Ruben Safir ruben-at-mrbrklyn.com=
writes:

If there is always a bias, why is there a while condition, let alone not=
an if condition. What situation were you anticipating?

These logic-gate perceptrons have two inputs, so you should be able to d=
eclare:

vector<float> inputs =3D {0, 1};

But the weight vector is three values:

vector<float> weights =3D {0.1, -0.2, 0.3};

The last one is what we call the bias. You can think of it as the weight=
attached to a =E2=80=9Cpseudo=E2=80=9D input which is always one. So what =
the loop does is to make that =E2=80=9Cpseudo=E2=80=9D input real:

while(input.size() < p.weights.size()) {
  input.push_back(1);   // Add pseudo input that's always 1
}

Now, since the input.size() matches the weights.size(), we can call the =
dot product to do the calculation:

dot_product(p.weights, input)

It=E2=80=99s just the =E2=80=9Crobustness principle=E2=80=9D: be strict =
in what you send and tolerant in what you accept. In the feed_forward funct=
ion I=E2=80=99m receiving the input vector. I=E2=80=99m sending it to the d=
ot_product function. I=E2=80=99m tolerant if the input is too small, and I =
augment it with ones (pseudo inputs) until there are enough inputs that the=
dot_product will work.

BTW, it only has to append the first time through the feed_forward. Ther=
eafter, if I continue to use the same input vector then the size will alrea=
dy be correct and the loop body is never executed.

Or if you just want to declare the inputs to have three values, then it =
just amounts to a check that you did that, and doesn=E2=80=99t have to appe=
nd:

vector<float> inputs =3D {0, 1, 1};  // The last 1 is the =
pseudo input for bias

CL