Big-O notation

Roy Smith roy at panix.com
Wed Apr 16 14:50:42 CEST 2003


Peter van Kampen <news at datatailors.xs4all.nl> wrote:
> I think I almost 'get it'. Except who or what decides what Big-O a
> certain algorithm has. Is that by 'hunch' or by running benchmarks or
> are there still other ways to recognize 'Big-O's

You do it by analysis.  Basicly, you find the "inner loop", i.e. the bit 
of core code that gets run over and over while the algorithm does its 
work.  Then, you figure out (by looking at the code) how many times that 
inner loop will run given that the input is a given size.

Sometimes, in a high level language like Python, there are implied loops 
in what look like atomic operations.  This can obscure the real 
algorithmic complexity of a program, unless you know enough to look 
deeper.  Take the classic Python example:

s1 = ""
for s2 in stringList:
   s1 += s2

On the surface, it looks like this should be an O(N) program, where N is 
the number of strings in stringList.  Certainly, for N strings, the loop 
gets executed N times.  But, the string addition has an couple of 
inherent loops of its own.  The most obvious is that each character in 
s2 has to be copied someplace, so it really looks like this:

s1 = ""
for s2 in stringList:
   for c in s2:
      s1 += c

Now it looks like it takes O(k*N) steps, where N is again the number of 
strings in stringList, and k is the average number of characters in a 
string.

Here's the first tricky part -- for the purpose of big-O notation, the k 
is meaningless.  It's essentially a constant factor.  For a given 
universe of strings (say, words in a dictionary), k is not going to 
change as you add more strings.  Sure, it's going to have minor 
variations as the average bops around, but once you've got a statisticly 
valid sample of input strings, it'll pretty much stay put.  k may be a 
function of the kinds of strings you're giving it (k might be bigger for 
German compared to English), but it's not a function of the number of 
strings you're feeding the program.  For big-O purposes, O(k*N) where k 
is a constant is the same as O(N).  What we're trying to describe is the 
shape of the growth curve, not the scaling factors.

And, here's the second tricky part -- since strings in Python are 
immutable, the way you do s1 += s2 is to figure out how long each of the 
two components are, allocate a new string big enough to hold the sum of 
those, and copy the two original strings into the new one.  So, let's 
tear that apart a bit and see what it looks like, complexity wise.

We need to figure out how long each component string is.  On the 
surface, that's O(N), where N is the length of the string.  In C, it 
looks something like:

length = 0;
while (*string != NULL) {
   length++;
   string++;
}

This takes 2*N steps (it does two additions on each of N trips through 
the loop), but big-O says that O(2*N) is the same as O(N).

Fortunately for us, Python doesn't need to do that.  Python stores 
strings in a way where the length is pre-computed and stored.  All it 
really needs to do to get the length is something like "return 
string.length", which is a constant-time operation (i.e. it takes the 
same amount of time regardless of the length of the string).  That's 
called O(1).

So, we could rewrite the whole thing as something like:

s1 = ""
for s2 in stringList:
   len1 = len(s1)
   len2 = len(s2)
   newLen = len1 + len2
   temp = allocate buffer for newLen characters
   for c in s1:
      temp += c
   for c in s2:
      temp += c
   s1 = temp

Now, we analyse the complexity of each bit of that.  It does the outer 
loop N times, but how much work is involved in each iteration of the 
outer loop?  Each of the len()'s we already figured out is constant 
time, so they don't add anything complexity-wise.  Adding the two 
lengths is also constant time.

For the sake of argument, I'm going to assume that memory allocation is 
constant time, although I don't really know that for sure.  To really do 
the analysis right, I'd have to know the details of how Python memory 
management works, and I don't, but I'm reasonably confident that it's 
not going to be a determining factor here.

Now we come to the first interesting part.  The "for c in s1" loop takes 
as many iterations as s1 is long.  Well, how long is s1?  It varies as 
the program progresses, but after we've processed M strings, it's 
approximately k*M characters long (where k is again the average string 
length).  We again toss out the constant k, and come up with the 
complexity of this particular loop being O(M).  But, what's M?  It 
varies from 1 to N, and on average, it's N/2.  But, the constant factor 
of 1/2 doesn't mean anything to big-O, so O(M) = O(N/2) = O(N).

The "for c in s2" loop we've already discussed and decided it was O(k), 
which is the same as O(1), i.e. constant time.

And finally, the rebinding of s1 to the new temporary string is constant 
time.  One minor hitch is that this causes the old s1 to fall out of 
scope and become a candidate for garbage collection.  Again, without 
knowing the details of Python's memory management, I can't say for sure 
what the algorithmic complexity of freeing a string is, but for now I'm 
going to assume it's O(1).

So, overall, we do an O(N) inner loop (plus a lot of constant-time stuff 
that we're ignoring), O(N) times, giving us an overall algorithmic 
complexity of O(N^2).  This is often called quadratic behavior, and is 
generally considered evil and something to be avoided if at all possible.

How evil?  Well, I read recently that the new Python 2.3 release is 
supposed to be 10% to 20% faster than 2.2.  That sounds impressive (and, 
don't get me wrong, it is impressive), but it's bupkis compared to 
algorithm tuning.  Imagine you've got a program which runs in O(N^2) 
time.  If you could find a way to reduce it to O(N), for an input set of 
just 10 items, you would have speeded it up by a factor of 10!  Compared 
to upgrading your Python interpreter, you did 50-100 times better tuning 
the algorithm.  Now try to consider the implications of going from 
O(N^2) to O(N) with 25,000 input items!

BTW, this is perhaps one of the few arguments against teaching Python 
(or any high-level language) as a first programming language.  So much
algorithmicly complex stuff gets pushed down into atomic language 
constructs that there's no longer much of a correlation between code 
complexity and algorithmic complexity.  This, of course, is what makes 
high level languages so useful in the first place, but I fear we may be 
training a new generation of programmers for whom algorithmic complexity 
is not as familiar a concept as it used to be.

I'm not saying we should still be teaching C (or assembler) as a first 
language.  On the other hand, I suspect analysis of algorithms probably 
needs to be emphasized more (and earlier) in the curriculum.  When I 
first learned this stuff, it was mostly just a codification and rigorous 
analysis of concepts I'd already explored experimentally in the normal 
course of writing C programs.  I suspect now it's rather completely new 
ground, and may seem rather esoteric and theoretical and not 
particularly relevant to real-world problems like designing java 
animiations for web pages :-)




More information about the Python-list mailing list