How does an oblivious Turing machine work?

Question

I am reading the book Computational Complexity: A Modern Approach and I am having problems understanding oblivious Turing machines .

An oblivious Turing machine (TM) is such a TM that the movement of its heads is entirely determined by the length of the input . That is, the TM is oblivious to its input. So far so good.

But one of the excercises is to prove the following theorem:

If a language L is decidable in time T(n) 
then there exists an oblivious TM that decides L in time O(T(n)^2). 

It is obvious that the oblivious TM must not operate on the original input of L but at some coded version. That is, the gist of the theorem is the coding of a bitstring to an integer (length of the input of the oblivious TM). But if one would want to code the set of possible inputs of L (bitstrings) to an integer, one would run into very high numbers fairly quickly since there are 2^n bitstrings of length n .

Am I understanding the problem correctly? How do you prove the theorem?

Answer 1

Here I suggest you read this paper. It is a rather interesting and wonderful paper that will give you a proof at a lower time bound that requested. (I think you should be able to finagle it to be O(N^2) or you could conclude that O(N*log(N)) is technically O(N^2) but following this proof directly may upset your professor. I imagine he intends for you to approach it differently.

Edit:

The original link to the paper no longer works. Here is another publicly posted one.

http://www-dev.ccs.neu.edu/home/viola/classes/papers/PippengerF-Oblivious.pdf

"Relations Among Complexity Measures" by Michael J. Fischer and Nicholas Pippenger, 1979.