When it comes to algorithms, you frequently have to evaluate problems like this:

Let $x$ be an $n$-bit integer. For each of the following questions, give your answer as a function of $n$.

Or a question like this:

[given algorithm] Assume that the subtraction takes $O(n)$ time on an $n$-bit number.

What does "let $x$ be an $n$-bit integer" really mean? It is just the amount of bits reserved for a random int variable $x$? How does the $n$-bit number relate to the big $O$ of $n$ notation?


We usually represent numbers as finite sequence of digits. In base-$2$ each digit is called bit and has value $0$ or $1$. So we write each number $a$ as $\sum_{k=0}^n a_i2^i$, where $a_i$ are digits and assuming $a_n\neq0$ number $a$ is called $n$-bit number. Notice that if $a$ is $n$-bit number then $2^{n-1}\leq a<2^n$. So number of bits of number $a$ is $O(\log a)$.

If algorithm works on individual digits, then the complexity is dependant on length of given number (by length I mean number of bits). For instance how does addition algorithm works? You want to add $a$ and $b$, first you add $a_0$ and $b_0$, write down the result, if necessary carry $1$, continue for next bit and so on. As a result you do $n$ bit additions, so adding two $n$-bit numbers has complexity $O(n)$.

So basicly, saying that algorithm takes time $O(n)$ for $n$-digit number, means that it is linear with respect to the length of given number. And also that it takes time $O(\log n)$ with respect to the number itself.

  • $\begingroup$ ohh its starting to make sense now, see i was unaware that n bit binary is n additions cause you have to add the bits one by one (seeing as i do most adding in decimal notation and there i dont even think about it since its so elementary). what is the $O(log n)$ notation you speak of, i don't get it. $\endgroup$ – notamathwiz Oct 24 '13 at 21:47
  • $\begingroup$ @notamathwiz Notice that when you multiply a number by 2, its length increases by 1. So if number $n$ has $k$ digits, then $k=O(\log n)$ and if algorithm takes time $O(k)$ (linear with respect to number of bits), then it is also running in time $O(\log n)$ (with respect to that number). $\endgroup$ – Adam Oct 24 '13 at 22:41

It asks you how does the space/time of algorithm evaluation growth with $n$. O(n) means linear dependency. $O(n^2)$ is square dependency, $O(e^n)$ is exponential.

  • $\begingroup$ i understand that part, im just trying to grasp the deeper most fundamental meaning of the n-bit int. $\endgroup$ – notamathwiz Oct 24 '13 at 20:56
  • $\begingroup$ It sounds like you are asking about fundamental dependency between a and b in $a = f(b)$ for abstract function $f$. $\endgroup$ – Val Oct 24 '13 at 20:59
  • $\begingroup$ im question is geared more toward understanding what you said about "the growth of n" what does that really mean, does that mean that the more data we input the more n growths? how does that relate to n? say we have a data set that has 10 integers and n-bit is 32-bit does that mean that the space we need to allocate is 320 bits? that's what im trying to understand. $\endgroup$ – notamathwiz Oct 24 '13 at 21:05
  • 2
    $\begingroup$ O(n) means that you may need cO(n) bits in your computer. Your computer must have size cn "bits" for n-bit numbers and c*m "bits" for m-bit arguments. It is not about datasets. It about a single parameter. If you have m numbers, n bits each, then you'll have some O(m,n). Likewise O(n) it is not always linear. $\endgroup$ – Val Oct 24 '13 at 21:20
  • $\begingroup$ particluarly for 10x32 this means that you can add your 10 numbers together one by one, in O(10) time using 32-bit computer (space) or do the addition in parallel in O(m,n)~log(10) time using O(m,n)~10x32 computer. $\endgroup$ – Val Oct 25 '13 at 6:02

$n$ is used for the sake of scaling things in terms of bits. If you didn't know the size of $x$, then lots of complexities would have to be expressed as a log base 2 as you don't know how many bits $x$ has unless it is expressed this way. Consider the question of how many addition operations have to be done on $a$ and $b$ where the addition is done base 2 though the numbers are in base 10? First, you want to know how long is each number in base 2, which could be seen as $n_a$ and $n_b$ which makes the answer the maximum of these values plus one as if the numbers are the same length there may be a carry bit to factor into things here.

The intention of Big O is to understand in the worst case how does the complexity grow relative to the size of the input, often denoted as $n$. Sorting would be one of the classic problems that has been studied to great lengths in terms of the number of comparisons that have to be done in order to sort a list.

Just to give a couple of examples, first consider the BubbleSort algorithm where the algorithm is to compare a pair of elements, and if a pair change order, start all over again at the beginning. So, for example consider this list of numbers that are to be sorted in descending order:

2 5 3 7

Comparing 2 and 5, these aren't in the proper order so we reverse these 2 elements and start again.

5 2 3 7

Comparing 5 and 2, these are in the proper order. Next, we compare 2 and 3 which aren't in the right order and have to be reversed and then we start again.

5 3 2 7

Comparing 5 and 3, these are in the proper order. Next, we compare 3 and 2 which are also in the proper order. Next, 2 and 7 are compared and have to be reversed.

5 3 7 2

Comparing 5 and 3, these are in the right order. Next, 3 and 7 are compared and have to be reversed.

5 7 3 2

Comparing 5 and 7, these aren't in the right order and so we reverse these now.

7 5 3 2

Now, the comparisons all work out and the list is sorted. Notice all the comparisons that had to be done which in the worst case will be $O(n^2)$ as if the list is in reverse order, there are a quadratic number of comparisons to be done.

In contrast, take that list and apply MergeSort:

2 5 3 7

After the initial split and sort each half, as the idea is divide and conquer here:

5 2 7 3

Now, merge the lists:

7 5 3 2

Merge is better in overall complexity in the worst case as the divisions reduce the comparisons to being $O(n \log n)$ which is slightly better.

In your example, if the number of repetitions is constant, then the big O stays the same. However, if the loop was for each bit within a loop for each bit within a loop for each bit, then the complexity is likely to be $O(n^3)$ unless the iterator for the loop is incrementing in a non linear fashion.

  • $\begingroup$ what about a $O(n^2)$? say you have a loop that cycles two times and in the loop you add two x integers that are n-bit. does that mean the big Oh notation is the one above? $\endgroup$ – notamathwiz Oct 24 '13 at 21:50

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