So, i would like to know the time complexity of the following codes:
x = (float) rand() / rand(); // T(4)
while (x >= 0.01) // T(?)
{
x *= 0.8; // T(?) x T(2)
}
Assuming that all the basic operations are perfomed once, is the best case T(1) - constant time? Since, that might only happen when the random x generated is <= 0.01.
What about the average case? Is it T(?) x T(1) / 2?
Thanks a lot!
Worst case, if the second call to rand() returns 0 and the first call doesn't, you get a floating point division by zero, and if you are using standard IEEE 754 arithmetic, the result is +infinity. In that case, the loop will run forever.
If you changed your code to exclude that case, and exclude the case that rand () might return a 128 bit integer, then for any implementation the size of x is limited, so the number of multiplications by 0.8 is limited, so the runtime is O (1).
This answer refers to a version of the question in which $x$ is sampled by dividing two random numbers.
As mentioned by Rick Decker's answer, given $x$, we can approximate the running time by $O(\max(\log x,1))$. Assuming that rand returns a random number in $[0,1]$, the running time should be proportional (up to an additive constant) to
$$
\int_0^1 \int_0^1 \max(\log \tfrac{x}{y},0) \, dx \, dy.
$$
Let us start by computing the inner integral:
$$
\int_0^1 \max(\log \tfrac{x}{y},0) \, dx =
\int_y^1 \log \tfrac{x}{y} \, dx = (1-y)\log y + \int_y^1 \log x \, dx = \\
-(1-y)\log y + \left. x(\log x-1) \right|_y^1 = -(1-y)\log y-1-y(\log y-1) = \\
y-\log y-1.
$$
Integrating this over $y$, we get
$$
\int_0^1 (y-\log y-1) \, dy =
\left. \tfrac{1}{2} y^2 - y\log y \right|_0^1 = \frac{1}{2}.
$$
This shows that in this idealized setting, the expected running time is $O(1)$ (rather than infinity, which could also have been the case).
x is generated as $\infty$ and then the program does not halt. But in all (constant amount) other cases the runtime is constant. Then average time is also constant (while worst case is $\infty$)? And in his code rand() returns integers in some interval, I think.
– rus9384
Jul 13 '17 at 16:55
rand returns an integer from $1$ to $N$ (or from $0$ to $N$), in the large $N$ limit we get the same result, since rand/rand is the same as (rand/N)/(rand/N), and rand/N converges in the large $N$ limit to the uniform distribution over $[0,1]$.
– Yuval Filmus
Jul 13 '17 at 17:04
rand() returns an integer in some range. Thus there is a non-zero probability it returns 0, and there is a non-zero probability that x is infinity and the loop never terminates. If a discrete random variable has a non-zero probability of being infinity, then its expected value is infinite. So I think gnasher729 is right, and technically the average running time is infinite.
– D.W.
Jul 13 '17 at 17:39
In general, the time taken by this snippet is mainly governed by how many times the loop iterates. In other words, how many times will you need to multiply $x$ by $0.8$ to get a result less than $0.01$?
In other words, for a fixed $x$, what $n$ value will ensure that $(0.8)^nx<0.01$?
rand()is bounded by constant number in your case. Then, initialxalso is bounded and answer is simple. Ah, well, there is one case when this program does not halt. – rus9384 Jul 13 '17 at 16:13rand(), so knowledge of any particular programming language is not needed to be able to answer the question. Is this code fragment in C? Not everyone here may know C. We want questions to be language-independent and to be understandable even to people who don't know a particular programming language. Thank you! – D.W. Jul 13 '17 at 17:41x. – D.W. Jul 14 '17 at 05:38x = (rand() + 1) / (rand() + 1), it would have $O(1)$ complexity. – rus9384 Jul 14 '17 at 07:50