Reputation: 136
I posted this question a few hours ago but I think I deleted it! Really sorry... I am working on Project Euler Problem 17.
Although there are other more obvious solutions, as a learning exercise, I approached the problem intending to solve it using recursion. I had also hoped that certain pieces of the code might later be usable in other contexts. The problem description itself is in the docstring at the top of the code, for those that are unfamiliar.
Here is the code in question:
"""If the numbers 1 to 5 are written out in words:
one, two, three, four, five
then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words,
how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two)
contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of
"and" when writing out numbers is in compliance with British usage.
"""
import collections
import re
SMALLS = [(0, "zero"),
(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")]
MULTS_OF_TEN = [(20, "twenty"),
(30, "thirty"),
(40, "forty"),
(50, "fifty"),
(60, "sixty"),
(70, "seventy"),
(80, "eighty"),
(90, "ninety")]
HUNDRED = [(100, "hundred")]
BIGS = [(1000, "thousand"),
(10**6, "million"),
(10**9, "billion")]
# other bigs: trillion, quadrillion, quintillion, sextillion, septillion, octillion, nonillion, decillion,
# undecillion, duodecillion, tredecillion, quattuordecillion, quindecillion, sexdecillion,
# septendecillion, octodecillion, novemdecillion, vigintillion
SMALLS = collections.OrderedDict(reversed(SMALLS))
MULTS_OF_TEN = collections.OrderedDict(reversed(MULTS_OF_TEN))
HUNDRED = collections.OrderedDict(reversed(HUNDRED))
BIGS = collections.OrderedDict(reversed(BIGS))
def int_to_words(num, follows_hundreds=False):
"""Retuns the text-equivelent of num, using recursion for distinct
pieces.
"""
def do_chunk(n,
num_text_map,
include_quotient=True,
is_hundreds=False):
for x in num_text_map:
quotient = n // x
remainder = n % x
if n == x: return num_text_map[x]
if quotient and remainder:
quotient_text = (int_to_words(quotient) + " " + num_text_map[x]) \
if include_quotient else \
(num_text_map[x])
remainder_text = int_to_words(remainder, follows_hundreds=True) \
if is_hundreds else \
int_to_words(remainder)
return quotient_text + " " + remainder_text
elif quotient:
quotient_text = (int_to_words(quotient) + " " + num_text_map[x]) \
if include_quotient else \
(num_text_map[x])
return quotient_text
return False
result = do_chunk(num, BIGS)
if result: return result
result = do_chunk(num, HUNDRED, is_hundreds=True)
if result: return result
result = do_chunk(num, MULTS_OF_TEN, include_quotient=False)
if result and follows_hundreds: return "and " + result
if result: return result
result = do_chunk(num, SMALLS)
if result and follows_hundreds: return "and " + result
if result: return result
def count_letters(string):
looking_for = "[a-z]"
instances = re.findall(looking_for, string)
return len(instances)
def tally_letters(start, end):
total_letters = 0
for i in range(start, end + 1):
total_letters += count_letters(int_to_words(i))
return total_letters
And here is the output of the program, compared against the expected solution.
>>> answer = tally_letters(1, 1000)
>>> assert answer == 21124
Traceback (most recent call last):
File "<pyshell#3>", line 1, in <module>
assert answer == 21124
AssertionError
>>> answer
1: 21118
It confounds me that I'm off by a difference of 6. Thanks in advance for the help.
Upvotes: 3
Views: 371
Reputation: 353059
I think if someone gave me nine balloons, and then gave me one more, I would say that I had ten balloons. On the other hand, if I had ninety-nine balloons, and then I received one more, I would say "I have one hundred balloons", not "I have hundred balloons":
>>> int_to_words(10)
'ten'
>>> int_to_words(100)
'hundred'
>>> int_to_words(1000)
'thousand'
len("one")*2 == 6
.
Upvotes: 7