On Monday, October 5, 2015 at 2:33:50 PM UTC-4, puzzler wrote: > > You're not using the combinatorics library as efficiently as you could > be. Here's the best strategy for generating all the team combinations with > the same number of players: >
Is a combinatorics library even needed for this? (-> (for [a teams b teams :when (= (count a) (count b))] #{a b}) (hash-set)) Should produce all distinct pairs with equal player counts, given that "teams" is bound to a collection of teams, each of which is a collection of player objects, and "seq" works on the former collection and "count" on the latter ones. If each team is something more complex (e.g. a map with a :players key) things get only slightly more complicated (e.g. changing (count a) to (count (:players a))). -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en --- You received this message because you are subscribed to the Google Groups "Clojure" group. To unsubscribe from this group and stop receiving emails from it, send an email to clojure+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.