Miscellanea
April 26, 2011
We have three short exercises today.
FizzBuzz: Looking back over past exercises, I was surprised to find that we haven’t done this classic interview question. You are to write a function that displays the numbers from 1 to an input parameter n, one per line, except that if the current number is divisible by 3 the function should write “Fizz” instead of the number, if the current number is divisible by 5 the function should write “Buzz” instead of the number, and if the current number is divisible by both 3 and 5 the function should write “FizzBuzz” instead of the number. For instance, if n is 20, the program should write 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, 17, Fizz, 19, and Buzz on twenty successive lines.
Prime Words: Consider that a word consisting of digits and the letters A through Z can represent an integer in base 36, where the digits represent their base-10 counterparts, A is a decimal 10, B is a decimal 11, and so on, until Z is a decimal 35. For instance, PRAXIS36 = P36 × 365 + R36 × 364 + A36 × 363 + X36 × 362 + I36 × 361 + S36 × 360 = 25 × 365 + 27 × 364 + 10 × 363 + 33 × 362 + 18 × 361 + 28 × 360 = 25 × 60466176 + 27 × 1679616 + 10 × 46656 + 33 × 1296 + 18 × 36 + 28 × 1 = 1557514036. You are to write a function that takes a base-36 number as input and returns true if the number is prime and false if the number is composite.
Split A List: You are to write a function that takes an input list and returns two lists, the first half of the input list and the second half of the input list. If the input list has an odd number of elements, it is your choice in which half to place the center element. You are only permitted to scan the list once.
Your task is to write the three functions described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
My try in REXX
numeric digits 30 call fizzbuzz 20 call prime_words 'PRAXIS' call split_a_list 'This is a list' call split_a_list 'This is a second list' exit fizzbuzz: procedure parse arg nbr do i = 1 to nbr aus = '' if i//3 == 0 then aus = 'Fizz' if i//5 == 0 then aus = strip(aus)||'Buzz' if length(aus) > 0 then say aus else say i end return prime_words: procedure parse arg nbr base36 = '123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ' sum = 0 exp = 0 rnbr = reverse(nbr) do i = 1 to length(rnbr) z = substr(rnbr,i,1) p = pos(z,base36) erg = p * (36 ** exp) sum = sum + erg exp = exp + 1 end aus = nbr '(Decimal:'sum')' if is_prime(sum) then say aus 'is prime' else say aus 'is not prime' return split_a_list: procedure parse arg text w = words(text) m = w % 2 say 'List 1: ' subword(text,1,m) say 'List 2: ' subword(text,m+1) return is_prime: procedure parse arg nbr do x = 2 to nbr-1 if nbr//x == 0 then return 0 end return 1[…] today’s Programming Praxis exercise, our goal is to write three fucntions: FizzBuzz, a function to […]
My Haskell solution (see http://bonsaicode.wordpress.com/2011/04/26/programming-praxis-miscellanea/ for a version with comments):
import Data.Foldable (toList) import Data.Numbers.Primes import Data.Sequence (ViewL(..), (|>), fromList, viewl, empty) fizzbuzz :: Integral a => a -> IO () fizzbuzz n = mapM_ (putStrLn . f) [1..n] where f n = case (mod n 3, mod n 5) of (0, 0) -> "FizzBuzz" (0, _) -> "Fizz" (_, 0) -> "Buzz" _ -> show n isPrimeWord :: String -> Bool isPrimeWord = isPrime . sum . zipWith (*) (iterate (* 36) 1) . reverse . map (\c -> maybe 0 id . lookup c $ zip (['0'..'9'] ++ ['A'..'Z']) [0..]) splitList :: [a] -> ([a], [a]) splitList = f True (empty, empty) where f _ (l,r) [] = (toList l, toList r) f True (l,r) (x:xs) = f False (l, r |> x) xs f False (l,r) (x:xs) = f True ((\(h :< t) -> (l |> h, t |> x)) $ viewl r) xsmy implementation in c
