6.001, Spring Semester, 2005--Quiz II Part 1:(25 points) We are going to explore a new kind of data structure, called a cycle. You can think of this as a kind of circular list: indeed we are going to represent a cy cle as a loop of cons cells, where the car of each cell point s to a value, and the cdr of each cell points to the next element in the circular sequence. To create a cycle from a list, we can use (define (list->cycle lst) et-cdr! (last lst) lst) (define (last lst) (if (null? lst) (error "not long enough") (if (null? (cdr lst)) (last (cdr lst))))) (define test-cycle (list->cycle (a b c g))) To see what this struct ure looks like, you might find it convenient to draw a box-and-pointer Associated wit h a cycle, we have several operat ions head will ret urn the value stored at the cell in the cy cle pointed to by its argument, e. g (head test-cycle)will return the value a rotate-left willrot ate t he cycle one element to the left, e.g,(head (rotate-left test-cycle)) will return the value b; (you can think of a cycle as a circular list in which the arranged in clo ckwise sequence, the head is at the top, and"left means rot at ing the sequence counter-clockwise) rotate-right will rot ate the cycle one element to the right, in analogy to rot ate-left, e. g (head (rot ate-right test-cy cle)) will ret urn t he value￾
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