CS61A_sp95_f - CS61A Spring 1995 Final CS61A Spring 1995...

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CS61A, Spring 1995 Final Question 1 (5 points): (a) Write a function prefix-to-infix that takes a Scheme arithmetic expression as its argument and returns a list containing the equivalent expression in the ordinary arithmetic notation with operators between the operands, like this: > (prefix-to-infix '(+ (* 2 3) (- 7 4))) ((2 * 3) + (7 - 4)) > (prefix-to-infix '(* (remainder 9 2) 5) ((9 remainder 2) * 5) You may assume that every function in the argument expression has exactly two arguments. (b) The procedure you wrote in part (a) carries out a tree reordering; in each sublist, the three elements are rearranged from the original order (0 1 2) to the new order (1 0 2) . That is, what used to be element number 1 now comes first; what used to be element number 0 now comes second, and what used to be number 2 remains third. We can represent this ordering by the list (1 0 2) . We'd like to generalize this by writing a procedure that takes an ordering as an additional argument, so that we could say (define (prefix-to-infix tree) (tree-reorder '(1 0 2) tree)) Here's an example of a tree-reorder that isn't a prefix-to-infix : > (tree-reorder '(2 1 2 1) '((a b c) (d e f) (g h i))) ((i h i h) (f e f e) (i h i h) (f e f e)) What follows is a partial implementation; your job is to fill in the blank. (define (tree-reorder ordering tree) (if (atom? tree) tree (map _______________________________________________________ ordering))) Assume that no number in the ordering is bigger than the length of any sublist; no error checking is needed. Question 2 (5 points): You are given a possibly infinite stream of lists of numbers. In the following example, the notation {...} represents a stream, while (...) represents a list: {(0 1 0 0 3) (1 2 3 0 4 2) (0 0 0 5 0) () (3) ...}
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