Adding Fibonacci Sequence (recursive) - in Racket - for issue #15

Algorithms/armstrong-Number.rkt

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+#lang racket
+
+;Armstrong number is a number that is equal to the sum of cubes of its digits. For example 0, 1, 153, 370, 371 and 407 are the Armstrong numbers.
+;
+;Let's try to understand why 153 is an Armstrong number.
+;
+;153 = (1*1*1)+(5*5*5)+(3*3*3)  
+;where:  
+;(1*1*1)=1  
+;(5*5*5)=125  
+;(3*3*3)=27  
+;So:  
+;1+125+27=153
+
+;; I got help from this article
+;; https://www.javatpoint.com/armstrong-number-in-c#:~:text=Armstrong%20number%20is%20a%20number,153%20is%20an%20Armstrong%20number.&text=Let's%20try%20to%20understand%20why%20371%20is%20an%20Armstrong%20number.
+
+;; number -> boolean
+;; code
+;(define (armstrong-Number n) false)   ; stub
+
+(define (armstrong-Number n)
+  (local [(define (armstrong-sum n)                            ;; helper function to find the sum, of cubes, of digits, of n
+            (cond [(zero? n) 0]                                ;; basecase to stop structural recirsion                               
+                  [else
+                   (+                                          ;; add them all
+                    (expt (modulo n 10) 3)                     ;; get    the last digit of n by modulo 10 of n
+                    (armstrong-sum (quotient n 10))) ]))]      ;; remove the last digit of n by quotient 10 of n - integer division by 10
+    (= (armstrong-sum n) n)))                                  ;; compare the sum of cube of digits of n, and n
+
+
+
+
+;; test examples
+;(equal? (armstrong-Number 0) true)
+;(equal? (armstrong-Number 1) true)
+;(equal? (armstrong-Number 9) false)
+;(equal? (armstrong-Number 153) true)
+;(equal? (armstrong-Number 100) false)
+;(= 153 (+ (expt 1 3) (expt 5 3) (expt 3 3)))
+;

Mathematics/fibonacci/fib.rkt

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+#lang racket
+
+;; The Fibonacci Sequence is the series of numbers:
+;; 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
+
+;; integer -> list of integers
+;; building intution for fibn - inner helper function
+;; fn = fn-1 + fn-2 where fn < 2 = n
+;; f0 = 0, f1 = 1  ; basecases
+;; f2 = f1 + f0 = 1 + 0 = 1
+;; f3 = f2 + f1 = 1 + 1
+;; f8 = f7 + f6 = 13 + 8 = 21
+
+;; code
+;(define (fib n) (list 0)) ; stub
+
+(define (fib n)
+  (local [(define (fibn n)                             ;; inner helper function to calculate (fib n)
+            (cond [(= n 0) 0]
+                  [(= n 1) 1]
+                  [else
+                   (+                                  ;; add fn-1 to fn-2
+                    (fibn (- n 1))
+                    (fibn (- n 2)))]))]                ;; self reference for structital recursion
+    (build-list (add1 n) fibn)))
+
+;; test examples for fibn
+;(equal? (fibn 0) 0)                                   ;; base case for structural recursion
+;(equal? (fibn 1)  1)                                  ;; base case for structural recursion
+;(equal? (fibn 8) (+ (fibn 7) (fibn 6)))               ;; intution
+;(equal? (fibn 8) 21)
+;
+
+;; test examples for fib
+;(equal? (fib 0) (cons 0 empty))                  ;; base case for structural recursion
+;(equal? (fib 1) (list 0 1))                ;; base case for structural recursion
+;(equal? (fib 8) (list 0 1 1 2 3 5 8 13 21))

problems/palindrome/isPalindrome?.rkt

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+#lang racket
+
+;Build a function that checks weather a given number is a palindrome or not without using a strings. Implement it using any language.
+;
+;For example:
+;input: 12321
+;output: true
+;
+; input: 12344321
+; output: True
+
+;; intution
+;; n and reverse n will be same for a palindrome
+;; will use reverse-number (see pr #341 for issue #35)
+
+;; I got the Idea from link below
+;; https://www.integers.co/questions-answers/is-12344321-a-palindrome-number.html
+
+
+
+;; code
+(define (isPalinDrome n)
+  (local [
+          (define (reverse-n n acc)                               ; using an decerasing accumulator
+            (cond [(= (order-of-magnitude n) 0) n]                ; basecase, used to stop recursion
+                  [else
+                   (+ 
+                    (* (modulo n 10) (expt 10 acc))               ; get the right-most digit of n, multiply it with 10^x
+                    (reverse-n (quotient n 10) (sub1 acc)))]))
+
+          (define (reverse-number num)
+            (reverse-n num (order-of-magnitude num)))
+          ]
+    (= (reverse-number n) n)))
+
+; test examples
+;(equal? (isPalinDrome 9) true) 
+;(equal? (isPalinDrome 12349321) false)
+;(equal? (isPalinDrome 12321) true)