Finding the extremum of convex function by three-part method

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In machine learning, it is a common problem to find the extremum of convex function, such as gradient descent method, Newton method, and so on, today we introduce a three-way method to find the extremum problem of a convex function.

For a convex function such as $f (x), x\in [left,right]$, where LM and RM are divided by interval [left,right], we find that if f (LM) <f (RM), then the horizontal axis x of the point with the lowest function value must be in [LEFT,RM] Between. If x is between [Rm,right], it appears that there is a lower point around RM, which is obviously impossible. Similarly, when F (LM) >f (RM), the maximum horizontal x must be within the range of [Lm,right].

By using this property, we can approximate the target point at the same time as we reduce the interval, and get the extremum.

As an example, the title derives from HTTP://HIHOCODER.COM/CONTEST/HIHO40/PROBLEM/1, such as a parabolic y=ax^2+bx+c and a point P (x, y) in a Cartesian coordinate system, and a point p to the shortest distance d of the parabola, Which -200≤a,b,c,x,y≤200. Our other pivot represents the parabolic symmetry pumping, which can be found when x>pivot, we can take left = Pivot,right = inf, and vice versa =-inf, right = pivot, whose distance exactly satisfies the convex function. the shortest distance we ask for D is exactly the extremum of this convex function.

The code is as follows:

#include <stdlib.h>#include<stdio.h>#include<string.h>#include<limits.h>#include<iostream>#include<cmath>using namespacestd;DoubleA, B, C, x, y;Const DoubleMAX =100000;DoubleDisDoubleX) {    DoubleY = a*x*x+b*x+C; returnsqrt ((x-x) * (x-x) + (y-y) * (yY));}DoubleSolveDoubleLDoubleR) {    DoubleLM = L + (r-l)/3; DoubleRM = R-(r-l)/3; DoubleLMD =dis (LM); DoubleRMD =Dis (RM); if(Fabs (LMD-RMD) <0.0001 )    {        returnLMD; }    if(LMD >RMD) {        returnSolve (LM, R); }    Else    {        returnsolve (l, RM); }}intMainintargcChar*argv[]) {     while(cin>>a>>b>>c>>x>>y) {DoublePivot =-b/(2*a); DoubleL =0, r =0; if(Pivot <x) {L= pivot +0.0001; R=MAX; }        Else{L= -MAX; R= Pivot-0.0001; }        Doubleres =Solve (L, R); printf ("%.3lf\n", RES); }}

Finding the extremum of convex function by three-part method