Algorithm complexity analysis

Notes and exercises

  • The goal of algorithm analysis is to study the efficiency of an algorithm in a language and machine-independent way.

  • The two most important tools for this are (1) the RAM model of computation and (2) the asymptotic analysis of worst-case complexity.

  • The Random Access Machine (RAM) model of computation is a simple model of a generic computer that is based on three main assumptions: (1) each simple operation takes exactly one time step, (2) loops and subroutines are considered composites of all simple operations they perform, and (3) memory access from cache and RAM takes one time unit. None of these hold in practice, but the model is extremely useful because it captures the essence of algorithm behaviour while being very simple to work with.

  • Best, worst, and average-case complexity are functions defined by the minimum, maximum, and average number of steps taken for any instance of size n of the input string. (Think about a graph with n on the x-axis and number of steps on the y-axis, with number of steps for each instance of a problem of size n forming columns of dots with increasing variation as n – and thus the number of possible instances – increases. The three functions trace the lowest, highest, and middle dots at each input size n. See Fig 2.1 in ADM.)

  • Using these functions to analyse algorithms is impractical, however, because they are not smooth and require lots of detail about the algorithm and its implementation.

  • Big O notation ignores such details and focuses on the essentials to capture the rate at which runtime (or space requirements) grow as a function of the input size (the letter O is used because the growth rate of a function is also called its order). In essence, this means only focusing on the higest order term and ignoring constants (which depend on things like hardware and programming language used to run the algorithm).

  • A function $f(n)$ is $O(g(n))$ if there exist constants $c$ and $n_0$ such that $f(n) \leq cg(n)$ for any $n > n_0$. Intuitively, this means that $f(n)$ grows no faster than $cg(n)$ above a certain input size. For example: $T(n) = 2n^2 + 3n$ is $O(n^2)$, since $5n^2 \geq 2n^2 + 3n$ for all positive values of $n$.

  • Amortised worse-case complexity takes into account that the running time of a given operation in an algorithm may take a very long or a very short time depending on the situation, and averages those different running times of the operation in a sequence over that sequence. Adding an element to an array that is dynamically resized takes $O(1)$ time until the array is full, when the array needs to create a new array of twice its original size, copy all elements over to the new array, and add the new element, which takes $O(n)$ time. Average worst-case complexity averages these runtimes to find that pushing elements onto a dynamically resized array takes: $\frac{nO(1) + O(n)}{n + 1} = O(1)$, constant time. (Source)


  1. Is $2^{n+1} = O(2^n)$?

  2. Is $(x + y)^2 = O(x^2 + y^2)$?

  3. What’s the time complexity of $f(n) = min(n, 100)$?


  1. The way to go is to start from the definition. The statement is true if there is a $c$ and $n_0$ for which $c2^n \geq 2^{n+1}$ for $n > n_0$. The key is to rewrite the right hand side to $c2^n \geq 2 \times 2^n$, which makes it obvious that the statement holds whenever $c \geq 2$.

  2. Starting from the definition, the statement is true if there exist constants $c$ and $n_0$ for which $c(x^2 + y^2) \geq (x + y)^2$ for $n > n_0$. Expanding the right hand side, we get $c(x^2 + y^2) \geq x^2 + 2xy + y^2$. Ignoring the middle term, the statement holds for $c = 1$; considering only the middle term, we see that it is largest when $x = y$, in which case the statement holds for $c = 2$. Thus, $3(x^2 + y^2) \geq (x + y)^2$, so the statement is true.

  3. I reflexively answered $n$. Thinking for a moment (an embarassingly long one, admittedly), I realised that $n$ here refers not to the length of an array but to a single number. So the operation is $O(1)$.