chaining multi-parameter functions

Question

Suppose I have two functions, f:X->Y and g:Y*Y->Z. I want to make a third function, h(a, b) = g(f(a), f(b)).

h a b = g (f a) (f b)

Is there any way to write it like h(a, b) = g*f (a, b)?

And what if h(a,b,c,d) = g2*g1*f2*f1 (a,b,c,d), where g_i takes 2 args?

your two examples are out of sync. The first should be g (f1 a) (f2 b) or the second g (f a) (f b) (f c) (f d). — Will Ness
– Will Ness, Commented Feb 18, 2013 at 18:38
You can write h = \a b -> g (f a) (f b) as h = (. f) . g . f, but you shouldn't (readability). — Daniel Fischer
– Daniel Fischer, Commented Feb 18, 2013 at 18:40
in your new 2nd example, you mean f functions taking ` arg, and g funcs taking 2? Then you need a 3rd level. Could you write it out in full exactly? — Will Ness
– Will Ness, Commented Feb 18, 2013 at 18:59

gspr · Accepted Answer · 2013-02-18 18:39:50Z

10

Searching Hoogle for functions with the right signature reveals on from Data.Function. According to its documentation,

g `on` f

seems to be what you want.

answered Feb 18, 2013 at 18:39

gspr

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Chris Taylor · Accepted Answer · 2013-02-18 18:50:06Z

7

The on combinator (in Data.Function, as pointed out by gspr in another answer) is defined by

g `on` f = \x y -> g (f x) (f y)

Which would allow you to write

h = g `on` f

You can make higher-dimensional generalizations of this, for example

g `on3` f = \x y z -> g (f x) (f y) (f z)

g `on4` f = \w x y z -> g (f w) (f x) (f y) (f z)

So that you could write

h = g `on3` f

There may be a way to write on3 and on4 in terms of on, but if there is I can't see it at the moment.

edited Feb 18, 2013 at 18:50

answered Feb 18, 2013 at 18:44

Chris Taylor

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Matt Fenwick · Accepted Answer · 2013-02-18 19:38:25Z

6

You may also find Arrows interesting. Here's one way to do it:

h g f a b = uncurry g ((f *** f) (a, b))

Which is equivalent to your example (except that g and f are not free) and on. Using:

definition of *** for functions:
```
(***) f g ~(x,y) = (f x, g y)
```
definition of uncurry:
```
uncurry f p =  f (fst p) (snd p)
```

And substituting them into the original equation:

h g f a b = uncurry g (f a, f b) (used *** definition)
h g f a b = g (f a) (f b) (used uncurry definition)

answered Feb 18, 2013 at 19:38

Matt Fenwick

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