我正在尝试最小化能量函数以找到Tammes问题的解决方案。我的代码通过使用梯度流方法工作，我可以改变我的变量t的值（能量函数的功率）来估计解决方案。
Tammes的问题是t趋于无穷大的结果，显然我希望t尽可能大。
问题在于我增加了我的解决方案（计算的能量）会失去准确性。
我认为这与pow（）函数以及它用如此大的整数计算的事实有关。
我的代码如下：
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <iomanip>
#include <fstream>
#include <time.h>
#include <stdlib.h>
#include <sstream>
using namespace std;
#define PI 3.14159265358979323846
int main()
{
int a,b,c,d,f,i,j,k,m,n,s,t,Success,Fails;
double p,theta,phi,Time,Averagetime,Energy,energy,Distance,Length,DotProdForce,
Forcemagnitude,ForceMagnitude[201],Force[201][4],E[1000001],En[501],Epsilon[4],Ep,
x[201][4],new_x[201][4],y[201][4],A[201],alpha[201][201],degree,bestalpha[501];
clock_t t1,t2;
t1=clock();
t=1;
while(t<1001){
n=2;
while(n<51){
cout << "N=" << n << "\n";
b=1;
Time=0.0;
while(b<101){
clock_t t3,t4;
t3=clock();
if(n>200){
cout << n << " is too many points for me :-( \n";
exit(0);
}
srand((unsigned)time(0));
for (i=1;i<=n;i++){
x[i][1]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
x[i][2]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
x[i][3]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
Length=sqrt(pow(x[i][1],2)+pow(x[i][2],2)+pow(x[i][3],2));
for (k=1;k<=3;k++){
x[i][k]=x[i][k]/Length;
}
}
Energy=0.0;
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
Energy=Energy+1.0/pow(Distance,t);
}
}
for(i=1;i<=n;i++){
y[i][1]=x[i][1];
y[i][2]=x[i][2];
y[i][3]=x[i][3];
}
m=100;
if (m>100){
cout << "The m="<< m << " loop is inefficient...lessen m \n";
exit(0);
}
a=1;
while(a<m){
for (i=1;i<=n;i++){
x[i][1]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
x[i][2]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
x[i][3]=((rand()*1.0)/(1.0*RAND_MAX)-0.5)*2.0;
Length=sqrt(pow(x[i][1],2)+pow(x[i][2],2)+pow(x[i][3],2));
for (k=1;k<=3;k++){
x[i][k]=x[i][k]/Length;
}
}
energy=0.0;
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
energy=energy+1.0/pow(Distance,t);
}
}
if(energy<Energy)
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
Energy=energy;
y[i][j]=x[i][j];
}
}
else
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
energy=Energy;
x[i][j]=y[i][j];
}
}
a=a+1;
}
/*
ostringstream String1;
String1 << "BestTamrandompoints_" << t << "_" << n; //add number
ofstream File1 (String1.str().c_str());
File1 << "Energy=" << Energy << "\n";
for(i=1;i<=n;i++){
File1 << x[i][1] << " " << x[i][2] << " " << x[i][3] << "\n";
}
File1.close();
*/
E[0]=Energy;
a=1;
s=0;
f=0;
Forcemagnitude=1.0;
if(n>80 && t<11)
p=0.005;
else if(n>60 && t<11)
p=0.01;
else if(n>40 && t<11)
p=0.02;
else if(n>25 && t<11)
p=0.05;
else if(n>40 && t>11)
p=0.002;
else if(n>30 && t>11)
p=0.005;
else if(n>20 && t>11)
p=0.01;
else if(n>5 && t>11)
p=0.025;
else
p=0.05;
while(Forcemagnitude>0.000001 && a<1000001){
for(i=1;i<=n;i++){
Force[i][1]=0.0;
Force[i][2]=0.0;
Force[i][3]=0.0;
}
for(i=1;i<=n;i++){
for(j=1;j<i;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
Force[i][1]=Force[i][1]+((x[i][1]-x[j][1])/(pow((Distance),3)));
Force[i][2]=Force[i][2]+((x[i][2]-x[j][2])/(pow((Distance),3)));
Force[i][3]=Force[i][3]+((x[i][3]-x[j][3])/(pow((Distance),3)));
}
for (j=i+1;j<=n;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
Force[i][1]=Force[i][1]+((x[i][1]-x[j][1])/(pow((Distance),3)));
