scheme

I have some code to find the maximum height and replace it with the associated name. There are separate lists for height and names, each the same length and non-empty. I can solve this using structural recursion but have to change it into accumulative, and I am unsure how to do that. All the examples I have seen are confusing me. Is anybody able to turn the code into one using accumulative recursion? (define (tallest names heights) (cond [(empty? names) heights] [(> (first heights) (first (rest heights))) (cons (first names) (tallest (rest (rest names)) (rest (rest heights))))] [else (tallest

How can I increment a value in scheme with closure? I'm on lecture 3A in the sicp course. (define (sum VAL) // how do I increment VAL everytime i call it? (lambda(x) (* x x VAL))) (define a (sum 5)) (a 3) Use set! for storing the incremented value. Try this: (define (sum VAL) (lambda (x) (set! VAL (add1 VAL)) (* x x VAL))) Because VAL was enclosed at the time the sum procedure was called, each time you call a it'll "remember" the previous value in VAL and it'll get incremented by one unit. For example: (define a (sum 5)) ; VAL = 5 (a 3) ; VAL = 6 => 54 ; (* 3 3 6) (a 3) ; VAL = 7 => 63 ; (* 3

The problem is to: Write a function (encode L) that takes a list of atoms L and run-length encodes the list such that the output is a list of pairs of the form (value length) where the first element is a value and the second is the number of times that value occurs in the list being encoded. For example: (encode '(1 1 2 4 4 8 8 8)) ---> '((1 2) (2 1) (4 2) (8 3)) This is the code I have so far: (define (encode lst) (cond ((null? lst) '()) (else ((append (list (car lst) (count lst 1)) (encode (cdr lst))))))) (define (count lst n) (cond ((null? lst) n) ((equal? (car lst) (car (cdr lst))) (count

I am new to scheme and having a hard time with using car and cdr. I have an AST string literal in ast. (define ast '(program ((assign (var i int) (call (func getint void int) ())) (assign (var j int) (call (func getint void int) ())) (while (neq (var i int) (var j int)) ((if (gt (var i int) (var j int)) ((assign (var i int) (minus (var i int) (var j int)))) ((assign (var j int) (minus (var j int) (var i int))))))) (call (func putint int void) ((var i int))))) ) I know car returns the head of ast. So (car ast) returns 'program. I am confused how to use car and cdr to get strings from ast such

I'm looking to create a function that returns a list of 'n' functions each of which increments the input by 1, 2, 3... n respectively. I use DrRacket to try this out. A sample of expected outcome : > (map (lambda (f) (f 20)) (func-list 5)) (21 22 23 24 25) I'm able to write this down in a static-way : > (define (func-list num) > (list (lambda (x) (+ x 1)) (lambda (x) (+ x 2)) (lambda (x) (+ x 3)) (lambda (x) (+ x 4)) (lambda (x) (+ x 5))) [Edit] Also that a few restrictions are placed on implementation : Only 'cons' and arithmetic operations can be used The func-list should take as input only

I need a procedure which takes a list and checks if an element is part of that list even when the list contains lists. So far, I've written this: (define (element-of-set? element set) (cond ((null? set) #f) ((eq? element (car set)) #t) (else (element-of-set? element (cdr set)))))) But, if you have a list that looks like this: (a (a b b (c b) 3) 5 5 (e s) (s e s)) then my written element-of-set? does not recognize that 3 is a part of this list. What am I doing wrong? You're not recurring down the car of your set argument. If you evaluate (element-of-set? 3 '(a (a b b (c b) 3) 5 5 (e s) (s e s))

tScheme novice question: I need to determine if a list contains an even or odd amount of atoms using recursion. I know the easy way is to get list length and determine if it is even or odd, but I would like to see hows it's done using recursion. (oddatom (LIST 'x 'y 'v 'd 'r 'h 'y)) should return #t , while (oddatom '((n m) (f p) l (u k p))) should return #f appreciate the help. Here's my version of the solution: (define (oddatom? lst) (let recur ((odd #f) (x lst)) (cond ((null? x) odd) ((pair? x) (recur (recur odd (car x)) (cdr x))) (else (not odd))))) Joshua Taylor I like Chris Jester-Young

I am defining a function binomial(n k) (aka Pascal's triangle) but am getting an error: application: not a procedure; expected a procedure that can be applied to arguments given: 1 arguments...: 2 I don't understand the error because I thought this defined my function: (define (binomial n k) (cond ((or (= n 0) (= n k)) 1) (else (+ (binomial(n) (- k 1))(binomial(- n 1) (- k 1)))))) In Scheme (and Lisps in general), parentheses are placed before a procedure application and after the final argument to the procedure. You've done this correctly in, e.g., (= n 0) (= n k) (- k 1) (binomial(- n 1) (-