C# 运用多元一次方程解决问题
高斯消元运用
最近项目涉及到一个需求,需要把指定数量的多个商品,混合装入到多个不同型号的箱子中(每种型号的箱子装入商品的种类和个数是固定的)。这就涉及到解多元一次方程
- 针对多元一次方程一般用高斯消元去处理,当消元后仍有不能消掉的元 则需要解多元一次方程
1. 将数据转换为二维数组,消元代码如下
- 样本数据如下
***
| 2, 3, 4, 3, 21 | => | 2, 3, 4, 3, 21 | => | 2, 3, 4, 3 , 21 || 3, 1, 6, 2, 17 | => | 0, 7, 0, 5, 29 | => | 0, 7, 0, 5 , 29 |
| 1, 3, 2, 1, 12 | => | 0, -3,0, 1, -3 | => | 0, 0, 0, -22, -66 |
***
var rowLength = arr.GetLength(0);//行数
var colLength = arr.GetLength(1);//列数for (int mainCol = 0; mainCol < colLength – 1; mainCol++) //按照列,一列一列的消元
{
var mainRow = mainCol; //主元行=主元列for (int row = mainRow + 1; row < rowLength; row++) // 从主元的下一行开始 行循环
{
if (row >= (colLength – 1))
break;
/
n1X + n2Y = a
m1X + m2Y = b
=> n1m1X + n2m1Y = am1
=> m1n1X + m2n1Y = bn1
/
//如果主元参数为0( 找出此列不为0的行) 运用行相加 变换0参数
if (arr[mainRow, mainCol] == 0)
{
for (int rowi = mainRow + 1; rowi < rowLength; rowi++)
{
if (arr[rowi, mainCol] != 0)
{
for (int colj = mainCol; colj < colLength; colj++)
{
arr[mainRow, colj] += arr[rowi, colj];
}
break;
}
}
}if (arr[row, mainCol] == 0)
continue; //当前行 此列已经是0 继续消元下一行var m = GetMinCommonMultiple(arr[mainCol, mainCol], arr[row, mainCol]);
int factorMain = m / arr[mainCol, mainCol];//主列的因子
int factor = m / arr[row, mainCol]; //待消元列因子for (int colk = mainCol; colk < colLength; colk++)
{
arr[row, colk] = arr[mainCol, colk] factorMain – arr[row, colk] * factor;
}
}
}
2. 消元后,找出仍然无法消除的元,解一元多次不定方程
- 通过消元 最终可以确定 a[3]=3 ,a[1]=2 a[0]和a[2]解有多个 需要解不定方程思路是将 多元方程切割成二元方程然后穷举求解。 列入3x+2y+4z+6a=18, 令 w=2y+4z+6a,再令 2w=2(y+2z+3a) ,带入公式 得 3x+2w=16,穷举求出一个解 x=2 w=6 ,然后 再解 y+2z+3a=6。 同样的方法再次处理
- 代码如下
/// <summary>
/// 解不定方程
/// </summary>
public static bool ResoveIndefiniteEquation(int[] arr, int value, int startIndex, ref int[] result, ref int count)
{
count++;//递归计数,防止无限递归if (arr == null || arr.Length == 0)
return false;if (arr.Length == 1)
{
var reValue = value / arr[0];
if (value % arr[0] == 0 && reValue >= MinResove && reValue <= MaxResove)
{
result[startIndex] = reValue;
return true;
}
return false;
}//获取第二部分元的最大公约数
var commonDivisor = arr[1];
for (int i = 1; i < (arr.Length – 1); i++)
{
commonDivisor = GetCommonDivisor(commonDivisor, arr[i + 1]);
}for (int i = MinResove; i <= MaxResove; i++)
{
var currValue = value – i * arr[0];
if (currValue % commonDivisor == 0)
{
var newArr = new int[arr.Length – 1];
Array.Copy(arr, 1, newArr, 0, arr.Length – 1);result[startIndex] = i;
var currArr = ResoveIndefiniteEquation(newArr, currValue, startIndex + 1, ref result, ref count);
if (currArr)
return true;if (count > 1000000)
{
Console.WriteLine(“递归太深无法求解”);
return false;
}
}
}
return false;
}
3.完整代码
{
/// <summary>
/// 不定方程解的最小值
/// </summary>
private const int MinResove = 1;
/// <summary>
/// 不定方程解的最大值
/// </summary>
private const int MaxResove = 100;
/// <summary>
/// 利用高斯消元求解一元多次方程组
/// </summary>
/// <param name=“arr“></param>
public static int?[] ResoveGauss(int[,] arr)
{
if (arr == null) throw new ArgumentNullException(“arr”);
var rowLength = arr.GetLength(0);//行数
var colLength = arr.GetLength(1);//列数
if (colLength < 2) throw new IndexOutOfRangeException(“arr.GetLength(1)的值必须大于等于2”);
var result = new int?[colLength – 1];
#if DEBUG
Console.WriteLine(“解一元多次方程:”);
PrintArr(arr);
#endif
for (int mainCol = 0; mainCol < colLength – 1; mainCol++) //按照列,一列一列的消元
{
var mainRow = mainCol; //主元行=主元列
for (int row = mainRow + 1; row < rowLength; row++) // 从主元的下一行开始 行循环
