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Postings from May 2005
On Fri, May 20, 2005 at 10:14:26PM +0000, mark.a.biggar@comcast.net wrote: > > > Mark A. Biggar wrote: > > > Well the identity of % is +inf (also right side only). > > > > I read $n % any( $n..Inf ) == $n. The point is there's no > > unique right identity and thus (Num,%) disqualifies for a > > Monoid. BTW, the above is a nice example where a junction > > needn't be preserved :) > > If as usual the definition of a right identity value e is that a op e = a for all a, > then only +inf works. Besdies you example should have been; > $n % any (($n+1)..Inf), $n % $n = 0. > > > > E.g. if X<Y is left associative and returns Y when true then ... > > > > Sorry, is it the case that $x = $y < $z might put something else > > but 0 or 1 into $x depending on the order relation between $y and $z? > > Which is one reason why I siad that it might not make sense to define the chaining ops in terms of the associtivity of the binary ops, But as we are interested in what [<] over the empty list shoud return , the identity (left or right) of '<' is unimportant as I think that should return false as there is nothing to be less then anything else. Note that defaulting to undef therefore works in that case. The identity operand is -inf for < and <=, and +inf for > and >=. A chained relation < (>, <=, >=) is then taken to mean monotonically increasing (decreasing, non-decreasing, non-increasing), and an empty list, like a one element list, is always in order. --Thread Previous | Thread Next