Outline for May 7, 1997
- Greetings and Felicitations
- HRU result and proof
- General result (HRU Theorem): Whether a given state of a given protection
system is safe for a given generic right is undecidable
- Proof approach: reduce halting problem to safety problem
- Proof: T is an arbitrary Turing machine; encode T's state as
an ACM and the head moves as commands of the protection system
T in
state q has scanned k cells; cells k+1, ... are blank.
Represent q as an ACM with k subjects and objects meeting the
following:
- If cell s of tape contains symbol X, then
A[s, s] contains the right X
- A[s, s+1] contains the right Own;
this induces an ordering on the subjects
according to the symbols on the tape; note
Own is above main diagonal
- A[k, k] contains the right End
(signalling the farthest right the head has moved)
- If tape head at cell s, A[s, s] contains the right
q
Initially, q0 is an ACM with one subject and one object
corresponding to the first tape cell.
Move: d(q, X) = (p, Y, L)
means in state q and head scanning symbol X,
changes state to p, overwrites X with Y,
and moves head left (R would be right).
Moves represented as commands; example:
above move is:
command cqX(s, s')
if Own in A[s, s'] and
q in A[s', s'] and
X in A[s', s']
then
delete q from A[s', s']
delete X from A[s', s']
enter Y into A[s', s']
enter p into A[s', s']
end.
where s' is current
position of tape head and s is square to left. A move to the right has
to handle the End case; 2 different commands, appropriately
conditioned
Now, if T reaches its final state qf, then right
qf leaked into some position of ACM;
equivalently, if T halts, the
right qf is leaked.
So if the safety question is decideable, so is the
halting problem, which is false; so the safety problem is undecidable.
- Take-Grant: rules, initial and terminal spans, bidirectionality of
t and g given subject-only graph
Notes by Michael Clifford:
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Department of Computer Science
University of California at Davis
Davis, CA 95616-8562
Page last modified on 4/4/97