Given the following function:
Function f(n,m)
if n == 0 or m == 0: return 1
return f(n-1, m) + f(n, m-1)
What's the runtime compexity of f? I understand how to do it quick and dirty, but how to properly characterize it? Is it O(2^(m*n))?
This is an instance of Pascal's triangle: every element is the sum of the two elements just above it, the sides being all ones.
So f(n, m) = (n + m)! / (n! . m!).
Now to know the number of calls to f required to compute f(n, m), you can construct a modified Pascal's triangle: instead of the sum of the elements above, consider 1 + sum (the call itself plus the two recursive calls).
Draw the modified triangle and you will quickly convince yourself that this is exactly 2.f(n, m) - 1.
You can obtain the asymptotic behavior of the binomial coefficients from Stirling's approximation. http://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas
f(n, m) ~ (n + m)^(n + m) / (n^n . m^m)
The runtime of f(n, m) is in O(f(n, m)). This is easily verified by the following observation:
Function g(n, m):
if n=0 or m=0: return 1
return g(n-1, m) + g(n, m-1) + 1
The function f is called equally often as g. Furthermore, the function g is called exactly g(n, m) times to evaluate the result of g(n, m). Likewise, the function f is called exactly g(n, m) = 2*f(n, m)-1 times in order to evaluate the result of f(n, m).
As @Yves Daoust points out in his answer, f(n, m) = (n + m)!/(n!*m!), therefore you get a non-recursive runtime of O((n+m)!/(n!*m!)) for f.
f(n, 0) = 1
f(0, m) = 1
f(n, m) = f(n - 1, m) + f(n, m - 1)
The values look like a Pascal triangle to me:
n 0 1 2 3 4 ..
m
0 1 1 1 1 1 ..
1 1 2 3 4
2 1 3 6
3 1 4 .
4 1 .
. .
. .
The values of Pascal's triangle can be expressed as binomial coefficients. Translating the coordinates one gets the solution for f:
f(n, m) = (n + m)
( m )
= (n + m)! / (m! (n + m - m)!)
= (n + m)! / (n! m!)
which is a nice term symmetric in both arguments n and m. (Final term first given by @Yves Daoust in this discussion)
The recursion equation of f can be derived by using the symmetry of the binomial coefficients and Pascal's Rule
f(n, m) = (n + m)
( n )
= (n + m)
( m )
= ((n + m) - 1) + ((n + m) - 1)
( m ) ( m - 1 )
= ((n - 1) + m) + (n + (m - 1))
( m ) ( m )
= f(n - 1, m) + f(n, m - 1)
The "number of calls of f" calculation function F is similar to f, we just have to add the call to f itself and the two recursive calls:
F(0, m) = F(n, 0) = 1, otherwise
F(n, m) = 1 + F(n - 1, m) + F(n, m - 1)
(Given first by @blubb in this discussion).
If we write it down, we get another triangle scheme:
1 1 1 1 1 ..
1 3 5 7
1 5 11
1 7 .
1 .
.
.
Comparing the triangles value by value, one guesses
F(n, m) = 2 f(n, m) - 1 (*)
(Result first suggested by @blubb in this discussion)
We get
F(0, m) = 2 f(0, m) - 1 ; using (*)
= 1 ; yields boundary condition for F
F(n, 0) = 2 f(n, 0) - 1
= 1
as it should and examining the otherwise clause, we see that
F(n, m) = 2 f(n, m) - 1 ; assumption
= 2 ( f(n - 1, m) + f(n, m - 1) ) - 1 ; definition f
= 1 + (2 f(n - 1, m) - 1) + (2 f(n, m - 1) - 1) ; algebra
= 1 + F(n - 1, m) + F(n, m - 1) ; 2 * assumption
Thus if we use (*) and the otherwise clause for f, the otherwise clause for F results.
As the finite difference equation and the start condition for F hold, we know it is F (uniqueness of the solution).
Now on calculating / estimating the values of F (i.e. the runtime of your algorithm).
As
F = 2 f - 1
we see that
O(F) = O(f).
So the runtime of this algorithm is
O( (n + m)! / (n! m!) )
(Result first given by @Yves Daoust in this discussion)
Using the Stirling approximation
n! ~= sqrt(2 pi n) (n / e)^n
one can get a form without hard to calculate factorials. One gets
f(n, m) ~= 1/(2 pi) sqrt((n+m) / (n m)) [(n + m)^(n + m)] / (n^n m^m)
thus arriving at
O( sqrt((n + m) / (n m)) [(n + m)^(n + m)] / (n^n m^m) )
(Use of Stirling's formula first suggested by @Yves Daoust in this discussion)