Force[i][2]=Force[i][2]+((x[i][2]-x[j][2])/(pow((Distance),3)));
Force[i][3]=Force[i][3]+((x[i][3]-x[j][3])/(pow((Distance),3)));
}
}
for(i=1;i<=n;i++){
DotProdForce=Force[i][1]*x[i][1]+Force[i][2]*x[i][2]+Force[i][3]*x[i][3];
y[i][1]=x[i][1];
y[i][2]=x[i][2];
y[i][3]=x[i][3];
Force[i][1]=Force[i][1]-DotProdForce*y[i][1];
Force[i][2]=Force[i][2]-DotProdForce*y[i][2];
Force[i][3]=Force[i][3]-DotProdForce*y[i][3];
x[i][1] = y[i][1]+p*(Force[i][1]);
x[i][2] = y[i][2]+p*(Force[i][2]);
x[i][3] = y[i][3]+p*(Force[i][3]);
Length=sqrt(pow(x[i][1],2)+pow(x[i][2],2)+pow(x[i][3],2));
for (j=1;j<=3;j++){
x[i][j]=x[i][j]/Length;
}
}
energy=0.0;
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
energy=energy+1.0/pow(Distance,t);
}
}
E[a]=energy;
for(i=1;i<=n;i++){
ForceMagnitude[i]=pow((pow(Force[i][1],2)+pow(Force[i][2],2)
+pow(Force[i][3],2)),0.5);
}
cout << "E[" << a << "]=" << E[a] << "\n";
cout << fixed << setprecision(40) << "Energy=" << Energy << " energy=" << energy << "\n";
if (energy<Energy)
Energy=energy,s=s+1;
else
energy=Energy,f=f+1,p=(9.5*p)/10;
for(i=1;i<=n-1;i++){
if(ForceMagnitude[i]<ForceMagnitude[i+1])
ForceMagnitude[i]=ForceMagnitude[i+1];
else
ForceMagnitude[i+1]=ForceMagnitude[i];
}
Ep=sqrt(pow(E[a]-E[a-1],2));
if (Ep<0.00000000000000000000000000000000000000000000000000000000001){
ForceMagnitude[n]=0.000001;
cout << "BROKEN \n";
}
else
ForceMagnitude[n]=ForceMagnitude[n];
Forcemagnitude=ForceMagnitude[n];
cout << fixed << setprecision(60) << "FM=" << Forcemagnitude << "\n";
cout << "Energy=" << Energy << "\n";
a=a+1;
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
Distance=sqrt(pow(x[i][1]-x[j][1],2)+pow(x[i][2]-x[j][2],2)
+pow(x[i][3]-x[j][3],2));
degree=(180/PI);
alpha[i][j]=degree*acos((2.0-pow(Distance,2))/2.0);
}
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
cout << "alpha[" << i << "][" << j << "]=" << alpha[i][j] << "\n";
}
}
for(i=1;i<=n-1;i++){
for(j=i+1;j<=n-1;j++){
if(alpha[i][j]>alpha[i][j+1])
alpha[i][j]=alpha[i][j+1];
else
alpha[i][j+1]=alpha[i][j];
}
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
cout << "alpha[" << i << "][" << j << "]=" << alpha[i][j] << "\n";
}
}
for(i=1;i<=n-2;i++){
if(alpha[i][n]>alpha[i+1][n])
alpha[i][n]=alpha[i+1][n];
else
alpha[i+1][n]=alpha[i][n];
}
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
cout << "alpha[" << i << "][" << j << "]=" << alpha[i][j] << "\n";
}
}
bestalpha[b]=alpha[n-1][n];
En[b]=Energy;
b=b+1;
t4=clock();
float diff ((float)t4-(float)t3);
float seconds = diff / CLOCKS_PER_SEC;
Time = Time + seconds;
}
Averagetime = Time/(b-1);
cout << fixed << setprecision (4) << "Average Time: " << Averagetime << "(s) \n";
cout << fixed << setprecision(10) << "Energy=" << Energy << "\n";
cout << "s=" << s << " f=" << f << "\n";
Success=Success+s;
Fails=Fails+f;
cout << "Successes=" << Success << " Failures=" << Fails << "\n";
cout << setprecision(5) << "Success Rate=" << ((Success*1.0)/((Success*1.0)+(Fails*1.0)))*100.0 << "\n";
t2=clock();
float diff ((float)t2-(float)t1);
float seconds = diff / CLOCKS_PER_SEC;
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
y[i][j]=x[i][j];
}
}
a=1;
d=0;
c=0;
while(a<2){
theta=x[a][2]/x[a][3];
theta=d*PI+atan(theta);
for(i=1;i<=n;i++){
new_x[i][1]=x[i][1];
