if n < 2
    return false
while n != 1
    if n % 2 != 0
        return false
    n = n/2
return true

The loop will terminate when n is odd. If n = 1, true is returned and that means that n0 was a power of 2. Otherwise it's false.

I'm new at this and struggle with the proving the power of 2 part for before and during iterations and after the loop terminates. Also what does a loop invariant look like for a true vs false return on a loop?

Is it reasonable to say something like:

at the every iteration i, 1 <= n <= n0/(2^i)
true returned when n = 1 and n0 is a power of 2
false returned when n is odd and n > 1

Is it possible to even make strong loop invariant here without referencing the initial value of n?

  • $\begingroup$ The loop invariant is that $n_0/n$ is a power of 2, where $n_0$ is the original value of $n$. The loop terminates since $n$ always decreases. $\endgroup$ Sep 27, 2018 at 2:41
  • $\begingroup$ Thanks for responding to me. How were you able to find that pattern? Is it just from experience and practice? $\endgroup$
    – weztex
    Sep 27, 2018 at 4:18
  • $\begingroup$ I'm not sure what's the difference between experience and practice. In college and beyond, you should never solve 100 exercises of the form "3+4=?". A few should suffice. Your problem-solving skills improve with time, and experience with one thing could help with another. $\endgroup$ Sep 27, 2018 at 4:20

1 Answer 1


One loop invariant is "$n_0/n$ is a power of 2 and $n \geq 1$", where $n_0$ is the original value of $n$.

The loop invariant holds initially (before the first iteration of the loop) since $n_0 = n$, and $n_0 \geq 2$. As for what happens in the loop, there are several cases:

  • If $n$ is odd then $n = 2k+1$ for some $k$; since $n > 1$ (otherwise the loop would have terminated), moreover $n \geq 3$, and so $n_0 = 2^t (2k+1)$ for some $t \geq 0$ and $k \geq 1$, showing that $n_0$ isn't a power of 2.
  • If $n$ is even then the value of $n$ in the next iteration is $n' = n/2 \geq 1$, and so $n_0/n' = 2n_0/n$ is also a power of 2.

If the loop terminates naturally then necessarily $n = 1$, and so the loop invariant guarantees that $n_0 = n_0/n$ is a power of 2.

Finally, since $n/2 < n$, the value of $n$ keeps decreasing; since always $n \geq 1$, it cannot decrease forever, i.e., the loop terminates. In fact, the running time is $\Theta(\log n)$.


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