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Advanced Iterators

❝ Great fleas have little fleas upon their backs to bite ’em,
And little fleas have lesser fleas, and so ad infinitum. ❞
— Augustus De Morgan

Diving In

GOYA + INGRES + RENOIR + SARGENT = ARTISTS. Or, to put it another way, 8643 + 598712 + 719657 + 2378190 = 3705202. Am I speaking in tongues? No, it's just a puzzle. Here, let me line it up for you.

GOYA + INGRES + RENOIR + SARGENT = ARTISTS
8643 + 598712 + 719657 + 2378190 = 3705202

G = 8
O = 6
Y = 4
A = 3
I = 5
N = 9
R = 7
E = 1
S = 2
T = 0

Puzzles like this are called cryptarithms or alphametics. The letters spell out actual words, but if you replace each letter with a digit from 0–9, it also “spells” an arithmetic equation. The trick is to figure out which letter maps to each digit. All the occurrences of each letter must map to the same digit, no digit can be repeated, and no “word” can start with the digit 0.

The most well-known alphametic puzzle is SEND + MORE = MONEY.

In this chapter, we'll dive into an incredible Python program originally written by Raymond Hettinger. This program solves alphametic puzzles in just 14 lines of code.

[download alphametics.py]

import re
import itertools

def solve(puzzle):
    words = re.findall('[A-Z]+', puzzle.upper())
    unique_characters = {c for c in ''.join(words)}
    assert len(unique_characters) <= 10
    first_letters = {word[0] for word in words}
    n = len(first_letters)
    sorted_characters = ''.join(first_letters) + \
        ''.join(unique_characters - first_letters)
    characters = tuple(ord(c) for c in sorted_characters)
    digits = tuple(ord(c) for c in '0123456789')
    zero = digits[0]
    for guess in itertools.permutations(digits, len(characters)):
        if zero not in guess[:n]:
            equation = puzzle.translate(dict(zip(characters, guess)))
            if eval(equation):
                return equation

if __name__ == '__main__':
    import sys
    for puzzle in sys.argv[1:]:
        print(puzzle)
        solution = solve(puzzle)
        if solution:
            print(solution)

you@localhost:~$ python3 alphametics.py "GOYA + INGRES + RENOIR + SARGENT = ARTISTS"
GOYA + INGRES + RENOIR + SARGENT = ARTISTS
8643 + 598712 + 719657 + 2378190 = 3705202
you@localhost:~$ python3 alphametics.py "I + LOVE + YOU == DORA"
I + LOVE + YOU == DORA
1 + 2784 + 975 == 3760
you@localhost:~$ python3 alphametics.py "SEND + MORE == MONEY"
SEND + MORE == MONEY
9567 + 1085 == 10652

Finding all occurrences of a pattern

FIXME

>>> import re
>>> re.findall('[A-Z]+', 'SEND + MORE == MONEY')
['SEND', 'MORE', 'MONEY']

FIXME

Finding the unique items in a sequence

Set comprehensions make it trivial to find the unique items in a sequence. [FIXME-not sure if I'm going to cover set comprehensions in an earlier chapter; if not, this is certainly an abrupt and inadequate introduction to the topic.]

>>> a_list = ['a', 'c', 'b', 'a', 'd', 'b']
>>> {c for c in a_list}                      ①
{'a', 'c', 'b', 'd'}
>>> a_string = 'EAST IS EAST'
>>> {c for c in a_string}                    ②
{'A', ' ', 'E', 'I', 'S', 'T'}
>>> words = ['SEND', 'MORE', 'MONEY']
>>> ''.join(words)                           ③
'SENDMOREMONEY'
>>> {c for c in ''.join(words)}              ④
{'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}

Given a list of several strings, a set comprehension with the identity function will return a set of unique strings from the list. This makes sense if you think of it like a for loop. Take the first item from the list, put it in the set. Second. Third. Fourth — wait, that's in the set already, so it only gets listed once. Fifth. Sixth — again, a duplicate, so it only gets listed once. The end result? All the unique items in the original list, without any duplicates. The original list doesn't even need to be sorted first.
The same technique works with strings, since a string is just a sequence of characters.
Given a list of strings, ''.join(a_list) concatenates all the strings together into one.
So, given a list of strings, this set comprehension returns all the unique characters across all the strings, with no duplicates.

