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Maximizing the area of a polygon Description Demonstrates how to solve a nonlinear problem in Java Further explanation of this example: 'Xpress NonLinear Reference Manual'
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Polygon.java import com.dashoptimization.DefaultMessageListener; import com.dashoptimization.IntHolder; import com.dashoptimization.XPRSconstants; import com.dashoptimization.XPRSenumerations.ObjSense; import com.dashoptimization.XPRSprob; import com.dashoptimization.XPRSprob.MapFunction; import com.dashoptimization.XPRSprob.MapFunctor; import java.util.ArrayList; /** Code example how to build nonlinear formulae from strings and token sequences. <pre> Xpress Optimizer Examples ========================= Maximize the area of polygon of N vertices and diameter of 1 The position of vertices is indicated as (rho,theta) coordinates where rho denotes the distance to the base point (vertex with number N) and theta the angle from the x-axis. (c) 2021-2024 Fair Isaac Corporation </pre> Polygon example: maximise the area of an N sided polygon *** Demonstrating how to build nonlinear formulae from strings and token sequences *** <pre> Variables: rho : 0..N-1 ! Distance of vertex from the base point theta : 0..N-1 ! Angle from x-axis Objective: (sum (i in 1..N-2) (rho(i)*rho(i-1)*sin(theta(i)-theta(i-1)))) * 0.5 Constraints: Vertices in increasing degree order: theta(i) >= theta(i-1) +.0001 : i = 1..N-2 Boundary conditions: theta(N-1) <= Pi 0.1 <= rho(i) <= 1 : i = 0..N-2 Third side of all triangles <= 1 rho(i)^2 + rho(j)^2 - rho(i)*rho(j)*2*cos(theta(j)-theta(i)) <= 1 : i in 0..N-3, j in i..N-2 </pre> */ public final class Polygon { public static void main(String[] args) { try (XPRSprob prob = new XPRSprob(null)) { prob.addMessageListener(new DefaultMessageListener()); // Number of sides of the Polygon int nSide = 5; // Theta int[] theta = prob.varArray('C', nSide - 1, 0.0, Math.PI, i -> String.format("THETA%d", i + 1)); // Rho int[] rho = prob.varArray('C', nSide - 1, 0.01, 1.0, i -> String.format("RHO%d", i + 1)); // Objective function. We build the objective function as // a formula in infix notation. See below for submitting a // formula as string. // Tokens are always integers, while values may be integers // (for example operator or delimiter constants) or double // values (actual numbers). That is why the `val` list has // elements of type Number. ArrayList<Integer> tok = new ArrayList<Integer>(); ArrayList<Number> val = new ArrayList<Number>(); for (int i = 1; i < nSide - 1; ++i) { if ( tok.size() > 0 ) { tok.add(XPRSconstants.TOK_OP); val.add(XPRSconstants.OP_PLUS); } tok.add(XPRSconstants.TOK_COL); // RHO(i) val.add(rho[i]); tok.add(XPRSconstants.TOK_OP); // * val.add(XPRSconstants.OP_MULTIPLY); tok.add(XPRSconstants.TOK_COL); // RHO(i-1) val.add(rho[i-1]); tok.add(XPRSconstants.TOK_OP); // * val.add(XPRSconstants.OP_MULTIPLY); tok.add(XPRSconstants.TOK_IFUN); // SIN val.add(XPRSconstants.IFUN_SIN); tok.add(XPRSconstants.TOK_LB); // ( val.add(XPRSconstants.TOK_LB); tok.add(XPRSconstants.TOK_COL); // THETA(i) val.add(theta[i]); tok.add(XPRSconstants.TOK_OP); // - val.add(XPRSconstants.OP_MINUS); tok.add(XPRSconstants.TOK_COL); // THETA(i-1) val.add(theta[i-1]); tok.add(XPRSconstants.TOK_RB); // ) val.add(XPRSconstants.TOK_RB); tok.add(XPRSconstants.TOK_OP); // * val.add(XPRSconstants.OP_MULTIPLY); tok.add(XPRSconstants.TOK_CON); // 0.5 val.add(0.5); } tok.add(XPRSconstants.TOK_EOF); val.add(0.0); // Since nonlinear objectives cannot be directly expressed in Xpress, we maximize a free // variable objx and constrain this variable to be equal to the nonlinear objective. int objx = prob.addCol(1.0, XPRSconstants.MINUSINFINITY, XPRSconstants.PLUSINFINITY); int objeq = prob.addRow(new int[]{objx}, new double[]{-1.0}, 'E', 0.0); prob.nlpChgFormula(objeq, 0, tok.stream().mapToInt(i -> i).toArray(), val.stream().mapToDouble(d -> d.doubleValue()).toArray()); prob.chgObjSense(ObjSense.MAXIMIZE); // Vertices in increasing degree order for (int i = 0; i < nSide - 2; ++i) prob.addRow(new int[]{ theta[i], theta[i+1] }, new double[]{ -1.0, 1.0 }, 'G', 0.001); // Third side of all triangles <= 1 for (int i = 1; i < nSide - 1; i++) { for (int j = i + 1; j < nSide; j++) { int row = prob.addRow(new int[0], new double[0], 'L', 1.0); prob.nlpChgFormulaStr(row, String.format("RHO%d ^ 2 + RHO%d ^ 2 - RHO%d * RHO%d * 2 * COS ( THETA%d - THETA%d )", i, j, i, j, j, i)); } } IntHolder solvestatus = new IntHolder(); IntHolder solstatus = new IntHolder(); prob.optimize("s", solvestatus, solstatus); System.out.printf("solvestatus: %s%n", solvestatus); System.out.printf("solstatus: %s%n", solstatus); System.out.printf("Solution value: %f%n", prob.attributes().getObjVal()); double[] x = new double[prob.attributes().getCols()]; prob.getSolution(null, x, 0, x.length - 1); for (int i = 0; i < rho.length; ++i) System.out.printf("RHO%d = %f%n", i + 1, x[rho[i]]); for (int i = 0; i < theta.length; ++i) System.out.printf("THETA%d = %f%n", i + 1, x[theta[i]]); } } } | |||||||||||
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