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Difficulty level: ♦♦♦♦♢
❝ Great fleas have little fleas upon their backs to bite ’em,
And little fleas have lesser fleas, and so ad infinitum. ❞
— Augustus De Morgan
Just as regular expressions put strings on steroids, the itertools module puts iterators on steroids. But first, I want to show you a classic puzzle.
HAWAII + IDAHO + IOWA + OHIO == STATES
510199 + 98153 + 9301 + 3593 == 621246
H = 5
A = 1
W = 0
I = 9
D = 8
O = 3
S = 6
T = 2
E = 4Puzzles 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.
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.
import re
import itertools
def solve(puzzle):
words = re.findall('[A-Z]+', puzzle.upper())
unique_characters = set(''.join(words))
assert len(unique_characters) <= 10, 'Too many letters'
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 can run the program from the command line. On Linux, it would look like this. (These may take some time, depending on the speed of your computer, and there is no progress bar. Just be patient!)
you@localhost:~/diveintopython3/examples$ python3 alphametics.py "HAWAII + IDAHO + IOWA + OHIO == STATES" HAWAII + IDAHO + IOWA + OHIO = STATES 510199 + 98153 + 9301 + 3593 == 621246 you@localhost:~/diveintopython3/examples$ python3 alphametics.py "I + LOVE + YOU == DORA" I + LOVE + YOU == DORA 1 + 2784 + 975 == 3760 you@localhost:~/diveintopython3/examples$ python3 alphametics.py "SEND + MORE == MONEY" SEND + MORE == MONEY 9567 + 1085 == 10652
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The first thing this alphametics solver does is find all the letters (A–Z) in the puzzle.
>>> import re
>>> re.findall('[0-9]+', '16 2-by-4s in rows of 8') ①
['16', '2', '4', '8']
>>> re.findall('[A-Z]+', 'SEND + MORE == MONEY') ②
['SEND', 'MORE', 'MONEY']
re module is Python’s implementation of regular expressions. It has a nifty function called findall() which takes a regular expression pattern and a string, and finds all occurrences of the pattern within the string. In this case, the pattern matches sequences of numbers. The findall() function returns a list of all the substrings that matched the pattern.
Here’s another example that will stretch your brain a little.
>>> re.findall(' s.*? s', "The sixth sick sheikh's sixth sheep's sick.")
[' sixth s', " sheikh's s", " sheep's s"]
Surprised? The regular expression looks for a space, an s, and then the shortest possible series of any character (.*?), then a space, then another s. Well, looking at that input string, I see five matches:
The sixth sick sheikh's sixth sheep's sick.The sixth sick sheikh's sixth sheep's sick.The sixth sick sheikh's sixth sheep's sick.The sixth sick sheikh's sixth sheep's sick.The sixth sick sheikh's sixth sheep's sick.But the re.findall() function only returned three matches. Specifically, it returned the first, the third, and the fifth. Why is that? Because it doesn’t return overlapping matches. The first match overlaps with the second, so the first is returned and the second is skipped. Then the third overlaps with the fourth, so the third is returned and the fourth is skipped. Finally, the fifth is returned. Three matches, not five.
This has nothing to do with the alphametics solver; I just thought it was interesting.
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Sets make it trivial to find the unique items in a sequence.
>>> a_list = ['The', 'sixth', 'sick', "sheik's", 'sixth', "sheep's", 'sick'] >>> set(a_list) ① {'sixth', 'The', "sheep's", 'sick', "sheik's"} >>> a_string = 'EAST IS EAST' >>> set(a_string) ② {'A', ' ', 'E', 'I', 'S', 'T'} >>> words = ['SEND', 'MORE', 'MONEY'] >>> ''.join(words) ③ 'SENDMOREMONEY' >>> set(''.join(words)) ④ {'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}
set() 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. Fifth — wait, that’s in the set already, so it only gets listed once, because Python sets don’t allow duplicates. Sixth. Seventh — 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.''.join(a_list) concatenates all the strings together into one.The alphametics solver uses this technique to build a set of all the unique characters in the puzzle.
unique_characters = set(''.join(words))This list is later used to assign digits to characters as the solver iterates through the possible solutions.
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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 <module> AssertionError >>> assert 2 + 2 == 5, "Only for very large values of 2" ③ Traceback (most recent call last): File "<stdin>", line 1, in <module> AssertionError: Only for very large values of 2
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.False, the assert statement will raise an AssertionError.AssertionError is raised.Therefore, this line of code:
assert len(unique_characters) <= 10, 'Too many letters'…is equivalent to this:
if len(unique_characters) > 10:
raise AssertionError('Too many letters')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 can not possibly have a solution.
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A generator expression is like a generator function without the function.
