I'll do my best to explain it here on simple terms, but be warned that this topic takes my students a couple of months to finally grasp. You can find more information on the Chapter 2 of the Data Structures and Algorithms in Java book.
There is no mechanical procedure that can be used to get the BigOh.
As a "cookbook", to obtain the BigOh from a piece of code you first need to realize that you are creating a math formula to count how many steps of computations get executed given an input of some size.
The purpose is simple: to compare algorithms from a theoretical point of view, without the need to execute the code. The lesser the number of steps, the faster the algorithm.
For example, let's say you have this piece of code:
int sum(int* data, int N) {
int result = 0; // 1
for (int i = 0; i < N; i++) { // 2
result += data[i]; // 3
}
return result; // 4
}
This function returns the sum of all the elements of the array, and we want to create a formula to count the computational complexity of that function:
Number_Of_Steps = f(N)
So we have f(N), a function to count the number of computational steps. The input of the function is the size of the structure to process. It means that this function is called such as:
Number_Of_Steps = f(data.length)
The parameter N takes the data.length value. Now we need the actual definition of the function f(). This is done from the source code, in which each interesting line is numbered from 1 to 4.
There are many ways to calculate the BigOh. From this point forward we are going to assume that every sentence that doesn't depend on the size of the input data takes a constant C number computational steps.
We are going to add the individual number of steps of the function, and neither the local variable declaration nor the return statement depends on the size of the data array.
That means that lines 1 and 4 takes C amount of steps each, and the function is somewhat like this:
f(N) = C + ??? + C
The next part is to define the value of the for statement. Remember that we are counting the number of computational steps, meaning that the body of the for statement gets executed N times. That's the same as adding C, N times:
f(N) = C + (C + C + ... + C) + C = C + N * C + C
There is no mechanical rule to count how many times the body of the for gets executed, you need to count it by looking at what does the code do. To simplify the calculations, we are ignoring the variable initialization, condition and increment parts of the for statement.
To get the actual BigOh we need the Asymptotic analysis of the function. This is roughly done like this:
- Take away all the constants
C.
- From
f() get the polynomium in its standard form.
- Divide the terms of the polynomium and sort them by the rate of growth.
- Keep the one that grows bigger when
N approaches infinity.
Our f() has two terms:
f(N) = 2 * C * N ^ 0 + 1 * C * N ^ 1
Taking away all the C constants and redundant parts:
f(N) = 1 + N ^ 1
Since the last term is the one which grows bigger when f() approaches infinity (think on limits) this is the BigOh argument, and the sum() function has a BigOh of:
O(N)
There are a few tricks to solve some tricky ones: use summations whenever you can.
As an example, this code can be easily solved using summations:
for (i = 0; i < 2*n; i += 2) { // 1
for (j=n; j > i; j--) { // 2
foo(); // 3
}
}
The first thing you needed to be asked is the order of execution of foo(). While the usual is to be O(1), you need to ask your professors about it. O(1) means (almost, mostly) constant C, independent of the size N.
The for statement on the sentence number one is tricky. While the index ends at 2 * N, the increment is done by two. That means that the first for gets executed only N steps, and we need to divide the count by two.
f(N) = Summation(i from 1 to 2 * N / 2)( ... ) =
= Summation(i from 1 to N)( ... )
The sentence number two is even trickier since it depends on the value of i. Take a look: the index i takes the values: 0, 2, 4, 6, 8, ..., 2 * N, and the second for get executed: N times the first one, N - 2 the second, N - 4 the third... up to the N / 2 stage, on which the second for never gets executed.
On formula, that means:
f(N) = Summation(i from 1 to N)( Summation(j = ???)( ) )
Again, we are counting the number of steps. And by definition, every summation should always start at one, and end at a number bigger-or-equal than one.
f(N) = Summation(i from 1 to N)( Summation(j = 1 to (N - (i - 1) * 2)( C ) )
(We are assuming that foo() is O(1) and takes C steps.)
