Attempted Solution to Five Thirty Eight Riddler Express from 20171215

538_riddler_20171215.py

#!/usr/bin/env python3

"""
This script attempts to solve the Riddler express problem from
20171215: https://fivethirtyeight.com/features/please-help-me-i-have-a-number-problem/

This script is super brute force and not really useful, although it does work

Potential improvement would be to target numbers that are likely to have
longer words like "seven", "three" or a teen number
"""

num2words = {
    1: 'One', 
    2: 'Two', 
    3: 'Three', 
    4: 'Four', 
    5: 'Five',
    6: 'Six', 
    7: 'Seven',
    8: 'Eight', 
    9: 'Nine', 
    10: 'Ten',
    11: 'Eleven', 
    12: 'Twelve', 
    13: 'Thirteen',
    14: 'Fourteen',
    15: 'Fifteen', 
    16: 'Sixteen', 
    17: 'Seventeen', 
    18: 'Eighteen',
    19: 'Nineteen', 
    20: 'Twenty', 
    30: 'Thirty', 
    40: 'Forty',
    50: 'Fifty', 
    60: 'Sixty', 
    70: 'Seventy', 
    80: 'Eighty',
    90: 'Ninety', 
    0: 'Zero'}

zero_words = {
    0:'',
    1:" Thousand ",
    2:" Million ",
    3:" Billion ",
    4:" Trillion ",
    5:" Quadrillion ",
    6:" Quintillion ",
    7:" Sextillion ",
    8:" Septillion ",
    9:" Octillion ",
    10:" Nonillion ",
    11:" Decillion ",
    12:" Undecillion ",
    13:" Duodecillion ",
    14:" Tredecillion ",
    15:" Quattuordecillion "}

# convert number to "count speak" when number is less than 20
def less_than_twenty(n):
    return num2words[n]

# convert number to "count speak" when number is between 20 and 100
def twenty_to_hundred(n):
    tens = (n // 10) * 10
    ones = n - tens
    if ones > 0:
        return num2words[tens] + ' ' + num2words[ones]
    else:
        return num2words[tens]

# convert number to "count speak" when number between 100 and 1000
def hundred_to_thousand(n):
    return num2words[(n // 100)] + ' Hundred '

# convert any number between 0 and 1000
def under_thousand(n):
    if n < 20:
        return less_than_twenty(n)
    elif n >= 20 and n < 100:
        return twenty_to_hundred(n)
    else:
        tens_ones = int(str(n)[1:])
        if tens_ones < 20:
            return hundred_to_thousand(n) + less_than_twenty(tens_ones)
        else:
            return hundred_to_thousand(n) + twenty_to_hundred(tens_ones)

# convert a number of any length to words (as said by count from Sesame Street)
def number_to_words(n):
    nstr = str(n)
    # fancy list comprehension to write number n like [1,234,567]
    nums_as_list = [int(nstr[::-1][i:i+3][::-1]) for i in range(0, len(nstr), 3)][::-1]
    result = []
    for i, s in enumerate(nums_as_list):
        ri = len(nums_as_list) - i - 1
        result.append(under_thousand(s) + zero_words[ri])
    return ''.join(result)

# find the longest number The Count could "tweet" (including the exclamation point)
def longest_twitter_number(limit):
    n = 111337337333737373737377 # random guess at starting number
    len_n = 3
    while len_n < limit:
        n += 1
        len_n = len(number_to_words(n))

    return n - 1

if __name__ == '__main__':
    limit = 280
    print(longest_twitter_number(limit))