Approximate computations

Transcription

1 Living with floting-point numers Stndrd normlized representtion (sign + frction + exponent): Approximte computtions Rnges of vlues: Representtions for:, +, +0, 0, NN (not numer) Jordi Cortdell Deprtment of Computer Science Be creful when operting with rel numers: doule x, y; cin >> x >> y; // cout.precision(20); cout << x + y << endl; // Single nd doule-precision FP numers flot: doule: 1 sign e 1023 or 1 sign i 2 i 2 e 1023 i=1 Introduction to Progrmming Dept. CS, UPC 2 Compring floting-point numers Comprisons: = + c; if ( == c) Allow certin tolernce for equlity comprisons: if (expr1 == expr2) // Wrong! // my e flse if (s(expr1 expr2) < ) // Ok! Not every numer cn e represented. Exmple: 0.15 is stored s Introduction to Progrmming Dept. CS, UPC 3 Introduction to Progrmming Dept. CS, UPC 4

2 Root of continuous function Root of continuous function Bolzno s theorem: Let f e rel-vlued continuous function. Let nd e two vlues such tht < nd f() f() < 0. Then, there is vlue c [,] such tht f(c)=0. Design function tht finds root of continuous function f in the intervl [, ] ssuming the conditions of Bolzno s theorem re fulfilled. Given precision ( ), the function must return vlue c such tht the root of f is in the intervl [c, c+ ]. c c Introduction to Progrmming Dept. CS, UPC 5 Root of continuous function Strtegy: nrrow the intervl [, ] y hlf, checking whether the vlue of f in the middle of the intervl is positive or negtive. Iterte until the width of the intervl is smller. Introduction to Progrmming Dept. CS, UPC 6 Root of continuous function // Pre: f is continuous, < nd f()*f() < 0. // Returns c [,] such tht root exists in the // intervl [c,c+ ]. // Invrint: root of f exists in the intervl [,] Introduction to Progrmming Dept. CS, UPC 7 Introduction to Progrmming Dept. CS, UPC 8

3 Root of continuous function Root of continuous function doule root(doule, doule, doule epsilon) { while ( > epsilon) { doule c = ( + )/2; if (f() f(c) <= 0) = c; else = c; return ; // A recursive version doule root(doule, doule, doule epsilon) { if ( <= epsilon) return ; doule c = ( + )/2; if (f() f(c) <= 0) return root(,c,epsilon); else return root(c,,epsilon); Introduction to Progrmming Dept. CS, UPC 9 The Newton-Rphson method Introduction to Progrmming Dept. CS, UPC 10 The Newton-Rphson method A method for finding successively pproximtions to the roots of rel-vlued function. The function must e differentile. Introduction to Progrmming Dept. CS, UPC 11 Introduction to Progrmming Dept. CS, UPC 12

4 The Newton-Rphson method Squre root (using Newton-Rphson) Clculte Find the zero of the following function: where Recurrence: source: s_method Introduction to Progrmming Dept. CS, UPC 13 Squre root (using Newton-Rphson) // Pre: >= 0 // Returns x such tht x 2 - < doule squre_root(doule ) { doule x = 1.0; // Mkes n initil guess // Itertes using the Newton-Rphson recurrence while (s(x x ) >= epsilon) x = 0.5 (x + /x); return x; Introduction to Progrmming Dept. CS, UPC 15 Introduction to Progrmming Dept. CS, UPC 14 Squre root (using Newton-Rphson) Exmple: squre_root(1024.0) x Introduction to Progrmming Dept. CS, UPC 16

5 Approximting definite integrls Approximting definite integrls There re vrious methods to pproximte definite integrl: The pproximtion is etter if severl intervls re used: The trpezoidl method pproximtes the re with trpezoid: Introduction to Progrmming Dept. CS, UPC 17 Approximting definite integrls Introduction to Progrmming Dept. CS, UPC 18 Approximting definite integrls // Pre: >=, n > 0 // Returns n pproximtion of the definite integrl of f // etween nd using n intervls. doule integrl(doule, doule, int n) { doule h = ( )/n; h doule s = 0; for (int i = 1; i < n; ++i) s = s + f( + i h); return (f() + f() + 2 s) h/2; Introduction to Progrmming Dept. CS, UPC 19 Introduction to Progrmming Dept. CS, UPC 20

