Firstly, hi all! It has been quite a long time since I last posted or
committed anything, but I'm still here!. :-)
Then, congratulations for all the great progresses fop is making!
And finally, concerning the keeps ...
Andreas L. Delmelle wrote:
[inserting penalties with higher value to represent numeric keeps]
This should steer the line-breaking algorithm in the right direction
to satisfy all keep constraints, IIC. The only big difference
compared to an auto keep-constraint, if I judge correctly, would then
be that we would somehow have to use penalties to represent all legal
break-opportunities. Instead of glues being considered as feasible
breakpoints, they would always be preceded by a zero-width penalty
having a value corresponding to the keep-constraint governing the
base FO.
I'm not sure the "steering capability" of penalty values would be enough
to get the prescribed result [section 4.8 "Keeps and breaks" (particularly
the last paragraph)]: the algorithm could still prefer violating a keep
with force = N to satisfy some keeps with force < N, as, IIRC, the
demerits ultimately depends much more on the necessary stretch / shrink
than on the penalty value.
I think that the breaking algorithm could be performed one time for each
distinct force value. Something like this:
lastConfirmedBreaks = ... the set of breaking points considering only
"always" keeps
ArrayList forceLevelList = findForceLevels(sequence); // in the reversed
order
int forceLevelIndex = 0;
boolean tryAgain = true;
while (tryAgain && forceLevelIndex <= forceLevelList.size() - 1) {
revisedSequence = setPenaltyValue(seq, lastConfirmedBreaks,
forceLevelList.get(i), HIGH_PENALTY_VALUE);
... compute the set of breaking points for revisedSequence
if (... they are still acceptable) {
lastConfirmedBreaks = ... these ones
i ++;
} else {
tryAgain = false
}
}
- in the sequence, keeps having force = "always" would be represented by
+INF penalties;
- keeps with numeric force start with a 0 penalty value;
- the method setPenaltyValue() sets those with the given force a high
value, and those with greater force (which have not been chosen as breking
points, this is why we must pass the computed breaks too) are set to +INF
so we are sure they would not be violated
If this approach is correct, the key point would be how to decide whether
or not the computed set of breaks is "still acceptable" ...
Hope this helps ...
Luca
