1364 lines
38 KiB
Python
1364 lines
38 KiB
Python
from __future__ import generators
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tutorial_tests = """
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Let's try a simple generator:
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>>> def f():
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... yield 1
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... yield 2
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>>> for i in f():
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... print i
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1
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2
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>>> g = f()
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>>> g.next()
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1
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>>> g.next()
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2
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"Falling off the end" stops the generator:
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>>> g.next()
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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File "<stdin>", line 2, in g
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StopIteration
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"return" also stops the generator:
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>>> def f():
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... yield 1
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... return
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... yield 2 # never reached
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...
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>>> g = f()
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>>> g.next()
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1
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>>> g.next()
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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File "<stdin>", line 3, in f
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StopIteration
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>>> g.next() # once stopped, can't be resumed
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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StopIteration
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"raise StopIteration" stops the generator too:
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>>> def f():
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... yield 1
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... raise StopIteration
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... yield 2 # never reached
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...
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>>> g = f()
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>>> g.next()
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1
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>>> g.next()
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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StopIteration
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>>> g.next()
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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StopIteration
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However, they are not exactly equivalent:
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>>> def g1():
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... try:
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... return
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... except:
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... yield 1
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...
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>>> list(g1())
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[]
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>>> def g2():
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... try:
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... raise StopIteration
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... except:
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... yield 42
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>>> print list(g2())
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[42]
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This may be surprising at first:
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>>> def g3():
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... try:
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... return
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... finally:
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... yield 1
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...
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>>> list(g3())
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[1]
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Let's create an alternate range() function implemented as a generator:
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>>> def yrange(n):
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... for i in range(n):
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... yield i
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...
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>>> list(yrange(5))
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[0, 1, 2, 3, 4]
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Generators always return to the most recent caller:
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>>> def creator():
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... r = yrange(5)
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... print "creator", r.next()
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... return r
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...
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>>> def caller():
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... r = creator()
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... for i in r:
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... print "caller", i
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...
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>>> caller()
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creator 0
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caller 1
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caller 2
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caller 3
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caller 4
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Generators can call other generators:
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>>> def zrange(n):
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... for i in yrange(n):
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... yield i
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...
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>>> list(zrange(5))
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[0, 1, 2, 3, 4]
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"""
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# The examples from PEP 255.
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pep_tests = """
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Specification: Return
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Note that return isn't always equivalent to raising StopIteration: the
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difference lies in how enclosing try/except constructs are treated.
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For example,
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>>> def f1():
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... try:
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... return
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... except:
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... yield 1
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>>> print list(f1())
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[]
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because, as in any function, return simply exits, but
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>>> def f2():
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... try:
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... raise StopIteration
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... except:
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... yield 42
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>>> print list(f2())
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[42]
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because StopIteration is captured by a bare "except", as is any
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exception.
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Specification: Generators and Exception Propagation
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>>> def f():
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... return 1/0
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>>> def g():
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... yield f() # the zero division exception propagates
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... yield 42 # and we'll never get here
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>>> k = g()
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>>> k.next()
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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File "<stdin>", line 2, in g
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File "<stdin>", line 2, in f
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ZeroDivisionError: integer division or modulo by zero
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>>> k.next() # and the generator cannot be resumed
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Traceback (most recent call last):
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File "<stdin>", line 1, in ?
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StopIteration
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>>>
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Specification: Try/Except/Finally
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>>> def f():
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... try:
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... yield 1
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... try:
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... yield 2
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... 1/0
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... yield 3 # never get here
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... except ZeroDivisionError:
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... yield 4
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... yield 5
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... raise
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... except:
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... yield 6
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... yield 7 # the "raise" above stops this
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... except:
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... yield 8
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... yield 9
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... try:
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... x = 12
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... finally:
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... yield 10
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... yield 11
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>>> print list(f())
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[1, 2, 4, 5, 8, 9, 10, 11]
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>>>
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Guido's binary tree example.
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>>> # A binary tree class.
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>>> class Tree:
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...
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... def __init__(self, label, left=None, right=None):
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... self.label = label
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... self.left = left
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... self.right = right
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...
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... def __repr__(self, level=0, indent=" "):
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... s = level*indent + `self.label`
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... if self.left:
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... s = s + "\\n" + self.left.__repr__(level+1, indent)
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... if self.right:
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... s = s + "\\n" + self.right.__repr__(level+1, indent)
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... return s
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...
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... def __iter__(self):
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... return inorder(self)
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>>> # Create a Tree from a list.
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>>> def tree(list):
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... n = len(list)
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... if n == 0:
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... return []
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... i = n / 2
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... return Tree(list[i], tree(list[:i]), tree(list[i+1:]))
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>>> # Show it off: create a tree.
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>>> t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
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>>> # A recursive generator that generates Tree leaves in in-order.
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>>> def inorder(t):
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... if t:
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... for x in inorder(t.left):
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... yield x
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... yield t.label
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... for x in inorder(t.right):
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... yield x
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>>> # Show it off: create a tree.
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... t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
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... # Print the nodes of the tree in in-order.
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... for x in t:
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... print x,
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A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
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>>> # A non-recursive generator.
