Time complexity and recurrence relations

Problem 1

1.1

Consider the following function of $n$ :

\begin{aligned} &n^2 \quad \sum_{i=0}^{n}5\cdot{i} \quad n^3\cdot \sqrt{\frac{1}{n^3}} \quad n^2 + 2^n \quad (\Pi_{i=1}^{9}i) \quad (\sum_{i=0}^{\log_2(n)}2^i) + 1 \quad 7^{\ln{(n)}} \\ &-\ln{(\frac{1}{n})} \quad \ln{(2^n)} \quad 10 \quad n\log_2{(n^7)} \quad \sqrt{n^4} \quad n^n \quad 5n \\ \end{aligned}

Group the above functions that have identical growth rate and order these groups on increasing growth. Hence:

If you place functions $f_1(n)$ and $f_2(n)$ in the same group, then we must have $f_1(n) = \Theta{(f_2(n))}$ ;
If you place function $f_1(n)$ in a group ordered before the group in which you place function $f_2(n)$ , then we must have $f_1(n) = \mathcal{O}{(f_2(n))} \land f_1(n) \neq \Omega{(f_2(n))}$ .

Solution:

Theorem 3.25 states the following:

\lim_{n \to \infty} \frac{f(n)}{g(n)} \space \text{is defined and is} \space \begin{cases} \infty & \space \text{then} \space f(n) = \Omega(g(n)); \\ c \text{, with c > 0 a constant} & \space \text{then} \space f(n) = \Theta(g(n)); \\ 0 & \space \text{then} \space f(n) = \mathcal{O}(g(n)); \end{cases}

The following are the rules of thumb to determine the order of growth of functions:

\begin{align} c \cdot f(n) &= \Theta(f(n)) \\ \log_{a}(n) &= \Theta(\log_{b}(n)) \\ \lim_{n \to \infty} \frac{n^c}{n^{c+d}} &= \lim_{n \to \infty} \frac{1}{n^d} = 0 \rightarrow n^c = \mathcal{O}(n^{c+d}) \\ \log_{2}(n)^c &= \mathcal{O}(n^d) \space \forall c > 0, d > 0 \\ n^c &= \mathcal{O}(d^{n}) \space \forall c > 0, d > 1 \\ d^{\frac{n}{u}} &= \mathcal{O}(c^{\frac{n}{v}}) \space \forall c \geq d \geq 1, u \geq v \geq 1 \\ \sum_{i=1}^{m}{c_{i} \cdot n^{d_i}} &= \mathcal{O}(n^{d_i}) \\ f(n) + g(n) &= \Theta(g(n)) \\ h(n) \cdot n(n) &= \mathcal{O}({h(n) \cdot g(n)}) \end{align}

Let’s start with analysing the functions that have the same growth rate:

$n^2$ : polynomial growth, $n^2 = \Theta(n^2)$ based on rule 3.
$\sum_{i=0}^{n}5\cdot{i}$ : polynomial growth, $\sum_{i=0}^{n}5\cdot{i} = 5 \cdot \frac{n(n+1)}{2} = \frac{5}{2}n^2 + \frac{5}{2}n$ , which is $\Theta(n^2)$ based on rule $7$ .
$n^3 \cdot \sqrt{\frac{1}{n^3}}$ : polynomial growth, $= n^3 \cdot \frac{1}{n^{\frac{3}{2}}} \rightarrow \Theta(n^{\frac{3}{2}})$ based on rule $9, 3$
$n^2 + 2^n$ : exponential growth, $=\Theta(2^n)$ based on rule $7$
$(\Pi_{i=1}^{9}i)$ : constant, $=1 \cdot 2 \dots 9$ based on rule 9
$\sum_{i=0}^{\log_2{(n)}}{2^i} + 1$ : linear, since the term is a geometrics series, such that $\sum_{i=0}^{\log_2{(n)}}{2^i} + 1 = \frac{1 \cdot (2^{\log_2(n)+1}-1)}{2-1} + 1 = 2 \cdot 2^{\log_2(n)} = 2n$ . Therefore based on rule $1$ we have $\Theta(n)$
$7^{\ln(n)}$ : polynomial, since $7^{\ln(n)} = n^{\ln(7)}$ . Apply rule $3$ we have $7^{\ln(n)} = \Theta(n^{\ln 7}) \rightarrow \mathcal{O}(n^2)$
$-\ln(\frac{1}{n})$ : logarithmic, since $-\ln (\frac{1}{n}) = \ln(n)$ . Apply rule $1$ yield $\Theta{(\ln{n})}$ .
$\ln (2^n)$ : linear, since $\ln(2^n) = n \cdot \ln_{2}$ and rule $1$ yield $\Theta{(n)}$
$10$ : constant
$n\log_2(n^7)$ : polynomial, since $n\log_{2}(n^7) = n \cdot 7\log_{2}{(n)} = 7n\log_{2}(n)$ . Based on rule $9$ and $1$ yield $\mathcal{O}(n^2)$
$\sqrt{n^4}$ : polynomial growth, $=n^2$ based on rule 3.
$n^n$ : super-exponential.
$5n$ : linear based on rule 1, gives $\Theta(n)$

Thus the order from increasing rate:

$10 \quad (\Pi_{i=1}^{9}i)$ (constant)
$-\ln(\frac{1}{n})$ (logarithmic)
$\ln (2^n) \quad 5n \quad \sum_{i=0}^{\log_2{(n)}}{2^i} + 1$ (linear)
$n^2 \quad \sum_{i=0}^{n}5\cdot{i} \quad n^3 \cdot \sqrt{\frac{1}{n^3}} \quad \sqrt{n^4} \quad n\log_2(n^7) \quad 7^{\ln(n)}$ (polynomial)
$n^2 + 2^n$ (exponential)
$n^n$ (super-exponential)

1.2

Consider the following recurrence

T(n) = \begin{cases} 7 & \text{if } n \leq 1;\\ 3T(n-2) & \text{if } n > 1. \end{cases}

Use induction to prove that $T(n) = f(n)$ with $f(n) = 7 \cdot 3^{[\frac{n}{2}]}$

Solution:

base case:

Given $T(n) = f(n) = 7 \cdot 3^{[\frac{n}{2}]}$ apply for $n=0$ and $n=1$ :

$n = 1 \rightarrow T(n) = 7 \cdot 3^{[\frac{1}{2}]} = 7 \cdot 3^0 = 7$
$n = 0 \rightarrow T(n) = 7 \cdot 3^{[\frac{0}{2}]} = 7 \cdot 3^0 = 7$

Thus the base case holds.

induction hypothesis:

Assume $T(n) = f(n)$ holds for $n > 1$ . We will proves that $f(n+2) = T(n+2)$ also holds.

\begin{align} f(n) &= T(n) \\ T(n) = f(n) &= 3T(n-2) = 3f(n-2) = 3 \cdot 7 \cdot 3^{[\frac{n-2}{2}]} \\ T(n+2) &= 3T(n) \end{align}

Therefore:

\begin{align} T(n+2) &= 3 \cdot 3 \cdot 7 \cdot 3^{[\frac{n-2}{2}]} = 7 \cdot 3^{[\frac{n-2}{2} + 2]} \\ T(n+2) &= 7 \cdot 3^{[\frac{n+2}{2}]} = f(n+2) \end{align}

Therefore the induction hypothesis holds.

conclusion:

$T(n) = f(n) = 7 \cdot 3^{[\frac{n}{2}]}$ is true for all $n \geq 0$ .

