Way calculation of a function in Haskell using composition -



Way calculation of a function in Haskell using composition -

i have function in haskell:

thrit::(a->a)->a->a thrit f= f.f.f

and want know how computer calculate expressions answer:

((thrit.thrit)(+1))0 9

((thrit thrit)(+1))0 27

let's expand expressions

((thrit . thrit) (+1)) 0 = (thrit (thrit (+1)) 0 = (thrit ((+1) . (+1) . (+1))) 0 = ((+1) . (+1) . (+1) . (+1) . (+1) . (+1) . (+1) . (+1) . (+1)) 0

which 9

((thrit thrit) (+1)) 0 = ((thrit . thrit . thrit) (+1)) 0 = (thrit (thrit (thrit (+1))) 0

you may have noticed (thrit (thrit (thrit (+1))) 0 3 times (thrit (thrit (+1)) 0

hence 27

haskell

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