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>>> def inorder(node):
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... stack = []
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... while node:
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... while node.left:
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... stack.append(node)
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... node = node.left
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... yield node.label
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... while not node.right:
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... try:
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... node = stack.pop()
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... except IndexError:
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... return
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... yield node.label
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... node = node.right
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>>> # Exercise the non-recursive generator.
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>>> for x in t:
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... print x,
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A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
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"""
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# Examples from Iterator-List and Python-Dev and c.l.py.
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email_tests = """
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The difference between yielding None and returning it.
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>>> def g():
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... for i in range(3):
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... yield None
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... yield None
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... return
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>>> list(g())
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[None, None, None, None]
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Ensure that explicitly raising StopIteration acts like any other exception
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in try/except, not like a return.
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>>> def g():
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... yield 1
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... try:
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... raise StopIteration
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... except:
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... yield 2
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... yield 3
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>>> list(g())
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[1, 2, 3]
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A generator can't be resumed while it's already running.
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>>> def g():
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... i = me.next()
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... yield i
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>>> me = g()
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>>> me.next()
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Traceback (most recent call last):
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...
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File "<string>", line 2, in g
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ValueError: generator already executing
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Next one was posted to c.l.py.
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>>> def gcomb(x, k):
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... "Generate all combinations of k elements from list x."
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...
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... if k > len(x):
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... return
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... if k == 0:
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... yield []
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... else:
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... first, rest = x[0], x[1:]
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... # A combination does or doesn't contain first.
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... # If it does, the remainder is a k-1 comb of rest.
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... for c in gcomb(rest, k-1):
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... c.insert(0, first)
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... yield c
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... # If it doesn't contain first, it's a k comb of rest.
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... for c in gcomb(rest, k):
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... yield c
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>>> seq = range(1, 5)
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>>> for k in range(len(seq) + 2):
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... print "%d-combs of %s:" % (k, seq)
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... for c in gcomb(seq, k):
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... print " ", c
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0-combs of [1, 2, 3, 4]:
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[]
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1-combs of [1, 2, 3, 4]:
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[1]
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[2]
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[3]
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[4]
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2-combs of [1, 2, 3, 4]:
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[1, 2]
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[1, 3]
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[1, 4]
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[2, 3]
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[2, 4]
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[3, 4]
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3-combs of [1, 2, 3, 4]:
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[1, 2, 3]
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[1, 2, 4]
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[1, 3, 4]
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[2, 3, 4]
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4-combs of [1, 2, 3, 4]:
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[1, 2, 3, 4]
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5-combs of [1, 2, 3, 4]:
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From the Iterators list, about the types of these things.
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>>> def g():
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... yield 1
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...
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>>> type(g)
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<type 'function'>
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>>> i = g()
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>>> type(i)
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<type 'generator'>
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>>> dir(i)
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['gi_frame', 'gi_running', 'next']
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>>> print i.next.__doc__
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next() -- get the next value, or raise StopIteration
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>>> iter(i) is i
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1
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>>> import types
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>>> isinstance(i, types.GeneratorType)
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1
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And more, added later.
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>>> i.gi_running
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0
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>>> type(i.gi_frame)
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<type 'frame'>
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>>> i.gi_running = 42
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Traceback (most recent call last):
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...
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TypeError: object has read-only attributes
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>>> def g():
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... yield me.gi_running
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>>> me = g()
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>>> me.gi_running
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0
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>>> me.next()
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1
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>>> me.gi_running
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0
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A clever union-find implementation from c.l.py, due to David Eppstein.
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Sent: Friday, June 29, 2001 12:16 PM
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To: python-list@python.org
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Subject: Re: PEP 255: Simple Generators
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>>> class disjointSet:
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... def __init__(self, name):
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... self.name = name
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... self.parent = None
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... self.generator = self.generate()
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...
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... def generate(self):
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... while not self.parent:
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... yield self
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... for x in self.parent.generator:
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... yield x
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...
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... def find(self):
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... return self.generator.next()
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...
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... def union(self, parent):
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... if self.parent:
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... raise ValueError("Sorry, I'm not a root!")
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... self.parent = parent
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...
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... def __str__(self):
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... return self.name
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>>> names = "ABCDEFGHIJKLM"
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>>> sets = [disjointSet(name) for name in names]
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>>> roots = sets[:]
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>>> import random
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>>> random.seed(42)
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>>> while 1:
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... for s in sets:
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... print "%s->%s" % (s, s.find()),
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... print
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... if len(roots) > 1:
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... s1 = random.choice(roots)
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... roots.remove(s1)
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... s2 = random.choice(roots)
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... s1.union(s2)
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... print "merged", s1, "into", s2
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... else:
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... break
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A->A B->B C->C D->D E->E F->F G->G H->H I->I J->J K->K L->L M->M
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merged D into G
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A->A B->B C->C D->G E->E F->F G->G H->H I->I J->J K->K L->L M->M
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merged C into F
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A->A B->B C->F D->G E->E F->F G->G H->H I->I J->J K->K L->L M->M
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merged L into A
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A->A B->B C->F D->G E->E F->F G->G H->H I->I J->J K->K L->A M->M
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merged H into E
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A->A B->B C->F D->G E->E F->F G->G H->E I->I J->J K->K L->A M->M
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merged B into E
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A->A B->E C->F D->G E->E F->F G->G H->E I->I J->J K->K L->A M->M
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merged J into G
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A->A B->E C->F D->G E->E F->F G->G H->E I->I J->G K->K L->A M->M
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merged E into G
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A->A B->G C->F D->G E->G F->F G->G H->G I->I J->G K->K L->A M->M
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merged M into G
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A->A B->G C->F D->G E->G F->F G->G H->G I->I J->G K->K L->A M->G
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merged I into K
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A->A B->G C->F D->G E->G F->F G->G H->G I->K J->G K->K L->A M->G
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merged K into A
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A->A B->G C->F D->G E->G F->F G->G H->G I->A J->G K->A L->A M->G
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merged F into A
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A->A B->G C->A D->G E->G F->A G->G H->G I->A J->G K->A L->A M->G
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merged A into G
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A->G B->G C->G D->G E->G F->G G->G H->G I->G J->G K->G L->G M->G
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"""
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# Fun tests (for sufficiently warped notions of "fun").
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fun_tests = """
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Build up to a recursive Sieve of Eratosthenes generator.
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>>> def firstn(g, n):
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... return [g.next() for i in range(n)]
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>>> def intsfrom(i):
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... while 1:
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... yield i
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... i += 1
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>>> firstn(intsfrom(5), 7)
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[5, 6, 7, 8, 9, 10, 11]
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>>> def exclude_multiples(n, ints):
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... for i in ints:
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... if i % n:
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... yield i
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>>> firstn(exclude_multiples(3, intsfrom(1)), 6)
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[1, 2, 4, 5, 7, 8]
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>>> def sieve(ints):
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... prime = ints.next()
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... yield prime
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... not_divisible_by_prime = exclude_multiples(prime, ints)
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... for p in sieve(not_divisible_by_prime):
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... yield p
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>>> primes = sieve(intsfrom(2))
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>>> firstn(primes, 20)
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
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Another famous problem: generate all integers of the form
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2**i * 3**j * 5**k
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in increasing order, where i,j,k >= 0. Trickier than it may look at first!
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Try writing it without generators, and correctly, and without generating
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3 internal results for each result output.
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>>> def times(n, g):
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... for i in g:
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... yield n * i
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>>> firstn(times(10, intsfrom(1)), 10)
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[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
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>>> def merge(g, h):
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... ng = g.next()
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... nh = h.next()
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... while 1:
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... if ng < nh:
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... yield ng
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... ng = g.next()
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... elif ng > nh:
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... yield nh
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... nh = h.next()
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... else:
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... yield ng
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... ng = g.next()
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... nh = h.next()
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The following works, but is doing a whale of a lot of redundant work --
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it's not clear how to get the internal uses of m235 to share a single
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generator. Note that me_times2 (etc) each need to see every element in the
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result sequence. So this is an example where lazy lists are more natural
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(you can look at the head of a lazy list any number of times).
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>>> def m235():
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... yield 1
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... me_times2 = times(2, m235())
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... me_times3 = times(3, m235())
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... me_times5 = times(5, m235())
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... for i in merge(merge(me_times2,
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... me_times3),
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... me_times5):
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... yield i
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Don't print "too many" of these -- the implementation above is extremely
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inefficient: each call of m235() leads to 3 recursive calls, and in
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turn each of those 3 more, and so on, and so on, until we've descended
|
|
enough levels to satisfy the print stmts. Very odd: when I printed 5
|
|
lines of results below, this managed to screw up Win98's malloc in "the
|
|
usual" way, i.e. the heap grew over 4Mb so Win98 started fragmenting
|
|
address space, and it *looked* like a very slow leak.
|
|
|
|
>>> result = m235()
|
|
>>> for i in range(3):
|
|
... print firstn(result, 15)
|
|
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24]
|
|
[25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80]
|
|
[81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192]
|
|
|
|
Heh. Here's one way to get a shared list, complete with an excruciating
|
|
namespace renaming trick. The *pretty* part is that the times() and merge()
|
|
functions can be reused as-is, because they only assume their stream
|
|
arguments are iterable -- a LazyList is the same as a generator to times().
|
|
|
|
>>> class LazyList:
|
|
... def __init__(self, g):
|
|
... self.sofar = []
|
|
... self.fetch = g.next
|
|
...
|
|
... def __getitem__(self, i):
|
|
... sofar, fetch = self.sofar, self.fetch
|
|
... while i >= len(sofar):
|
|
... sofar.append(fetch())
|
|
... return sofar[i]
|
|
|
|
>>> def m235():
|
|
... yield 1
|
|
... # Gack: m235 below actually refers to a LazyList.
|
|
... me_times2 = times(2, m235)
|
|
... me_times3 = times(3, m235)
|
|
... me_times5 = times(5, m235)
|
|
... for i in merge(merge(me_times2,
|
|
... me_times3),
|
|
... me_times5):
|
|
... yield i
|
|
|
|
Print as many of these as you like -- *this* implementation is memory-
|
|
efficient.
|
|
|
|
>>> m235 = LazyList(m235())
|
|
>>> for i in range(5):
|
|
... print [m235[j] for j in range(15*i, 15*(i+1))]
|
|
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24]
|
|
[25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80]
|
|
[81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192]
|
|
[200, 216, 225, 240, 243, 250, 256, 270, 288, 300, 320, 324, 360, 375, 384]
|
|
[400, 405, 432, 450, 480, 486, 500, 512, 540, 576, 600, 625, 640, 648, 675]
|
|
|
|
|
|
Ye olde Fibonacci generator, LazyList style.
|
|
|
|
>>> def fibgen(a, b):
|
|
...
|
|
... def sum(g, h):
|
|
... while 1:
|
|
... yield g.next() + h.next()
|
|
...
|
|
... def tail(g):
|
|
... g.next() # throw first away
|
|
... for x in g:
|
|
... yield x
|
|
...
|
|
... yield a
|
|
... yield b
|
|
... for s in sum(iter(fib),
|
|
... tail(iter(fib))):
|
|
... yield s
|
|
|
|
>>> fib = LazyList(fibgen(1, 2))
|
|
>>> firstn(iter(fib), 17)
|
|
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584]
|
|
"""
|
|
|
|
# syntax_tests mostly provokes SyntaxErrors. Also fiddling with #if 0
|
|
# hackery.
|
|
|
|
syntax_tests = """
|
|
|
|
>>> def f():
|
|
... return 22
|
|
... yield 1
|
|
Traceback (most recent call last):
|
|
...
|
|
SyntaxError: 'return' with argument inside generator (<string>, line 2)
|
|
|
|
>>> def f():
|
|
... yield 1
|
|
... return 22
|
|
Traceback (most recent call last):
|
|
...
|
|
SyntaxError: 'return' with argument inside generator (<string>, line 3)
|
|
|
|
"return None" is not the same as "return" in a generator:
|
|
|
|
>>> def f():
|
|
... yield 1
|
|
... return None
|
|
Traceback (most recent call last):
|
|
...
|
|
SyntaxError: 'return' with argument inside generator (<string>, line 3)
|
|
|
|
This one is fine:
|
|
|
|
>>> def f():
|
|
... yield 1
|
|
... return
|
|
|
|
>>> def f():
|
|
... try:
|
|
... yield 1
|
|
... finally:
|
|
... pass
|
|
Traceback (most recent call last):
|
|
...
|
|
SyntaxError: 'yield' not allowed in a 'try' block with a 'finally' clause (<string>, line 3)
|
|
|
|
>>> def f():
|
|
... try:
|
|
... try:
|
|
... 1/0
|
|
... except ZeroDivisionError:
|
|
... yield 666 # bad because *outer* try has finally
|
|
... except:
|
|
... pass
|
|
... finally:
|
|
... pass
|
|
Traceback (most recent call last):
|
|
...
|
|
SyntaxError: 'yield' not allowed in a 'try' block with a 'finally' clause (<string>, line 6)
|
|
|
|
But this is fine:
|
|
|
|
>>> def f():
|
|
... try:
|
|
... try:
|
|
... yield 12
|
|
... 1/0
|
|
... except ZeroDivisionError:
|
|
... yield 666
|
|
... except:
|
|
... try:
|
|
... x = 12
|
|
... finally:
|
|
... yield 12
|
|
... except:
|
|
... return
|
|
>>> list(f())
|
|
[12, 666]
|
|
|
|
>>> def f():
|
|
... yield
|
|
Traceback (most recent call last):
|
|
SyntaxError: invalid syntax
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... yield
|
|
Traceback (most recent call last):
|
|
SyntaxError: invalid syntax
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... yield 1
|
|
>>> type(f())
|
|
<type 'generator'>
|
|
|
|
>>> def f():
|
|
... if "":
|
|
... yield None
|
|
>>> type(f())
|
|
<type 'generator'>
|
|
|
|
>>> def f():
|
|
... return
|
|
... try:
|
|
... if x==4:
|
|
... pass
|
|
... elif 0:
|
|
... try:
|
|
... 1/0
|
|
... except SyntaxError:
|
|
... pass
|
|
... else:
|
|
... if 0:
|
|
... while 12:
|
|
... x += 1
|
|
... yield 2 # don't blink
|
|
... f(a, b, c, d, e)
|
|
... else:
|
|
... pass
|
|
... except:
|
|
... x = 1
|
|
... return
|
|
>>> type(f())
|
|
<type 'generator'>
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... def g():
|
|
... yield 1
|
|
...
|
|
>>> type(f())
|
|
<type 'None'>
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... class C:
|
|
... def __init__(self):
|
|
... yield 1
|
|
... def f(self):
|
|
... yield 2
|
|
>>> type(f())
|
|
<type 'None'>
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... return
|
|
... if 0:
|
|
... yield 2
|
|
>>> type(f())
|
|
<type 'generator'>
|
|
|
|
|
|
>>> def f():
|
|
... if 0:
|
|
... lambda x: x # shouldn't trigger here
|
|
... return # or here
|
|
... def f(i):
|
|
... return 2*i # or here
|
|
... if 0:
|
|
... return 3 # but *this* sucks (line 8)
|
|
... if 0:
|
|
... yield 2 # because it's a generator
|
|
Traceback (most recent call last):
|
|
SyntaxError: 'return' with argument inside generator (<string>, line 8)
|
|
"""
|
|
|
|
# conjoin is a simple backtracking generator, named in honor of Icon's
|
|
# "conjunction" control structure. Pass a list of no-argument functions
|
|
# that return iterable objects. Easiest to explain by example: assume the
|
|
# function list [x, y, z] is passed. Then conjoin acts like:
|
|
#
|
|
# def g():
|
|
# values = [None] * 3
|
|
# for values[0] in x():
|
|
# for values[1] in y():
|
|
# for values[2] in z():
|
|
# yield values
|
|
#
|
|
# So some 3-lists of values *may* be generated, each time we successfully
|
|
# get into the innermost loop. If an iterator fails (is exhausted) before
|
|
# then, it "backtracks" to get the next value from the nearest enclosing
|
|
# iterator (the one "to the left"), and starts all over again at the next
|
|
# slot (pumps a fresh iterator). Of course this is most useful when the
|
|
# iterators have side-effects, so that which values *can* be generated at
|
|
# each slot depend on the values iterated at previous slots.
|
|
|
|
def conjoin(gs):
|
|
|
|
values = [None] * len(gs)
|
|
|
|
def gen(i, values=values):
|
|
if i >= len(gs):
|
|
yield values
|
|
else:
|
|
for values[i] in gs[i]():
|
|
for x in gen(i+1):
|
|
yield x
|
|
|
|
for x in gen(0):
|
|
yield x
|
|
|
|
# That works fine, but recursing a level and checking i against len(gs) for
|
|
# each item produced is inefficient. By doing manual loop unrolling across
|
|
# generator boundaries, it's possible to eliminate most of that overhead.
|
|
# This isn't worth the bother *in general* for generators, but conjoin() is
|
|
# a core building block for some CPU-intensive generator applications.
|
|
|
|
def conjoin(gs):
|
|
|
|
n = len(gs)
|
|
values = [None] * n
|
|
|
|
# Do one loop nest at time recursively, until the # of loop nests
|
|
# remaining is divisible by 3.
|
|
|
|
def gen(i, values=values):
|
|
if i >= n:
|
|
yield values
|
|
|
|
elif (n-i) % 3:
|
|
ip1 = i+1
|
|
for values[i] in gs[i]():
|
|
for x in gen(ip1):
|
|
yield x
|
|
|
|
else:
|
|
for x in _gen3(i):
|
|
yield x
|
|
|
|
# Do three loop nests at a time, recursing only if at least three more
|
|
# remain. Don't call directly: this is an internal optimization for
|
|
# gen's use.
|
|
|
|
def _gen3(i, values=values):
|
|
assert i < n and (n-i) % 3 == 0
|
|
ip1, ip2, ip3 = i+1, i+2, i+3
|
|
g, g1, g2 = gs[i : ip3]
|
|
|
|
if ip3 >= n:
|
|
# These are the last three, so we can yield values directly.
|
|
for values[i] in g():
|
|
for values[ip1] in g1():
|
|
for values[ip2] in g2():
|
|
yield values
|
|
|
|
else:
|
|
# At least 6 loop nests remain; peel off 3 and recurse for the
|
|
# rest.
|
|
for values[i] in g():
|
|
for values[ip1] in g1():
|
|
for values[ip2] in g2():
|
|
for x in _gen3(ip3):
|
|
yield x
|
|
|
|
for x in gen(0):
|
|
yield x
|
|
|
|
# And one more approach: For backtracking apps like the Knight's Tour
|
|
# solver below, the number of backtracking levels can be enormous (one
|
|
# level per square, for the Knight's Tour, so that e.g. a 100x100 board
|
|
# needs 10,000 levels). In such cases Python is likely to run out of
|
|
# stack space due to recursion. So here's a recursion-free version of
|
|
# conjoin too.
|
|
# NOTE WELL: This allows large problems to be solved with only trivial
|
|
# demands on stack space. Without explicitly resumable generators, this is
|
|
# much harder to achieve. OTOH, this is much slower (up to a factor of 2)
|
|
# than the fancy unrolled recursive conjoin.
|
|
|
|
def flat_conjoin(gs): # rename to conjoin to run tests with this instead
|
|
n = len(gs)
|
|
values = [None] * n
|
|
iters = [None] * n
|
|
_StopIteration = StopIteration # make local because caught a *lot*
|
|
i = 0
|
|
while 1:
|
|
# Descend.
|
|
try:
|
|
while i < n:
|
|
it = iters[i] = gs[i]().next
|
|
values[i] = it()
|
|
i += 1
|
|
except _StopIteration:
|
|
pass
|
|
else:
|
|
assert i == n
|
|
yield values
|
|
|
|
# Backtrack until an older iterator can be resumed.
|
|
i -= 1
|
|
while i >= 0:
|
|
try:
|
|
values[i] = iters[i]()
|
|
# Success! Start fresh at next level.
|
|
i += 1
|
|
break
|
|
except _StopIteration:
|
|
# Continue backtracking.
|
|
i -= 1
|
|
else:
|
|
assert i < 0
|
|
break
|
|
|
|
# A conjoin-based N-Queens solver.
|
|
|
|
class Queens:
|
|
def __init__(self, n):
|
|
self.n = n
|
|
rangen = range(n)
|
|
|
|
# Assign a unique int to each column and diagonal.
|
|
# columns: n of those, range(n).
|
|
# NW-SE diagonals: 2n-1 of these, i-j unique and invariant along
|
|
# each, smallest i-j is 0-(n-1) = 1-n, so add n-1 to shift to 0-
|
|
# based.
|
|
# NE-SW diagonals: 2n-1 of these, i+j unique and invariant along
|
|
# each, smallest i+j is 0, largest is 2n-2.
|
|
|
|
# For each square, compute a bit vector of the columns and
|
|
# diagonals it covers, and for each row compute a function that
|
|
# generates the possiblities for the columns in that row.
|
|
self.rowgenerators = []
|
|
for i in rangen:
|
|
rowuses = [(1L << j) | # column ordinal
|
|
(1L << (n + i-j + n-1)) | # NW-SE ordinal
|
|
(1L << (n + 2*n-1 + i+j)) # NE-SW ordinal
|
|
for j in rangen]
|
|
|
|
def rowgen(rowuses=rowuses):
|
|
for j in rangen:
|
|
uses = rowuses[j]
|
|
if uses & self.used == 0:
|
|
self.used |= uses
|
|
yield j
|
|
self.used &= ~uses
|
|
|
|
self.rowgenerators.append(rowgen)
|
|
|
|
# Generate solutions.
|
|
def solve(self):
|
|
self.used = 0
|
|
for row2col in conjoin(self.rowgenerators):
|
|
yield row2col
|
|
|
|
def printsolution(self, row2col):
|
|
n = self.n
|
|
assert n == len(row2col)
|
|
sep = "+" + "-+" * n
|
|
print sep
|
|
for i in range(n):
|
|
squares = [" " for j in range(n)]
|
|
squares[row2col[i]] = "Q"
|
|
print "|" + "|".join(squares) + "|"
|
|
print sep
|
|
|
|
# A conjoin-based Knight's Tour solver. This is pretty sophisticated
|
|
# (e.g., when used with flat_conjoin above, and passing hard=1 to the
|
|
# constructor, a 200x200 Knight's Tour was found quickly -- note that we're
|
|
# creating 10s of thousands of generators then!), and is lengthy.
|
|
|
|
class Knights:
|
|
def __init__(self, m, n, hard=0):
|
|
self.m, self.n = m, n
|
|
|
|
# solve() will set up succs[i] to be a list of square #i's
|
|
# successors.
|
|
succs = self.succs = []
|
|
|
|
# Remove i0 from each of its successor's successor lists, i.e.
|
|
# successors can't go back to i0 again. Return 0 if we can
|
|
# detect this makes a solution impossible, else return 1.
|
|
|
|
def remove_from_successors(i0, len=len):
|
|
# If we remove all exits from a free square, we're dead:
|
|
# even if we move to it next, we can't leave it again.
|
|
# If we create a square with one exit, we must visit it next;
|
|
# else somebody else will have to visit it, and since there's
|
|
# only one adjacent, there won't be a way to leave it again.
|
|
# Finelly, if we create more than one free square with a
|
|
# single exit, we can only move to one of them next, leaving
|
|
# the other one a dead end.
|
|
ne0 = ne1 = 0
|
|
for i in succs[i0]:
|
|
s = succs[i]
|
|
s.remove(i0)
|
|
e = len(s)
|
|
if e == 0:
|
|
ne0 += 1
|
|
elif e == 1:
|
|
ne1 += 1
|
|
return ne0 == 0 and ne1 < 2
|
|
|
|
# Put i0 back in each of its successor's successor lists.
|
|
|
|
def add_to_successors(i0):
|
|
for i in succs[i0]:
|
|
succs[i].append(i0)
|
|
|
|
# Generate the first move.
|
|
def first():
|
|
if m < 1 or n < 1:
|
|
return
|
|
|
|
# Since we're looking for a cycle, it doesn't matter where we
|
|
# start. Starting in a corner makes the 2nd move easy.
|
|
corner = self.coords2index(0, 0)
|
|
remove_from_successors(corner)
|
|
self.lastij = corner
|
|
yield corner
|
|
add_to_successors(corner)
|
|
|
|
# Generate the second moves.
|
|
def second():
|
|
corner = self.coords2index(0, 0)
|
|
assert self.lastij == corner # i.e., we started in the corner
|
|
if m < 3 or n < 3:
|
|
return
|
|
assert len(succs[corner]) == 2
|
|
assert self.coords2index(1, 2) in succs[corner]
|
|
assert self.coords2index(2, 1) in succs[corner]
|
|
# Only two choices. Whichever we pick, the other must be the
|
|
# square picked on move m*n, as it's the only way to get back
|
|
# to (0, 0). Save its index in self.final so that moves before
|
|
# the last know it must be kept free.
|
|
for i, j in (1, 2), (2, 1):
|
|
this = self.coords2index(i, j)
|
|
final = self.coords2index(3-i, 3-j)
|
|
self.final = final
|
|
|
|
remove_from_successors(this)
|
|
succs[final].append(corner)
|
|
self.lastij = this
|
|
yield this
|
|
succs[final].remove(corner)
|
|
add_to_successors(this)
|
|
|
|
# Generate moves 3 thru m*n-1.
|
|
def advance(len=len):
|
|
# If some successor has only one exit, must take it.
|
|
# Else favor successors with fewer exits.
|
|
candidates = []
|
|
for i in succs[self.lastij]:
|
|
e = len(succs[i])
|
|
assert e > 0, "else remove_from_successors() pruning flawed"
|
|
if e == 1:
|
|
candidates = [(e, i)]
|
|
break
|
|
candidates.append((e, i))
|
|
else:
|
|
candidates.sort()
|
|
|
|
for e, i in candidates:
|
|
if i != self.final:
|
|
if remove_from_successors(i):
|
|
self.lastij = i
|
|
yield i
|
|
add_to_successors(i)
|
|
|
|
# Generate moves 3 thru m*n-1. Alternative version using a
|
|
# stronger (but more expensive) heuristic to order successors.
|
|
# Since the # of backtracking levels is m*n, a poor move early on
|
|
# can take eons to undo. Smallest square board for which this
|
|
# matters a lot is 52x52.
|
|
def advance_hard(vmid=(m-1)/2.0, hmid=(n-1)/2.0, len=len):
|
|
# If some successor has only one exit, must take it.
|
|
# Else favor successors with fewer exits.
|
|
# Break ties via max distance from board centerpoint (favor
|
|
# corners and edges whenever possible).
|
|
candidates = []
|
|
for i in succs[self.lastij]:
|
|
e = len(succs[i])
|
|
assert e > 0, "else remove_from_successors() pruning flawed"
|
|
if e == 1:
|
|
candidates = [(e, 0, i)]
|
|
break
|
|
i1, j1 = self.index2coords(i)
|
|
d = (i1 - vmid)**2 + (j1 - hmid)**2
|
|
candidates.append((e, -d, i))
|
|
else:
|
|
candidates.sort()
|
|
|
|
for e, d, i in candidates:
|
|
if i != self.final:
|
|
if remove_from_successors(i):
|
|
self.lastij = i
|
|
yield i
|
|
add_to_successors(i)
|
|
|
|
# Generate the last move.
|
|
def last():
|
|
assert self.final in succs[self.lastij]
|
|
yield self.final
|
|
|
|
if m*n < 4:
|
|
self.squaregenerators = [first]
|
|
else:
|
|
self.squaregenerators = [first, second] + \
|
|
[hard and advance_hard or advance] * (m*n - 3) + \
|
|
[last]
|
|
|
|
def coords2index(self, i, j):
|
|
assert 0 <= i < self.m
|
|
assert 0 <= j < self.n
|
|
return i * self.n + j
|
|
|
|
def index2coords(self, index):
|
|
assert 0 <= index < self.m * self.n
|
|
return divmod(index, self.n)
|
|
|
|
def _init_board(self):
|
|
succs = self.succs
|
|
del succs[:]
|
|
m, n = self.m, self.n
|
|
c2i = self.coords2index
|
|
|
|
offsets = [( 1, 2), ( 2, 1), ( 2, -1), ( 1, -2),
|
|
(-1, -2), (-2, -1), (-2, 1), (-1, 2)]
|
|
rangen = range(n)
|
|
for i in range(m):
|
|
for j in rangen:
|
|
s = [c2i(i+io, j+jo) for io, jo in offsets
|
|
if 0 <= i+io < m and
|
|
0 <= j+jo < n]
|
|
succs.append(s)
|
|
|
|
# Generate solutions.
|
|
def solve(self):
|
|
self._init_board()
|
|
for x in conjoin(self.squaregenerators):
|
|
yield x
|
|
|
|
def printsolution(self, x):
|
|
m, n = self.m, self.n
|
|
assert len(x) == m*n
|
|
w = len(str(m*n))
|
|
format = "%" + str(w) + "d"
|
|
|
|
squares = [[None] * n for i in range(m)]
|
|
k = 1
|
|
for i in x:
|
|
i1, j1 = self.index2coords(i)
|
|
squares[i1][j1] = format % k
|
|
k += 1
|
|
|
|
sep = "+" + ("-" * w + "+") * n
|
|
print sep
|
|
for i in range(m):
|
|
row = squares[i]
|
|
print "|" + "|".join(row) + "|"
|
|
print sep
|
|
|
|
conjoin_tests = """
|
|
|
|
Generate the 3-bit binary numbers in order. This illustrates dumbest-
|
|
possible use of conjoin, just to generate the full cross-product.
|
|
|
|
>>> for c in conjoin([lambda: iter((0, 1))] * 3):
|
|
... print c
|
|
[0, 0, 0]
|
|
[0, 0, 1]
|
|
[0, 1, 0]
|
|
[0, 1, 1]
|
|
[1, 0, 0]
|
|
[1, 0, 1]
|
|
[1, 1, 0]
|
|
[1, 1, 1]
|
|
|
|
For efficiency in typical backtracking apps, conjoin() yields the same list
|
|
object each time. So if you want to save away a full account of its
|
|
generated sequence, you need to copy its results.
|
|
|
|
>>> def gencopy(iterator):
|
|
... for x in iterator:
|
|
... yield x[:]
|
|
|
|
>>> for n in range(10):
|
|
... all = list(gencopy(conjoin([lambda: iter((0, 1))] * n)))
|
|
... print n, len(all), all[0] == [0] * n, all[-1] == [1] * n
|
|
0 1 1 1
|
|
1 2 1 1
|
|
2 4 1 1
|
|
3 8 1 1
|
|
4 16 1 1
|
|
5 32 1 1
|
|
6 64 1 1
|
|
7 128 1 1
|
|
8 256 1 1
|
|
9 512 1 1
|
|
|
|
And run an 8-queens solver.
|
|
|
|
>>> q = Queens(8)
|
|
>>> LIMIT = 2
|
|
>>> count = 0
|
|
>>> for row2col in q.solve():
|
|
... count += 1
|
|
... if count <= LIMIT:
|
|
... print "Solution", count
|
|
... q.printsolution(row2col)
|
|
Solution 1
|
|
+-+-+-+-+-+-+-+-+
|
|
|Q| | | | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | |Q| | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | | | |Q|
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | |Q| | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | |Q| | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | | |Q| |
|
|
+-+-+-+-+-+-+-+-+
|
|
| |Q| | | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | |Q| | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
Solution 2
|
|
+-+-+-+-+-+-+-+-+
|
|
|Q| | | | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | |Q| | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | | | |Q|
|
|
+-+-+-+-+-+-+-+-+
|
|
| | |Q| | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | | | |Q| |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | |Q| | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| |Q| | | | | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
| | | | |Q| | | |
|
|
+-+-+-+-+-+-+-+-+
|
|
|
|
>>> print count, "solutions in all."
|
|
92 solutions in all.
|
|
|
|
And run a Knight's Tour on a 10x10 board. Note that there are about
|
|
20,000 solutions even on a 6x6 board, so don't dare run this to exhaustion.
|
|
|
|
>>> k = Knights(10, 10)
|
|
>>> LIMIT = 2
|
|
>>> count = 0
|
|
>>> for x in k.solve():
|
|
... count += 1
|
|
... if count <= LIMIT:
|
|
... print "Solution", count
|
|
... k.printsolution(x)
|
|
... else:
|
|
... break
|
|
Solution 1
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 1| 58| 27| 34| 3| 40| 29| 10| 5| 8|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 26| 35| 2| 57| 28| 33| 4| 7| 30| 11|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 59|100| 73| 36| 41| 56| 39| 32| 9| 6|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 74| 25| 60| 55| 72| 37| 42| 49| 12| 31|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 61| 86| 99| 76| 63| 52| 47| 38| 43| 50|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 24| 75| 62| 85| 54| 71| 64| 51| 48| 13|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 87| 98| 91| 80| 77| 84| 53| 46| 65| 44|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 90| 23| 88| 95| 70| 79| 68| 83| 14| 17|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 97| 92| 21| 78| 81| 94| 19| 16| 45| 66|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 22| 89| 96| 93| 20| 69| 82| 67| 18| 15|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
Solution 2
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 1| 58| 27| 34| 3| 40| 29| 10| 5| 8|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 26| 35| 2| 57| 28| 33| 4| 7| 30| 11|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 59|100| 73| 36| 41| 56| 39| 32| 9| 6|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 74| 25| 60| 55| 72| 37| 42| 49| 12| 31|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 61| 86| 99| 76| 63| 52| 47| 38| 43| 50|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 24| 75| 62| 85| 54| 71| 64| 51| 48| 13|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 87| 98| 89| 80| 77| 84| 53| 46| 65| 44|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 90| 23| 92| 95| 70| 79| 68| 83| 14| 17|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 97| 88| 21| 78| 81| 94| 19| 16| 45| 66|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
| 22| 91| 96| 93| 20| 69| 82| 67| 18| 15|
|
|
+---+---+---+---+---+---+---+---+---+---+
|
|
"""
|
|
|
|
__test__ = {"tut": tutorial_tests,
|
|
"pep": pep_tests,
|
|
"email": email_tests,
|
|
"fun": fun_tests,
|
|
"syntax": syntax_tests,
|
|
"conjoin": conjoin_tests}
|
|
|
|
# Magic test name that regrtest.py invokes *after* importing this module.
|
|
# This worms around a bootstrap problem.
|
|
# Note that doctest and regrtest both look in sys.argv for a "-v" argument,
|
|
# so this works as expected in both ways of running regrtest.
|
|
def test_main():
|
|
import doctest, test_generators
|
|
if 0: # change to 1 to run forever (to check for leaks)
|
|
while 1:
|
|
doctest.master = None
|
|
doctest.testmod(test_generators)
|
|
print ".",
|
|
else:
|
|
doctest.testmod(test_generators)
|
|
|
|
# This part isn't needed for regrtest, but for running the test directly.
|
|
if __name__ == "__main__":
|
|
test_main()
|