Hofstadter's Sentences

hofstadter_sentences

#!/usr/bin/python
# -*- coding: utf-8 -*-
# Metamagical Themas, Douglas R Hofstadter
# Self-Referential Sentences: a follow up
#
# The noted logician Raphael Robinson submitted a playful puzzle
# in the self documenting genre. Readers are asked to complete
# the following sentence:
#
# "In this sentence, the number of occurrences of 0 is _, of 1
# is _, of 2 is _, of 3 is _, of 4 is_, of 5 is _, of 6 is _,
# of 7 is _, of 8 is _, and of 9 is _."
#
# Each blank is to be filled with a numeral of one or more digits,
# written in decimal notation. Robinson states that there are
# exactly two solutions.
#
# esouthren, Jan 2021
#   I found one solution; can you find the other?
#

origin = {0:1, 1:1, 2:1, 3:1, 4:1, 5:1, 6:1, 7:1, 8:1, 9:1}

def is_valid_sentence(input):
    for key in input:
        char_count = get_char_count(input, key)

        if char_count != input[key]:
            return False
    return True


def get_char_count(input, key):
    count = 0
    for key2 in input:
        for x in [int(d) for d in str(key2)]:
            if x == key:
                count += 1
        for x in [int(d) for d in str(input[key2])]:
            if x == key:
                count += 1
    return count


def print_sentence(i):
    print 'In this sentence, the number of occurrences of 0 is {},of 1 is {}, of 2 is {}, of 3 is {}, of 4 is {}, of 5 is {}, of 6 is{}, of 7 is {}, of 8 is {}, and of 9 is {}.'.format(i[0], i[1],i[2],i[3],i[4],i[5],i[6],i[7],i[8],i[9]))


foundAnswer = False
while not foundAnswer:
    for key in origin:
        if is_valid_sentence(origin):
            print_sentence(origin)
            foundAnswer = True
            break

        x = get_char_count(origin, key)
        if x != origin[key]:
            origin[key] = x