The alphametics solver uses this technique to get a list of all the unique characters in the puzzle.

unique_characters = {c for c in ''.join(words)}

This list is later used to assign digits to characters as the solver iterates through the possible solutions.

Making assertions

Like many programming languages, Python has an assert statement. Here's how it works.

>>> assert 1 + 1 == 2  ①
>>> assert 1 + 1 == 3  ②
Traceback (most recent call last):
  File "<stdin>", line 1, in 
AssertionError

The assert statement is followed by any valid Python expression. In this case, the expression 1 + 1 == 2 evaluates to True, so the assert statement does nothing.
However, if the Python expression evaluates to False, the assert statement will raise an AssertionError.

Therefore, this line of code:

assert len(unique_characters) <= 10

…is equivalent to…

if len(unique_characters) > 10:
    raise AssertionError

But a bit easier to read and write.

The alphametics solver uses this exact assert statement to bail out early if the puzzle contains more than ten unique letters. Since each letter is assigned a unique digit, and there are only ten digits, a puzzle with more than ten unique letters is unsolvable.

Generator objects

FIXME

>>> unique_characters = {'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}
>>> gen = (ord(c) for c in unique_characters)
>>> gen
<generator object <genexpr> at 0x00BADC10>
>>> next(gen)
69
>>> next(gen)
68
>>> tuple(ord(c) for c in unique_characters)
(69, 68, 77, 79, 78, 83, 82, 89)

FIXME

Calculating Permutations… The Lazy Way!

First of all, what the heck are permutations? Permutations are a mathematical concept. (There are actually several definitions, depending on what kind of math you're doing. Here I'm talking about combinatorics, but if that doesn't mean anything to you, don't worry about it. As always, Wikipedia is your friend.)

The idea is that you take a list of things (could be numbers, could be letters, could be dancing bears) and find all the possible ways to split them up into smaller lists. All the smaller lists have the same size, which can be as small as 1 and as large as the total number of items. Oh, and nothing can be repeated. Mathematicians say things like "let's find the permutations of 3 different items taken 2 at a time," which means you have a sequence of 3 items and you want to find all the possible ordered pairs.

>>> import itertools                              ①
>>> perms = itertools.permutations([1, 2, 3], 2)  ②
>>> next(perms)                                   ③
(1, 2)
>>> next(perms)
(1, 3)
>>> next(perms)
(2, 1)                                            ④
>>> next(perms)
(2, 3)
>>> next(perms)
(3, 1)
>>> next(perms)
(3, 2)
>>> next(perms)                                   ⑤
Traceback (most recent call last):
  File "<stdin>", line 1, in 
StopIteration

The itertools module has all kinds of fun stuff in it, including a permutations() function that does all the hard work of finding permutations.
The permutations() function takes a sequence (here a list of three integers) and a number, which is the number of items you want in each smaller group. The function returns an iterator, which you can use in a foor loop or any old place that iterates. Here I'll step through the iterator manually to show all the values.
The first permutation of [1, 2, 3] taken 2 at a time is (1, 2).
Note that permutations are ordered: (2, 1) is different than (1, 2).
That's it! Those are all the permutations of [1, 2, 3] taken 2 at a time. Pairs like (1, 1) and (2, 2) never show up, because they contain repeats so they aren't valid permutations. When there are no more permutations, the iterator raises a StopIteration exception.

The permutations() function doesn't have to take a list. It can take any sequence — even a string.

>>> import itertools
>>> perms = itertools.permutations('ABC', 3)  ①
>>> next(perms)
('A', 'B', 'C')                               ②
>>> next(perms)
('A', 'C', 'B')
>>> next(perms)
('B', 'A', 'C')
>>> next(perms)
('B', 'C', 'A')
>>> next(perms)
('C', 'A', 'B')
>>> next(perms)
('C', 'B', 'A')
>>> next(perms)
Traceback (most recent call last):
  File "<stdin>", line 1, in 
StopIteration
>>> list(itertools.permutations('ABC', 3))    ③
[('A', 'B', 'C'), ('A', 'C', 'B'),
 ('B', 'A', 'C'), ('B', 'C', 'A'),
 ('C', 'A', 'B'), ('C', 'B', 'A')]

A string is just a sequence of characters. For the purposes of finding permutations, the string 'ABC' is equivalent to the list ['A', 'B', 'C'].
The first permutation of the 3 items ['A', 'B', 'C'], taken 3 at a time, is ('A', 'B', 'C'). There are five other permutations — the same three characters in every conceivable order.
Since the permutations() function always returns an iterator, an easy way to debug permutations is to pass that iterator to the built-in list() function to see all the permutations immediately.

