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Algorithms - Lab 3, Problem 1

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

Group 7 consists of:

**Mazdak Farrokhzad**901011-0279

twingoow@gmail.com

Program: IT

**Niclas Alexandersson**920203-0111

nicale@student.chalmers.se

Program: IT

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

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

[->, >=stealth’,shorten >=1pt,auto,node distance=3.5cm,semithick] [.\(C(1)\) \(C(0)\) ] [.\(C(2)\) \(C(0)\) [.\(C(1)\) \(C(0)\) ] ] [.\(C(3)\) \(C(0)\) [.\(C(1)\) \(C(0)\) ] [.\(C(2)\) \(C(0)\) [.\(C(1)\) \(C(0)\) ] ] ] ]

\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.

[>=stealth,shorten >=2pt,thick]

= [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:

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

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

The complexity of this algorithm is:

\[\begin{aligned} T(n) &= c_0 + \sum_{i=1}^{n-1} c_1\\ &= c_0 + c_1 + c_1 n\\ &\in \mathcal{O}(n) \end{aligned}\]

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