Notquitetrivial

Bx Examples Repository

Title: notQuiteTrivial

Version: 0.1

Type: Precise

Overview

Invented to illustrate a point about Least Change: almost the simplest case where there can be any question about what the best consistency restoration is.

Models

Let $B = \{true, false\}$ and let $M = B$, $N = B \times B$.

Consistency

$m\in M$ is consistent with $n = (s,t) \in N$ iff $m = s$.

Consistency Restoration

Forward

An obvious choice is

$\overrightarrow{R}(m,(s,t)) = (m,t)$

Backward

No choice if we wish to be correct and hippocratic.

$\overleftarrow{R}(m,(s,t)) = s$

Properties

Correct, hippocratic

Variants

$\overrightarrow{R}(m,(s,t)) = (s,t)$ if $m=s$, otherwise $(m,\neg t)$.

Discussion

Is the version of $\overrightarrow{R}$ presented in the main body, or in the Variants section, better from a Least Change point of view? Most people's intuition probably favours the version in the main body: why would we modify the second component of $n \in N$ if we did not have to? But this depends on how well the presentation of this example matches the real-world situation it comes from. Suppose, for example, that in $n = (s,t)$, $s$ records the presence or absence of a pebble - or a chocolate - in one pot, while $t$ records the presence or absence of a pebble or chocolate in another pot. The model space permits items to be destroyed, or created, but we might regard destroying or creating an item as a bigger change than moving one existing item from one pot to the other. If so, we might prefer the variant. The point is that even in so simple a case, the idea of "the" least change consistency restoration may not be quite as straightforward as one might assume.

References

None yet

Author(s)

Perdita Stevens

Reviewer(s)

None yet

Comments

This is where any member of the wiki can comment.

Unless otherwise stated, the content of this page is licensed under GNU Free Documentation License.