Given a k-tape Turing machine can we reduce the number of tapes without
too large an increase in time and space (memory)? Not just an esoteric
question, tape reduction plays an important role in time and space
hierarchies and creating efficient reductions.
For space we have an easy result: Every k-tape s(n)-space bounded
Turing machine can be simulated by a 1-tape machine in O(s(n))
space. Create a single supertape with separate "tracks" for
each of the original tapes and add markers for the locations of the
heads on each of these tapes.
For deterministic and nondeterministic time, this constructions yields
a t2(n)-time 1-tape simulation of a k-tape t(n)-time
machine and this is the best you can do (consider { x#x |
x∈Σ*}). We can do much better if we reduce k
tapes to 2 tapes.
We can simulate any nondeterministic t(n)-time k-tape machine in
nondeterministic O(t(n)) time on a 2-tape machine. Roughly the
simulation guesses every step of the transition function on one tape
and uses the other tape to verify the transition function on each tape
of the original machine one tape at a time.
For deterministic time we can only achieve a weaker result: Every
t(n)-time k-tape machine has a 2-tape simulation using O(t(n)log t(n))
time. The proof (due to Hennie and Stearns) is quite involved and I
won't give it here. The proof has a nice side effect: The construction
creates an oblivious machine where the head movements depend only on
the input length. One can use this fact to show that we can simulate
any t(n)-time Turing machines with O(t(n)log t(n))-size bounded fan-in
circuits and reduce any t(n)-time nondeterministic computation to
satisfiability questions of size O(t(n)log t(n)).
Lance, there was a post you made about a prospective theory grad student's view of the different theory groups which I cannot seem to find anymore. Has it been censored because it was sensitive?