On Tuesday 08 August 2006 14:29, Jon LaBadie wrote:
> >> -- The most recently used "tapecycle" number of tapes is in A.
> >> -- Any remaining tapes are in I. The single least recently used
> >>    of these is also in P.
>
> When I first read it, I was in a mindset of # tapes in rotation
> matches tapecycle and I applied the last sentence to that situation,
> i.e. the least recently used of tapecycle # of tapes is in P.  Which
> would have to be true, otherwise you'd never cycle through them again.

Unless you have used fewer than tapecycle number of tapes, in which case the 
only tapes in P are new tapes.

> I failed to consider you meant single most recently used of A+I is in P.

Any tape in P is never in A. Consider that P is a subset of I, and A and I are 
disjoint sets.

> If at least tapecycle tapes have been used, would it be correct to
> say P is all unused, labeled tapes plus 1 previously used tape?

Yes, unless you have set some tapes as no-reuse.

Cheers,

--Ian
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