In Haskell, given that:

foldr fg [] = g foldr f g (x:xs) fx (foldr f g xs) Using foldr write the function group: : (a->a->Bool)-> [a]->[[a]] which, given a predicate p::a->a->Bool and a list, breaks the list into a series of (longest possible) sublists in which any two consecutive elements satisfy the predicate (in particular grouping the empty list will give the list containing the empt list). For example, suppose that the predicate nbr determines whether two integers differ by at most one, then group nbr [] = [0] group nbr [2,1,3,4,5,5,4,7,4,3, 3] [[2,1], [3,4,5,5,4],[7], [4,3,3]] foldr fg [] = g foldr f g (x:xs) fx (foldr f g xs) Using foldr write the function group: : (a->a->Bool)-> [a]->[[a]] which, given a predicate p::a->a->Bool and a list, breaks the list into a series of (longest possible) sublists in which any two consecutive elements satisfy the predicate (in particular grouping the empty list will give the list containing the empt list). For example, suppose that the predicate nbr determines whether two integers differ by at most one, then group nbr [] = [0] group nbr [2,1,3,4,5,5,4,7,4,3, 3] [[2,1], [3,4,5,5,4],[7], [4,3,3]]