如果我在循环中执行矩阵中的乘法错误 [英] Multiplication error in matrices if I do it in a loop
问题描述
这是我的功能
This is my function
void clear_up(MatrixXd A, int dim) //to set all initial values of the matrix to 0
{
for(int i=0;i<dim;i++)
{ for(int j=0;j<dim;j++)
{ A(i,j)=0; } }
}
int sumrow(MatrixXd A,int row, int dim) //adding up elements of a particular row
{
int sum=0;
for(int i=0;i<dim;i++)
{
sum=sum+A(row,i);
}
return sum;
}
MatrixXd generate_B(MatrixXd A, int dim) //A is a symmetric matrix with diagonal elements 0 and other elements either 0 or 1
{
cout<<"A \n"<<A<<"\n";
MatrixXd B(dim,dim), inter1(dim,dim), inter2(dim,dim), D(dim,dim);
clear_up(B,dim); clear_up(inter1,dim); clear_up(inter2,dim); clear_up(D,dim);
for(int i=0;i<dim;i++)
{
for(int j=0;j<dim;j++)
{
if(i==j)
{
D(i,j)= 1/sqrt(sumrow(A,i,dim)); //D is thus a diagonal matrix
}
else { D(i,j) = 0; }
}
}
cout<<"D \n"<<D<<"\n";
for(int a=0;a<dim;a++)
{
for(int b=0;b<dim;b++)
{
for(int c=0;c<dim;c++)
{
inter1(a,b)+= A(a,c)*D(c,b); //inter1 = A*D
}
}
}
cout<<"inter1 \n"<<inter1<<"\n";
for(int x=0;x<dim;x++)
{
for(int y=0;y<dim;y++)
{
for(int z=0;z<dim;z++)
{
inter2(x,y)+= D(x,z)*inter1(z,y); //inter2 = D*A*D
}
}
}
cout<<"inter2 \n"<<inter2<<"\n";
for(int X=0; X<dim; X++)
{ for(int Y=0; Y<dim; Y++)
{ if(X==Y) { B(X,Y) = 1 - inter2(X,Y); }
else { B(X,Y) = 0 - inter2(X,Y); }
} //B = I - D*A*D (I = identity matrix)
}
cout<<"B \n"<<B<<"\n";
clear_up(inter1,dim); clear_up(inter2,dim); clear_up(D,dim);
return B;
}
A是对称矩阵，对角元素为0，非对角元素为0或1.数学上，B应该是具有对角元素1的对称矩阵，它是1次迭代。但是，当我在循环中运行它时，一些矩阵是对称的，而有些矩阵则不对称。可能是什么原因？
我的尝试：
尝试将其拆分为不同的功能。还尝试清除垃圾值。
A is a symmetric matrix with diagonal elements as 0 and non diagonal elements as 0 or 1. Mathematically, B should be a symmetric matrix with diagonal elements 1, which it is for 1 iteration. However, when I run it in a loop, some matrices are symmetric whereas some are not. What could be the reason?
What I have tried:
Tried splitting it into different functions. Also tried clearing up junk values.
推荐答案
编译并不意味着你的代码是对的！ ：笑：
将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。
所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。
首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：
Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.
So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.
Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input Expected output Actual output
1 2 1
2 4 4
3 6 9
4 8 16然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。
所以，你可以查看代码和很明显，它在某处：
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
private int Double(int value)
{
return value * value;
}
一旦你知道可能出现的问题，就开始使用调试器找出原因。在你的行上设一个断点：
Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on your line:
cout<<"A \n"<<A<<"\n";
并运行你的应用程序。在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。
如果没有，为什么不呢？它有何不同？
这是一项非常值得开发的技能，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！
是的，我可能会告诉你问题是什么 - 但这并不难做到，并且你将同时学到一些非常值得的东西！
and run your app. Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
Yes, I could probably tell you what "the problem" is - but it's not difficult to do this yourself, and you will learn something really worthwhile at the same time!
但是当我在一个循环中运行它时，一些矩阵是对称的，而另一些则不是。可能是什么原因？
However, when I run it in a loop, some matrices are symmetric whereas some are not. What could be the reason?
严重吗？你甚至不敢举一个问题的例子。
最小值是一个调用序列重现问题的例子。
有一个工具可以让你看到你的代码在做什么，它的名字是调试器。掌握调试器不是可选的，它对任何程序员都是强制性的，也不例外。
它也是一个很好的学习工具，因为它向你展示了现实，你可以看到哪个期望与现实相符。
当你不明白你的代码在做什么或为什么它做它的作用时，答案是调试器。
使用调试器查看你的代码是什么代码正在做。只需设置断点并查看代码执行情况，调试器允许您逐行执行第1行并在执行时检查变量，这是一个令人难以置信的学习工具。
调试器 - 维基百科，免费的百科全书 [ ^ ]
掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]
使用Visual Studio 2010进行基本调试 - YouTube [ ^ ]
调试器在这里向您展示您的代码正在做什么以及您的任务是与它应该做的比较。
调试器中没有魔法，它没有发现错误，它只是帮助你。当代码没有达到预期的效果时，你就会接近一个错误。
Serious ? you did not even dared to give an example of the problem.
The minimum is an example with calling sequence to reproduce the problem.
There is a tool that allow you to see what your code is doing, its name is debugger. Mastering the debugger is not optional, it is mandatory for any programmer, no exception.
It is also a great learning tool because it show you reality and you can see which expectation match reality.
When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.
Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.
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