We have a problem here: when i takes the value N / 2 + 1 upwards, the inner Summation ends at a negative number! That's impossible and wrong. We need to split the summation in two, being the pivotal point the moment i takes N / 2 + 1.
f(N) = Summation(i from 1 to N / 2)( Summation(j = 1 to (N - (i - 1) * 2)) * ( C ) ) + Summation(i from 1 to N / 2) * ( C )
Since the pivotal moment i > N / 2, the inner for won't get executed, and we are assuming a constant C execution complexity on its body.
Now the summations can be simplified using some identity rules:
- Summation(w from 1 to N)( C ) = N * C
- Summation(w from 1 to N)( A (+/-) B ) = Summation(w from 1 to N)( A ) (+/-) Summation(w from 1 to N)( B )
- Summation(w from 1 to N)( w * C ) = C * Summation(w from 1 to N)( w ) (C is a constant, independent of
w)
- Summation(w from 1 to N)( w ) = (N * (N + 1)) / 2
Applying some algebra:
f(N) = Summation(i from 1 to N / 2)( (N - (i - 1) * 2) * ( C ) ) + (N / 2)( C )
f(N) = C * Summation(i from 1 to N / 2)( (N - (i - 1) * 2)) + (N / 2)( C )
f(N) = C * (Summation(i from 1 to N / 2)( N ) - Summation(i from 1 to N / 2)( (i - 1) * 2)) + (N / 2)( C )
f(N) = C * (( N ^ 2 / 2 ) - 2 * Summation(i from 1 to N / 2)( i - 1 )) + (N / 2)( C )
=> Summation(i from 1 to N / 2)( i - 1 ) = Summation(i from 1 to N / 2 - 1)( i )
f(N) = C * (( N ^ 2 / 2 ) - 2 * Summation(i from 1 to N / 2 - 1)( i )) + (N / 2)( C )
f(N) = C * (( N ^ 2 / 2 ) - 2 * ( (N / 2 - 1) * (N / 2 - 1 + 1) / 2) ) + (N / 2)( C )
=> (N / 2 - 1) * (N / 2 - 1 + 1) / 2 =
(N / 2 - 1) * (N / 2) / 2 =
((N ^ 2 / 4) - (N / 2)) / 2 =
(N ^ 2 / 8) - (N / 4)
f(N) = C * (( N ^ 2 / 2 ) - 2 * ( (N ^ 2 / 8) - (N / 4) )) + (N / 2)( C )
f(N) = C * (( N ^ 2 / 2 ) - ( (N ^ 2 / 4) - (N / 2) )) + (N / 2)( C )
f(N) = C * (( N ^ 2 / 2 ) - (N ^ 2 / 4) + (N / 2)) + (N / 2)( C )
f(N) = C * ( N ^ 2 / 4 ) + C * (N / 2) + C * (N / 2)
f(N) = C * ( N ^ 2 / 4 ) + 2 * C * (N / 2)
f(N) = C * ( N ^ 2 / 4 ) + C * N
f(N) = C * 1/4 * N ^ 2 + C * N
And the BigOh is:
O(N²)
There's a function in the standard-library for this: itertools.permutations.
import itertools
list(itertools.permutations([1, 2, 3]))
If for some reason you want to implement it yourself or are just curious to know how it works, here's one nice approach, taken from http://code.activestate.com/recipes/252178/:
def all_perms(elements):
if len(elements) <=1:
yield elements
else:
for perm in all_perms(elements[1:]):
for i in range(len(elements)):
# nb elements[0:1] works in both string and list contexts
yield perm[:i] + elements[0:1] + perm[i:]
A couple of alternative approaches are listed in the documentation of itertools.permutations. Here's one:
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = range(n)
cycles = range(n, n-r, -1)
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
And another, based on itertools.product:
def permutations(iterable, r=None):
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
for indices in product(range(n), repeat=r):
if len(set(indices)) == r:
yield tuple(pool[i] for i in indices)
Best Answer
If the values are discrete and the number of distinct values isn't too high, you could just accumulate the number of times each value occurs in a histogram, then find the median from the histogram counts (just add up counts from the top and bottom of the histogram until you reach the middle). Or if they're continuous values, you could distribute them into bins - that wouldn't tell you the exact median but it would give you a range, and if you need to know more precisely you could iterate over the list again, examining only the elements in the central bin.