How to solve the equation of binary logic operation, where all unknowns are 0 or 1

For example, I have the following equations, linear equations and nonlinear equations, in which the unknowns can only take 0 or 1, and it is based on this operation rule,

for Addition operation：

1+1=0，1+0=1，0+1=1，0+0=0，

For multiplication operation：

0*0=0，0*1=0，1*0=0，1*1=1

so how to solve the values of all unknowns?

x1+x2+x3+(x4+1)*(x5+x6+1)=x1*x2, 
x2+x3+x4+(x5+1)*(x6+x1+1)=x2*x3, 
x3+x4+x5+(x6+1)*(x1+x2+1)=x3*x4, 
x4+x5+x6+(x1+1)*(x2+x3+1)=x4*x5, 
x5+x6+x1+(x2+1)*(x3+x4+1)=x5*x6, 
x6+x1+x2+(x3+1)*(x4+x5+1)=x6*x1, 

and in fact, the original question consists of 416 equations and 416 variables including form

c0...c31 and x1...x384 which should be only 0 or 1

as i have attached in the

equations.txt

[dcydhb dcydhb]

i am sorry there is no tried codes as i really don't konw how to solve and set the

binary logic operation.

[Walter Roberson]

the first operation is XOR. The second operation is AND.

AND can be implemented as multiplication. XOR can be implemented as addition modulo 2.

[dcydhb dcydhb]

thanks for your tips but how to realized them in matlab?for example i can solve linear equation with 'EquationToMatrix 'and 'fsolve', 'fzero' in nolinear but how to operate with binary logic operation?

[Bruno Luong]

You can try brute force search, there is only 2^6 = 64 possible combination.

[dcydhb dcydhb]

i want to say that the equation i have posted is just an example equation and my original equation consists of 300 unknowns so

i need other way rahher than brute force search.

[Walter Roberson]

https://www.mathworks.com/help/comm/ref/gf.html

gf(x,1) appears to implement the required mathematics.

Answer 1

[Bruno Luong]

[x1,x2,x3,x4,x5,x6]=ndgrid(0:1);
x1 = x1(:);
x2 = x2(:);
x3 = x3(:);
x4 = x4(:);
x5 = x5(:);
x6 = x6(:);

b = [x1+x2+x3+(x4+1).*(x5+x6+1)-x1.*x2, ...
x2+x3+x4+(x5+1).*(x6+x1+1)-x2.*x3, ...
x3+x4+x5+(x6+1).*(x1+x2+1)-x3.*x4, ...
x4+x5+x6+(x1+1).*(x2+x3+1)-x4.*x5, ...
x5+x6+x1+(x2+1).*(x3+x4+1)-x5.*x6, ...
x6+x1+x2+(x3+1).*(x4+x5+1)-x6.*x1 ];
b=mod(b,2);
idx = find(all(b==0,2));
for i=1:length(idx)
    k=idx(i);
    fprintf('x1=%d,x2=%d,x3=%d,x4=%d,x5=%d,x6=%d\n', x1(k), x2(k), x3(k), x4(k), x5(k), x6(k));
end

Result

x1=1,x2=0,x3=0,x4=0,x5=0,x6=0
x1=0,x2=1,x3=0,x4=0,x5=0,x6=0
x1=0,x2=0,x3=1,x4=0,x5=0,x6=0
x1=0,x2=0,x3=0,x4=1,x5=0,x6=0
x1=0,x2=0,x3=0,x4=0,x5=1,x6=0
x1=0,x2=0,x3=0,x4=0,x5=0,x6=1
x1=1,x2=1,x3=1,x4=1,x5=1,x6=1

[dcydhb dcydhb]

thanks for your answer and i was wonder when the unknowns and equations increased to 400, is this method still an effective way to solve the equation?

[Bruno Luong]

Of course not, since 2^400 is beyond any possible computer capacity.

[dcydhb dcydhb]

so is there any other way?

[Walter Roberson]

mathematically, your systems are non-linear, since they have functions of variables being multiplied by functions of variables. In what you posted, the maximum degree is 2, which makes the system "quadratic", and there are techniques for dealing with quadratic functions. Does this extend in general to your equations, that you never "and" more than two variables together ?

[dcydhb dcydhb]

i am sorry what does your 'and' mean, i have just attached the original equations and i think may you can just explore the original equations and thanks a lot!