>>> unique_characters = {'E', 'D', 'M', 'O', 'N', 'S', 'R', 'Y'}
>>> gen = (ord(c) for c in unique_characters) ①
>>> gen ②
<generator object <genexpr> at 0x00BADC10>
>>> next(gen) ③
69
>>> next(gen)
68
>>> tuple(ord(c) for c in unique_characters) ④
(69, 68, 77, 79, 78, 83, 82, 89)
next(gen) returns the next value from the iterator.tuple(), list(), or set(). In these cases, you don’t need an extra set of parentheses — just pass the “bare” expression ord(c) for c in unique_characters to the tuple() function, and Python figures out that it’s a generator expression.☞Using a generator expression instead of a list comprehension can save both CPU and RAM. If you’re building an list just to throw it away (e.g. passing it to
tuple()orset()), use a generator expression instead!
Here’s another way to accomplish the same thing, using a generator function:
def ord_map(a_string):
for c in a_string:
yield ord(c)
gen = ord_map(unique_characters)The generator expression is more compact but functionally equivalent.
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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 <module> StopIteration
itertools module has all kinds of fun stuff in it, including a permutations() function that does all the hard work of finding permutations.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 for loop or any old place that iterates. Here I’ll step through the iterator manually to show all the values.[1, 2, 3] taken 2 at a time is (1, 2).(2, 1) is different than (1, 2).[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 <module>
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')]
'ABC' is equivalent to the list ['A', 'B', 'C'].['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.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.⁂
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')]
itertools.product() function returns an iterator containing the Cartesian product of two sequences.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', encoding='utf-8')) 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']
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. (Strings also have an lstrip() method to strip leading whitespace, and a strip() method which strips both.)sorted() function takes a list and returns it sorted. By default, it sorts alphabetically.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) ② >>> for name_length, name_iter in groups: ③ ... 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
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.list() function “exhausted” the iterator, i.e. you’ve already generated every item in the iterator to make the list. There’s no “reset” button on an iterator; you can’t just start over once you’ve exhausted it. If you want to loop through it again (say, in the upcoming for loop), you need to call itertools.groupby() again to create a new iterator.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.☞The
itertools.groupby()function only works if the input sequence is already sorted by the grouping function. In the example above, you grouped a list of names by thelen()function. That only worked because the input list was already sorted by length.
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)]
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.)zip() function does something prosaic that turns out to be extremely useful: it takes any number of sequences and returns an iterator which returns tuples of the first items of each sequence, then the second items of each, then the third, and so on.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.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'}
zip function will create a pairing of letters and digits, in order.dict() function to create a dictionary that uses letters as keys and their associated digits as values. (This isn’t the only way to do it, of course. You could use a dictionary comprehension to create the dictionary directly.) 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.
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Python strings have many methods. You learned about some of those methods in the Strings chapter: lower(), count(), and format(). Now I want to introduce you to a powerful but little-known string manipulation technique: the translate() method.
>>> translation_table = {ord('A'): ord('O')} ①
>>> translation_table ②
{65: 79}
>>> 'MARK'.translate(translation_table) ③
'MORK'
ord() function returns the ASCII value of a character, which, in the case of A–Z, is always a byte from 65 to 90.translate() method on a string takes a translation table and runs the string through it. That is, it replaces all occurrences of the keys of the translation table with the corresponding values. In this case, “translating” MARK to MORK.What does this have to do with solving alphametic puzzles? As it turns out, everything.
>>> characters = tuple(ord(c) for c in 'SMEDONRY') ① >>> characters (83, 77, 69, 68, 79, 78, 82, 89) >>> guess = tuple(ord(c) for c in '91570682') ② >>> guess (57, 49, 53, 55, 48, 54, 56, 50) >>> translation_table = dict(zip(characters, guess)) ③ >>> translation_table {68: 55, 69: 53, 77: 49, 78: 54, 79: 48, 82: 56, 83: 57, 89: 50} >>> 'SEND + MORE == MONEY'.translate(translation_table) ④ '9567 + 1085 == 10652'
alphametics.solve() function.
itertools.permutations() function in the alphametics.solve() function.
alphametics.solve() function does inside the for loop.
translate() method of the original puzzle string. This converts each letter in the string to the corresponding digit (based on the letters in characters and the digits in guess). The result is a valid Python expression, as a string.That’s pretty impressive. But what can you do with a string that happens to be a valid Python expression?
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This is the final piece of the puzzle (or rather, the final piece of the puzzle solver). After all that fancy string manipulation, we’re left with a string like '9567 + 1085 == 10652'. But that’s a string, and what good is a string? Enter eval(), the universal Python evaluation tool.
>>> eval('1 + 1 == 2')
True
>>> eval('1 + 1 == 3')
False
>>> eval('9567 + 1085 == 10652')
True
But wait, there’s more! The eval() function isn’t limited to boolean expressions. It can handle any Python expression and returns any datatype.
>>> eval('"A" + "B"')
'AB'
>>> eval('"MARK".translate({65: 79})')
'MORK'
>>> eval('"AAAAA".count("A")')
5
>>> eval('["*"] * 5')
['*', '*', '*', '*', '*']
But wait, that’s not all!
