Fast exponentiation algorithm - How to arrive at it?

I've been learning about fast exponentiation when I found this algorithm:

int expo(int a, int b) {
  int result = 1;


  while (b) {
    if (b%2==1) {
      result *= a;
    }
    b /= 2;
    a *= a;
  }

  return result;

}

This is supposed to compute $a^b$ in logarithmic time, but I couldn't understand or find anywhere a way to arrive at this procedure, other than by noting that (this particular formula didn't help me in understanding why the algorithm is correct):

$$
x^n =
\begin{cases}
1 & \text{if $n=0$} \\
\frac{1}{x}^{-n} & \text{if $n<0$} \\

x \cdot \left( x^{\frac{n-1}{2}} \right)^2 & \text{if $n$ is odd} \\
\left( x^{\frac{n}{2}} \right)^2 & \text{if $n$ is even}
\end{cases}
$$

From what I understand, the transition from this particular identity to the actual algorithm is quite obvious, but I honestly don't get it and I've worked by hand quite a few examples for this algorithm.

Can anyone explain?

Answer

Write $b = b_0 + b_1 2 + b_2 2^2 + ... + b_n 2^n$ where $b_k$ are the binary digits of the representation of $b$ in base $2$. Note that $n$ varies logarithmically with $b$. Then:

$$
a^b = a^{b_0} \cdot (a^2)^{b_1} \cdot (a^{2^2})^{b_2} \cdot ... \cdot (a^{2^n})^{b_n}
$$

The terms where bits $b_k = 0$ evaluate to $1$ and can be skipped. The terms with $b_k = 1$ are of the form $a^{2^k}$ and can be calculated by repeatedly squaring $a$ $k$ times. This is precisely what the posted code does.