A Study to Generate a Weak Order from a Partially Ordered Set, Taken. . .
71
Hav (x, X 1 ) < Hav (x, X) <| X 2 | +Hav (x, X 1 )
(12a)
Hav (y, X 2 ) < Hav (y, X) <| X 1 | +Hav (y, X 2 )
(12b)
because as an extremal case the subposet based on X 1 can once be completely
below the subposet (X 2 , <) or completely above (X 2 , <). Hence: When Dom(X 1 , X 2 )
< 0.5 then the role of the separability matrix is overwhelming (because the sum of
Dom- and Sep-matrices is bounded, due to the finite number of comparabilities and
incomparabilities and the Eqs. 8 and 9 fail. Therefore it is needed that the poset, to
be considered, has more comparabilities than incomparabilities. This is the reason,
why instead of 20 sites (the real example) only the 10 first sites were selected.
Summarizing: From a methodological point of view, we want to check, as to how
far a deviation of Dom(X 1 , X 2 ) from 1 can lead to acceptable results.
3 Results
3.1 Randomly Generated Datasets
In order to test as to how far deviations of Dom(X 1 , X 2 ) from 1 lead to errors
in the estimation of Hav, 22 smaller datasets (each of 10 objects) were randomly
generated. For each object x of these artificial data sets the Hav-value based on the
scheme given in Eq. 12 was calculated, HavDom(x) and the exact value, Havexact,
based on the lattice theoretical method presented by De Loof et al. (2006, 2011,
2012). The deviation was calculated:
Eps(x) :=| H axexact (x)–H avDom(x) |
(13)
For each dataset a final value epsav was determined:
epsav :=
Eps(x)/n with n =| X |
(14)
the quantity epsav being the average error related to any single object. In Fig. 1 the
scatterplot, together with the regression equation is shown.
Figure 1 confirms that the deviations epsav will be rather large, when Dom(X 1 ,
X 2 ) becomes small values. It is clear that the way, how the partitioning of X into
two subsets X 1 and X 2 is selected, plays an important role. However, aiming at an
efficient method for the calculation of Hav, the principles were:
71
Hav (x, X 1 ) < Hav (x, X) <| X 2 | +Hav (x, X 1 )
(12a)
Hav (y, X 2 ) < Hav (y, X) <| X 1 | +Hav (y, X 2 )
(12b)
because as an extremal case the subposet based on X 1 can once be completely
below the subposet (X 2 , <) or completely above (X 2 , <). Hence: When Dom(X 1 , X 2 )
< 0.5 then the role of the separability matrix is overwhelming (because the sum of
Dom- and Sep-matrices is bounded, due to the finite number of comparabilities and
incomparabilities and the Eqs. 8 and 9 fail. Therefore it is needed that the poset, to
be considered, has more comparabilities than incomparabilities. This is the reason,
why instead of 20 sites (the real example) only the 10 first sites were selected.
Summarizing: From a methodological point of view, we want to check, as to how
far a deviation of Dom(X 1 , X 2 ) from 1 can lead to acceptable results.
3 Results
3.1 Randomly Generated Datasets
In order to test as to how far deviations of Dom(X 1 , X 2 ) from 1 lead to errors
in the estimation of Hav, 22 smaller datasets (each of 10 objects) were randomly
generated. For each object x of these artificial data sets the Hav-value based on the
scheme given in Eq. 12 was calculated, HavDom(x) and the exact value, Havexact,
based on the lattice theoretical method presented by De Loof et al. (2006, 2011,
2012). The deviation was calculated:
Eps(x) :=| H axexact (x)–H avDom(x) |
(13)
For each dataset a final value epsav was determined:
epsav :=
Eps(x)/n with n =| X |
(14)
the quantity epsav being the average error related to any single object. In Fig. 1 the
scatterplot, together with the regression equation is shown.
Figure 1 confirms that the deviations epsav will be rather large, when Dom(X 1 ,
X 2 ) becomes small values. It is clear that the way, how the partitioning of X into
two subsets X 1 and X 2 is selected, plays an important role. However, aiming at an
efficient method for the calculation of Hav, the principles were: