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69 lines
4.2 KiB
Org Mode
69 lines
4.2 KiB
Org Mode
1 year ago
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* Asymptotic Notations
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** Omega Notation [ $\Omega$ ]
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+ It is used to shown the lower bound of the algorithm.
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+ For any positive integer $n_0$ and a positive constant $c$, we say that, $f(n) = \Omega (g(n))$ if \[ f(n) \ge c.g(n) \ \ \forall n \ge n_0 \]
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+ So growth rate of $g(n)$ should be less than or equal to growth rate of $f(n)$
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*Note* : If $f(n) = O(g(n))$ then $g(n) = \Omega (f(n))$
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** Theta Notation [ $\theta$ ]
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+ If is used to provide the asymptotic *equal bound*.
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+ $f(n) = \theta (g(n))$ if there exists a positive integer $n_0$ and a positive constants $c_1$ and $c_2$ such that \[ c_1 . g(n) \le f(n) \le c_2 . g(n) \ \ \forall n \ge n_0 \]
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+ So the growth rate of $f(n)$ and $g(n)$ should be equal.
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*Note* : So if $f(n) = O(g(n))$ and $f(n) = \Omega (g(n))$, then $f(n) = \theta (g(n))$
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** Little-Oh Notation [o]
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+ The little o notation defines the strict upper bound of an algorithm.
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+ We say that $f(n) = o(g(n))$ if there exists positive integer $n_0$ and positive constant $c$ such that, \[ f(n) < c.g(n) \ \ \forall n \ge n_0 \]
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+ Notice how condition is <, rather than $\le$ which is used in Big-Oh. So growth rate of $g(n)$ is strictly greater than that of $f(n)$.
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** Little-Omega Notation [ $\omega$ ]
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+ The little omega notation defines the strict lower bound of an algorithm.
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+ We say that $f(n) = \omega (g(n))$ if there exists positive integer $n_0$ and positive constant $c$ such that, \[ f(n) > c.g(n) \ \ \forall n \ge n_0 \]
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+ Notice how condition is >, rather than $\ge$ which is used in Big-Omega. So growth rate of $g(n)$ is strictly less than that of $f(n)$.
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* Comparing Growth rate of funtions
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** Applying limit
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To compare two funtions $f(n)$ and $g(n)$. We can use limit
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\[ \lim_{n\to\infty} \frac{f(n)}{g(n)} \]
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+ If result is 0 then growth of $g(n)$ > growth of $f(n)$
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+ If result is $\infty$ then growth of $g(n)$ < growth of $f(n)$
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+ If result is any finite number (constant), then growth of $g(n)$ = growth of $f(n)$
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*Note* : L'Hôpital's rule can be used in this limit.
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** Using logarithm
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Using logarithm can be useful to compare exponential functions. When comaparing functions $f(n)$ and $g(n)$,
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+ If growth of $\log(f(n))$ is greater than growth of $\log(g(n))$, then growth of $f(n)$ is greater than growth of $g(n)$
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+ If growth of $\log(f(n))$ is less than growth of $\log(g(n))$, then growth of $f(n)$ is less than growth of $g(n)$
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+ When using log for comparing growth, comaparing constants after applying log is also required. For example, if functions are $2^n$ and $3^n$, then their logs are $n.log(2)$ and $n.log(3)$. Since $log(2) < log(3)$, the growth rate of $3^n$ will be higher.
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+ On equal growth after applying log, we can't decide which function grows faster.
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** Common funtions
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Commonly, growth rate in increasing order is
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\[ c < c.log(log(n)) < c.log(n) < c.n < n.log(n) < c.n^2 < c.n^3 < c.n^4 ... \]
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\[ n^c < c^n < n! < n^n \]
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Where $c$ is any constant.
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* Properties of Asymptotic Notations
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** Big-Oh
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+ *Product* : \[ Given\ f_1 = O(g_1)\ \ and\ f_2 = O(g_2) \implies f_1 f_2 = O(g_1 g_2) \] \[ Also\ f.O(g) = O(f g) \]
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+ *Sum* : For a sum of two functions, the big-oh can be represented with only with funcion having higer growth rate. \[ O(f_1 + f_2 + ... + f_i) = O(max\ growth\ rate(f_1, f_2, .... , f_i )) \]
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+ *Constants* : For a constant $c$ \[ O(c.g(n)) = O(g(n)) \], this is because the constants don't effect the growth rate.
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** Properties
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# Taken from https://www.youtube.com/watch?v=pmGau4xHjFM&ab_channel=UnacademyComputerScience (Analysis of an Algorithm - 2 | L 2 | Algorithms | Infinity Batch | GATE 2022 CS/IT | Ankush Sir)
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[[file:./imgs/asymptotic-notations-properties.png]]
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+ *Reflexive* : $f(n) = O(f(n)$ and $f(n) = \Omega (f(n))$ and $f(n) = \theta (f(n))$
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+ *Symmetric* : If $f(n) = \theta (g(n))$ then $g(n) = \theta (f(n))$
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+ *Transitive* : If $f(n) = O(g(n))$ and $g(n) = O(h(n))$ then $f(n) = O(h(n))$
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+ *Transpose* : If $f(n) = O(g(n))$ then we can also conclude that $g(n) = \Omega (f(n))$ so we say Big-Oh is transpose of Big-Omega and vice-versa.
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+ *Antisymmetric* : If $f(n) = O(g(n))$ and $g(n) = O(f(n))$ then we conclude that $f(n) = g(n)$
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+ *Asymmetric* : If $f(n) = \omega (g(n))$ then we can conclude that $g(n) \ne \omega (f(n))$
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