6 Monte Crlo methods Algorithms tht use repeted genertion of rndom numers to perform numericl computtions. The methods often rely on the existence of n lgorithm tht genertes rndom numers uniformly distriuted over n intervl. In C++ we cn use rnd(), tht genertes numers in the intervl [0, RAND_MAX) Introduction to Progrmming Dept. CS, UPC 21 Approximting Approximting Let us pick rndom point within the unit squre. Q: Wht is the proility for the point to e inside the circle? A: The proility is /4 Algorithm: Generte n rndom points in the unit squre Count the numer of points inside the circle (n in ) Approximte /4 n in /n Introduction to Progrmming Dept. CS, UPC 22 Approximting #include <cstdli> // Pre: n is the numer of generted points // Returns n pproximtion of using n rndom points doule pprox_pi(int n) { int nin = 0; doule rndmx = doule(rand_max); for (int i = 0; i < n; ++i) { doule x = rnd()/rndmx; doule y = rnd()/rndmx; if (x x + y y < 1.0) nin = nin + 1; return 4.0 nin/n; n , , , ,000, ,000, ,000, ,000,000, Introduction to Progrmming Dept. CS, UPC 23 Introduction to Progrmming Dept. CS, UPC 24

7 Generting rndom numers in n intervl Monte Crlo pplictions: exmples Assume rnd() genertes rndom nturl numer r in the intervl [0, R). Domin Intervl Rndom numer R [0,1) Τ r R R [0, ) rτr R [, ) + rτr Z [0, ) r mod Z [, ) + r mod ( ) Note: Be creful with integer divisions when delivering rel numers (enforce rel division). Determine the pproximte numer of lttice points in sphere of rdius r centered in the origin. Determine the volume of 3D region R defined s follows: A point P = (x, y, z) is in R if nd only if x 2 + y 2 + 2z nd 3x 2 + y 2 + z And mny other ppliction domins: Mthemtics, Computtionl Physics, Finnces, Simultion, Artificil Intelligence, Gmes, Computer Grphics, etc. Introduction to Progrmming Dept. CS, UPC 25 Exmple: intersection of two odies Determine the volume of 3D region R defined s follows: A point P = (x, y, z) is in R if nd only if x 2 + y 2 + 2z nd 3x 2 + y 2 + z The intersection of the two odies is inside rectngulr cuoid C, with center in the origin, such tht: x 2 50 y z 2 50 The volume of the cuoid is: Introduction to Progrmming Dept. CS, UPC 26 Exmple: intersection of two odies // Returns n estimtion of the volume of the // intersection of two odies with n rndom smples. doule volume_intersection(int n) { int nin = 0; doule s50 = sqrt(50)/rand_max; // scling for numers in [0,sqrt(50)) doule s10 = 10.0/RAND_MAX; // scling for numers in [0,10) // Generte n rndom smples for (int i = 0; i < n; ++i) { // Generte rndom point inside the cuoid doule x = s50 rnd(); doule y = s10 rnd(); doule z = s50 rnd(); // Check whether the point is inside the intersection if (x x + y y + 2 z z <= 100 nd 3 x x + y y + z z <= 150) ++nin; Volume C = = 4000 return nin/n; Introduction to Progrmming Dept. CS, UPC 27 Introduction to Progrmming Dept. CS, UPC 28

8 Exmple: intersection of two odies Volume Summry Approximte computtions is resort when no exct solutions cn e found numericlly. Intervls of tolernce re often used to define the level of ccurcy of the computtion. Rndom smpling methods cn e used to sttisticlly estimte the result of some complex prolems. numer of smples Introduction to Progrmming Dept. CS, UPC 29 Introduction to Progrmming Dept. CS, UPC 30