人口增长模式与预期不符

Question

我试图制作一个细菌模型只是为了好玩，我正在使用 pow(a, b) 函数作为计算人口增长的函数。当细菌数量达到其环境可以提供的最大食物单位数量时，由于个体之间的竞争，它会下降 70%。我将结果保存到 txt 文件中，以便稍后绘制。我遇到的问题是人口正确振荡直到我达到 t = 900 左右的繁殖周期数，然后人口默认为 0。代码如下，希望大家不要介意用葡萄牙语写的变量名和函数名。

bool
check_aliemento (unsigned long int *pop)
{
    if (*pop >= MAX_ALIMENTO) return false;
    return true;
}

unsigned long int
replicaBacteria (unsigned long int *popInit, unsigned int tempo_t, double taxa)
{
    unsigned long int nextPop = round ((*popInit) * 
                static_cast<double> (pow (1 + taxa, tempo_t))); 
                //I'm almost sure that the problem happens in this pow() function
    while (! check_aliemento (&nextPop))
    {
        nextPop = (0.7 * nextPop);
    }
    return nextPop;
}



   int
main ( int argc, char** argv )
{
    unsigned long int a = 2;
    ofstream myfile;
    myfile.open ( "C:\Users\Pedro\Desktop\values.txt" );
    for ( unsigned int i; i < 1000; i ++ )
    {
        unsigned long int pop = replicaBacteria ( &a, i, 0.05 );
        myfile << pop << " ==> time = " << i;
        myfile << "\r\n";
    }
    myfile.close ( );
    return 0;
}

示例输出：

8080 ==> time = 872
8484 ==> time = 873
8909 ==> time = 874
9354 ==> time = 875
9822 ==> time = 876
7219 ==> time = 877
7580 ==> time = 878
7958 ==> time = 879
8357 ==> time = 880
8775 ==> time = 881
9214 ==> time = 882
9675 ==> time = 883
7110 ==> time = 884
7466 ==> time = 885
7839 ==> time = 886
8232 ==> time = 887
8643 ==> time = 888
9075 ==> time = 889
9529 ==> time = 890
7003 ==> time = 891
7354 ==> time = 892
7721 ==> time = 893
8108 ==> time = 894
8513 ==> time = 895
0 ==> time = 896
0 ==> time = 897
0 ==> time = 898
0 ==> time = 899
0 ==> time = 900

Answer 1

你溢出了 long int nextPop。这是因为你在计算人口增长时，就好像它在 896 代人中是不间断的，然后才减少人口，直到它低于你的魔法极限（你说这里是 10,000）。

解决这个问题的一种方法是在每次迭代后保留人口并将其增长 5%，然后在必要时将结果削减到 70%。

所以你想要 pop = replicaBacteria ( pop, 0.05 ); 之类的东西而不是 unsigned long int pop = replicaBacteria ( &a, i, 0.05 );（在循环之前声明变量），然后 replicaBacteria 应该简单地将其输入乘以 5% 并缩放到如有必要，70% 的结果。

编辑以下代码以生成可以为我编译和运行的代码

像这样：

#include <cmath>
#include <fstream>
#include <iostream>

#define MAX_ALIMENTO 10000

double
replicaBacteria (double popInit, double taxa)
{
    double nextPop = popInit * (1 + taxa); 
    while (nextPop >= MAX_ALIMENTO)
    {
        nextPop = (0.7 * nextPop);
    }
    return nextPop;
}



   int
main ( int argc, char** argv )
{
    std::ofstream myfile( "/tmp/out.txt" );
    double pop = 2;
    unsigned int i;
    for ( i=0; i < 1000; i ++ )
    {
        pop = replicaBacteria ( pop, 0.05 );
        myfile << round(pop) << " ==> time = " << i;
        myfile << "\r\n";
    }
    myfile.close ( );
    return 0;
}

输出：

2 ==> time = 0
2 ==> time = 1
2 ==> time = 2
2 ==> time = 3
3 ==> time = 4
3 ==> time = 5
3 ==> time = 6
3 ==> time = 7
3 ==> time = 8
3 ==> time = 9
3 ==> time = 10
4 ==> time = 11
...
8925 ==> time = 990
9371 ==> time = 991
9840 ==> time = 992
7232 ==> time = 993
7594 ==> time = 994
7974 ==> time = 995
8372 ==> time = 996
8791 ==> time = 997
9230 ==> time = 998
9692 ==> time = 999

Answer 2

pow (1 + taxa, tempo_t)

显然 long int 不能容纳 2^900 所以你在这里看到的是整数溢出使用可以容纳大值的数据类型

unsigned long long

Answer 3

即使是 unsigned long long int 也不能容纳。当 tempo_t > 895 时

unsigned long long int nextPop = round ((*popInit) * 
                    static_cast<double> (pow (1 + taxa, tempo_t)));

此表达式的计算结果为零，因此您必须减少细菌的寿命。

Answer 4

稍微改变一下函数以了解数字何时变为零：

unsigned long int
replicaBacteria (unsigned long int *popInit, unsigned int tempo_t, double taxa)
{
   double r = round ((*popInit) * static_cast<double> (pow (1 + taxa, tempo_t))); 
   std::cout << r << std::endl;
   unsigned long int nextPop = r;
   std::cout << nextPop << std::endl;

   //I'm almost sure that the problem happens in this pow() function
   while (! check_aliemento (&nextPop))
   {
      nextPop = (0.7 * nextPop);
   }
   return nextPop;
}

我看到以下输出：

1.84269e+19
18426916303946758144
8513 ==> time = 895
1.93483e+19
0
0 ==> time = 896
2.03157e+19
0
0 ==> time = 897

这是在 64 位 Linux 机器上使用 g++ 4.8.2。由于您看到从 896 开始为零，因此可以安全地假设您的编译器存在完全相同的问题。 unsigned long int能表示的最大数小于1.93483e+19.

人口增长模式与预期不符

Population growth model not corresponding to expectations

c++

c

modeling