package algs91; // section 6.5
import stdlib.*;
/* ***********************************************************************
 *  Compilation:  javac Simplex.java
 *  Execution:    java Simplex
 *
 *  Given an M-by-N matrix A, an M-length vector b, and an
 *  N-length vector c, solve the  LP { max cx : Ax <= b, x >= 0 }.
 *  Assumes that b >= 0 so that x = 0 is a basic feasible solution.
 *
 *  Creates an (M+1)-by-(N+M+1) simplex tableaux with the
 *  RHS in column M+N, the objective function in row M, and
 *  slack variables in columns M through M+N-1.
 *
 *************************************************************************/

public class Simplex {
	private static final double EPSILON = 1.0E-10;
	private final double[][] a;   // tableaux
	private final int M;          // number of constraints
	private final int N;          // number of original variables

	private final int[] basis;    // basis[i] = basic variable corresponding to row i
	// only needed to print out solution, not book

	// sets up the simplex tableaux
	public Simplex(double[][] A, double[] b, double[] c) {
		M = b.length;
		N = c.length;
		a = new double[M+1][N+M+1];
		for (int i = 0; i < M; i++)
			for (int j = 0; j < N; j++)
				a[i][j] = A[i][j];
		for (int i = 0; i < M; i++) a[i][N+i] = 1.0;
		for (int j = 0; j < N; j++) a[M][j]   = c[j];
		for (int i = 0; i < M; i++) a[i][M+N] = b[i];

		basis = new int[M];
		for (int i = 0; i < M; i++) basis[i] = N + i;

		solve();

		// check optimality conditions
		assert check(A, b, c);
	}

	// run simplex algorithm starting from initial BFS
	private void solve() {
		while (true) {

			// find entering column q
			int q = bland();
			if (q == -1) break;  // optimal

			// find leaving row p
			int p = minRatioRule(q);
			if (p == -1) throw new Error("Linear program is unbounded");

			// pivot
			pivot(p, q);

			// update basis
			basis[p] = q;
		}
	}

	// lowest index of a non-basic column with a positive cost
	private int bland() {
		for (int j = 0; j < M + N; j++)
			if (a[M][j] > 0) return j;
		return -1;  // optimal
	}

	// index of a non-basic column with most positive cost
	private int dantzig() {
		int q = 0;
		for (int j = 1; j < M + N; j++)
			if (a[M][j] > a[M][q]) q = j;

		if (a[M][q] <= 0) return -1;  // optimal
		else return q;
	}

	// find row p using min ratio rule (-1 if no such row)
	private int minRatioRule(int q) {
		int p = -1;
		for (int i = 0; i < M; i++) {
			if (a[i][q] <= 0) continue;
			else if (p == -1) p = i;
			else if ((a[i][M+N] / a[i][q]) < (a[p][M+N] / a[p][q])) p = i;
		}
		return p;
	}

	// pivot on entry (p, q) using Gauss-Jordan elimination
	private void pivot(int p, int q) {

		// everything but row p and column q
		for (int i = 0; i <= M; i++)
			for (int j = 0; j <= M + N; j++)
				if (i != p && j != q) a[i][j] -= a[p][j] * a[i][q] / a[p][q];

		// zero out column q
		for (int i = 0; i <= M; i++)
			if (i != p) a[i][q] = 0.0;

		// scale row p
		for (int j = 0; j <= M + N; j++)
			if (j != q) a[p][j] /= a[p][q];
		a[p][q] = 1.0;
	}

	// return optimal objective value
	public double value() {
		return -a[M][M+N];
	}

	// return primal solution vector
	public double[] primal() {
		double[] x = new double[N];
		for (int i = 0; i < M; i++)
			if (basis[i] < N) x[basis[i]] = a[i][M+N];
		return x;
	}

	// return dual solution vector
	public double[] dual() {
		double[] y = new double[M];
		for (int i = 0; i < M; i++)
			y[i] = -a[M][N+i];
		return y;
	}


	// is the solution primal feasible?
	private boolean isPrimalFeasible(double[][] A, double[] b) {
		double[] x = primal();

		// check that x >= 0
		for (int j = 0; j < x.length; j++) {
			if (x[j] < 0.0) {
				StdOut.println("x[" + j + "] = " + x[j] + " is negative");
				return false;
			}
		}

		// check that Ax <= b
		for (int i = 0; i < M; i++) {
			double sum = 0.0;
			for (int j = 0; j < N; j++) {
				sum += A[i][j] * x[j];
			}
			if (sum > b[i] + EPSILON) {
				StdOut.println("not primal feasible");
				StdOut.println("b[" + i + "] = " + b[i] + ", sum = " + sum);
				return false;
			}
		}
		return true;
	}

	// is the solution dual feasible?
	private boolean isDualFeasible(double[][] A, double[] c) {
		double[] y = dual();

		// check that y >= 0
		for (int i = 0; i < y.length; i++) {
			if (y[i] < 0.0) {
				StdOut.println("y[" + i + "] = " + y[i] + " is negative");
				return false;
			}
		}

		// check that yA >= c
		for (int j = 0; j < N; j++) {
			double sum = 0.0;
			for (int i = 0; i < M; i++) {
				sum += A[i][j] * y[i];
			}
			if (sum < c[j] - EPSILON) {
				StdOut.println("not dual feasible");
				StdOut.println("c[" + j + "] = " + c[j] + ", sum = " + sum);
				return false;
			}
		}
		return true;
	}

	// check that optimal value = cx = yb
	private boolean isOptimal(double[] b, double[] c) {
		double[] x = primal();
		double[] y = dual();
		double value = value();

		// check that value = cx = yb
		double value1 = 0.0;
		for (int j = 0; j < x.length; j++)
			value1 += c[j] * x[j];
		double value2 = 0.0;
		for (int i = 0; i < y.length; i++)
			value2 += y[i] * b[i];
		if (Math.abs(value - value1) > EPSILON || Math.abs(value - value2) > EPSILON) {
			StdOut.println("value = " + value + ", cx = " + value1 + ", yb = " + value2);
			return false;
		}

		return true;
	}

	private boolean check(double[][]A, double[] b, double[] c) {
		return isPrimalFeasible(A, b) && isDualFeasible(A, c) && isOptimal(b, c);
	}

	// print tableaux
	public void show() {
		StdOut.println("M = " + M);
		StdOut.println("N = " + N);
		for (int i = 0; i <= M; i++) {
			for (int j = 0; j <= M + N; j++) {
				StdOut.format("%7.2f ", a[i][j]);
			}
			StdOut.println();
		}
		StdOut.println("value = " + value());
		for (int i = 0; i < M; i++)
			if (basis[i] < N) StdOut.println("x_" + basis[i] + " = " + a[i][M+N]);
		StdOut.println();
	}


	public static void test(double[][] A, double[] b, double[] c) {
		Simplex lp = new Simplex(A, b, c);
		StdOut.println("value = " + lp.value());
		double[] x = lp.primal();
		for (int i = 0; i < x.length; i++)
			StdOut.println("x[" + i + "] = " + x[i]);
		double[] y = lp.dual();
		for (int j = 0; j < y.length; j++)
			StdOut.println("y[" + j + "] = " + y[j]);
	}

	public static void test1() {
		double[][] A = {
				{ -1,  1,  0 },
				{  1,  4,  0 },
				{  2,  1,  0 },
				{  3, -4,  0 },
				{  0,  0,  1 },
		};
		double[] c = { 1, 1, 1 };
		double[] b = { 5, 45, 27, 24, 4 };
		test(A, b, c);
	}


	// x0 = 12, x1 = 28, opt = 800
	public static void test2() {
		double[] c = {  13.0,  23.0 };
		double[] b = { 480.0, 160.0, 1190.0 };
		double[][] A = {
				{  5.0, 15.0 },
				{  4.0,  4.0 },
				{ 35.0, 20.0 },
		};
		test(A, b, c);
	}

	// unbounded
	public static void test3() {
		double[] c = { 2.0, 3.0, -1.0, -12.0 };
		double[] b = {  3.0,   2.0 };
		double[][] A = {
				{ -2.0, -9.0,  1.0,  9.0 },
				{  1.0,  1.0, -1.0, -2.0 },
		};
		test(A, b, c);
	}

	// degenerate - cycles if you choose most positive objective function coefficient
	public static void test4() {
		double[] c = { 10.0, -57.0, -9.0, -24.0 };
		double[] b = {  0.0,   0.0,  1.0 };
		double[][] A = {
				{ 0.5, -5.5, -2.5, 9.0 },
				{ 0.5, -1.5, -0.5, 1.0 },
				{ 1.0,  0.0,  0.0, 0.0 },
		};
		test(A, b, c);
	}



	// test client
	public static void main(String[] args) {

		try                           { test1();             }
		catch (ArithmeticException e) { e.printStackTrace(); }
		StdOut.println("--------------------------------");

		try                           { test2();             }
		catch (ArithmeticException e) { e.printStackTrace(); }
		StdOut.println("--------------------------------");

		try                           { test3();             }
		catch (ArithmeticException e) { e.printStackTrace(); }
		StdOut.println("--------------------------------");

		try                           { test4();             }
		catch (ArithmeticException e) { e.printStackTrace(); }
		StdOut.println("--------------------------------");


		int M = Integer.parseInt(args[0]);
		int N = Integer.parseInt(args[1]);
		double[] c = new double[N];
		double[] b = new double[M];
		double[][] A = new double[M][N];
		for (int j = 0; j < N; j++)
			c[j] = StdRandom.uniform(1000);
		for (int i = 0; i < M; i++)
			b[i] = StdRandom.uniform(1000);
		for (int i = 0; i < M; i++)
			for (int j = 0; j < N; j++)
				A[i][j] = StdRandom.uniform(100);
		Simplex lp = new Simplex(A, b, c);
		StdOut.println(lp.value());
	}

}