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Jorge Llamb�as, On 25/08/2012 22:17:
On Sat, Aug 25, 2012 at 5:12 PM, And Rosta<and.rosta@hidden.email> wrote:I'm a bit lost by this. The way it works is that the parsed predicate and the incoming predicate each have an ordered list of arguments. You take the first argument off the incoming list and, starting at the start of the parsed list, work your way through the parsed list until you can merge an item with the current item from the incoming list. Once you've merged it, you then take the next item of the incoming list and continue along the parsed list till you can merge, and so forth.Continue along? Does that mean that the second argument of the incoming list can't be merged with one from the parsed list that was passed by the first argument?
Yes, but the order of the inflections marking arguments of the incoming is dictated not by argument 'place' (participant role) (which is encoded by the type of inflection rather than its relative order) but by the order in which they merge and interleave with arguments of the parsed predicate.
N(n,m) = Sum from i=1 to min(n,m) of n!m! / i!(n-i)!(m-i)! For four sorts of correspondence, it gets more complicated.Think of it this way: You have n men and m women, each may enter into monogamous heterosexual marriage or may remain single. How many different marriage patterns are there?That's the above N(n,m)
Righto. I don't understand well enough to actually do sums with it, but that doesn't matter.
But this assumes that the second woman can still choose one of the men discarded by the first woman.
She can in that before the marriages take place, the women line themselves up in such a way as to each be able to get the man they want, even tho they can't get men discarded by the women in front of them. --And.