 
Summary: Com S 631: Lower bounds and Separation Results
Lecture 4 Scribe: Ankit Agrawal
1. Time Vs. Space
A turing machine can visit at most t(n) cells in t(n) time. This implies that any deter
ministic t(n)time bounded machine can be simulated by a t(n)space bounded machine.
Can we improve this space bound aymptoticaly? In other words, is there a language in
DTIME(t(n)) that takes less than t(n) space? We will show the following theorem.
Theorem 1. DTIME t(n) DSPACE t(n)/ log t(n) .
Using space hierarchy, we have the following straightforward corollary.
Corollary 1. DTIME t DSPACE t
Proof. We know by space hierarchy theorem that DSPACE t(n)/ log t(n) DSPACE t(n) ,
which implies, DTIME t(n) DSPACE t/ log t DSPACE t .
To prove the above theorem, we will consider block respecting Turing machines. Suppose
M is a t(n) time bounded ktape turing machine. Let 1 b(n) t(n)/2 and a(n) =
t(n)/b(n). Divide the computation of M into a(n) time segments where each segment has
b(n) steps. Since M os t(n) time bounded, it visits atmost t(n)cells on each tape. Now
divide each tape into a(n) segments and thus each segment has b(n) cells.
Definition 1. M is b(n)block respecting if every tape head of M crosses a block boundary
at time c.b(n), where c is an integer.
We will use the following claim without proof.
