修正单纯形法·优化算法实现·Java

修正单纯性法

代码例如以下：

舍去了输入转化的内容，主要包括算法关键步骤。

public class LPSimplexM {

	private static final double inf = 1e9;

	private int n;							// 约束个数
	private double[][] A;					// 输入函数參数
	private double[] b;					// 约束值
	private double[] c;					// 目标函数系数
	private double Z;						// 目标值
	private void InitF() {					// 初始化 First
		/* problem 1.
		 * max Z = 5*x1 + 4*x2;			  min Z = -5*x1 - 4*x2;
		 * x1   + 3*x2 <= 90;				x1 + 3*x2 +  x3				= 90;
		 * 2*x1 + x2   <= 80;		=>	  2*x1 +   x2 	    + x4		= 80;
		 * x1   + x2   <= 45;				x1 +   x2   		  + x5 	= 45;
		 * xi >= 0
		 */
		/* problem 2.
		 * min Z = -7*x1 - 12*x2;
		 * 9*x1 + 4*x2  + x3          = 360;
		 * 4*x1 + 5*x2       +x4 	  = 200;
		 * 3*x1 + 10*x2			 + x5 = 300;
		 *   xi >= 0
		 */
		n = 3;
		A = new double[n+1][n+1];
		b = new double[n+1];
//		c = new double[n<<1];
		Z = inf;
		// 约束初始条件
		A[1][1] = 1;	A[1][2] = 3;
		A[2][1] = 2;	A[2][2] = 1;
		A[3][1] = 1;	A[3][2] = 1;
		// 条件值
		// problem 1.
//		b[1] = 90;
//		b[2] = 80;
//		b[3] = 45;
		// problem 2.
		b[1] = 360;
		b[2] = 200;
		b[3] = 300;
//		for(int i = 1; i <= n; i++)System.out.println("b[" + i + "] = " + b[i]);
		// 目标函数系数
//		c[1] = -5;	c[2] = -4;
	}

	int m;
	private double[][] p;
	private double[][] e, oe;
	private double[][] E, oE;
	private double[] X;
	private boolean[] G;
	private int[] S;
	private void InitS() {
		m = 2;
		p = new double[n+1][n+m+1];
		e = new double[n+1][n+1];
		oe = new double[n+1][n+1];
		E = new double[n+1][n+1];
		oE = new double[n+1][n+1];
		X = new double[n+1];
		G = new boolean[n+m+1];
		S = new int[n+1];

		c = new double[n+m+1];
		// problem 1.
//		c[1] = -5;	c[2] = -4;	c[3] = 0;	c[4] = 0;	c[5] = 0;
		// problem 2.
		c[1] = -7;	c[2] = -12;	c[3] = 0;	c[4] = 0;	c[5] = 0;
		// problem 1.
//		p[1][1] = 1;	p[1][2] = 3;	p[1][3] = 1;	p[1][4] = 0;	p[1][5] = 0;
//		p[2][1] = 2;	p[2][2] = 1;	p[2][3] = 0;	p[2][4] = 1;	p[2][5] = 0;
//		p[3][1] = 1;	p[3][2] = 1;	p[3][3] = 0;	p[3][4] = 0;	p[3][5] = 1;
		// problem 2.
		p[1][1] = 9;	p[1][2] = 4;	p[1][3] = 1;	p[1][4] = 0;	p[1][5] = 0;
		p[2][1] = 4;	p[2][2] = 5;	p[2][3] = 0;	p[2][4] = 1;	p[2][5] = 0;
		p[3][1] = 3;	p[3][2] = 10;	p[3][3] = 0;	p[3][4] = 0;	p[3][5] = 1;

		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				if(i == j)E[i][j] = oE[i][j] = 1;
				else E[i][j] = oE[i][j] = 0;

		for(int i = 1; i <= n; i++)
			X[i] = b[i];

		G[1] = false;	G[2] = false;	G[3] = true;	G[4] = true;	G[5] = true;