Other Fun Stuff in the `itertools` Module

>>> import itertools
>>> list(itertools.product('ABC', '123'))   ①
[('A', '1'), ('A', '2'), ('A', '3'), 
 ('B', '1'), ('B', '2'), ('B', '3'), 
 ('C', '1'), ('C', '2'), ('C', '3')]
>>> list(itertools.combinations('ABC', 2))  ②
[('A', 'B'), ('A', 'C'), ('B', 'C')]

The itertools.product() function returns an iterator containing the Cartesian product of two sequences.
The itertools.combinations() function returns an iterator containing all the possible combinations of the given sequence of the given length. This is like the itertools.permutations() function, except combinations don't include items that are duplicates of other items in a different order. So itertools.permutations('ABC', 2) will return both ('A', 'B') and ('B', 'A') (among others), but itertools.combinations('ABC', 2) will not return ('B', 'A') because it is a duplicate of ('A', 'B') in a different order.

[download favorite-people.txt]

>>> names = list(open('examples/favorite-people.txt'))  ①
>>> names
['Dora\n', 'Ethan\n', 'Wesley\n', 'John\n', 'Anne\n',
'Mike\n', 'Chris\n', 'Sarah\n', 'Alex\n', 'Lizzie\n']
>>> names = [name.rstrip() for name in names]           ②
>>> names
['Dora', 'Ethan', 'Wesley', 'John', 'Anne',
'Mike', 'Chris', 'Sarah', 'Alex', 'Lizzie']
>>> names = sorted(names)                               ③
>>> names
['Alex', 'Anne', 'Chris', 'Dora', 'Ethan',
'John', 'Lizzie', 'Mike', 'Sarah', 'Wesley']
>>> names = sorted(names, key=len)                      ④
>>> names
['Alex', 'Anne', 'Dora', 'John', 'Mike',
'Chris', 'Ethan', 'Sarah', 'Lizzie', 'Wesley']

This idiom returns a list of the lines in a text file.
Unfortunately (for this example), the list(open(filename)) idiom also includes the carriage returns at the end of each line. This list comprehension uses the rstrip() string method to strip trailing whitespace from each line.
The sorted() function takes a list and returns it sorted. By default, it sorts alphabetically.
But the sorted() function can also take a function as the key parameter, and it sorts by that key. In this case, the sort function is len(), so it sorts by len(each item). Shorter names come first, then longer, then longest.

What does this have to do with the itertools module? I'm glad you asked.

…continuing from the previous interactive shell…
>>> import itertools
>>> groups = itertools.groupby(names, len)  ①
>>> groups
<itertools.groupby object at 0x00BB20C0>
>>> list(groups)
[(4, <itertools._grouper object at 0x00BA8BF0>),
 (5, <itertools._grouper object at 0x00BB4050>),
 (6, <itertools._grouper object at 0x00BB4030>)]
>>> groups = itertools.groupby(names, len)
>>> for name_length, name_iter in groups:    ②
...     print('Names with {0:d} letters:'.format(name_length))
...     for name in name_iter:
...         print(name)
... 
Names with 4 letters:
Alex
Anne
Dora
John
Mike
Names with 5 letters:
Chris
Ethan
Sarah
Names with 6 letters:
Lizzie
Wesley

The itertools.groupby() function takes a sequence and a key function, and returns an iterator that generates pairs. Each pair contains the result of key_function(each item) and another iterator containing all the items that shared that key result.
In this example, given a list of names sorted by length, itertools.groupby(names, len) will put all the 4-letter names in one iterator, all the 5-letter names in another iterator, and so on. The groupby() function is completely generic; it could group strings by first letter, numbers by their number of factors, or any other key function you can think of.

Are you watching closely?

