Reading Working Papers in Linguistics 4 (2000) - The University of ...

OPACITY AND SYMPATHY THEORY
successively eliminate candidates according to their non-violations of the
constraints, taking into consideration the constraint ranking.
To that end, we successively define sets (using subscripts to denote
candidates and constraints) containing the discarded candidates. Starting
from the highest ranked constraint (say constraint 1), we eliminate all
candidates violating it. D 1 contains all candidates violating constraint 1,
while there is at least one candidate k (not belonging to the sets of the
discarded candidates F 1 ), obeying constraint 1. D 2 contains all candidates
violating constraint 2, the second highest-ranked constraint, while there is
at least one candidate k (not belonging to the sets of the discarded
candidates F 1 and D 1 ), obeying it. D 3 contains all candidates violating
constraint 3, the third highest-ranked constraint, while there is at least one
candidate k, which obeys it and which does not belong to the sets of
discarded candidates F 1 , D 1 and D 2 . This process is carried on until all
candidates but one have been discarded and can be written symbolically
as
D
D
D
D
1
2
3
n
=
=
=
=
{ i :Ιi,
1 = 0 ∧ ∃ k ∈ F1
−{ i} with Ik,1
= 1 }
{ i :Ιi,
2 = 0 ∧ ∃ k ∈ F1
−{ i}
− D1
with Ik,2
= 1 }
{ i :Ι = 0 ∧ ∃ k ∈ F −{ i}
− D − D with I = 1 }
i, 3
1
...
{ i :Ι = 0 ∧ ∃ k ∈ F − D − D − D ... − D with I = 1 }
i,n
1
1
1
2
At the end of the procedure, the optimal candidate is then going to be
OPT = F1 − D1
− D2
− D3...
− D n = { i opt }.
OPT should have only one candidate (if otherwise, i.e. if OPT = ∅ or it
contains more than one candidate, either the constraint ranking is
incorrect or perhaps more constraints need to be added). The discarding
of candidates from F 1 is performed in a simultaneous process by EVAL
in the OT mechanism.
Let us demonstrate this in tableau 1, which I cite here again
slightly modified as tableau 5. de(e2 and des2 are not the first candidates
to be ruled out due to their violating the highest ranked constraint. Instead
de( and des2 are out first because they violate the ❀constraint, ❀MAX–V.
In the first step of our algorithm, this constraint acts as though it is the
highest ranked (undominated) one. Once de( and des2 have been
discarded, ❀MAX–V is deprived of its special powers and now acts as a
dominated constraint. In the second step de(e2 is the next candidate to be
2
3
k,3
n−1
k,
n
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