Consider the binary search problem on a sorted array containing $n$ integers on 16 bits. Everybody agrees that the binary search needs $O(\log(n))$ time, because it makes at worst $O(\log(n))$ steps. But: at each step it needs to calculate a midpoint. The first midpoint is $[n/2]$. The next midpoint could be $[n/2]+[n/4]$. If the searched number is placed in the second half of the array, the midpoint at each step is no less than $[n/2]$. Each calculation of a new midpoint involves at least a number with $O(\log(n))$ bits, hence each midpoint calculation requires $O(\log(n))$. I obtain a more precise complexity of $O(\log(n)^2)$. Where is the error?

The same problem could appear on countless algorithms that use arrays or matrices. Take the most well-known dynamic programming algorithm for the knapsack problem. Everybody agrees it takes $O(nW)$ time (e.g., the wikipedia article on the knapsack problem), where $n$ is the number of items and $W$ is the capacity. But at each step, it needs to compute a difference of weights/capacities, which should account for an additional factor of $O(\log(W))$. Where is the error?

Where/how does this $O(\log(n))$ complexity factor disappear? If we had needed it, it would have appeared in many algorithms.

ps. A very similar problem arises in the questions asked here and here, thanks Ariel, Raphael and ryan for pointing this out.

  • $\begingroup$ Your statement "hence each midpoint calculation requires $O(log(n))$" is not true. In binary search $mid = (left+right)/2$ is $O(1)$. $\endgroup$
    – fade2black
    Commented Jul 22, 2017 at 21:07
  • $\begingroup$ Thanks for this quick reply. Don't forget that $left$, $right$ and $mid$ might be written on $O(\log(n))$ bits, e.g., $left=[n/2]$, $right=n$, and $mid=[([n/2]+n)/2]$. What machine can add numbers on $O(\log(n))$ bits in $O(1)$? $\endgroup$ Commented Jul 22, 2017 at 21:21
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    $\begingroup$ If you are using RAM computation model then as fade2black stated $\mathcal O(1)$ is the cost of array access and primitive operations involved. It seems that you use different model than one with the conclusion given hence the discrepancy. Also keep in mind that numbers are bounded, it is fair enough to treat it as constant. In the modern processors these operations are performed in constant number of cycles. When arithmetic exceeds the built-in size then it is no longer constant. $\endgroup$
    – Evil
    Commented Jul 22, 2017 at 21:22
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    $\begingroup$ This logarithmic factor should indeed appear unless you're dealing with a RAM machine. See Yuval's answer here cs.stackexchange.com/a/75960/27055 $\endgroup$
    – Ariel
    Commented Jul 22, 2017 at 21:30
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    $\begingroup$ See also our reference question on the matter. $\endgroup$
    – Raphael
    Commented Jul 23, 2017 at 10:33

2 Answers 2


You're not wrong, you're just using a different cost model.

Typically there are two:

  • Uniform cost model - assigns a constant cost to every machine operation regardless of size of numbers.

  • Logarithmic cost model - assigns a cost proportional to the number of bits involved for each machine operation.

Under uniform cost, a binary search takes constant operations to find the next index to search at, we get a recurrence relations like so: $$\begin{align} T(1) & = O(1)\\[0.5em] T(n) & = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + O(1)\\[0.5em] & = \Theta(\log n) \end{align}$$

Under logarithmic cost, we can assume it takes $f(n)$ operations to find the next midpoint and we get a recurrence like so: $$\begin{align} T(1) & = O(1)\\[0.5em] T(n) & = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + f(n)\\[0.5em] \end{align}$$ For binary search, typically $f(n)$ will be no more than computing $\lfloor \frac{l + r}{2} \rfloor$ which would take $\log_2 n$ operations for the addition, and to divide by $2$ we could do a bit shift taking $\log_2 n$ operations (see here). Which brings us to: $$\begin{align} T(n) & = T\left( \left\lfloor \frac{n}{2} \right\rfloor \right) + \Theta(\log n)\\[0.5em] & = \Theta(\log^2 n) \end{align}$$ These logarithmic costs are just for addition and shifting though. If you were to do other $n$-bit operations then $f(n)$ could change. For example, $n$-bit multiplication and exponentiation could be larger than $\Omega(n)$ and you would have to adjust you complexity accordingly under logarithmic cost.

  • $\begingroup$ Thanks for the reply; it is clear. Could you please expand on how the uniform and log cost models relate to the various (Transdichotomous or word) RAM machines out there? $\endgroup$ Commented Jul 23, 2017 at 11:50
  • $\begingroup$ @DanielPorumbel, perhaps JeffE's answer here might help for understanding it in context of RAM machines. Also this wiki article might help. $\endgroup$
    – ryan
    Commented Jul 23, 2017 at 19:04

Usually people neglect time it takes to perform simple math operations on numbers and assume them to be $O(1)$. In practice, computers work with words, so we assuming bounded number. And it uses summator to add , which is extremely fast, comparing to other $O(log(n))$ operations.

Indeed, if you go really deep, and you have unbounded numbers $n,m$, Turing machine needs $O(log(n)+log(n))$ to add they assuming they are not in unary. With your array problem, I'd suggest looking carefully what is your input size and write complexity very carefully even for math operations, I expect them to be neglecting small. Like size of your array versus bound on numbers in the array and so on.

  • $\begingroup$ Yes, but in my case, the number of digits $\log(n)$ is not bounded so a computer that works with words can not put $\log(n)$ bits on a word. It needs to use something else, and not a fast summator. Binary search is just an example, other algorithms might even do some multiplications, e.g., knapsack dynamic programming, where we consider the profit of using $k$ times item $i$, something like profit$(w,i)=max($profit$(w,i-1), $profit$(w-k\cdot w_i,i-1)+k\cdot p_i)$. Regarding the Turing Machine, it is slower than the RAM machine. The input size for binary search is simple: 16$\cdot n$. $\endgroup$ Commented Jul 22, 2017 at 21:37

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