I don't understand the condition from FOR row 16 to row 26. Can explain each row form if row to the end please.

syms x y z lambda
dist = x.^2+(y-1).^2+z.^2;       % distance of an abitary point (x,y,z) from the point(0,1,0) 
S = z.^2-(x.*y)-1 == 0;          % surface constraint  z^2 = xy+1    ------>   z^2-xy-1 == 0
L = dist + lambda * lhs(S);      % Langrange function
dL_dx = diff(L,x) == 0;          % derivative of L with respect to x
dL_dy = diff(L,y) == 0;          % derivative of L with respect to y
dL_dz = diff(L,z) == 0;          % derivative of L with respect to z
dL_dlambda = diff(L,lambda) == 0;      % derivative of L with respect to lambda
system = [dL_dx; dL_dy; dL_dz; dL_dlambda];    % build the system of equations
[x_val, y_val, z_val,lambda_val] = solve(system, [x y z lambda], 'Real', true);    % solve the system of equations and display the results 
results_numeric = double([x_val, y_val, z_val]);    % show results in a vector of data type double
[rows columns]=size(results_numeric);         % getting number of rows and columns
fprintf('points on the surface S that are closest to the point (0, 1, 0) are :\n');
fprintf('   x         y         z\n');
fprintf('--------------------------\n');
for i=1:rows
    fprintf('(');
    for j=1:columns
        if j < columns
            fprintf("%.4f , ",results_numeric(i,j));
        else 
            fprintf("%.4f",results_numeric(i,j));
        end
    end
    fprintf(")\n");
end

[Stephen23]

The FOR loop and IF can be simplified, assuming that there is always one or more columns:

for ii = 1:rows
    fprintf("(%.4f",results_numeric(ii,1));
    fprintf(",%.4f",results_numeric(ii,2:end));
    fprintf(")\n");
end

Answer 1

[Voss]

Lines 16 through 26:

for i=1:rows
    fprintf('(');
    for j=1:columns
        if j < columns
            fprintf("%.4f , ",results_numeric(i,j));
        else 
            fprintf("%.4f",results_numeric(i,j));
        end
    end
    fprintf(")\n");
end

are there to display results_numeric in the command window.

The idea is to print each row of results_numeric on its own line, enclosed in parentheses, with commas separating each number in the row.

% for each row:
for i=1:rows
    % first print an open parenthesis, "(":
    fprintf('(');
    % then for each element of this row:
    for j=1:columns
        if j < columns
            % if it's not the last element, print it with
            % 4 decimal points of precision ("%.4f") followed by
            % space-comma-space (" , "):
            fprintf("%.4f , ",results_numeric(i,j));
        else 
            % if it is the last element of the row, just print the number
            % (without the space-comma-space afterwards):
            fprintf("%.4f",results_numeric(i,j));
        end
    end
    % now print the close parenthesis ")"
    % and a newline character ("\n") so that whatever 
    % is printed next will begin on the next line:
    fprintf(")\n");
end

[Ngo Ha Gia Bao]

I am still not clear about how does if work like j and i work to caculate the clostest point to the surface . Can you explain more clearly about this please .Thank you very much

[Voss]

j and i iterate over the columns and rows of results_numeric, which is the solution to the system of equations set up in the previous lines.

In other words, the system is set up first, then solve solves it, then the results are displayed in the command window using fprintf iterating over each row and column of results_numeric.