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I was trying to formulate some problem. I want to find a relation between a floor and ceiling function. Suppose the Property 1 satisfies that it has $\lfloor \frac{n}{2} \rfloor$ number of $X$. Then the Property 2 satisfies that it has $k = \lceil \frac{d}{2} \rceil$ where $d = \lfloor \frac{n}{2} \rfloor$. I tried to put $k = \lceil \frac{n}{4} \rceil$ but it is contradicting for some values of $n$. Like if I take $n = 9$, then

$d = \lfloor \frac{9}{2} \rfloor$ $\Rightarrow d = 4$ and thus $k = 2$.

But if I take $k = \lceil \frac{n}{4} \rceil$ then $k = \lceil \frac{9}{4} \rceil$ $\Rightarrow k = 3$

which contradicts. Is there any way to find the relation between $n$ and $k$ directly. Kindly help. My data is given below for different values of $n$.

enter image description here

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Starting from $n=0$ you have

$$d=0,0,1,1,2,2,3,3,4,4,5,5,6,6,\cdots$$ $$k=0,0,1,1,1,1,2,2,2,2,3,3,3,3,\cdots$$

In the last row, you see that the length of the runs is $4$, but the first run has only two elements.

Hence,

$$k=\left\lfloor\frac{n+2}4\right\rfloor$$


Alternatively,

$$d=\left\lfloor\frac n2\right\rfloor\equiv n=2d+d'$$ where $d'=0,1$.

$$k=\left\lceil\frac d2\right\rceil\equiv d=2k-k'$$ where $k'=0,1$.

Then

$$n=4k-2k'+d'=4k+k''$$ where $k''=-2,-1,0,1$.

By shifting,

$$n+2=4k+k'''$$ where $k'''=0,1,2,3$ and

$$k=\left\lfloor\frac{n+2}4\right\rfloor.$$

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  • $\begingroup$ I have added an image. I hope that is useful. $\endgroup$ – monalisa May 19 '16 at 6:32
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    $\begingroup$ @monalisa: hem, I had already written it together with the solution. $\endgroup$ – Yves Daoust May 19 '16 at 6:32
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Yves Daoust's way is probably easiest but, just in case, here's a more systematic way of solving it.

\begin{align*} k &= \left \lceil \frac{d}{2} \right \rceil \\ &= \left \lfloor \frac{d+1}{2} \right \rfloor \\ &= \left\lfloor \frac{\lfloor n/2\rfloor + 1}{2} \right\rfloor \\ &= \left\lfloor \frac{n/2 + 1}{2} \right\rfloor \\ &= \left\lfloor \frac{n + 2}{4} \right\rfloor. \end{align*} In the first step I used the identity $$\left\lceil \frac{j}{m} \right\rceil =\left\lfloor \frac{j+m-1}{m} \right\rfloor$$ which holds for all integers $j, m$ with positive $m$. For the third step, it turns out that $$\left\lfloor\frac{\lfloor x \rfloor + j}{m}\right\rfloor = \left\lfloor\frac{x + j}{m}\right\rfloor$$ also holds for all integers $j, m$ with positive $m$, for all real $x$. The same holds for ceilings, but we cannot remove floors from inside ceilings or vice versa, which is why the first step was necessary.

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