Solving Recurrence relation: T(n) = 3T(n/5) + lgn * lgn

Question

Consider the following recurrence
T(n) = 3T(n/5) + lgn * lgn

What is the value of T(n)?

(A) Theta(n ^ log_5{3})
(B) Theta(n ^ log_3{5})
(c) Theta(n Log n )
(D) Theta( Log n )

Answer is (A)

My Approach :

lgn * lgn = theta(n) since c2lgn < 2*lglgn < c1*lgn for some n>n0

Above inequality is shown in this picture for c2 = 0.1 and c1 = 1

log_5{3} < 1,

Hence by master theorem answer has to be theta(n) and none of the answers match. How to solve this problem??

Answer 1

Your claim that lg n * lg n = Θ(n) is false. Notice that the limit of (lg n) ² / n tends toward 0 as n goes to infinity. You can see this using l'Hopital's rule:

lim _{n → ∞} (lg n) ² / n

= lim _{n → ∞} 2 lg n / n

= lim _{n → ∞} 2 / n

= 0

More generally, using similar reasoning, you can prove that lg n = o(n ^ε ) for any ε > 0.

Let's try to solve this recurrence using the master theorem. We see that there are three subproblems of size n / 5 each, so we should look at the value of log ₅ 3. Since (lg n) ² = o(n ^{log ₅ 3} ), we see that the recursion is bottom-heavy and can conclude that the recurrence solves to O(n ^{log ₅ 3} ), which is answer (A) in your list up above.

Hope this helps!

Answer 2

To apply we should check the relation between 我们应该检查它们之间的关系

n ^{log ₅ (3)} ~= n ^0.682 and (lg(n)) ²

Unfortunately lg(n) ² != 2*lg(n): it is lg(n ² ) that's equal to 2*lg(n)

Also, there is a big difference, in Master Theorem, if f(n) is O(n ^{log _b (a)-ε} ), or instead Θ(n ^{log _b a} ): if the former holds we can apply case 1, if the latter holds case 2 of the theorem.

With just a glance, it looks highly unlikely (lg(n)) ² = Ω(n ^0.682 ), so let's try to prove that (lg(n)) ² = O(n ^0.682 ), ie:

∃ n ₀ , c ∈ N ⁺ , such that for n>n ₀ , (lg(n)) ² < c * n ^0.682

Let's take the square root of both sides (assuming n > 1, the inequality holds)

lg(n) < c ₁ * n ^0.341 , (where c ₁ = sqrt(c))

now we can assume, that lg(n) = log ₂ (n) (otherwise the multiplicative factor could be absorbed by our constant - as you know constant factors don't matter in asymptotic analysis) and exponentiate both sides:

2 ^lg(n) < 2 ^{c ₂ * n 0.341} <=> n < 2 ^{c ₂ * n 0.341} <=> n < (n ^{2 0.341} ) ^{c ₂} <=> n < (n ^{2 0.341} ) ^{c ₂} <=> n < (n ^1.266 ) ^{c ₂}

which is immediately true choosing c ₂ = 1 and n ₀ = 1

Therefore, it does hold true that f(n) = O(n ^{log _b (a)-ε} ), and we can apply case 1 of the , and conclude that: 情况1，并得出结论：

T(n) = O(n ^{log ₅ 3} )

Same result, a bit more formally.