我正在閱讀在線書籍「計算分類理論」http://www.cs.man.ac.uk/~david/categories/book/book.pdf,本書中存在問題2.10的一些問題。特別是隨着powerset的定義。StandardML中一組套件上類型衝突的問題
abstype 'a Set = set of 'a list
with val emptyset = set([])
fun is_empty(set(s)) = length(s)=0
fun singleton(x) = set([x])
fun disjoint_union(set(s),set(nil))=set(s) |
disjoint_union(set(s),set(t::y))=
if list_member(t,s) then disjoint_union(set(s),set(y))
else disjoint_union(set(t::s),set(y))
fun union(set(s),set(t)) = set(append(s,t))
fun member(x,set(l)) = list_member(x,l)
fun remove(x,set(l)) = set(list_remove(x,l))
fun singleton_split(set(nil)) = raise empty_set
| singleton_split(set(x::s)) =(x,remove(x,set(s)))
fun split(s) = let val (x,s') = singleton_split(s) in (singleton(x),s') end
fun cardinality(s) = if is_empty(s) then 0 else
let val (x,s') = singleton_split(s) in 1 + cardinality(s') end
fun image(f)(s) = if is_empty(s) then emptyset else
let val (x,s') = singleton_split(s) in
union(singleton(f(x)),image(f)(s')) end
fun display(s)= if is_empty(s) then [] else
let val (x,s') = singleton_split(s) in x::display(s') end
fun cartesian(set(nil),set(b))=emptyset |
cartesian(set(a),set(b)) = let val (x,s') = singleton_split(set(a))
in union(image(fn xx => (x,xx))(set(b)),cartesian(s',set(b))) end
fun powerset(s) =
if is_empty(s) then singleton(emptyset)
else let
val (x,s') = singleton_split(s)
val ps'' = powerset(s')
in union(image(fn t => union(singleton(x),t))(ps''),ps'') end
end
的冪函數是從答案在附錄D我再創建一個集的冪給出:
val someset=singleton(3); (*corresponds to the set {3}*)
val powerset=powerset(someset); (* should result in {{},{3}} *)
val cardinality(someset); (*returns 1*)
val cardinality(powerset); (*throws an error*)
! Type clash: expression of type
! int Set Set
! cannot have type
! ''a Set
爲什麼我可以計算一個整數集的基數,但不一組整數?難道我做錯了什麼?
有沒有一種方法來定義抽象的相等性設置類型,以便list_remove將能夠使用int Set?即我可以強制抽象類型來支持平等嗎? –