2331Concept150-0038150-0038.xmlBig-\mathcal {O}Sometimes, we wish to simplify exact bounds, ignoring linear factors. To do this, we use big-\mathcal {O} notation.Let X be a set and let f, g : X \to \mathbb {N}. We say that f \in \mathcal {O}(g) when there exist constants a, b : \mathbb {N} such that f \le ag + b, i.e. \forall x : X, f(x) \le ag(x) + b.We write \mathcal {O}(g) for the set of all functions f bounded by g, i.e. \mathcal {O}(g) = \{f : X \to \mathbb {N} \mid f \in \mathcal {O}(g) \}.Traditionally, X = \mathbb {N} and function inputs are assumed to be named n: for example, \mathcal {O}(n^2) is syntactic sugar for \mathcal {O}(n \mapsto n^2).2330150-002Z150-002Z.xmlAsymptotic analysis479Concept150-0038150-0038.xmlBig-\mathcal {O}Sometimes, we wish to simplify exact bounds, ignoring linear factors. To do this, we use big-\mathcal {O} notation.Let X be a set and let f, g : X \to \mathbb {N}. We say that f \in \mathcal {O}(g) when there exist constants a, b : \mathbb {N} such that f \le ag + b, i.e. \forall x : X, f(x) \le ag(x) + b.We write \mathcal {O}(g) for the set of all functions f bounded by g, i.e. \mathcal {O}(g) = \{f : X \to \mathbb {N} \mid f \in \mathcal {O}(g) \}.Traditionally, X = \mathbb {N} and function inputs are assumed to be named n: for example, \mathcal {O}(n^2) is syntactic sugar for \mathcal {O}(n \mapsto n^2).482Example150-003B150-003B.xmlBig-\mathcal {O} boundsWe have n^2 \in \mathcal {O}(n^2 + n + 3). 480Proof#199unstable-199.xml150-003B By the definition, we must choose a, b : \mathbb {N} such that n^2 \le a(n^2 + n + 3) + b. Let a = 1 and b = 0; we have n^2 \le n^2 + n + 3 immediately. We have n^2 + n + 3 \in \mathcal {O}(n^2). 481Proof#200unstable-200.xml150-003B By the definition, we must choose a, b : \mathbb {N} such that n^2 + n + 3 \le an^2 + b. Let a = 2 and b = 3; we have n^2 + n + 3 \le 2n^2 + 3 = n^2 + n^2 + 3, since n^2 \le n^2, n \le n^2, and 3 \le 3. 483Theorem150-0039150-0039.xmlProperties of big-\mathcal {O}\mathcal {O} is a preorder: \mathcal {O} is reflexive: for all f, we have f \in \mathcal {O}(f). \mathcal {O} is transitive: for all f,g,h, if f \in \mathcal {O}(g) and g \in O(h), then f \in O(h). If f \in \mathcal {O}(g) and f' \in \mathcal {O}(g'), then f + f' \in \mathcal {O}(g + g'), where (f + f')(x) = f(x) + f'(x). For all f,g : X \to \mathbb {N}, we have \mathcal {O}(f + g) = \mathcal {O}(\max (f, g)). For all a, b : \mathbb {N} and f : X \to \mathbb {N}, we have \mathcal {O}(f) = \mathcal {O}(af + b), where (af + b)(x) = a \cdot f(x) + b by definition.509Concept150-003A150-003A.xmlCommon big-\mathcal {O} classesThe following classes are distinct, ordered by inclusion from top to bottom: Class Common Name \mathcal {O}(1) constant \mathcal {O}(\log n) logarithmic \mathcal {O}(n) linear \mathcal {O}(n \log n) quasilinear/log-linear \mathcal {O}(n^2) quadratic \mathcal {O}(n^3) cubic \mathcal {O}(2^n) exponential 510Example150-003C150-003C.xmlBig-\mathcal {O} bounds for list reverse algorithmsRecall the bounds for revSlow, \frac {1}{2}(n^2 + n), and for rev, n.We have a tight big-\mathcal {O} bound W_\texttt {revSlow}(n) \in \mathcal {O}(n^2). We can prove it using properties of big-\mathcal {O}: \begin {aligned} \frac {1}{2}(n^2 + n) &\in \mathcal {O}\left (\frac {1}{2}(n^2 + n)\right ) \\ &= \mathcal {O}(n^2 + n) \\ &= \mathcal {O}(\max (n^2, n)) \\ &= \mathcal {O}(n^2) \end {aligned} In your analyses, you need not justify your asymptotic bounds formally in this way; for example, you are welcome to immediately state that \frac {1}{2}(n^2 + n) \in \mathcal {O}(n^2) without justification. We also have a tight big-\mathcal {O} bound W_\texttt {rev}(n) \in \mathcal {O}(n), by reflexivity.536Table150-003E150-003E.xmlSolutions to common recurrencesIn the below table, we include solutions to common recurrences, where we let T(0) = c_0. Recurrence T(n) = \cdots Exact Solution \mathcal {O}(-) T(n-1) + c_1 c_0 + c_1n \mathcal {O}(n) 2T(n-1) + c_1 c_02^n + c_1\left (2^n-1\right ) \mathcal {O}(2^n) T(n-1) + c_1n + c_2 c_0 + c_1\frac {n(n+1)}{2} + c_2n \mathcal {O}(n^2) 2T(n-1) + c_1n + c_2 c_02^n + c_1\left (2^{n+1}-n-2\right ) + c_2(2^n - 1) \mathcal {O}(2^n) T(n-1) + c_1\log _2(n) + c_2 c_0 + c_1\log (n!) + c_2n \mathcal {O}(n \log n) 2329Example150-003B150-003B.xmlBig-\mathcal {O} boundsWe have n^2 \in \mathcal {O}(n^2 + n + 3). 480Proof#199unstable-199.xml150-003B By the definition, we must choose a, b : \mathbb {N} such that n^2 \le a(n^2 + n + 3) + b. Let a = 1 and b = 0; we have n^2 \le n^2 + n + 3 immediately. We have n^2 + n + 3 \in \mathcal {O}(n^2). 481Proof#200unstable-200.xml150-003B By the definition, we must choose a, b : \mathbb {N} such that n^2 + n + 3 \le an^2 + b. Let a = 2 and b = 3; we have n^2 + n + 3 \le 2n^2 + 3 = n^2 + n^2 + 3, since n^2 \le n^2, n \le n^2, and 3 \le 3.