From: Dan Starr <dvstarr@ea...>  20041030 14:03:50

Concerning setdifference ... If it is known that both sets are sorted, the calculation can proceed very quickly indeed. Consider, apart from hashing, to write #'sortedsetdiff, which presumes sorting, and then use this on sorted sets. Intuitively, since decent sorting takes place in (* n (log n) time, and #'sortedsetdiff is onepass, this could be a favorable solution for larger sets. Might compete favorably with hashing. It would require,however, two predicates: A :test parameter and a :pecedes parameter, so you might wind up inferring #'< from #'= and so forth. That could get sticky. Just a thought. Devious Dan 