Sum Square Digits Sequence
April 27, 2018
Here is our solution, with help from the Standard Prelude; we short-circuit the loop by stopping as soon as any member of the loop is found:
(define (ssds n) ; sum square digits sequence
(let loop ((n n) (ns (list)))
(if (or (= n 1) (member n '(4 16 37 58 89 145 42 20)))
(reverse (cons n ns))
(let ((x (sum (map square (digits n)))))
(loop x (cons n ns))))))
Here are two examples:
> (ssds 19) (19 82 68 100 1) > (ssds 18) (18 65 61 37)
You can run the program at https://ideone.com/Ro26ys. There is more to the topic of Happy Numbers than we have discussed here; oeis.org is a good place to learn more.
Programming Praxis 
Although the problem is ill-defined (there is a case of a number being neutral, i.e. neither happy or sad), we’ll look at the case where the number is happy if and only if the end result is 1. Here is my take with Julia.
global sad = [4, 16, 37, 58, 89, 145, 42, 20];
function DigitsOf(n::Int64)
digits = split(string(n), “”)
z = length(digits)
Z = Array{Int64}(z)
end
function IsHappy(n::Int64)
D, nd = DigitsOf(n)
S = [n]
end
Whatever the case, creative problems like this can make someone quite happy, for sure. Happy weekend everyone!
Perl6 solution with a simple loop:
#!perl6 sub is-happy(Int:D $n --> Bool) { $n == 1; } sub is-sad(Int:D $n --> Bool) { ?(4, 16, 37, 58, 89, 145, 42, 20).first: * == $n; } sub ssds(Int:D $n --> List) { my @seq = (($n),); repeat { @seq.push([+] @seq[* - 1].polymod(10 xx *).map({ $_ * $_ })); } while !(is-happy(@seq[* - 1]) or is-sad(@seq[* - 1])); @seq; } say ssds(19); # [19 82 68 100 1] say ssds(18); # [18 65 61 37]Here is my Golang solution, with a disclaimer that I’m just learning Go and so far don’t like it very much:
package main import ( "bufio" "fmt" "math" "os" "strconv" "strings" ) func main() { reader := bufio.NewReader(os.Stdin) for { fmt.Println("Enter a positive number >= 1, or negative number to quit: ") text, _ := reader.ReadString('\n') i, err := strconv.Atoi(strings.TrimSpace(text)) if err != nil { fmt.Println("Please enter a number") continue } if i <= 1 { fmt.Println("Bye") break } typ, sequence := happyOrSad(i) fmt.Printf("%d is %s and the sequence is %v\n", i, typ, sequence) } } func happyOrSad(num int) (string, []int) { seen := make(map[int]bool) seq := []int{num} for num != 1 && !seen[num] { seen[num] = true cur := num num = 0 for cur > 0 { num += int(math.Pow(float64(cur%10), float64(2))) cur /= 10 } seq = append(seq, num) } var typ string if num == 1 { typ = "happy" } else { typ = "sad" } return typ, seq }Here is a pastebin of the above code (thought WordPress formatted Go these days).
Here’s a Haskell solution. (I’m calling the sequence a “Porges sequence”.)
import Data.Bool (bool) import Data.List (intercalate, unfoldr) import Data.Tuple (swap) -- The infinite "Porges" sequence starting at the given number. porges :: Integral a => a -> [a] porges = iterate (sum . map (\i -> i * i) . digits) -- The prefix of the Porges sequence starting at 'n' and ending on the first -- repeated number. finitePorges :: Integral a => a -> [a] finitePorges n = let (is, j:_) = break (\i -> i == 1 || i `elem` cyc) $ porges n in is ++ bool (take (length cyc + 1) (porges j)) [j] (j == 1) showPorges :: (Integral a, Show a) => a -> String showPorges n = intercalate ", " (map show $ finitePorges n) ++ ", ..." -- The cycle into which a Porges sequence might fall. cyc :: Integral a => [a] cyc = [4, 16, 37, 58, 89, 145, 42, 20] -- The list of a number's base-10 digits, from least to most significant. digits :: Integral a => a -> [a] digits = unfoldr (\i -> bool Nothing (Just . swap $ i `quotRem` 10) (i /= 0)) main :: IO () main = do putStrLn $ showPorges 1 putStrLn $ showPorges 18 putStrLn $ showPorges 19Python 3 version, one function to generate the sequence, terminating with either a known sad number or 1, and one function to check the ‘happiness’ of a number:
I used mathematical operations to find the sum of squared digits, rather than casting to string, although Python does make that easy :)
def happy_seq(n): terminators = {1, 4, 16, 37, 58, 89, 145, 42, 20} seq = [n] while seq[-1] not in terminators: n, r = seq[-1], 0 while n: r, n = r + (n%10)**2, n // 10 seq.append(r) return seq def is_happy(n): return happy_seq(n)[-1] == 1Here’s a solution in C.
#include <stdio.h> #include <stdlib.h> int main(int argc, char* argv[]) { if (argc != 2) { fprintf(stderr, "Usage: sum_square <NUMBER>\n"); return EXIT_FAILURE; } int num = atoi(argv[1]); int terminals[] = {1,4,16,37,58,89,145,42,20}; size_t n_terminals = sizeof(terminals) / sizeof(int); while (1) { printf("%d\n", num); for (size_t i = 0; i < n_terminals; ++i) { if (num == terminals[i]) goto exit; } int sum = 0; while (num != 0) { int digit = num % 10; sum += digit * digit; num /= 10; } num = sum; } exit: return EXIT_SUCCESS; }Example Usage:
@Zach, the problem mentions and links to a proof that all natural numbers are either “happy” or “sad”. I think this contradicts there being any “neutral” numbers.
Zach -> Zack