Super combinator
From HaskellWiki
(Rewrite to explain what's really going on.) 
m (Formatting fix) 

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So these are supercombinators: 
So these are supercombinators: 

−  * <code>0<code> 
+  * <code>0</code> 
* <code>\x y > x + y</code> 
* <code>\x y > x + y</code> 

* <code>\f > f (\x > x + x)</code> 
* <code>\f > f (\x > x + x)</code> 

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A supercombinator which is not a lambda abstraction (i.e. n=0) is called a [[Constant applicative form]]. 
A supercombinator which is not a lambda abstraction (i.e. n=0) is called a [[Constant applicative form]]. 

−  Any Haskell program can be converted into supercombinators using [[Lambda lifting]]. 
+  Any lambda calculus expression (or, indeed, Haskell program) can be converted into supercombinators using [[Lambda lifting]]. 
[[Category:Glossary]] 
[[Category:Glossary]] 
Revision as of 03:52, 1 February 2010
A super combinator is either a constant, or a Combinator which contains only super combinators as subexpressions.
To get a fuller idea of what a supercombinator is, it may help to use the following equivalent definition:
Any lambda expression is of the form \x1 x2 .. xn > E
, where E is not a lambda abstraction and n≥0. (Note that if the expression is not a lambda abstraction, n=0.) This is a supercombinator if and only if:
 the only free variables in E are x1..xn, and
 every lambda abstraction in E is a supercombinator.
So these are supercombinators:

0

\x y > x + y

\f > f (\x > x + x)
These are not combinators, let alone supercombinators, because in each case, the variable y occurs free:

\x > y

\x > y + x
This is a combinator, but not a supercombinator, because the inner lambda abstraction is not a combinator:

\f g > f (\x > g x 2)
A supercombinator which is not a lambda abstraction (i.e. n=0) is called a Constant applicative form.
Any lambda calculus expression (or, indeed, Haskell program) can be converted into supercombinators using Lambda lifting.