 8
19 3 intesection of Pyamid function 9
20 Visualize intesection 0
21 Algoithm Analysis: Convegence a = l a = l
22 Algoithm Analysis: Convegence Eponential decay of infinity nom esiduals.
23 Compleity satisfy l ma ( ε, and l ( + l ( ε, since l ( 0 = l ( 0 =, whee log ( / ε = log (/ ε +. 3
24 Numeical Tests: Pyamid asis function Tests Pyamid Function defined as P h ( = min(,ma( h,0, whee Pyamid asis function P h : [0,] fo and h [0,] 4
25 Numeical Tests: Pyamid asis function h P Plots of fo seveal values of and h 5
26 Numeical Tests: Pyamid asis function 6
27 Numeical Tests: 3Pyamid Tests Tests 3-imensional Pyamid function P whee given the distinct integes i i i, L,, L, i (, = ma( P 3, Tested the algoithm on the functions f ( = ( P s h i i (, P s non empty susets (, L, P ( fo all pais of S i j and h (, 3 j. S i i of {, L,3}, ε = e 4. 7
28 Numeical Tests: 4Pyamid Tests 8
29 Comple as(c 9
30 Comple angle 30
31 Numeical Tests Tests algoithm on the functions Aveage atio of a test s function evaluations to log (/ ε + = 9 : Total nume of tests satisfying the asolute eo citeion:,776. Aveage atio of a test s function evaluations to log (/ ε + = 9, fo satisfying the asolute eo citeion:0.5. Minimum nume of function evaluations achieved y a test:. 3
32 Futue wo Plan algoithm wos fo any dimension d. Compleity in lowe ound O( c( dlog(/ ε, whee c( d is polynomialin d. Investigate the esticted function class may have finite compleity in the asolute citeion. 3
33 Than You Questions? 33
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