[Algorithm] Detailed explanation of Gaussian elimination

Original link (strongly recommend Amway)

0.Prerequisite knowledge

• Know how to solve a system of linear equations in three variables
• Have hands, have brains

1.Representation and storage of answers

First solve a system of equations:

2x+3y+5z=31
x-4y -z=-6
4x+2y-5z=9

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We write this system of equations into a table format that can be read by machines :

2 3 5 31
1 -4 -1 -6
4 2 -5 9

The first column represents $The\; coefficient\; of\; x$ , the second column represents$The\; coefficient\; of\; y$ ...
Notethat the extra fourth column is the specific value of the answer.
The first row represents equation 1, the second column represents equation 2...

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Because we always use $x,y,z$ means$The\; number\; n$ is inconvenient, so we use$a_1,a_2...a_n$To represent.

That is, the meaning of the above table is:

第$i(1\le i\le Column\; n)$ represents ai a_iin each formula$a_i$Coefficient of n $n+Column\; 1$ represents the specific valueof each formula. (Because these are ordinary values ​​when stored, readersmust distinguish them clearly)

第$i(1\le i\le Line\; n)$ represents the$i$ formulas.

Then throughout the text, use $a\left[i\right]\left[j\right]$ Display No.iiLine$i$ , jjValue/coefficient of column$j$ .

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2. Gaussian elimination idea

The name is so high-end, but the method is very crude. The specific operation is: addition, subtraction and elimination + substitution and elimination .

First, follow the coefficient matrix above:

2 3 5 31
1 -4 -1 -6
4 2 -5 9

Then start Gaussian elimination.

enumerationii $i$ ,sequentially eliminate unknowns $a_i$ 。

Then enumerate $j$ , find aminimum absolute valuethat is not$0$ 的 $a[j][i]$ 。

Find the minimum $a\left[j\right]\left[i\right]$的$After\; j$ , add the$The\; j$ formula and the$i$ expression exchange

enum $j$ ，设 $mul=a[i][i]/a[j][i]$ ，
将 $j$ expression multiplied by$mul$ and minus$i$ formula

Let’s analyze why we do this layer by layer.
First, enumerate $i$ , need toeliminate the$i$ unknown numbers.
Then, select$j$ use$a\left[j\right]\left[i\right]$ is the smallest, which is actuallynot necessary(I will explain why later). exchange$j$ formula and$i$ formula.
Later, enumerate each$j$ , willjja [ i ] [ i ] / a [ j ] [ i ] a[i][i]/a[j][i$]$$a\left[i\right]\left[i\right]/a\left[j\right]\left[i\right]$ , because jj needs to be$j$ formulaiiThe coefficient of term$i$ is adjustedto$i$ shikiko$The\; coefficients\; of\; i$ terms are the same;
change$j$ minusiiThe purpose of formula$i$ is to eliminate the iiThe coefficient of item$i$ .

Why choose the smallest absolute value instead of $0$ 的 $a\left[j\right]\left[i\right]$的$Where\; is\; j$ ? Becausea [ j ] [ i ] a[j][i]smallestabsolute value$a\left[j\right]\left[i\right]$ can roughlyformmore divisible situations, and this can avoid.
Cannot take$0$ is of course.

If you still don’t understand? Then simulate it layer by layer.
Let’s take this system of equations as an example:

2 3 5 31
1 -4 -1 -6
4 2 -5 9

First, eliminate the first coefficient.

1 -4 -1 -6
2 3 5 31
4 2 -5 9

Find the minimum $For\; the\; formula\; a\left[1\right]$ , adjust to the first one;

1 -4 -1 -6
1 3/2 5/2 31/2
4 2 -5 9

The second persimmon $\times 0.5$ 。

1 -4 -1 -6
0 11/2 7/2 43/2
4 2 -5 9

Subtract the first from the second equation.

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By analogy, the coefficient matrix of the final answer is: (not aligned, a bit ugly)

4 0 0 20
0 -4.5 0 -9
0 0 7.611 22.833

Then the final $n+1$ coefficient divided by$i$ rowiiThe value in column$i\; is\; enough.$

3. The code you want to see.

P3389 [Template] Gaussian elimination method

The comments are a bit sparse, but just understandable.

#include<bits/stdc++.h>
#define RI register int
using namespace std;typedef long long LL;const int inf=1073741823;int FL,CH;template<typename T>void in(T&a){
    
    for(a=0,FL=1,CH=getchar();!isdigit(CH);CH=getchar())FL=(CH=='-')?-1:1;for(;isdigit(CH);CH=getchar())a=a*10+CH-'0';a*=FL;}template<typename T,typename ...Args>void in(T&a,Args&...args){
    
    in(a);in(args...);}
const int maxn=110;const double eps=1e-6;
double a[maxn][maxn];
int n;
int main()
{
    
    
	cin>>n;
	for(RI i=1;i<=n;++i){
    
    
		for(RI j=1;j<=n+1;++j)scanf("%lf",&a[i][j]);}
	//输入系数矩阵
	for(RI rp,i=1;i<=n;++i)
	{
    
    
		rp=i;
		for(RI j=i+1;j<=n;++j)
		if(fabs(a[j][i])>fabs(a[rp][i]))rp=j;
		if(fabs(a[rp][i])<=eps)
		return printf("No Solution"),0;
		//寻找绝对值最小的 j
		//这个特殊一点，要判无解
		if(i!=rp)swap(a[rp],a[i]);
		//交换 j 和 i
		double div=a[i][i];
		for(RI j=1;j<=n;++j)
		if(j!=i){
    
    
			div=a[j][i]/a[i][i];
			//计算 mul 值
			for(RI k=1;k<=n+1;++k)
				a[j][k]-=a[i][k]*div;
			//乘上，再减去
			//高斯消元算法的精髓
		}
	}
	for(RI i=1;i<=n;++i)
		printf("%.2lf\n",a[i][n+1]/a[i][i]);//回代，并输出答案
	return 0;
}