#include <stdio.h> #include <math.h> void FizzBuzz(int n) { int i; for(i = 1; i <= n; i++) { /* if divisible by three */ if((i % 3) == 0 && (i % 5) == 0) printf("FizzBuzz\n"); else if((i % 3) == 0) printf("Fizz\n"); else if((i % 5) == 0) printf("Buzz\n"); else printf("%d\n", i); } } int PrimeWords(char *word) { int i; long long val; long long sqt; /* convert the base 36 word to decimal */ for(i = 0, val = 0; word[i] != '\0'; i++) { if(word[i] >= '0' && word[i] <= '9') val = val * 36 + (word[i] - '0'); else val = val * 36 + (word[i] - 'A' + 10); } /* test primality */ if((val % 2) == 0) /* check divisibility by 2 */ return 0; if((val % 3) == 0) /* check divisibility by 2 */ return 0; /* check the divisibilty of val by 6k-1 or 6k+1 upto squre root of val */ sqt = (long long)sqrt((double)val); for(i = 1; ;i++) { int j; /* test for 6k-1 */ j = 6*i - 1; if(j > sqt) return 1; if((val % j) == 0) return 0; /* test for 6k+1 */ j = 6*i + 1; if(j > sqt) return 1; if((val % j) == 0) return 0; } } /* Split the nodes of the given list into front and back halves, and return the two lists using the reference parameters. If the length is odd, the extra node should go in the front list. */ void FrontBackSplit(struct Node* source, struct Node** frontRef, struct Node** backRef) { struct Node* singleJump = source; struct Node* doubleJump = source; if(singleJump == NULL || singleJump->next == NULL) { *frontRef = source; *backRef = NULL; return; } while(doubleJump->next != NULL && doubleJump->next->next != NULL) { singleJump = singleJump->next; doubleJump = doubleJump->next->next; } *frontRef = source; *backRef = singleJump->next; singleJump->next = NULL; }Solution in Python (for simple tests look at github):
def fizz_buzz(number): for i in range(1, number + 1): if i % 15 == 0: print 'FizzBuzz' elif i % 3 == 0: print 'Fizz' elif i % 5 == 0: print 'Buzz' else: print i def is_prime(number): if number == 1: return False if number == 2: return True if number % 2 == 0: return False for i in range(3, int(sqrt(number)) + 1, 2): if number % i == 0: return False return True def hexatridecimal_to_dec(test_str): pos = len(test_str) - 1 result = 0 for character in list(test_str): code = ord(character) if ord('0') <= code and code <= ord('1'): result += (code - ord('0')) * 36 ** pos elif ord('A') <= code and code <= ord('Z'): result += (code - ord('A') + 10) * 36 ** pos else: print ('illegal character ', character) pos -= 1 return result def split_list(first_node): split_at = first_node double_next = first_node if first_node is None: return (None, None) while double_next.next_node is not None and double_next.next_node.next_node is not None: double_next = double_next.next_node.next_node split_at = split_at.next_node prev_node = split_at split_at = split_at.next_node prev_node.next_node = None return (first_node, split_at)I have been trying to learn macros , so here is my code
using predicate dispatch paper, implemented using macros.
(module example3 "../method.rkt" (define (div-3 x) (zero? (remainder x 3))) (define (div-5 x) (zero? (remainder x 5))) (define (div-3-5 x) (and (div-3 x) (div-5 x))) (add-subclass div-3 (div-3-5 0)) (add-subclass div-5 (div-3-5 0)) (add-subclass div-3-5 (0)) (define-method fizbuz (n) (when: (div-5 n) (fizbuz (- n 1)) (printf "~a\n" "Buzz")) (when: (div-3 n) (fizbuz (- n 1)) (printf "~a\n" "Fizz")) (when: (div-3-5 n) (fizbuz (- n 1)) (printf "~a\n" "FizzBuzz"))) (define-method fizbuz (#:n 0) (printf "\n")) (define-method fizbuz (n) (fizbuz (- n 1)) (printf "~a\n" n) ) (fizbuz 20) )@Vikas Tandi – how about PrimeWords(“3”) and PrimeWords(“2”) :)
Oh and in my code instead of:
if ord('0') <= code and code <= ord('1'):should be
if ord('0') <= code and code <= ord('9'):My solution to “FizzBuzz”: write a generator “fizzer()” that accepts an argument list of the “fizz-buzz” parameters.
