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Chapter 9: Q16P (page 390)

Prove that TQBF鈭�SPACE(n1/3)

Short Answer

Expert verified

Using the space hierarchy theorem, we are going to solve the above problem.

Step by step solution

01

Defining the space hierarchy theorem

We can define the space hierarchy theorem as 鈥渋f g is a space-constructible (1n鈫�1g(n)can be computed in space O(gn),f(n)=O(gn) then SPACE(fn)鈭�SPACE(gn),

Thus, it is known from space hierarchy theorem that there exists a language L , which is solvable in linear space but not in sub-linear space.

02

Arriving at the contraction to prove the answer

鈥楲鈥� can be reduced TQBF in log space since, TQBF is space-complete. So, ifTQBF鈭�SPACE(n1/3) then role="math" localid="1657785362677" L鈭�SPACE((nc)1/3+log(n)),

Now, suppose if 0.33<1 / c , then there exists a contradiction

Hence, it can be said thatTQBF鈭�SPACE(n1/3)

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