Question

Prove the bound of T(n) = 2T(n/3) + \Theta(n) using guess-and-verification method. ====== ============= SAMPLE ANSWER ========================== Step 1: Guess T(n) =____ Step 2: Prove the guess. Step a: Assume that the bound holds for all By definition, for an appropriate choice of constant Step b: In particular, let yielding Step c: Substitute the inequality in the original recurrence, yielding The (part of inequality) is trivially satisfied. Step d: T(n) ? (simplified) ? (the target inequality) for (show appropriate constant) Step e: Prove that holds for boundary cases. Assume T(1) = \Theta(1) = 1. Then T()= and T()= Let n= , plug n in hypothesis yields does not hold true) for Let n= , plug n in hypothesis yields The hypothesis holds true (or The hypothesis holds true (or does not hold true) for show this one.) (If the previous boundary case holds, no need to Step f: The inductive hypothesis holds true for boundary cases for Conclude:

          Prove the bound of T(n) = 2T(n/3) + \Theta(n) using guess-and-verification method.
======
============= SAMPLE ANSWER ==========================
Step 1: Guess T(n) =____
Step 2: Prove the guess.
Step a: Assume that the bound
holds for all
By definition,
for an appropriate choice of constant
Step b: In particular, let
yielding
Step c: Substitute the inequality in the original recurrence, yielding
The
(part of inequality) is trivially satisfied.
Step d:
T(n) ?
(simplified) ?
(the target inequality) for
(show appropriate constant)
Step e: Prove that holds for boundary cases. Assume T(1) = \Theta(1) = 1.
Then T()=
and T()=
Let n=
, plug n in hypothesis yields
does not hold true) for
Let n=
, plug n in hypothesis yields
The hypothesis holds true (or
The hypothesis holds true (or
does not hold true) for
show this one.)
(If the previous boundary case holds, no need to
Step f: The inductive hypothesis holds true for boundary cases for
Conclude:

Prove the bound of T(n) = 2T(n/3) + Θ(n) using guess-and-verification method.
======
============= SAMPLE ANSWER ==========================
Step 1: Guess T(n) =
Step 2: Prove the guess.
Step a: Assume that the bound
holds for all
By definition,
for an appropriate choice of constant
Step b: In particular, let
yielding
Step c: Substitute the inequality in the original recurrence, yielding
The
(part of inequality) is trivially satisfied.
Step d:
T(n) ?
(simplified) ?
(the target inequality) for
(show appropriate constant)
Step e: Prove that holds for boundary cases. Assume T(1) = Θ(1) = 1.
Then T()=
and T()=
Let n=
, plug n in hypothesis yields
does not hold true) for
Let n=
, plug n in hypothesis yields
The hypothesis holds true (or
The hypothesis holds true (or
does not hold true) for
show this one.)
(If the previous boundary case holds, no need to
Step f: The inductive hypothesis holds true for boundary cases for
Conclude:

Added by Josep M.

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