		S[1] = 3;	S[2] = 4;	S[3] = 5;
	}

	public LPSimplexM() {
		InitF();
		InitS();
		AlgorithmProcess();
		solve();
	}

	private void AlgorithmProcess() {
		double[] coE = new double[n+1];	// c * E^-1
		double[] r = new double [n+m+1];	// c - c * E^-1 * p
		double[] oEp = new double[n+1];	// E^-1 * p;
		boolean flag = false;
		while(true) {
			// x = E^-1 * b
			for(int i = 1; i <= n; i++){
				X[i] = 0;
				for(int j = 1; j <= n; j++)
					X[i] += oE[i][j]*b[j];
			}
			// c * E^-1
			for(int i = 1; i <= n; i++){
				coE[i] = 0;
				for(int j = 1; j <= n; j++)
					coE[i] += c[S[j]]*oE[j][i];
			}
			// r = c - c * E^-1 * p     => min r' id -> k
			int k = -1;
			flag = false;
			for(int i = 1; i <= n+m; i++)if(!G[i]){
				double ans = 0;
				for(int j = 1; j <= n; j++)
					ans += coE[j]*p[j][i];
				r[i] = c[i] - ans;
				if(!flag && r[i] < 0){
					k = i;
					flag = true;
				}else if(flag && r[i] < r[k]){
					k = i;
				}
			}
			if(k == -1)return ;									// solution output 1（X, S 为最优解）
			// E^-1 * p;				=> min theta>0' id -> s
			int s = -1;
			flag = false;
			for(int i = 1; i <= n; i++){
				oEp[i] = 0;
				for(int j = 1; j <= n; j++){
					oEp[i] += oE[i][j]*p[j][k];
				}
				if(oEp[i] > 0){
					if(!flag){
						s = i;
						flag = true;
					}else if(flag && X[i]/oEp[i] < X[s]/oEp[s]){
						s = i;
					}
				}
			}
			if(!flag)return ; 										// no solution 1（无同意解）
			if(s == -1)return ;									// no solution 2（该问题有无解集）
			// p[s] = p[k], 形成新的矢量基 E
			G[S[s]] = false;	G[k] = true;
			S[s] = k;
//			System.out.println("k = " + k + "; s = " + s);
			for(int i = 1; i <= n; i++){
				p[i][k] = -1.0*oEp[i]/oEp[s];
				if(i == s)
					p[i][k] = 1/oEp[s];
			}
			for(int i = 1; i <= n; i++){
				int id = S[i];
				for(int j = 1; j <= n; j++){
					if(i == s){
						e[j][i] = p[j][k];
					}else{
						if(j == i){
							e[j][i] = 1;
						}else{
							e[j][i] = 0;
						}
					}
				}
			}
//			for(int i = 1; i <= n; i++){
//				for(int j = 1; j <= n; j++){
//					System.out.print(oE[i][j] + " ");
//				}System.out.println();
//			}System.out.println("{oE}");
//			for(int i = 1; i <= n; i++){
//				for(int j = 1; j <= n; j++){
//					System.out.print(e[i][j] + " ");
//				}System.out.println();
//			}System.out.println("{e}");
			for(int i = 1; i <= n; i++){
				for(int j = 1; j <= n; j++){
					oe[i][j] = 0;
					for(int t = 1; t <= n; t++){
						oe[i][j] += e[i][t]*oE[t][j];
					}
				}
			}
			for(int i = 1; i <= n; i++){
				for(int j = 1; j <= n; j++){
					oE[i][j] = oe[i][j];
					//System.out.print(oE[i][j] + " ");
				}//System.out.println();
			}//System.out.println();
		}
	}
	// 最优解输出
	private void solve() {
		Z = 0;
		for(int i = 1; i <= n; i++){
			int id = S[i];
			Z += c[id]*X[i];
			System.out.println(id + " : " + X[i] + " * " + -c[id]);
		}
		System.out.println("Z = " + -Z);
	}

	public static void main(String[] args) {
		new LPSimplexM();
	}
}

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