CS 3110 Recitation 20
Using the substituion and master methods
Using the substituion method
The substitution method is a condensed way of proving an asymptotic bound on a recurrence by induction. In
the substitution method, instead of trying to find an exact closedform solution, we only try to find a closed
form
bound
on the recurrence. This is often much easier than finding a full closedform solution, as there is
much greater leeway in dealing with constants.
The substitution method is a powerful approach that is able to prove upper bounds for almost all recurrences.
However, its power is not always needed; for certain types of recurrences, the master method (see below) can
be used to derive a tight bound with less work. In those cases, it is better to simply use the master method,
and to save the substitution method for recurrences that actually need its full power.
Note that the substitution method still requires the use of induction. The induction will always be of the same
basic form, but it is still important to state the property you are trying to prove, split into one or more base
cases and the inductive case, and note when the inductive hypothesis is being used.
Substitution method example
Consider the following reccurence relation, which shows up fairly frequently for some types of algorithms:
T
(1) = 1
T
(
n
) = 2
T
(
n
1) + c
1
By expanding this out a bit (using the "iteration method"), we can guess that this will be O(2
n
). To use the
substitution method to prove this bound, we now need to guess a closedform upper bound based on this
asymptotic bound. We will guess an upper bound of
k
2
n

b
, where
b
is some constant. We include the
b
in
anticipation of having to deal with the constant
c
1
that appears in the recurrence relation, and because it does
no harm. In the process of proving this bound by induction, we will generate a set of constraints on
k
and
b
,
and if
b
turns out to be unnecessary, we will be able to set it to whatever we want at the end.
Our property, then, is
T
(
n
) ≤
k
2
n

b
, for some two constants
k
and
b
. Note that this property logically implies
that
T
(
n
) is O(2
n
), which can be verified with reference to the definition of O.
Base case:
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 '07
 GIAMBATTISTA,A
 LG, Recurrence relation, 2k, Master Theorem, Ih, O. Base

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