Petite Chez Scheme Version 8.3 Copyright (c) 1985-2011 Cadence Research Systems (define (fib n) (let loop ((n n) (f2 1) (f1 1) (fs (list))) (if (zero? N) (reverse fs) (loop (- n 1) f1 (+ f2 f1) (cons f2 fs))))) (fib 50) (1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 1 8 93 111 5 215 92037 39059190909025).
Yes: (defun fibonacci (n &optional (a 0) (b 1) (acc )) (if (zerop n) (nreverse acc) (fibonacci (1- n) b (+ a b) (cons a acc)))) (fibonacci 5); (0 1 1 2 3) The logic behind it is that you need to know the two previous numbers to generate the next. A 0 1 1 2 3 5.
Fibonacci Series
One of the most fundamental concepts learnt by a novice programmer is generation of Fibonacci Series. Fibonacci Number in O(log n). I've made a basic program that output Fibonacci sequence for whatever length 'n'. Creating a list of Fibonacci numbers in Scheme? Lisp program (using Scheme. // Fibonacci Series using Dynamic Programming class fibonacci { static int fib(int n) { /* Declare an array to store Fibonacci numbers.
B 1 1 2 3 5 8. New-b 1 2 3 5 8 13.
Instead of returning just one result I accumulate all the a-s until n is zero. EDIT Without reverse it's a bit more inefficient: (defun fibonacci (n &optional (a 0) (b 1)) (if (zerop n) nil (cons a (fibonacci (1- n) b (+ a b))))) (fibonacci 5); (0 1 1 2 3). The program prints the nth number of Fibonacci series.
This program doesn't print anything. If you're seeing output, it's probably because you're calling it from the read-eval- print-loop (REPL), which reads a form, evaluates it, and then prints the result.
E.g., you might be doing: CL-USER (fibonacci 4) 2 If you wrapped that call in something else, though, you'll see that it's not printing anything: CL-USER (progn (fibonacci 4) nil) NIL As you've got this written, it will be difficult to modify it to print each fibonacci number just once, since you do a lot of redundant computation. For instance, the call to (fibonacci (- n 1)) will compute (fibonacci (- n 1)), but so will the direct call to (fibonacci (- n 2)) That means you probably don't want each call to fibonacci to print the whole sequence.
Fibonacci Series Program In C
If you do, though, note that (print x) returns the value of x, so you can simply do: (defun fibonacci(n) (cond ((eq n 1) 0) ((eq n 2) 1) ((print (+ (fibonacci (- n 1)) (fibonacci (- n 2))))))) CL-USER (progn (fibonacci 6) nil) 1 2 1 3 1 2 5 NIL You'll see some repeated parts there, since there's redundant computation. You can compute the series much more efficiently, however, by starting from the first two numbers, and counting up: (defun fibonacci (n) (do ((a 1 b) (b 1 (print (+ a b))) (n n (1- n))) ((zerop n) b))) CL-USER (fibonacci 6) 2 3 5 8 13 21. An option to keep the basic structure you used is to pass an additional flag to the function that tells if you want printing or not: (defun fibo (n printseq) (cond ((= n 1) (if printseq (print 0) 0)) ((= n 2) (if printseq (print 1) 1)) (T (let ((a (fibo (- n 1) printseq)) (b (fibo (- n 2) NIL))) (if printseq (print (+ a b)) (+ a b)))))) The idea is that when you do the two recursive calls only in the first you pass down the flag about doing the printing and in the second call instead you just pass NIL to avoid printing again.