Happy numbers using python base 8

Question

I have written a code that works for base 10
How to make it work for base 8??

SQUARE = dict([(c, int(c)**2) for c in "0123456789"])
def is_happy(n):
 s = set()
 while (n > 1) and (n not in s):
   s.add(n)
   n = sum(SQUARE[d] for d in str(n))
 return n == 1

a=is_happy(28)

@thefourtheye He's probably trying to find these: https://en.wikipedia.org/wiki/Happy_number — Thanatos, Mar 23 '14 at 05:56
Really, did you write that code? Or did you copy it from [here](https://en.wikipedia.org/wiki/Happy_number#Programming_example)? — devnull, Mar 23 '14 at 06:02

score 2 · Answer 1 · answered Mar 23 '14 at 06:01

2

Use the octal representation instead.

n = sum(SQUARE[d] for d in oct(n))

answered Mar 23 '14 at 06:01

Ignacio Vazquez-Abrams

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score 2 · Accepted Answer · edited May 23 '17 at 12:06

You just need to replace all the places in the algorithm where "base-10" is used with base 8. The only place this really is is when we turn the number into a string, so we can square each "digit". To determine digits, we need a base. Normally, we choose base 10, as your sample shows. Converting an integer to a string in an arbitrary base (or, in your case 8) has been answered here.

We might also adjust the lookup table SQUARE.

def to_base_8(n):
    digits = []
    while n > 0:
        digits.append(str(n % 8))
        n = n // 8
    return ''.join(reversed(digits))

SQUARE = dict([(c, int(c)**2) for c in "01234567"])
def is_happy(n):
 s = set()
 while (n > 1) and (n not in s):
   s.add(n)
   n = sum(SQUARE[d] for d in to_base_8(n))
 return n == 1

a=is_happy(28)

Happy numbers using python base 8

2 Answers2