>>> x = 5
>>> eval("x * 5") ①
25
>>> eval("pow(x, 2)") ②
25
>>> import math
>>> eval("math.sqrt(x)") ③
2.2360679774997898
eval() takes can reference global variables defined outside the eval(). If called within a function, it can reference local variables too.Hey, wait a minute…
>>> import subprocess
>>> eval("subprocess.getoutput('ls ~')") ①
'Desktop Library Pictures \
Documents Movies Public \
Music Sites'
>>> eval("subprocess.getoutput('rm /some/random/file')") ②
subprocess module allows you to run arbitrary shell commands and get the result as a Python string.It’s even worse than that, because there’s a global __import__() function that takes a module name as a string, imports the module, and returns a reference to it. Combined with the power of eval(), you can construct a single expression that will wipe out all your files:
>>> eval("__import__('subprocess').getoutput('rm /some/random/file')") ①
'rm -rf ~'. Actually there wouldn’t be any output, but you wouldn’t have any files left either.eval() is EVIL
Well, the evil part is evaluating arbitrary expressions from untrusted sources. You should only use eval() on trusted input. Of course, the trick is figuring out what’s “trusted.” But here’s something I know for certain: you should NOT take this alphametics solver and put it on the internet as a fun little web service. Don’t make the mistake of thinking, “Gosh, the function does a lot of string manipulation before getting a string to evaluate; I can’t imagine how someone could exploit that.” Someone WILL figure out how to sneak nasty executable code past all that string manipulation (stranger things have happened), and then you can kiss your server goodbye.
But surely there’s some way to evaluate expressions safely? To put eval() in a sandbox where it can’t access or harm the outside world? Well, yes and no.
>>> x = 5
>>> eval("x * 5", {}, {}) ①
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<string>", line 1, in <module>
NameError: name 'x' is not defined
>>> eval("x * 5", {"x": x}, {}) ②
25
>>> import math
>>> eval("math.sqrt(x)", {"x": x}, {}) ③
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<string>", line 1, in <module>
NameError: name 'math' is not defined
eval() function act as the global and local namespaces for evaluating the expression. In this case, they are both empty, which means that when the string "x * 5" is evaluated, there is no reference to x in either the global or local namespace, so eval() throws an exception.math module, you didn’t include it in the namespace passed to the eval() function, so the evaluation failed.Gee, that was easy. Lemme make an alphametics web service now!
>>> eval("pow(5, 2)", {}, {}) ①
25
>>> eval("__import__('math').sqrt(5)", {}, {}) ②
2.2360679774997898
pow(5, 2) works, because 5 and 2 are literals, and pow() is a built-in function.__import__() function is also a built-in function, so it works too.Yeah, that means you can still do nasty things, even if you explicitly set the global and local namespaces to empty dictionaries when calling eval():
>>> eval("__import__('subprocess').getoutput('rm /some/random/file')", {}, {})
Oops. I’m glad I didn’t make that alphametics web service. Is there any way to use eval() safely? Well, yes and no.
>>> eval("__import__('math').sqrt(5)",
... {"__builtins__":None}, {}) ①
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<string>", line 1, in <module>
NameError: name '__import__' is not defined
>>> eval("__import__('subprocess').getoutput('rm -rf /')",
... {"__builtins__":None}, {}) ②
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<string>", line 1, in <module>
NameError: name '__import__' is not defined
"__builtins__" to None, the Python null value. Internally, the “built-in” functions are contained within a pseudo-module called "__builtins__". This pseudo-module (i.e. the set of built-in functions) is made available to evaluated expressions unless you explicitly override it.__builtins__. Not __builtin__, __built-ins__, or some other variation that will work just fine but expose you to catastrophic risks.So eval() is safe now? Well, yes and no.
>>> eval("2 ** 2147483647",
... {"__builtins__":None}, {}) ①
__builtins__, you can still launch a denial-of-service attack. For example, trying to raise 2 to the 2147483647th power will spike your server’s CPU utilization to 100% for quite some time. (If you’re trying this in the interactive shell, press Ctrl-C a few times to break out of it.) Technically this expression will return a value eventually, but in the meantime your server will be doing a whole lot of nothing.In the end, it is possible to safely evaluate untrusted Python expressions, for some definition of “safe” that turns out not to be terribly useful in real life. It’s fine if you’re just playing around, and it’s fine if you only ever pass it trusted input. But anything else is just asking for trouble.
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To recap: this program solves alphametic puzzles by brute force, i.e. through an exhaustive search of all possible solutions. To do this, it…
re.findall() function
set() function
assert statement
itertools.permutations() function
translate() string method
eval() function
True…in just 14 lines of code.
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itertools moduleitertools — Iterator functions for efficient loopingMany thanks to Raymond Hettinger for agreeing to relicense his code so I could port it to Python 3 and use it as the basis for this chapter.
© 2001–11 Mark Pilgrim