How can I prove that this is the upper bound?

  • Thread starter Thread starter SpatialVacancy1
  • Start date Start date
  • Tags Tags
    Combinatorics
SpatialVacancy1
Messages
1
Reaction score
0
Combinatorics...evil problem!

Hi all,

I am working on my combinatorics homework. I have completed all of the problems but one. Here it goes:

------------------------------------------
Let S_1 and S_2 be two sets where |S_1| = m and |S_2| = r, for m,r in Z+ (positive integers), and the elements in each of S_1, S_2 are in ascending order. It can be shown that the elements in S_1 and S_2 can be merged into ascending order by making no more than m + r - 1 comparisons. Use this result to establish the following:

For n >= 0, Let S be a set with |S| = 2^n. Prove that the number of comparisons needed to place the elements of S in ascending order is bounded above by n * 2^n.
------------------------------------------

Please help! Due by 12:00 PM EST tomorrow the 3rd!
 
Physics news on Phys.org
I was able to produce the right result by breaking S into one element sets then combining pairs of them to create half as many sets of two elements and assuming it took the upper bound of m+r-1 comparisons to create each set. I then combined pairs of two element sets to create half as many sets of 4 elements, and so on until I had one set of 2^n elements (S), and summed the upper bounds of all of comparisons required to make each intermediate set and finally S.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top