# Introduction

This is a hand-in written by Group 7 (FIRE) for the Algorithms course (TIN092) at Chalmers University of Technology.

Group 7 consists of:

• 901011-0279

• twingoow@gmail.com

• Program: IT

• 920203-0111

• nicale@student.chalmers.se

• Program: IT

# a) Complexity

Our algorithm for calculating a numerical value is, in pseudocode:

 Computes the numerical value for C(n) for average quicksort complexity Input: n is the size of the problem. C(n): if n = 0 then: return 1 end if sum <- 0 for i: 0 .. n-1 do: sum <- sum + C(i) end for return 2/n * sum + n end 

The complexity of this algorithm in turn is:

\begin{aligned} T(n) = \begin{cases} c_0 & \text{if } n = 0\\ c_1 + \displaystyle\sum_{i = 0}^{n-1} T(i) & \text{if } n > 0 \end{cases}\end{aligned}

$$T(n-1), n > 1$$ is, using the recursive definition above: $T(n-1) = c_1 + \sum_{i = 0}^{n-2} T(i)$

Using this, we expand $$T(n)$$ and get:

\begin{aligned} T(n) &= c_1 + \sum_{i = 0}^{n-2} T(i) + T(n-1)\\ &= 2T(n-1)\end{aligned}

Thus we get:

\begin{aligned} T(n) = \begin{cases} c_0 & \text{if } n = 0\\ 2^{n-1}(c_0 + c_1) & \text{if } n > 0 \end{cases}\end{aligned}

Meaning $$T(n) \in \mathcal{O}(2^n)$$.

# b) Dynamic programming

The execution tree for $$C(4)$$ is:

\label{fig:execution-tree}

The optimized version - using dynamic programming - of the algorithm in a) becomes:

 Computes the numerical value for C(n) for average quicksort complexity Input: n is the size of the problem. Memory: initialize mem[n] with 0 C(n): if mem[n] = 0 then: if n = 0 then: mem[n] <- 1 end if sum <- 0 for i: 0 .. n-1 do: sum <- sum + C(i) end for mem[n] <- 2/n * sum + n end if return mem[n] end 

From the outset, it is obvious that the memory complexity is $$\mathcal{O}(n)$$

The cost of initializing the memory is: $$n$$ The algorithm will do $$n$$ fills.

= [draw,anchor=center,minimum size=3em, thin]

in 0,1,2,3,4 at (0,-) (m) $$m_\m$$; (5.9/1.5 - /1.5,-) – +(0,); ; in 1,2,3,4

(5.9/1.5 - /1.5, -) circle(0.15em);

in 1, ..., (5.9/1.5 - /1.5, -+ 0.95) – +(-0.5,0);

\label{fig:lookup-calls}

It will also do: $$i - 1$$ lookup for each $$i$$, so the total complexity is:

\begin{aligned} T(n) &= n + n + \sum_{i=0}^{n-1} \left[i-1\right]\\ &= 2n + \sum_{i=0}^{n-1} i + \sum_{i=0}^{n-1} 1\\ &= 3n + \frac{n(n-1)}{2}\\ &= \frac{5n + n^2}{2}\\ &\in \mathcal{O}(n^2) \end{aligned}