new_x[i][2]=x[i][2]*cos(theta)-x[i][3]*sin(theta);
new_x[i][3]=x[i][2]*sin(theta)+x[i][3]*cos(theta);
}
phi=new_x[a][1]/new_x[a][3];
phi=c*PI+atan(phi);
for(i=1;i<=n;i++){
x[i][1]=new_x[i][1]*cos(phi)-new_x[i][3]*sin(phi);
x[i][2]=new_x[i][2];
x[i][3]=new_x[i][1]*sin(phi)+new_x[i][3]*cos(phi);
}
Epsilon[1]=sqrt(pow((x[a][1]),2));
Epsilon[2]=sqrt(pow((x[a][2]),2));
Epsilon[3]=sqrt(pow((x[a][3]-1.0),2));
cout << "x[" << a << "][1]=" << x[a][1] << " x[" << a << "][2]=" << x[a][2] << " x[" << a << "][3]=" << x[a][3] << "\n";
if(Epsilon[1]<0.00001 && Epsilon[2]<0.00001 && Epsilon[3]<0.00001)
a=a+1;
else if(d==0 && c==0)
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
x[i][j]=y[i][j];
c=1;
}
}
else if(d==0 && c==1)
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
x[i][j]=y[i][j];
d=1;
c=0;
}
}
else if(d==1 && c==0)
for(i=1;i<=n;i++){
for(j=1;j<=3;j++){
x[i][j]=y[i][j];
c=1;
}
}
else if(d==1 && c==1)
break;
}
cout << "a=" << a << " d=" << d << " c=" << c << "\n";
ostringstream String2;
String2 << "TamPoints_" << t << "_" << n; //add number
ofstream File2 (String2.str().c_str());
for(i=1;i<=n;i++){
File2 << x[i][1] << " " << x[i][2] << " " << x[i][3] << "\n";
}
File2.close();
ostringstream String3;
String3 << "TamEnergies_" << t << "_" << n; //add number
ofstream File3 (String3.str().c_str());
for(i=1;i<a;i++){
File3 << fixed << setprecision (10) << E[i] << "\n";
}
File3.close();
ofstream File4 ("mypoints");
for(i=1;i<=n;i++){
File4 << x[i][1] << " " << x[i][2] << " " << x[i][3] << "\n";
}
File4.close();
ostringstream String4;
String4 << "TamInfo_" << t << "_" << n; //add number
ofstream File5 (String4.str().c_str());
File5 << "Iterations=" << a-1 << "\n";
File5 << "Successes=" << s << " Failures=" << f << "\n";
File5 << fixed << setprecision(20) << "Energy=" << Energy << "\n";
File5 << fixed << setprecision(5) << "Total run time: " << seconds << "(s) \n";
File5 << fixed << setprecision(5) << "Average run time: " << Averagetime << "(s) \n";
File5.close();
ostringstream String5;
String5 << "Tam%Energies_" << t << "_" << n; //add number
ofstream File6 (String5.str().c_str());
for(i=1;i<b;i++){
File6 << fixed << setprecision(20) << En[i] << "\n";
}
File6.close();
ostringstream String6;
String6 << "TamAngles_" << t << "_" << n << ".2"; //add number
ofstream File7 (String6.str().c_str());
for(i=1;i<b;i++){
File7 << fixed << setprecision(15) << bestalpha[i] << "\n";
}
File7.close();
cout << fixed << setprecision(5) << "Run time: " << seconds << "(s)" << "\n";
n=n+1;
}
if(t==1)
t=2;
else if(t==2)
t=5;
else if(t==5)
t=10;
else if(t==10)
t=25;
else if(t==25)
t=50;
else if(t==50)
t=100;
else if(t==100)
t=250;
else if(t==250)
t=500;
else if(t==500)
t=1000;
else
t=t+1;
}
return 0;
}
注意：我很欣赏我的代码非常粗糙，但是我的论文很快就会出现，因此我不能把时间花在漂亮上，我只需要它就可以了。我提前为缺乏功能而道歉，我对任何类型的编码都不熟悉，并且没有意识到这些都要被使用，直到我觉得花费我所有的时间来适应它会产生反作用。我确定我的问题是能量计算是：1 / pow（距离，t），因此t增加它们会失去准确性。代码适用于小t（至少高达100）但是对于t = 250，代码开始在准确性上失败。
有没有办法快速解决这个问题？我听说过bignum图书馆，但我对此一无所知，今晚应该阅读。感谢所有提前，A。
答案 0 :(得分：1)
你走在正确的轨道上 - 你需要任意精度整数。我过去曾使用the GMP library，但很久以前......祝你好运！
答案 1 :(得分：0)
如果你想任意大，你需要一个bignum库。
要根据您需要的大小进行快速修复，请尝试将您的数字声明为unsigned long long（如果它们始终为正数）或long long而不是{{1 （在此处了解详情：http://msdn.microsoft.com/en-us/library/s3f49ktz(v=vs.80).aspx）