Answer to Question #153397 in Discrete Mathematics for Sam

Question

For discrete structures there are n exams to check and there are k graders. To guarantee a high quality of grading every exam may be checked by any number of graders (but always at least by one grader). This means that summed all together the graders may make up to k ∗ n exam checks. To avoid this it is required that for each pair of graders there is at most 1 exam that they have both checked. Prove that this rule creates a much better bound of at most ((k +n) 3/2 + (k +n))/2 exam checks. Hint: Consider modeling the grading work as a graph. 

Answer 1

Let $N_p$ is number of pairs of graders, $N$ is number of exams checks.

$N_p=C^2_k=\frac{k!}{2(k-2)!}=\frac{k(k-1)}{2}$

We have, by the given rule: $N_p\geq n$

Maximum $N$will be in case when $N_p=n$

Then: $N=2n=k(k-1)$

We can prove the given formula by induction:

For $k=2$ :

$\frac{(2+1)(\sqrt{2+1}+1)}{2}>N=2$

Let

$(k+n)(\frac{\sqrt{k+n}+1}{2})=(\frac{k(k+1)}{2})(\frac{\sqrt{\frac{k(k+1)}{2}}+1}{2})\geq N=k(k-1)$

Then we have for $k+1$

$(k+1+\frac{k(k+1)}{2})(\frac{\sqrt{k+1+\frac{k(k+1)}{2}}+1}{2})=$

$=(k+1)(\frac{\sqrt{k+1+\frac{k(k+1)}{2}}+1}{2})+\frac{k(k+1)}{2}(\frac{\sqrt{k+1+\frac{k(k+1)}{2}}+1}{2})>N=k(k-1)+2k$

So, the given formula is the upper bound for the maximum of exam checks.

Also:

if the number of graders is increasing by 1, the the number of exam checks is increasing by $1+(k-2)=k-1$

and we get:

$N=k(k-1)+(k-1)<\frac{(k+n)^{3/2}+k+n}{2}+(k-1)$