Problem 2

Consider the following Count algorithm

Algorithm Count(L, v):
Pre: L is an array, v is a value
i, c := 0, 0
while i neq |L| do
  if L[i] = v then
    c := c + 1
 end if
 i := i + 1
end while
return c.
Post: return the number of copies of v in L

2.1

Provide an invariant for the while loop at Line 2

Solution:

0 \leq i \leq |L|, c = \sum_{j=0}^{i-1} \lbrack L \lbrack j \rbrack = v \rbrack

2.2

Provide a bound function for the while loop at Line 2

Solution:

f(i) = |L| - i

2.3

Prove that Count algorithm is correct.

Solution:

Line 1: i, c := 0, 0

$L \lbrack 0, i)$ with $i=0$ is $L \lbrack 0, 0)$
$L \lbrack 0,0)$ is empty, hence $c = \sum_{j=0}^{i-1} \lbrack L \lbrack j \rbrack = v \rbrack = 0$
bound function $f(i) = |L| - i$ starts at $|L|, |L| \geq 0$

Line 2: while i neq |L| do

bound function $f(i)$ stops at $0$
invariant still holds, with $i \neq |L|$

Now prove invariant holds again til reach the end of the m-th loop:

Line 3-5: if L[i] = v then case:

$L \lbrack i \rbrack = v$ hence $v \in L\lbrack 0,i \rbrack$
invariant still holds, with $i \neq |L|$ and $L\lbrack v \rbrack = v$
$i_{\text{new}} = i + 1$ hence $0 < i_{\text{new}} \leq |L|$ implies $0 \leq i*{\text{new}} \leq |L|$ and $f(i*{\text{new}}) = f(i) - 1$
$c_{\text{new}} = c + 1 = \sum_{j=0}^{i-1} \lbrack L \lbrack j \rbrack = v \rbrack + 1 = \sum_{j=0}^{i_{\text{new}}} \lbrack L \lbrack j \rbrack = v \rbrack$
$f(i)$ strictly decreases after each iteration, $i_{\text{new}} := i + 1$

Therefore the invariant still holds within the if statement.

L7: end while

$i = |L|$ hence $f(i) = 0$ , the loop stops
$c = \sum_{j=0}^{i-1} \lbrack L \lbrack j \rbrack = v \rbrack = \sum_{j=0}^{|L|-1} \lbrack L \lbrack j \rbrack = v \rbrack$

Therefore the invariant still holds at the end of the loop.

2.4

What is the runtime and memory complexity of Count algorithm?

Solution:

L1 implies 2 instructions
L2 implies 2 instructions $|L| + 1$ times,
L3-5 (if loop) implies 4 instructions $|L|$ times
L6 implies 2 instructions $|L|$ times

Therefore number of work is $5 + 8N$ , thus runtime complexity would be $\Theta(5+8N)$ .

Memory complexity is $\Theta(1)$ , since only 2 variables are used.

2.5

Provide an algorithm FastCount(L, v) operates on ordered lists L and computes the same results as Count(L, v) but with a $\mathcal{O}(\log_2(|L|))$

Solution:

Algorithm FastCount(L, v):
Pre: L is an ordered array, v is a value
function binarySearchFirst(L, v)
  low, high := 0, |L| - 1
  results := -1
  while low <= high do
    mid := (low + high) / 2
    if L[mid] < v then
      low := mid + 1
    else if L[mid] > v then
      high := mid - 1
    else
      results := mid
      high := mid - 1
  end while
  return results
end function
 
function binarySearchLast(L, v)
  low, high := 0, |L| - 1
  results := -1
  while low <= high do
    mid := (low + high) / 2
    if L[mid] < v then
      low := mid + 1
    else if L[mid] > v then
      high := mid - 1
    else
      results := mid
      low := mid + 1
  end while
  return result
end function
 
firstIndex := binarySearchFirst(L, v)
if firstIndex = -1 then
  return 0
end if
lastIndex := binarySearchLast(L, v)
return lastIndex - firstIndex + 1
Post: return the number of copies of v in L

Time complexity and recurrence relations

Étiquette

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Problem 1

1.1

1.2

base case:

induction hypothesis:

conclusion:

Problem 2

2.1

2.2

2.3

2.4

2.5

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