{
if (row >= (colLength – 1))
break;
/*
n1*X + n2*Y = a
m1*X + m2*Y = b
=> n1*m1*X + n2*m1Y = a*m1
=> m1*n1*X + m2*n1*Y = b*n1
*/
//如果主元参数为0( 找出此列不为0的行) 运用行相加 变换0参数
if (arr[mainRow, mainCol] == 0)
{
for (int rowi = mainRow + 1; rowi < rowLength; rowi++)
{
if (arr[rowi, mainCol] != 0)
{
for (int colj = mainCol; colj < colLength; colj++)
{
arr[mainRow, colj] += arr[rowi, colj];
}
break;
}
}
}
if (arr[row, mainCol] == 0)
continue; //当前行 此列已经是0 继续消元下一行
var m = GetMinCommonMultiple(arr[mainCol, mainCol], arr[row, mainCol]);
int factorMain = m / arr[mainCol, mainCol];//主列的因子
int factor = m / arr[row, mainCol]; //待消元列因子
for (int colk = mainCol; colk < colLength; colk++)
{
arr[row, colk] = arr[mainCol, colk] * factorMain – arr[row, colk] * factor;
}
}
#if DEBUG
if (rowLength > mainCol + 1)
PrintArr(arr);
#endif
}
//回代过程
//回代行 判断 行数是否大于列数减-1 取到可以计算结果的回代行
var backRow = rowLength > (colLength – 1) ? colLength – 1 : rowLength;
var backRowIndex = backRow – 1;
//回代行的结果值
var backValue = arr[backRowIndex, colLength – 1];
//判断消解行 有多少个变元 ,存储 变元的 索引
var listColIndex = new List<int>();
for (int col = 0; col < colLength – 1; col++)
{
if (arr[backRowIndex, col] != 0)
{
listColIndex.Add(col);
}
}
if (listColIndex.Count == 0)
return null;
//根据需要求解的个数 解不定方程处理
if (listColIndex.Count > 1)
{
var resoveArr = listColIndex.Select(x => arr[backRowIndex, x]).ToArray();
var resoveResultArr = new int[listColIndex.Count];
var count = 0;
var flag = ResoveIndefiniteEquation(resoveArr, backValue, 0, ref resoveResultArr, ref count);
#if DEBUG
Console.WriteLine();
Console.WriteLine(“–解不定方程”);
var sb = new StringBuilder();
for (int i = 0; i < colLength; i++)
{
sb.Append(“********”);
}
Console.WriteLine(sb.ToString());
Console.WriteLine($”方程: {string.Join(“,\t”, resoveArr)} = {backValue}“);
Console.WriteLine($”结果: {string.Join(“,\t”, resoveResultArr)}“);
Console.WriteLine($”计算: {string.Join(“,\t”, resoveResultArr.Select((x, index) => $”{x * resoveArr[index]}“))}“);
Console.WriteLine($”递归次数: {count}“);
Console.WriteLine(sb.ToString());
Console.WriteLine();
#endif
if (!flag)
return null;
for (int i = 0; i < resoveResultArr.Length; i++)
{
result[listColIndex[i]] = resoveResultArr[i];
}
}
if (listColIndex.Count == 1)
{
if (backValue % arr[backRowIndex, listColIndex[0]] == 0)
{
result[listColIndex[0]] = backValue / arr[backRowIndex, listColIndex[0]];
}
else
{
#if DEBUG
var number = backValue * 1.0 / arr[backRowIndex, listColIndex[0]];
Console.WriteLine($”第{backRow}行 {backValue}/{arr[backRowIndex, listColIndex[0]]} = {number}, 不能被被整除计算失败”);
#endif
return null;
}
}
for (int row = backRowIndex – 1; row >= 0; row—) //从倒数第二行开始往前迭代
{
if (arr[row, row] == 0)
continue;
int addResult = 0;
var currlist = new List<int>();
for (int j = row; j < colLength – 2; j++)//j=2 j 最大值为2,每行未知数可能不止一个,故需要遍历已知的未知数并代入
{
if (!result[j + 1].HasValue && arr[row, j + 1] != 0)
currlist.Add(j + 1);//把没有计算出结果的列的索引存入
else
addResult += result[j + 1].GetValueOrDefault() * arr[row, j + 1];//k代表计算的行,j+1代表的列,系数与解要对应,故都为 j+1
}
var calculateValue = arr[row, colLength – 1] – addResult;
//发现没有计算出结果的列 解不定方程
if (currlist.Count > 0)
{
currlist.Add(row);
var resoveArr = currlist.Select(x => arr[row, x]).ToArray();
var resoveResultArr = new int[currlist.Count];
var count = 0;
var flag = ResoveIndefiniteEquation(resoveArr, calculateValue, 0, ref resoveResultArr, ref count);
#if DEBUG