# "args" must be a zero or more of tuples ("word", modulo) def fizzer(*args): cnt = 0 while True: cnt += 1 out = "" for word, modulo in args: if cnt % modulo == 0: out += word if out == "": out = cnt yield outThis is how you call it:
# create a "fizz-buzz" generator fb = fizzer(("Fizz", 3), ("Buzz", 5)) for n, fizz in zip(range(25), fb): print(fizz, end=" ") print() # alternate way to call it fb = fizzer(("Fizz", 3), ("Buzz", 5)) for n in range(25): print(fb.__next__(), end=" ") print() # just for grins: create a "fizz-buzz-bang" generator fbb = fizzer(("Fizz", 3), ("Buzz", 5), ("Bang", 7)) for n, fizz in zip(range(25), fbb): print(fizz, end=" ") print()This is the output:
@arturasl – Thanks for pointing that out :(
This check should be added after decimal conversion
My Python solution, with multiple answers for each.
Some of it isn’t original, but I’m most proud of my list splitting and my use of Horner’s Method for
converting a word to a base 36 number (though Python’s
int()provides all the functionality we need,anyway).
Here’s ruby versions of each. FizzBuzz is unimaginative, prime words is OK, split_list works for arrays, but I feel like it could be better.
# # FizzBuzz (1..ARGV[0].to_i).each do |v| if v % 15 == 0 puts "FizzBuzz" elsif v % 5 == 0 puts "Buzz" elsif v % 3 == 0 puts "Fizz" else puts "#{v}" end end# Prime Words # Require mathn for the prime? method. require 'mathn' class String def prime_word? self.to_i(36).prime? end end puts "\n\nPrime Words:" praxis = "praxis" puts "praxis is #{praxis.prime_word? ? "prime" : "not prime"}" lisp = "lisp" puts "lisp is #{lisp.prime_word? ? "prime" : "not prime"}"# Split List (assumes we can't just do l.length). On an odd numbered list # it will put the extra in the first list. def split_list(l) split = size = 0 l.each_index do |current| split = current / 2 size = current end puts "split = #{split} size = #{size}" l1 = l[0..split] l2 = l[split+1..size] return l1, l2 end puts "\n\nSplit List:" list = %w[this is a list of six elements] l1, l2 = split_list(list) puts "first half = #{l1} second half = #{l2}" list = %w[this is a list of six] l1, l2 = split_list(list) puts "first half = #{l1} second half = #{l2}"@Graham – is_prime_word_horner will not work if you pass it word with digits:
>>> reduce(lambda x, y: 36 * x + ord(y) - 55, '123', 0) -7960 >>> int('123', 36) 1371Anyway neat solution, especially the usage of all() and reduce() q:-)
Here is an attempt in python
number = input(‘What number would you like to count to? ‘) + 1
whole = 0
fizzz = 0
buzzz = 0
def whole_number(num):
global whole
if num % 1 == 0:
whole = 1
else:
whole = 0
def fizz(num):
global fizzz
if num % 3 == 0:
fizzz = 1
else:
fizzz = 0
def buzz(num):
global buzzz
if num % 5 == 0:
buzzz = 1
else:
buzzz = 0
for N in range(1,number,1):
whole_number(N)
fizz(N)
buzz(N)
if fizzz is 1 and buzzz is 1:
print(‘FizzBuzz’)
elif buzzz is 1 and fizzz is 0:
print(‘Buzz’)
elif buzzz is 0 and fizzz is 1:
print(‘Fizz’)
else:
print(N)
Sorry I dont know how to retain the indentation and code look
Ryan Take a look at https://programmingpraxis.com/contents/howto-posting-source-code/. That should explain how to format your code. I was going to paste it in, but I’m pretty sure it wouldn’t do what I want.
