Asymptotic complexity has limitations, but it is still a useful tool analyzing and improving program performance. The use of bigO notation simplifies the task of analyzing performance. For OCaml, we assume that all the reductions performed during evaluation take constant time, including all arithmetic operations and pattern matching.
This assumption may be surprising if you think in terms of the substitution model. Substitution may require a lot of work, because a substituted variable may appear at many different places in the code. However, the OCaml implementation is closer to the environment model of evaluation, in which the actual substitutions are never done, but bindings are recorded in a separate environment for later lookup. Environments can be implemented so that all environment operations, including lookup, can be performed in constant time.
The following multiplication routine for nonnegative integers is correct, but not very efficient:
let rec times1 a b = if b = 0 then 0 else a + times1 a (b  1)
What is the order of growth of the time required by times1
as a
function of n, where n
is the magnitude of the parameter b
?
Note that the "size" of a number can be measured either in terms of its
magnitude or in terms of the number of digits (the space it takes to write the
number down). Often the number of digits is used, but here we use the magnitude.
Note that it takes only about log_{10} x digits to write down a
number of magnitude x, thus these two measures are very different.
We assume that all the primitive operations in the times1
function
(if
, +
, =
, and 
) and the overhead for function
calls take constant time. Thus if n=0,
the routine takes some constant time c_{2}.
If n>0, the time taken on an input of magnitude n
is at most some constant c_{1}
plus the time taken by the recursive call on n
1
. In other words, there are constants c_{1} and
c_{2} such that T(n) satisfies:
T(n) = T(n1) + c_{1} for n > 0 T(0) = c_{2}
This is a recurrence relation (or simply recurrence)
defining a function T(n). It simply states that the time
to multiply a number a by another number b of size n > 0
is the time required to multiply a by a number of size n
1 plus a
constant amount of work (the primitive operations performed).
The recurrence relation is an inductive definition of a function. This particular recurrence relation has a unique closedform solution that defines T(n) without any recursion. In order to determine the running time of recursive functions we will write recurrences that describe the running time, find closed form solutions to them, and express those solutions in terms of BigO notation.
Expanding out this particular recurrence gives some idea of how to convert it to closed form:
T(n) = T(n1) + c_{1} for n > 0 = T(n2) + c_{1} + c_{1} = T(n3) + c_{1} + c_{1} + c_{1} = ... = T(nk) + kc_{1} for 0 ≤ k ≤ n = ... = T(0) + nc_{1} for n ≥ 0 = c_{2} + c_{1}n for n ≥ 0
which is O(n).
More generally, we guess a closed form solution for a recurrence, and then prove by induction that it holds. Sometimes expanding out the recurrence for a few steps can help with the guessing.
Now consider a different procedure for multiplying two numbers:
let rec times2 a b = if b = 0 then 0 else if even b then times2 (a*2) (b/2) else a + times2 a (b1)
Again we want an expression for the running time in terms of n, the
magnitude of the parameter b
. We assume that double and half
operations are constant time (these could be done in constant time using
arithmetic shift) as well as the standard primitives. The recurrence relation
for this problem is more complicated than the previous one:
T(n) = T(n1) + c_{1} for n > 0 and n odd T(n) = T(n/2) + c_{2} for n > 0 and n even T(0) = c_{3}
We somehow need to figure out how often the first versus the second branch of this recurrence relation will be taken. It's easy if n is a power of two, i.e. if n = 2^{m} for some integer m. In this case, the second branch of will only get taken when n = 1, because 2^{m} is even except when m = 0, i.e. when n = 1. Thus, for this special case we can use the expansion technique above to get a closed form in terms of m:
T(2^{m}) = T(2^{m1}) + c_{2} for m > 0 = T(2^{m2}) + c_{2} + c_{2} = ... = T(2^{mm}) + mc_{2} = T(1) + mc_{2} = T(0) + c_{1} + mc_{2} = c_{3} + c_{1} + mc_{2}
which is O(log n). We can show that this is the case using induction.
