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aa:lab:notations [2016/11/02 18:02]
dalex
aa:lab:notations [2016/11/02 18:18] (current)
dalex
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 Let $math[f(n) = \log(n)] and $math[g(n) = n]\\ Let $math[f(n) = \log(n)] and $math[g(n) = n]\\
 We see that, for $math[c = 1, n_0 = 1 \Rightarrow 0 \le \log(n) \le n]\\ We see that, for $math[c = 1, n_0 = 1 \Rightarrow 0 \le \log(n) \le n]\\
-for $math[c = 10, n_0 = 64 \Rightarrow 0 \le 10\log(n) \le n]\\ +for $math[c = 10, n_0 = 64 \Rightarrow 0 \le 10 \log(n) \le n]\\ 
-for $math[c = 100, n_0 = 1024 \Rightarrow 0 \le 100\log(n) \le n]\\+for $math[c = 100, n_0 = 1024 \Rightarrow 0 \le 100 \log(n) \le n]\\
 etc.\\ etc.\\
 Thus, $math[g(n) \in \omega(f(n))] Thus, $math[g(n) \in \omega(f(n))]
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 Note that equations are not symmetric and should only be read from left to right. Consider: Note that equations are not symmetric and should only be read from left to right. Consider:
 $$\Theta(n) = O(n)$$ $$\Theta(n) = O(n)$$
-While it is true that, for any function in $math[\Theta(n)] there is a function equal to it in $math[O(n)],​ we can clearly see that there are functions in $math[O(n)] for which there is no correspondent in $math[\Theta(n)] (e.g. $math[f(n) = 1, $f(n) = \log(n)] etc.)+While it is true that, for any function in $math[\Theta(n)] there is a function equal to it in $math[O(n)],​ we can clearly see that there are functions in $math[O(n)] for which there is no correspondent in $math[\Theta(n)] (e.g. $math[f(n) = 1, f(n) = \log(n)] etc.)
  
 As a rule, each asymptotic notation on the left side of the equal sign should be read as an **universally quantified** function ($math[\forall f]) from that class and each asymptotic notation on the right should be read as an **existentially quantified** function ($math[\exists g]) from that class. As a rule, each asymptotic notation on the left side of the equal sign should be read as an **universally quantified** function ($math[\forall f]) from that class and each asymptotic notation on the right should be read as an **existentially quantified** function ($math[\exists g]) from that class.
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 $$\left(\frac{\omega(n^2)}{\Theta(n)}\right) = \Omega(n) + o(n)$$ $$\left(\frac{\omega(n^2)}{\Theta(n)}\right) = \Omega(n) + o(n)$$
  
-$math[\forall f \in \omega(n^2)\ and\ \forall g \in \Theta(n),\ \exists h \in \Omega(n)] and $math[\exists j \in o(n)]  such that+$math[\forall f \in \omega(n^2)\ and\ 
 + \forall g \in \Theta(n),\ \exists h \in \Omega(n)] and $math[\exists j \in o(n)]  such that
 $math[\left(\frac{f(n)}{g(n)}\right) = h(n) + j(n),\ \forall n \in \mathbb{R}^{+}] $math[\left(\frac{f(n)}{g(n)}\right) = h(n) + j(n),\ \forall n \in \mathbb{R}^{+}]
  
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   * $math[\log(n\cdot \log(n))\in\Theta(\log(n))]   * $math[\log(n\cdot \log(n))\in\Theta(\log(n))]
   * $math[\sqrt{n}\in\omega(\log(n))]   * $math[\sqrt{n}\in\omega(\log(n))]
-  * $math[f(n) + g(n) \in O(n\cdot\log(n))for $f(n)\in\Theta(n)and $g(n)\in O(n\cdot\log n)]+  * $math[f(n) + g(n) \in O(n\cdot\log(n))for $math[f(n)\in\Theta(n)and $math[g(n)\in O(n\cdot\log n)]
  
 ===== Exercises (syntactic sugars) ===== ===== Exercises (syntactic sugars) =====
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 ===== Practice ===== ===== Practice =====
 Prove/​disprove the following: Prove/​disprove the following:
-  * $math[f(n) = \Omega(\log(n))and $g(n)=O(n) \implies f(n)=\Omega(\log(g(n))] +  * $math[f(n) = \Omega(\log(n))and $math[g(n)=O(n) \implies f(n)=\Omega(\log(g(n))] 
-  * $math[f(n) = \Omega(\log(n))and $g(n)=O(n) \implies f(n)=\Theta(\log(g(n))] +  * $math[f(n) = \Omega(\log(n))and $math[g(n)=O(n) \implies f(n)=\Theta(\log(g(n))] 
-  * $math[f(n) = \Omega(g(n))and $g(n)=O(n^2) \implies \frac{g(n)}{f(n)}=O(n)]+  * $math[f(n) = \Omega(g(n))and $math[g(n)=O(n^2) \implies \frac{g(n)}{f(n)}=O(n)]
  
 Find two functions $math[f] and $math[g] such that: Find two functions $math[f] and $math[g] such that: