How to print all possible balanced parentheses for an expression

algorithmcatalanpuzzle

For example, with elements a,b,c,d, there are 5 possible ways to take neighboring elements and reduce them into a single element, where exactly two elements must be combined at a time (below represented by parentheses):

(((ab)c)d), ((a(bc))d), ((ab)(cd)), (a((bc)d)) and (a(b(cd)))

The first example multiplies a*b, then multiplies that product by c, then multiplies that product by d. The second example first multiplies b*c, then multiplies that product by a, then multiplies that product by d.

Any valid parenthesized expression of 2n elements will necessarily have n ( and n ) with the property that, reading from left to right, there are always at least as many ( as ).

I know that for n numbers, the number of ways is the (n-1)th Catalan number. But how does one accurately generate all the resulting groupings?

Thanks

(As an aside: There are over 160 equivalent formulations of this problem, all based on different combinatorial objects enumerated by the Catalan Numbers. For the most up to date list of these, see Richard Stanley's Catalan Addendum.)

Best Answer

Here is actual code in Python, using generators to avoid using too much memory.

#! /usr/bin/python

def parenthesized (exprs):
    if len(exprs) == 1:
        yield exprs[0]
    else:
        first_exprs = []
        last_exprs = list(exprs)
        while 1 < len(last_exprs):
            first_exprs.append(last_exprs.pop(0))
            for x in parenthesized(first_exprs):
                if 1 < len(first_exprs):
                    x = '(%s)' % x
                for y in parenthesized(last_exprs):
                    if 1 < len(last_exprs):
                        y = '(%s)' % y
                    yield '%s%s' % (x, y)

for x in parenthesized(['a', 'b', 'c', 'd']):
    print x

Related Solutions

Python – How to generate all permutations of a list

There's a function in the standard-library for this: itertools.permutations.

import itertools
list(itertools.permutations([1, 2, 3]))

If for some reason you want to implement it yourself or are just curious to know how it works, here's one nice approach, taken from http://code.activestate.com/recipes/252178/:

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                # nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]

A couple of alternative approaches are listed in the documentation of itertools.permutations. Here's one:

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

And another, based on itertools.product:

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

How to find list of possible words from a letter matrix [Boggle Solver]

My answer works like the others here, but I'll post it because it looks a bit faster than the other Python solutions, from setting up the dictionary faster. (I checked this against John Fouhy's solution.) After setup, the time to solve is down in the noise.

grid = "fxie amlo ewbx astu".split()
nrows, ncols = len(grid), len(grid[0])

# A dictionary word that could be a solution must use only the grid's
# letters and have length >= 3. (With a case-insensitive match.)
import re
alphabet = ''.join(set(''.join(grid)))
bogglable = re.compile('[' + alphabet + ']{3,}$', re.I).match

words = set(word.rstrip('\n') for word in open('words') if bogglable(word))
prefixes = set(word[:i] for word in words
               for i in range(2, len(word)+1))

def solve():
    for y, row in enumerate(grid):
        for x, letter in enumerate(row):
            for result in extending(letter, ((x, y),)):
                yield result

def extending(prefix, path):
    if prefix in words:
        yield (prefix, path)
    for (nx, ny) in neighbors(path[-1]):
        if (nx, ny) not in path:
            prefix1 = prefix + grid[ny][nx]
            if prefix1 in prefixes:
                for result in extending(prefix1, path + ((nx, ny),)):
                    yield result

def neighbors((x, y)):
    for nx in range(max(0, x-1), min(x+2, ncols)):
        for ny in range(max(0, y-1), min(y+2, nrows)):
            yield (nx, ny)

Sample usage:

# Print a maximal-length word and its path:
print max(solve(), key=lambda (word, path): len(word))

Edit: Filter out words less than 3 letters long.

Edit 2: I was curious why Kent Fredric's Perl solution was faster; it turns out to use regular-expression matching instead of a set of characters. Doing the same in Python about doubles the speed.

Best Answer

Related Solutions

Python – How to generate all permutations of a list

How to find list of possible words from a letter matrix [Boggle Solver]

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