# Prove that a boolean function computable in T(n) by a RAM machine is in DTIME(T(n)^2)

The question is exercise 1.9 from Arora-Barak's book Computational Complexity — A Modern Approach:

Define a RAM Turing machine to be a Turing machine that has random access memory. We formalize this as follows: The machine has an infinite array A that is initialized to all blanks. It accesses this array as follows. One of the machine's work tapes is designated as the address tape. Also the machine has two special alphabet symbols denoted by R and W and an additional state we denote by q_access. Whenever the machine enters q_access, if its address tape contains 'i'R (where 'i' denotes the binary representation of i) then the value A[i] is written in the cell next to the R symbol. If its tape contains 'i'Wa (where a is some symbol in the machine's alphabet) then A[i] is set to the value a.

Show that if a Boolean function $f$ is computable within time $T(n)$ (for some time constructible $T$) by a RAM TM, then is is in $\mathrm{DTIME}(T(n)^2)$.

The trivial solution by using an additional tape recording pairs (address,value) turns out to be in $\mathrm{DTIME}(T(n)^3)$, since that tape can be of size $O(T(n)^2)$ with $O(T(n))$ pairs while the address of each pair can be of size $O(T(n))$.

• How do you know the bound on address size? Couldn't my first write be to $2^{2^{T(n)}}$? And if you can bound it to $T(n)$, then the address size is $\log\left(T(n)\right)$, not $T(n)$. – Xodarap Jul 10 '12 at 19:35
• Since the address should be written in the tape by a $T(n)$-time Turing Machine, the size (i.e. string length) of the address can not exceed $O(T(n))$, the accessible address space is $O(2^{T(n)})$. – c c Jul 11 '12 at 4:59
• Note that Arora and Barak explicitly ask in their introduction for other not post answers to their questions. See also the policy about homework questions. – Kaveh Jul 12 '12 at 3:17
• Sorry for that. I just study the book by myself and get troubled in that question. I don't know if such $O(T(n)^2)$ simulation really exists or it is just a typo. If you know the answer, please email me in private to ccqmpux@gmail.com, and I'll then close the question. – c c Jul 12 '12 at 5:56
• You can find more in the first chapter of the handbook of theoretical computer science. – Kaveh Sep 16 '12 at 21:22

Since the address should be written in the tape by a $T(n)$-time Turing Machine, the size (i.e. string length) of the address can not exceed $O(T(n))$.
[The] tape can be of size $O(T(n)^2)$ with $O(T(n))$ pairs while the address of each pair can be of size $O(T(n))$.