x = n; y = 0

while x >= b:

   x = x DIV b

   y = y + 1

return y

This function takes in $n,b\in\mathbb{N}, n > 0, b > 1$, and returns $k\in\mathbb{N}$ such that $b^k\leq n \lt b^{k+1}$

So far I think it is:

$b^y \leq n$

$x \leq n$

but now I'm stuck

  • 2
    $\begingroup$ Why are you talking about 'the' loop invariant. Think about what statement you want to show true at the end of the program first. $\endgroup$ – Apiwat Chantawibul Jul 29 '17 at 22:09
  • $\begingroup$ The statement that I wanna show true is $b^k \leq n \lt b^{k+1}$, but I need a proper LI in order to do this $\endgroup$ – K Split X Jul 29 '17 at 23:13

Try finding a loop invariant of the following form: $$ f(x,y) \leq n < g(x,y). $$

For example, when $y=0$ you could choose $f(x,y) = x$ and $g(x,y) = x+1$, and in the end you want this invariant to imply $b^y \leq n < b^{y+1}$.

To get some intuition, note that if $n = b^k z + w$, where $0 \leq w < b^k$, then after $k$ iterations, we have $x = z$. This can help you formulate the loop invariant.

  • $\begingroup$ What is $z,w$ in this case? $\endgroup$ – K Split X Jul 30 '17 at 14:29
  • $\begingroup$ It's the result of dividing $n$ by $b^k$. $\endgroup$ – Yuval Filmus Jul 30 '17 at 15:42

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