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Chapter 9: Q16P (page 390)

Prove that TQBF∉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 “if g is a space-constructible (1n→1g(n)can be computed in space O(gn),f(n)=O(gn) then SPACE(fn)∅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

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

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

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

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Most popular questions from this chapter

Consider the following function pad:∑*×Ν→∑*#*that is defined as follows. Let PAD (s, l) = s#3, where j = max (0,l - m) and mis the length of s. Thus, pad (s, l)simply adds enough copies of the new symbol # to the end of s so that the length of the result is at least l. For any language A and function f:N→N, define the language pad(A, f) aspad(A,f)={pad(s,f(m))| where s∈Aand ‘m’ is the length of ‘s’ }. Prove that if A∈TIME(n6), then pad(A,n2)∈TIME(n3).

Give regular expressions with exponentiation that generate the following languages over the alphabet {0,1}.

a. All strings of length 500

b. All strings of length 500 or less

c. All strings of length 500 or more

d. All strings of length different than 500

e. All strings that contain exactly 500 1s

f. All strings that contain at least 500 1s

g. All strings that contain at most 500 1s

h. All strings of length 500 or more that contain a 0 in the 500th position

i. All strings that contain two 0s that have at least 500 symbols between them

Prove that NTIME(n)ΦPSPACE.

Show that the language MAX-CLIQUEfrom problem 7.48 isin.PSAT

Question: LetEREXI^

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