Strong induction has the same steps as ordinary induction, but the induction hypothesis is a little different:
We can formalize this process as an inference rule that gives us a way to prove a proposition ∀n.P(n) for some predicate P. Here are the rules for weak (ordinary) and strong induction respectively:
Weak induction:

Strong induction:

Note that the base case does not occur explicitly in the strong induction rule. However, it is still there implicitly: for n=0, you must prove ∀m (m < 0 ⇒ P(m)) ⇒ P(0), which reduces to P(0) since ∀m (m < 0 ⇒ P(m)) is vacuously true, as there are no m < 0.
In practice, it is often easier to prove asymptotic complexity bounds with strong induction than with weak induction, because you have a stronger induction hypothesis to work with, namely ∀m m < n ⇒ P(m) rather than P(n). However, the two rules are logically equivalent: one can prove the validity of the strong rule using the weak rule with the induction hypothesis P'(n) = ∀m (m < n ⇒ P(m)).
Let's return to the analysis of times2
. When
n is a power of 2, we have the following recurrence:
T(n) = T(n/2) + c_{2} for n > 1 and n a power of 2 T(1) = c_{4} where c_{4} = c_{3} + c_{1}
We show T(n) = c_{2} log n + c_{4} is the solution to this recurrence using strong induction.
Base case: T(1) = c_{4} = c_{2} log 1 + c_{4}.
Induction hypothesis: T(m) = c_{2} log m + c_{4} for all 1 ≤ m < n.
Inductive step:
T(n) = T(n/2) + c_{2}
= c_{2} log(n/2) + c_{4} + c_{2} by I.H.
= c_{2}(log n − log 2) + c_{4} + c_{2}
= c_{2} log n − c_{2} + c_{4} + c_{2}
= c_{2} log n + c_{4}.
Thus in the case of an input of size n which is a power of
2, the running time of times2
is O(log n).
In the general case, the running time of times2
is
also O(log n), but we need a slightly different recurrence to
make this easier to show. Note that it is not possible for the
parameter b
to be odd on two successive recursive calls
of times2
, because when it is odd, one is subtracted
making it even. Thus we can rewrite the original recurrence as:
T(n) = T((n−1)/2) + c_{1} + c_{2} for n > 1 and n odd T(n) = T(n/2) + c_{2} for n > 1 and n even T(1) = c_{4} T(0) = c_{3}
Note that this new occurrence is analogous to rewriting the code so that both recursive calls divide the input size in half:
let rec times3 a b = if b = 0 then 0 else if even b then times3 (a*2) (b/2) else a + times3 (a*2) ((b1)/2)
It is a little difficult to come up with a closed form solution to this recurrence. However, using strong induction, we can show that whatever it is, it is bounded asymptotically by a function of the form d log n + e for sufficiently large constants d and e, and that is good enough to show that the solution is O(log n). So we want to show by strong induction that T(n) ≤ d log n + e for some (as yet to be determined) constants d and e. To find suitable d and e, we'll just jump in and see what we need to make the proof work.
Base case: We want T(1) = c_{4} ≤ d log 1 + e. This will be true provided we choose e ≥ c_{4}.
Induction hypothesis: T(m) ≤ d log m + e for all 1 ≤ m < n.
Inductive step, n even:
T(n) = T(n/2) + c_{2}
≤ d log(n/2) + e + c_{2} by I.H.
= d log n − d + e + c_{2}
≤ d log n + e, provided we choose d ≥ c_{2}.
Inductive step, n odd:
T(n) = T((n−1)/2) + c_{1} + c_{2}
≤ d log((n−1)/2) + e + c_{1} + c_{2} by I.H.
= d log(n−1) − d + e + c_{1} + c_{2}
≤ d log n − d + e + c_{1} + c_{2}
≤ d log n + e, provided we choose d ≥ c_{1} + c_{2}.
Choosing d and e just large enough to satisfy all these conditions, we see that T(n) ≤ (c_{1} + c_{2}) log n + c_{4} for all n ≥ 1, therefore T(n) is O(log n).