racket

The question i'm trying to answer: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? Where am I going wrong? my prime? test seems to be the issue, but it works fine on relatively small numbers. However the prime? test gives a wrong answer with larger numbers. Is there an easier way to go about this? (define b 3) (define z 0) (define divides? (lambda (a b) (= (remainder a b) 0))) (define (prime? n) (cond ((or (= n 1) (= n 0)) false) ((even? n) false) ((= n 2) true) ((= n b) true) ((divides? n b) false) (else (and (set! b (+ b 1))

Suppose I have something like this: (define pair (cons 1 (lambda (x) (* x x)) If I want to return the front object of the pair I do this: (car pair) And it returns 1. However when the object is a procedure I don't get the exact description of it. In other words: (cdr pair) returns #<procedure> and not (lambda (x) (*x x)) . How do I fix this? Although there's no way to do this generally, you can rig up something to do it for procedures that you define. Racket struct s can define a prop:procedure that allows the struct to be applied (called) as a procedure. The same struct can hold a copy of

I am trying to understand recursion in scheme and i have a hard time doing the dry run for it, for example a simple fibonacci number problem can someone breakdown the steps in which the additions take place ? (define (fib n) (if (<= n 2) 1 (+ (fib (- n 1)) (fib (- n 2))))) If you're using Racket, as your tags indicate, then you have a built-in stepper. Enter the program into DrRacket, and click Step in the top-right menu: First step http://f.cl.ly/items/341k1X2c44422e220T3I/Screen%20Shot%202013-02-01%20at%208.00.27%20PM.png Then a stepper window will open up. Click Step over and over, and you

I got a list of lists in racket and have to transpose them. (: transpose ((list-of(list-of %a)) -> (list-of (list-of %a)))) (check-expect (transpose (list (list 1 2 3) (list 4 5 6))) (list (list 1 4) (list 2 5) (list 3 6))) (define transpose (lambda (xs) (cond ((empty? xs)empty) ((pair? xs)(make-pair (make-pair (first(first xs)) (make-pair (first(first(rest xs)))empty)) (transpose (rest(rest xs)))))))) That's my code at the moment. I think the problem is in the recursive call (correct me if I'm wrong please). The actual outcome is (list (list 1 4)) . The rest seems kinda ignored. It would

In standard Scheme it is possible to write (if (> x 2) (set! x (- x 1))) but this is not possible in Racket -- Racket's if always requires two arms. Why? Rationale The one-armed variant of if was removed from Racket to prevent bugs. In functional code one always uses the two-armed variant of if . (if test expr-on-true expr-on-false) Forgetting the second arm expr-on-false would not lead to a syntax-error, but to a runtime error (the expression would return #<void> ). To prevent these often occurring bugs in functional code, it was decided to introduce the form when for the one-armed variant of