Console.WriteLine();
Console.WriteLine($”—-发现第{row + 1}行 第({string.Join(“,”, currlist)})列 没有计算出结果,需要解不定方程”);
var sb = new StringBuilder();
for (int i = 0; i < colLength; i++)
{
sb.Append(“********”);
}
Console.WriteLine(sb.ToString());
Console.WriteLine($”方程: {string.Join(“,\t”, resoveArr)} = {calculateValue}“);
Console.WriteLine($”结果: {string.Join(“,\t”, resoveResultArr)}“);
Console.WriteLine($”计算: {string.Join(“,\t”, resoveResultArr.Select((x, index) => $”{x * resoveArr[index]}“))}“);
Console.WriteLine($”递归次数: {count}“);
Console.WriteLine(sb.ToString());
Console.WriteLine();
#endif
if (!flag)
return null;
for (int i = 0; i < resoveResultArr.Length; i++)
{
result[currlist[i]] = resoveResultArr[i];
}
}
else
{
if (calculateValue % arr[row, row] == 0)
{
result[row] = calculateValue / arr[row, row];//本行的未知数用本行最右边数-本行已知未知数代入系数之差 再除以本未知数系数
}
else
{
#if DEBUG
var number = calculateValue * 1.0 / arr[row, row];
Console.WriteLine($”第{row + 1}行 {calculateValue}/{arr[row, row]} = {number}, 不能被被整除计算失败”);
#endif
return null;
}
}
}
#if DEBUG
Console.WriteLine(“———-结果————“);
Console.Write($”结果: {string.Join(“,”, result)}“);
#endif
return result;
}
/// <summary>
/// 解不定方程
/// </summary>
public static bool ResoveIndefiniteEquation(int[] arr, int value, int startIndex, ref int[] result, ref int count)
{
count++;//递归计数,防止无限递归
if (arr == null || arr.Length == 0)
return false;
if (arr.Length == 1)
{
var reValue = value / arr[0];
if (value % arr[0] == 0 && reValue >= MinResove && reValue <= MaxResove)
{
result[startIndex] = reValue;
return true;
}
return false;
}
//获取第二部分元的最大公约数
var commonDivisor = arr[1];
for (int i = 1; i < (arr.Length – 1); i++)
{
commonDivisor = GetCommonDivisor(commonDivisor, arr[i + 1]);
}
for (int i = MinResove; i <= MaxResove; i++)
{
var currValue = value – i * arr[0];
if (currValue % commonDivisor == 0)
{
var newArr = new int[arr.Length – 1];
Array.Copy(arr, 1, newArr, 0, arr.Length – 1);
result[startIndex] = i;
var currArr = ResoveIndefiniteEquation(newArr, currValue, startIndex + 1, ref result, ref count);
if (currArr)
return true;
if (count > 1000000)
{
Console.WriteLine(“递归太深无法求解”);
return false;
}
}
}
return false;
}
private static void PrintArr(int[,] arr)
{
var rowLength = arr.GetLength(0);//行数
var colLength = arr.GetLength(1);//列数
var sb = new StringBuilder();
for (int i = 0; i < colLength; i++)
{
sb.Append(“——–“);
}
Console.WriteLine(sb.ToString());
for (int i = 0; i < rowLength; i++)
{
for (int j = 0; j < colLength; j++)
{
Console.Write(arr[i, j]);
var str = j == (colLength – 1) ? “” : “,”;
Console.Write($” {str}\t “);
}
Console.WriteLine();
}
}
/// <summary>
/// 求最大公约数
/// </summary>
/// <param name=“a“></param>
/// <param name=“b“></param>
/// <returns></returns>
public static int GetCommonDivisor(int a, int b)
{
if (a == 0 || b == 0) return 0;
if (Math.Abs(a) < Math.Abs(b))
{
var temp = a;
a = b;
b = temp;
}
return (a % b == 0) ? b : GetCommonDivisor(a % b, b);
}
/// <summary>
/// 求最小公倍数
/// </summary>
/// <param name=“a“></param>
/// <param name=“b“></param>
/// <returns></returns>
public static int GetMinCommonMultiple(int a, int b)
{
var commonDivisor = GetCommonDivisor(a, b);
if (commonDivisor == 0)
return 0;
return a * b / commonDivisor;
}
}
4. 运行结果
{
var arr = new int[,]
{
{ 2, 3, 4, 3, 21 },
{ 3, 1, 6, 2, 17 },
{ 1, 3, 2, 1, 12},
};
GaussHelper.ResoveGauss(arr);
Console.ReadKey();
}