6001 structure & interpretation of computer programs recitation 12/ november 4, 1997

topics mutable data structures finish circular buffer example from last time sample quiz problems patterns, representations, tree search, streams 2/24/2019 daniel jackson

circular buffer (from last time) model always contains at least one element values inserted to right of bar values removed from right of bar basic operations make-circular-buffer : T –> CB insert! : CB x T –> undefined remove! : CB –> T rotate-left! : CB –> undefined rotate-right! : CB –> undefined size : CB –> int implementation hints use only a singly-linked list all ops are 0(1) time except for rotate-right! and size 1 6 2 5 4 3 2/24/2019 daniel jackson

implementation notes represent each entry as a cons cell represent buffer as a cons cell whose car (cdr) is the cell to the left (right) of the bar code (define (make-cb v) (let ((cell (cons v nil))) (set-cdr! cell cell) (cons cell cell))) (define (insert! b v) (let ((new-cell (cons v (cdr b)))) (set-cdr! (car b) new-cell) (set-cdr! b new-cell))) (define (remove! b) (let ((old-cell (cdr b))) (set-cdr! (car b) (cdr old-cell)) (set-cdr! b (cdr old-cell)) (car old-cell))) 2/24/2019 daniel jackson

implementation, ctd (define (size b) (define (count cell) (if (eq? (car b) cell) 1 (inc (count (cdr cell))))) (count (cdr b))) ; a hidden operation: (prev c) is the cell that precedes c (define (prev c) (define (follow current) (let ((next (cdr current))) (if (eq? c next) current (follow next)))) 2/24/2019 daniel jackson

implementation, ctd (define (rotate-left! b) (set-car! b (cdr b)) (set-cdr! b (cddr b))) (define (rotate-right! b) (set-cdr! b (car b)) (set-car! b (prev (car b)))) 2/24/2019 daniel jackson

sample problems: pattern matching (1) Write a simplified version of (match pat dat) which takes iin two args, both lists, representing the pattern & the datum. The pattern has ?, ?variable & symbols. The datum has symbols. It returns #f if there is no dictionary or it returns the appropriate dictionary. You can assume you have a procedure (insert var val d) that returns a new dictionary obtainined by adding the pair (var, val) to d, unless var is already there, in which case it returns d if (var, val) is present and #f otherwise. You can also assume that you have primitive procedures var?, anon-var? that take a symbol and return true if the symbol is a variable, anonymous variable examples (match '(? ?a b c ?d) '(hi there b c hello)) ==> ((?a there) (?d hello)) (match '(? ?a b c ?d) '(hi there b bad hello))==> #f 2/24/2019 daniel jackson

sample problems: pattern matching (2) solution (define (match pat dat) (cond ((not (eq? (length pat) (length dat))) #f) ((null? pat) ‘()) (let ((d (match (cdr pat) (cdr dat)))) (if (not d) d (cond ((var? (car pat)) (insert (car pat) (car dat) d)) ((anon-var? (car pat)) d) ((eq? (car pat) (car dat)) d) (else #f)))))) 2/24/2019 daniel jackson

sample problems: pattern matching (3) Write a simplified version of (instantiate temp dict) which takes in two args, a template, which is a list of ?variables & symbols and dict which is a dictionary that better have all the ?variables mentioned in the template. It returns the template with the variables substituted appropriately. example (instantiate '(this ?is a ?test) '((?is hi) (?test hello) (?other ignored))) ==> (this hi a hello) solution (define (instantiate temp dict) (define (lookup v) (cadr (assq v dict))) (map lookup temp)) 2/24/2019 daniel jackson

sample problems: mutable bag Design a Data Structure for.... & implement (or provide a sketch of how it works) the following functions: Represent a mutable bag of elements. A bag is like a set with duplicates. Create empty bag - O(1) add element to bag - O(n) where n is number of distinct elements in bag return total # of elements - O(n) return # of distinct elements - O(n) delete an element - O(n) determine how many occurences of a particular element are in the bag - O(n) After you have something that works with those time bounds, see if you can improve (by changing slightly the data structure) the time bounds for return total # of elements & return # of distinct elements to O(1) Can you think of other useful operations on the bag? 2/24/2019 daniel jackson

sample problem: tree search a tree consists of a key, a value and a list of subtrees write a search procedure that takes a tree and a key and returns a list of all values associated with the key assume you have procedures empty?, key, value, next where (next t) returns the subtrees of t code (define (search t k) (define (dfs trees) (if (null? trees) nil (let ((t1 (car trees))) (if (eq? (key t1) k) (cons (val t1) (dfs (append (next t1) (cdr trees)))) (dfs (append (next t1) (cdr trees))))))) (dfs (list t))) 2/24/2019 daniel jackson

sample problem: streams (1) what does map-stream do? what is its type? implement it using the primitives car-stream, cdr-stream, cons-stream solution map-stream : (t1 -> t2) x stream(t1) –> stream(t2) (define (map-stream p s) (cons-stream (p (car-stream s)) (map-stream p (cdr-stream s)))) 2/24/2019 daniel jackson

sample problem: streams (2) take the tree from before define a procedure that converts a tree to a stream containing key/value pairs in depth first order. use this to implement depth-frst search code (define (streamify t k) (define (dfs trees) (if (null? trees) the-empty-stream (let ((t1 (car trees))) (cons-stream (cons (key t1) (val t1)) (dfs (append (next t1) (cdr trees))))))) (dfs (list t))) (define (search t k) (filter-stream (lambda (e) (eq? (key e) k)) (streamify t))) 2/24/2019 daniel jackson

sample problem: procedures with state write a procedure that takes a procedure p of one argument and returns another procedure, that behaves exactly like p except when presented with the argument ‘how-many, which causes it to return the number of times p has been called. example (define sc (counter square)) (sc 3) ==> 9 (sc 4) ==> 16 (sc ‘how-many) ==> 2 solution (define (counter p) (let ((ct 0)) (lambda (i) (if (eq? i ‘how-many) ct (begin (set! ct (inc ct)) (p i)))))) 2/24/2019 daniel jackson