>>> list(range(0, 3))
[0, 1, 2]
>>> list(range(10, 13))
[10, 11, 12]
>>> list(itertools.chain(range(0, 3), range(10, 13)))        ①
[0, 1, 2, 10, 11, 12]
>>> list(zip(range(0, 3), range(10, 13)))                    ②
[(0, 10), (1, 11), (2, 12)]
>>> list(zip(range(0, 3), range(10, 14)))                    ③
[(0, 10), (1, 11), (2, 12)]
>>> list(itertools.zip_longest(range(0, 3), range(10, 14)))  ④
[(0, 10), (1, 11), (2, 12), (None, 13)]

The itertools.chain() function takes two iterators and returns an iterator that contains all the items from the first iterator, followed by all the items from the second iterator. (Actually, it can take any number of iterators, and it chains them all in the order they were passed to the function.)
The zip() function does something prosaic that turns out to be extremely useful: it any number of sequences and returns an iterator with the first items of each sequence, then the second items of each, then the third, and so on.
The zip() function stops at the end of the shortest sequence. range(10, 14) has 4 items (10, 11, 12, and 13), but range(0, 3) only has 3, so the zip() function returns an iterator of 3 items.
On the other hand, the itertools.zip_longest() function stops at the end of the longest sequence, inserting None values for items past the end of the shorter sequences.

OK, that was all very interesting, but how does it relate to the alphametics solver? Here's how:

>>> characters = ('S', 'M', 'E', 'D', 'O', 'N', 'R', 'Y')
>>> guess = ('1', '2', '0', '3', '4', '5', '6', '7')
>>> tuple(zip(characters, guess))  ①
(('S', '1'), ('M', '2'), ('E', '0'), ('D', '3'),
 ('O', '4'), ('N', '5'), ('R', '6'), ('Y', '7'))
>>> dict(zip(characters, guess))   ②
{'E': '0', 'D': '3', 'M': '2', 'O': '4',
 'N': '5', 'S': '1', 'R': '6', 'Y': '7'}

Given a list of letters and a list of digits (each represented here as 1-character strings), the zip function will create a pairing of letters and digits, in order.
Why is that cool? Because that data structure happens to be exactly the right structure to pass to the dict() function to create a dictionary that uses letters as keys and their associated digits as values. Although the printed representation of the dictionary lists the pairs in a different order (dictionaries have no "order" per se), you can see that each letter is associated with the digit, based on the ordering of the original characters and guess sequences.

The alphametics solver uses this technique to create a dictionary that maps letters in the puzzle to digits in the solution, for each possible solution.

characters = tuple(ord(c) for c in sorted_characters)
digits = tuple(ord(c) for c in '0123456789')
...
for guess in itertools.permutations(digits, len(characters)):
    ...
    equation = puzzle.translate(dict(zip(characters, guess)))

But what is this translate() method? Ah, now you're getting to the really fun part.

A New Kind Of String Manipulation

FIXME

>>> characters = tuple(ord(c) for c in 'SMEDONRY')
>>> characters
(83, 77, 69, 68, 79, 78, 82, 89)
>>> digits = tuple(ord(c) for c in '0123456789')
>>> digits
(48, 49, 50, 51, 52, 53, 54, 55, 56, 57)
>>> guess = (49, 50, 48, 51, 52, 53, 54, 55)
>>> translation_table = dict(zip(characters, guess))
>>> translation_table
{68: 51, 69: 48, 77: 50, 78: 53, 79: 52, 82: 54, 83: 49, 89: 55}
>>> "SEND + MORE == MONEY".translate(translation_table)
'1053 + 2460 == 24507'

FIXME

>>> translation_table = {ord("A"): ord("O")}
>>> translation_table
{65: 79}
>>> 'MARK'.translate(translation_table)
'MORK'

FIXME

Evaluating Arbitrary Strings As Python Expressions

FIXME

Putting It All Together

To recap: this program solves alphametic puzzles by brute force, i.e. through an exhaustive search of all possible solutions. To do this, it…

Finds all the letters in the puzzle with the re.findall() function
Find all the unique letters in the puzzle with set comprehensions
Checks if there are more than 10 unique letters (meaning the puzzle is definitely unsolvable) with an assert statement
FIXME sorts the letters with a set difference operation
Converts the letters to their ASCII equivalents with a generator object
Calculates all the possible solutions with the itertools.permutations() function
Converts each possible solution to a Python expression with the translate() string method
Tests each possible solution by evaluating the Python expression with the eval() function
Returns the first solution that evaluates to True

…in just 14 lines of code.

Advanced Iterators

Diving In

Finding all occurrences of a pattern

Finding the unique items in a sequence

Making assertions

Generator objects

Calculating Permutations… The Lazy Way!

Other Fun Stuff in the itertools Module

A New Kind Of String Manipulation

Evaluating Arbitrary Strings As Python Expressions

Putting It All Together

Further Reading

Other Fun Stuff in the `itertools` Module