@arturasl: Thanks for catching that. I only thought to work with letters, basing my
ord(y) - 55off of the offset for the letters A through Z…Like I said, the use of Python’s integer
conversion is probably the preferred method—in this case, the only correct method :-)
—that I came up with. Thanks again!
A non-destructive solution without using reverse (in case that’s cheating):
(define (split xs)
(let loop ((xs xs) (rest xs))
(if (or (null? rest) (null? (cdr rest)))
(values ‘() xs)
(let-values (((first second) (loop (cdr xs) (cddr rest))))
(values (cons (car xs) first) second)))))
Err, I’ll try formatted:
(define (split xs) (let loop ((xs xs) (rest xs)) (if (or (null? rest) (null? (cdr rest))) (values '() xs) (let-values (((first second) (loop (cdr xs) (cddr rest)))) (values (cons (car xs) first) second)))))Here’s my 2 cents:
FizzBuzz
Thought I’d do something different than the standard if-elif-type answer.
from itertools import cycle, izip def fizzbuzz(n): fizzer = cycle(['','','Fizz']) buzzer = cycle(['','','','','Buzz']) counter = xrange(1,n+1) for f,b,c in izip(fizzer, buzzer, counter): print f+b if f or b else cPrime Words
isprime() is from a prior exercise
from itertools import count from operator import mul from primes import isprime from string import digits, ascii_uppercase # translation table base-36 -> decimal xlat = {c:n for n,c in enumerate(digits + ascii_uppercase)} def prime_words(s): digits = (xlat[c] for c in reversed(s.upper())) powers = (pow(36,e) for e in count()) return isprime(sum(imap(mul, digits, powers)))Split A List
Uses deques to efficiently add/pop items from either end.
Basic idea is add each item in the list to the end of the ‘back’ half. Every other time an item is added to the back, an item
is popped off the front of the ‘back’ half and added to the ‘front’ half.
from collections import deque def split_a_list(lst, front_heavy=True): '''returns front/back halves of a list. front_heavy determines which half is longer (heavy) if the list has an odd number of items. ''' front = deque() back = deque() for item in lst: back.append(item) if front_heavy: front.append(back.popleft()) front_heavy = not front_heavy return list(front), list(back)Tested in MIT Scheme
praxis.scm
;;https://programmingpraxis.com/2011/04/26/miscellanea/ ;FizzBuzz (define (enumerate-interval low high) (if (> low high) '() (cons low (enumerate-interval (+ low 1) high)))) (define (fizz-buzz n) (for-each (lambda (x) (display x) (newline)) (map fizz-buzz-func (enumerate-interval 1 20)))) (define (fizz-buzz-func n) (cond ((and (= (remainder n 3) 0) (= (remainder n 5) 0)) 'fizz-buzz) ((= (remainder n 3) 0) 'fizz) ((= (remainder n 5) 0) 'buzz) (else n))) (fizz-buzz 20) ;prime words ;prime? is from SICP (define (prime? n) (define (smallest-divisor n) (find-divisor n 2)) (define (find-divisor n test-divisor) (cond ((> (square test-divisor) n) n) ((divides? test-divisor n) test-divisor) (else (find-divisor n (+ test-divisor 1))))) (define (divides? a b) (= (remainder b a) 0)) (= n (smallest-divisor n))) (define (string-to-number-list s n) (let ((numbers "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ")) (cond ((= n (string-length s)) '()) (else (cons (cons (string-find-next-char numbers (string-ref s n)) (- (string-length s) (+ n 1))) (string-to-number-list s (+ n 1))))))) (define (prime-words? w) (prime? (fold-left + 0 (map (lambda (x) (* (car x) (expt 36 (cdr x)))) (string-to-number-list w 0))))) (prime-words? "LISP") ;#t (prime-words? "PRAXIS") ;#f ;split a list ;if length is odd, put the center element in first half (define (col first-half second-half) (cons first-half second-half)) (define (split-a-list l col) (define (split-a-list-internal l c cn col) (cond ((null? l) (col '() '())) ((< c cn) (split-a-list-internal (cdr l) (+ c 1) cn (lambda (first-half second-half) (col (cons (car l) first-half) second-half)))) (else (split-a-list-internal (cdr l) (+ c 1) cn (lambda (first-half second-half) (col first-half (cons (car l) second-half))))))) (split-a-list-internal l 0 (/ (length l) 2) col)) (define test-list (list 1 2 3 4 5)) ;return a list contains two lists (first-half and second-half) (split-a-list test-list col) ;output: ((1 2 3) 4 5)Here’s my Ruby solution:
def fizz_buzz(n) 1.upto(n) do |i| case when i % 3 == 0 && i % 5 == 0 puts "FizzBuzz" when i % 3 == 0 puts "Fizz" when i % 5 == 0 puts "Buzz" else puts i end end end def get_numeric_value(word) exp, sum = word.length - 1, 0 word.bytes{ |char| if char >= 0 && char <= 9 then sum += char * 36 ** exp else sum += (char - 55) * 36 ** exp end exp = exp - 1 } return sum end require 'mathn' def is_prime_word(word) return Prime.instance.prime?(get_numeric_value(word)) end def split_list(list) if list.length % 2 == 0 then return [list.slice(0, list.length / 2), list.slice(list.length / 2, list.length - 1)] else return [list.slice(0, list.length / 2 + 1), list.slice(list.length / 2 + 1, list.length - 1)] end end puts fizz_buzz(20) puts is_prime_word('PRAXIS') puts split_list([1,2,3,4]).to_s puts split_list([1,2,3]).to_sI feel I cheated a bit with my solution for the 3rd assignment because I’m not scanning the list in situ. Oh well… =)
Simple Haskell for FizzBuzz.
run n = mapM_ putStrLn $ take n $ zipWith (\a b -> if null b then show a else b) [1..] $ zipWith (++) (cycle ["","","Fizz"]) (cycle ["","","","","Buzz"])
Scheme.
In a table, I store a continuation that builds a list made of 1/ what’s on the stack, and 2/ what you give it, as well as the current rest of the list. Each time I scan one more element of the list, I put it on the stack (with CONS) and call myself recursively, after setting the continuation and the rest of the list in the table. When the list is empty, compute the middle index, fetch the continuation and rest at this time, save the rest somewhere, and call the continuation with ‘() to return the list up to the middle. Then assemble everything and you’re done. Lists of length 0 and 1 are edge cases.
(define (split l)
(define table (make-table))
(define cnt 0)
(define bottom '())
(define (aux l)
(if (null? l)
(case cnt
((0 1) '())
(else
(let* ((mid (floor (/ cnt 2)))
(val (table-ref table mid)))
(set! bottom (cdr val))
((car val) '()))))
(begin
(set! cnt (+ cnt 1))
(cons (car l)
(call/cc
(lambda (build-top)
(table-set! table cnt (cons build-top (cdr l)))
(aux (cdr l))))))))
(let ((res (aux l)))
(vector res bottom)))
Morally the same as above, but cleaner. Uses a hack (?) from gambit (4.2.8 in my test), the fact that ((lambda(x) x) (values a b)) will effectively return the two values.
(define (split l)
(define (aux l #!optional (table (make-table)) (cnt 1))
(if (null? l)
(let* ((mid (floor (/ cnt 2)))
(val (table-ref table mid (cons (lambda (x) x) '()))))
((car val) (values '() (cdr val))))
(call-with-values
(lambda ()
(call/cc
(lambda (build-top)
(table-set! table cnt (cons build-top (cdr l)))
(aux (cdr l) table (+ 1 cnt)))))
(lambda (val bot)
(values (cons (car l) val) bot)))))
(aux l))