Re: Discussion paper: what are the level 2 datums?
Arnold Neumaier wrote:
Nate Hayes wrote:
Arnold Neumaier wrote:
Nate Hayes wrote:
Arnold Neumaier wrote:
Anyway, one must be able to conclude from the decoration whether or
not
an interval is valid, and one needs at least operations extracting the
four trits v d c b from a tetrit representation.
Agreed.
Any interval with a "domain" decoration (F,F) is invalid. This is also
the "worst" state of a tetrit, i.e., by the propagation rules of
tetrits it is impossible that (F,F) could ever be "absorbed", "lost",
or "changed" into another state that is either better or worse. So it
is the perfect state to represent invalid constructions and/or
uninitialized interval variables.
The new domain tetrit provides the four states:
priority tetrit trit meaing
3 (T,F) + "is defined"
2 (T,T) 0 "possibly defined"
1 (F,T) - "is undefined"
0 (F,F) "nowhere defined, nowhere undefined"
The first three states and thier meaning are the same as the old trits.
But with trits, there was not an equivalent of the (F,F) state, which
essentially makes the distinction of "v" in the above example.
I'm fairly certain the only decorations IEEE 1788 needs are
"defined", "continuous", and "bounded".
We need all four. But we can dispense with the ohers.
If we were still using trits, then "v" might be needed. But since we
are now using tetrits I believe "v" is not necessary.
Does the interpretational difference matter when there are only 10
different combinations that are useful? With tetrits it shoudn't be
significantly more.
Tetrits simplify and reduce the number of "unusable" combinations.
In place of 3^4=81 we have now 4^3=64 combinations.
A slight reduction, but still far off the 10 or so useful ones.
Right. But keep in mind, too, that "defined and continous" will only be a
single bit (and not a tetrit) if Motion 22 passes. So between the
"domain" and "defined and continous" decorations that is only 2*4=8
combinations, of which 3 would be useless. However, there is no way to
represent 5 states in fewer than 3 bits of information, anyways.
So one can probably map the trit formulation and the tetrit
formulation equivalently to 10 (or 16) abstract decoration objects
that are just organized differently by the two formulations.
Yes.
It would be nice to have an explicit such correspondence.
So how do the 5 or 8 combinations of Motion 22 and the 10 trit
combinations
v d c b | #cases
- 0 0 0 | 1
+ - 0 0 | 1
+ 0 0 0- | 2
+ + +0 +0- | 6
v=valid, d=-defined, c=continuous, b=bounded
+ (True), - (False), and 0 (no claim),
correspond to each other?
Let me just concentrate on v, d, and c for now.
Below are the states represented in terms of the "domain" tetrit (Motion 18)
and a "defined and continuous" bit (Motion 22 + Amendment), as well as thier
v, d, and c equivalents as you describe them above:
domain d&c v d c meaning
(T,F) T + + + "is defined and continuous"
(T,T) T + 0 + useless
(F,T) T + - + useless
(F,F) T - 0 + useless
(T,F) F + + 0 "is defined"
(T,T) F + 0 0 "possibly defined"
(F,T) F + - 0 "is undefined"
(F,F) F - 0 0 "nowhere defined, nowhere undefined"
The table shows that IEEE 1788 only needs the "domain" tetrit (as defined in
Motion 18) and a "defined and continuous" bit (as defined in Motion 22 +
Amendment) to efficiently encode all of the desired states otherwise
represented by the v, d and c trits. Furthermore, this encoding only
requires a total of 8 combinations, of which only 3 are useless. So it is
the most storage-efficient method, since it is not possible to represent the
5 valid states in fewer than 3 bits, anyways.
Nate Hayes