Difference between revisions of "TADM2E 2.1"

Latest revision as of 18:23, 11 September 2014

2-1 ans

This loop can be expressed as the sum:

$ \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=1}^{j}1 $

Reducing this, sum by sum from the rhs:

$ \begin{align} &\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=1}^{j}1 =\\ &\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}j =\\ &\sum_{i=1}^{n-1}\left(\sum_{j=1}^{n}j - \sum_{j=1}^{i}j\right) =\\ &\sum_{i=1}^{n-1}\left(\frac{n(n+1)}{2} - \frac{i(i+1)}{2}\right) =\\ &\frac{1}{2}\sum_{i=1}^{n-1}n^2+n-i^2-i =\\ &\frac{1}{2}\left((n-1)n^2 + (n-1)n - \left(\frac{n(n+1)(2n+1)}{6} - n^2\right) - \left(\frac{n(n+1)}{2} - n\right)\right) =\\ &f(n) = \frac{n(n(n+1))}{2} - \frac{n(n+1)(2n+1)}{12} - \frac{n(n+1)}{4} \end{align} $

n=1 gives zero; and order is O((n^3)/3)

Alternative Derivation

$ \begin{align} &mystery(n)=\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\sum_{k=1}^{j} 1\\ &=\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}j\\ \end{align} $

Taking the following well-known formula for the sum of the integers $ 1 $ to $ n $:-

$ \sum_{x=1}^{n}x = \frac{1}{2}n(n + 1) $

And noting that:-

$ \sum_{j=i+1}^{n}j = \sum_{j=1}^{n}j - \sum_{j=1}^{i}j $

We continue our derivation thus:-

$ \begin{align} &=\sum_{i=1}^{n-1}(\frac{1}{2}n(n+1) - \frac{1}{2}i(i+1))\\ &=\frac{1}{2}n(n-1)(n+1) - \frac{1}{2}\sum_{i=1}^{n-1}(i^2 + i)\\ \end{align} $

Given our formula for the sum of integers, substituting $ n-1 $ for $ n $ gives us:-

$ \sum_{x=1}^{n-1}x=\frac{1}{2}(n-1)((n-1)+1)=\frac{1}{2}n(n-1) $

Thus continuing with our derivation:-

$ \begin{align} &=\frac{1}{2}n(n-1)(n+1)- \frac{1}{2}\frac{1}{2}n(n-1) - \frac{1}{2}\sum_{i=1}^{n-1}i^2\\ &=\frac{1}{2}n(n-1)(n + 1 - \frac{1}{2}) - \frac{1}{2}\sum_{i=1}^{n-1}i^2\\ &=\frac{1}{2}n(n-1)(n + \frac{1}{2}) - \frac{1}{2}\sum_{i=1}^{n-1}i^2\\ \end{align} $

Let us consider the well-known formula for the sum of the squares of the integers between $ 1 $ and $ n $:-

$ \sum_{x=1}^{n}x^2=\frac{1}{6}n(n+1)(2n+1) $

As before, let us substitute $ n-1 $ for $ n $:-

$ \sum_{x=1}^{n-1}x^2=\frac{1}{6}(n-1)((n-1)+1)(2(n-1)+1)=\frac{1}{6}n(n-1)(2n-1) $

Substituting this into our derivation gives us:-

$ \begin{align} &=\frac{1}{2}n(n-1)(n + \frac{1}{2}) - \frac{1}{2}\frac{1}{6}n(n-1)(2n - 1)\\ &=\frac{1}{2}n(n-1)(n + \frac{1}{2} - \frac{1}{6}(2n-1))\\ &=\frac{1}{2}n(n-1)(n + \frac{1}{2} - \frac{1}{3}n + \frac{1}{6})\\ &=\frac{1}{2}n(n-1)(\frac{2}{3}n + \frac{2}{3})\\ &=\frac{1}{3}n(n-1)(n+1) \end{align} $

Thus the solution is $ mystery(n)=\frac{1}{3}n(n-1)(n+1) $. This is the equivalent to the solution given by the previous derivation.

The Big-Oh complexity of this function (ignoring constants) is O(n^3), as the RAM model dictates that each iteration of the increment of $ r $ is a time-step. We can ignore the first and last line's contribution to running time for the purposes of Big-Oh, as they do not contribute to the growth of the time taken with input size.

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