From 8ffac10bf733f5f75bf85750562710c1cd60cf0a Mon Sep 17 00:00:00 2001
From: braxtonhall
Date: Mon, 24 Oct 2022 15:14:18 -0700
Subject: Add Ron

---
 entries/rxg/fib.rkt | 132 ++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 132 insertions(+)
 create mode 100644 entries/rxg/fib.rkt

(limited to 'entries/rxg/fib.rkt')

diff --git a/entries/rxg/fib.rkt b/entries/rxg/fib.rkt
new file mode 100644
index 0000000..61373cd
--- /dev/null
+++ b/entries/rxg/fib.rkt
@@ -0,0 +1,132 @@
+#lang racket
+
+;; fib.rkt - A variety of iterative implementations of the Fibonacci Sequence
+;; All are based on accumulators
+
+;; Natural -> Natural
+;; produce the n'th Fibonacci number
+(define (fib-iter n)
+  (cond
+    [(= n 0) 1]
+    [(= n 1) 1]
+    [else   
+     (let-values ([(n-2 n-1)
+                   ;; i-2 = fib (i-2)
+                   ;; i-1 = fib (i-1)
+                   (for/fold ([i-2 1] [i-1 1])
+                             ([i (in-range 2 n)])
+                     (values i-1 (+ i-2 i-1)))])
+       (+ n-2 n-1))]))
+
+
+(define (fib-iter2 n)
+  (cond
+    [(= n 0) 1]
+    [(= n 1) 1]
+    [else   
+     ;; i-2 = fib (i-2)
+     ;; i-1 = fib (i-1)
+     (let loop ([i-2 1] [i-1 1] [i 2])
+       (if (= i n)
+           (+ i-2 i-1)
+           (loop i-1 (+ i-2 i-1) (add1 i))))]))
+
+(define (fib-iter3 n)
+  (cond
+    [(= n 0) 1]
+    [(= n 1) 1]
+    [else   
+     ;; i-2 = fib (i-2)
+     ;; i-1 = fib (i-1)
+     (do ([i-2 1 i-1]
+          [i-1 1 (+ i-2 i-1)]
+          [i 2 (add1 i)])
+       [(= i n)
+        (+ i-2 i-1)])]))
+
+;; New variant of for with accumulators and a final expression in terms of
+;; the accumulators
+(define-syntax (for/acc stx)
+  (syntax-case stx ()
+    [(_ ([id* init*] ...)
+        (clauses ...)
+        body
+        result)
+     (with-syntax ([original stx])
+     #'(let-values ([(id* ...)
+                   (for/fold/derived original
+                                     ([id* init*] ...)
+                                     (clauses ...)
+                                     body)])
+         result))]))
+
+(define (fib-iter4 n)
+  (cond
+    [(= n 0) 1]
+    [(= n 1) 1]
+    [else   
+     ;; i-2 = fib (i-2)
+     ;; i-1 = fib (i-1)
+     (for/acc ([i-2 1] [i-1 1])
+              ([i (in-range 2 n)])
+              (values i-1 (+ i-2 i-1))
+              (+ i-2 i-1))]))
+
+;; New variant of for with accumulators and a final expression in terms of
+;; the accumulators
+(define-syntax (for/do stx)
+  (syntax-case stx ()
+    [(_ ([id* init* step*] ...)
+        (clauses ...)
+        result)
+     (with-syntax ([original stx])
+     #'(let-values ([(id* ...)
+                   (for/fold/derived original
+                                     ([id* init*] ...)
+                                     (clauses ...)
+                                     (values step* ...))])
+         result))]))
+
+(define (fib-iter5 n)
+  (cond
+    [(= n 0) 1]
+    [(= n 1) 1]
+    [else   
+     ;; i-2 = fib (i-2)
+     ;; i-1 = fib (i-1)
+     (for/do ([i-2 1 i-1] [i-1 1 (+ i-2 i-1)])
+             ([i (in-range 2 n)])
+             (+ i-2 i-1))]))
+
+
+;; Fibonacci's problem, as described by Greg Rawlins in "Compared to What":
+;;   Suppose you have a pair of rabbits and suppose every month each pair
+;;   bears a new pair that from the second month on becomes productive.
+;;   how many pairs of rabbits will you have in a year?
+
+;; Analysis:  
+;; - At time 0 you have 1 unproductive pair: 1 pair, 0 productive pairs
+;; - Each month, each productive pair produces an unproductive pair
+;; - Each month, last months unproductive pairs transition to productive
+;; - how many pairs are there at time step 12?
+
+;; The following function solves the problem *directly* as a
+;; structural recursion over natural numbers with two
+;; accumulators (for lost context (fertile) and result-so-far (total))
+
+;; Natural -> Natural
+;; produce the solution to Fibonacci's problem after n months
+(define (fib-rabbit n0)
+  ;; Accumulator: total is Natural
+  ;; Invariant: total pairs of rabbits after (- n0 n) months
+  ;; Accumulator: fertile is Natural
+  ;; Invariant: productive pairs of rabbits after (- n0 n) months
+  (local [(define (fib-acc fertile total n)
+            (cond [(zero? n) total]
+                  [else 
+                   (fib-acc total      ;; next month, all will be productive
+                            (+         ;; next months pairs include:
+                             total     ;; - this months pairs plus
+                             fertile)  ;; - offspring from productive pairs
+                            (sub1 n))]))] 
+    (fib-acc 0 1 n0)))
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