Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to reduce operands to logical scalar values.

Hi, I get the following error with the given code,

"Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to reduce operands to logical scalar values."

This performs a minimization using the active set method. But seemingly, it stops over this operator problem.

Any ideas welcome!

% Define symbolic variables
syms x1 x2 x3 lambda1 lambda3

% Objective function
f = x1^2 + x2^2 + x3^2 - x1 + 2*x2 - 4*x3;

% Constraints
g1 = -x1 + x2 + 1;  % -x1 + x2 >= -1 is equivalent to g1 <= 0
g2 = x1 + x2 + x3 + 3;  % x1 + x2 + x3 >= -3 is equivalent to g2 <= 0
g3 = x3;  % x3 >= 0 is equivalent to g3 <= 0

% Initial point
x0 = [0; 0; 0];

% Iterative process
x = x0;
activeSet = [];  % Initially empty active set

maxIter = 10;  % Maximum number of iterations
tol = 1e-6;  % Tolerance for convergence

for iter = 1:maxIter
    % Compute gradients of the objective function and constraints
    grad_f = gradient(f, [x1, x2, x3]);
    grad_g1 = gradient(g1, [x1, x2, x3]);
    grad_g2 = gradient(g2, [x1, x2, x3]);
    grad_g3 = gradient(g3, [x1, x2, x3]);
    
    % Evaluate gradients at the current point
    grad_f_val = double(subs(grad_f, {x1, x2, x3}, x'));
    grad_g1_val = double(subs(grad_g1, {x1, x2, x3}, x'));
    grad_g2_val = double(subs(grad_g2, {x1, x2, x3}, x'));
    grad_g3_val = double(subs(grad_g3, {x1, x2, x3}, x'));
    
    % Check KKT conditions to determine active set
    if abs(grad_g1_val) < tol && subs(g1, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 1];  % Constraint 1 is active
    end
    
    if abs(grad_g2_val) < tol && subs(g2, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 2];  % Constraint 2 is active
    end
    
    if abs(grad_g3_val) < tol && subs(g3, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 3];  % Constraint 3 is active
    end
    
    % Solve the subproblem using the active set
    A_eq = [];
    b_eq = [];
    A_ineq = [];
    b_ineq = [];
    lb = [];
    ub = [];
    options = optimoptions('quadprog', 'Display', 'off');
    
    if any(activeSet == 1)
        A_ineq = [A_ineq; -1, 1, 0];
        b_ineq = [b_ineq; -1];
    end
    
    if any(activeSet == 2)
        A_ineq = [A_ineq; 1, 1, 1];
        b_ineq = [b_ineq; -3];
    end
    
    if any(activeSet == 3)
        A_ineq = [A_ineq; 0, 0, 1];
        b_ineq = [b_ineq; 0];
    end
    
    [x_new, ~, exitflag] = quadprog(eye(3), -grad_f_val', A_ineq, b_ineq, A_eq, b_eq, lb, ub, [], options);
    
    % Check convergence
    if norm(x_new - x) < tol
        break;
    end
    
    % Update x and active set
    x = x_new;
    activeSet = unique(activeSet);
end

% Display results
disp(['Optimal point: x = [', num2str(x'), ']']);
disp(['Objective function value: ', num2str(double(subs(f, {x1, x2, x3}, x')))]);

Answer 1

[Torsten]

    if all(abs(grad_g1_val) < tol) && subs(g1, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 1];  % Constraint 1 is active
    end
    
    if all(abs(grad_g2_val) < tol) && subs(g2, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 2];  % Constraint 2 is active
    end
    
    if all(abs(grad_g3_val) < tol) && subs(g3, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 3];  % Constraint 3 is active
    end
    

instead of

    if abs(grad_g1_val) < tol && subs(g1, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 1];  % Constraint 1 is active
    end
    
    if abs(grad_g2_val) < tol && subs(g2, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 2];  % Constraint 2 is active
    end
    
    if abs(grad_g3_val) < tol && subs(g3, {x1, x2, x3}, x') <= 0
        activeSet = [activeSet, 3];  % Constraint 3 is active
    end

[Sergio]

Thanks @Torsten!