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tutorial_tests = """
Let's try a simple generator:
>>> def f():
... yield 1
... yield 2
>>> for i in f():
... print i
1
2
>>> g = f()
>>> g.next()
1
>>> g.next()
2
"Falling off the end" stops the generator:
>>> g.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in g
StopIteration
"return" also stops the generator:
>>> def f():
... yield 1
... return
... yield 2 # never reached
...
>>> g = f()
>>> g.next()
1
>>> g.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 3, in f
StopIteration
>>> g.next() # once stopped, can't be resumed
Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
"raise StopIteration" stops the generator too:
>>> def f():
... yield 1
... raise StopIteration
... yield 2 # never reached
...
>>> g = f()
>>> g.next()
1
>>> g.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
>>> g.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
However, they are not exactly equivalent:
>>> def g1():
... try:
... return
... except:
... yield 1
...
>>> list(g1())
[]
>>> def g2():
... try:
... raise StopIteration
... except:
... yield 42
>>> print list(g2())
[42]
This may be surprising at first:
>>> def g3():
... try:
... return
... finally:
... yield 1
...
>>> list(g3())
[1]
Let's create an alternate range() function implemented as a generator:
>>> def yrange(n):
... for i in range(n):
... yield i
...
>>> list(yrange(5))
[0, 1, 2, 3, 4]
Generators always return to the most recent caller:
>>> def creator():
... r = yrange(5)
... print "creator", r.next()
... return r
...
>>> def caller():
... r = creator()
... for i in r:
... print "caller", i
...
>>> caller()
creator 0
caller 1
caller 2
caller 3
caller 4
Generators can call other generators:
>>> def zrange(n):
... for i in yrange(n):
... yield i
...
>>> list(zrange(5))
[0, 1, 2, 3, 4]
"""
# The examples from PEP 255.
pep_tests = """
Specification: Yield
Restriction: A generator cannot be resumed while it is actively
running:
>>> def g():
... i = me.next()
... yield i
>>> me = g()
>>> me.next()
Traceback (most recent call last):
...
File "<string>", line 2, in g
ValueError: generator already executing
Specification: Return
Note that return isn't always equivalent to raising StopIteration: the
difference lies in how enclosing try/except constructs are treated.
For example,
>>> def f1():
... try:
... return
... except:
... yield 1
>>> print list(f1())
[]
because, as in any function, return simply exits, but
>>> def f2():
... try:
... raise StopIteration
... except:
... yield 42
>>> print list(f2())
[42]
because StopIteration is captured by a bare "except", as is any
exception.
Specification: Generators and Exception Propagation
>>> def f():
... return 1//0
>>> def g():
... yield f() # the zero division exception propagates
... yield 42 # and we'll never get here
>>> k = g()
>>> k.next()
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in g
File "<stdin>", line 2, in f
ZeroDivisionError: integer division or modulo by zero
>>> k.next() # and the generator cannot be resumed
Traceback (most recent call last):
File "<stdin>", line 1, in ?
StopIteration
>>>
Specification: Try/Except/Finally
>>> def f():
... try:
... yield 1
... try:
... yield 2
... 1//0
... yield 3 # never get here
... except ZeroDivisionError:
... yield 4
... yield 5
... raise
... except:
... yield 6
... yield 7 # the "raise" above stops this
... except:
... yield 8
... yield 9
... try:
... x = 12
... finally:
... yield 10
... yield 11
>>> print list(f())
[1, 2, 4, 5, 8, 9, 10, 11]
>>>
Guido's binary tree example.
>>> # A binary tree class.
>>> class Tree:
...
... def __init__(self, label, left=None, right=None):
... self.label = label
... self.left = left
... self.right = right
...
... def __repr__(self, level=0, indent=" "):
... s = level*indent + repr(self.label)
... if self.left:
... s = s + "\\n" + self.left.__repr__(level+1, indent)
... if self.right:
... s = s + "\\n" + self.right.__repr__(level+1, indent)
... return s
...
... def __iter__(self):
... return inorder(self)
>>> # Create a Tree from a list.
>>> def tree(list):
... n = len(list)
... if n == 0:
... return []
... i = n // 2
... return Tree(list[i], tree(list[:i]), tree(list[i+1:]))
>>> # Show it off: create a tree.
>>> t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
>>> # A recursive generator that generates Tree labels in in-order.
>>> def inorder(t):
... if t:
... for x in inorder(t.left):
... yield x
... yield t.label
... for x in inorder(t.right):
... yield x
>>> # Show it off: create a tree.
>>> t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ")
>>> # Print the nodes of the tree in in-order.
>>> for x in t:
... print x,
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
>>> # A non-recursive generator.
>>> def inorder(node):
... stack = []
... while node:
... while node.left:
... stack.append(node)
... node = node.left
... yield node.label
... while not node.right:
... try:
... node = stack.pop()
... except IndexError:
... return
... yield node.label
... node = node.right
>>> # Exercise the non-recursive generator.
>>> for x in t:
... print x,
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
"""
# Examples from Iterator-List and Python-Dev and c.l.py.
email_tests = """
The difference between yielding None and returning it.
>>> def g():
... for i in range(3):
... yield None
... yield None
... return
>>> list(g())
[None, None, None, None]
Ensure that explicitly raising StopIteration acts like any other exception
in try/except, not like a return.
>>> def g():
... yield 1
... try:
... raise StopIteration
... except:
... yield 2
... yield 3
>>> list(g())
[1, 2, 3]
Next one was posted to c.l.py.
>>> def gcomb(x, k):
... "Generate all combinations of k elements from list x."
...
... if k > len(x):
... return
... if k == 0:
... yield []
... else:
... first, rest = x[0], x[1:]
... # A combination does or doesn't contain first.
... # If it does, the remainder is a k-1 comb of rest.
... for c in gcomb(rest, k-1):
... c.insert(0, first)
... yield c
... # If it doesn't contain first, it's a k comb of rest.
... for c in gcomb(rest, k):
... yield c
>>> seq = range(1, 5)
>>> for k in range(len(seq) + 2):
... print "%d-combs of %s:" % (k, seq)
... for c in gcomb(seq, k):
... print " ", c
0-combs of [1, 2, 3, 4]:
[]
1-combs of [1, 2, 3, 4]:
[1]
[2]
[3]
[4]
2-combs of [1, 2, 3, 4]:
[1, 2]
[1, 3]
[1, 4]
[2, 3]
[2, 4]
[3, 4]
3-combs of [1, 2, 3, 4]:
[1, 2, 3]
[1, 2, 4]
[1, 3, 4]
[2, 3, 4]
4-combs of [1, 2, 3, 4]:
[1, 2, 3, 4]
5-combs of [1, 2, 3, 4]:
From the Iterators list, about the types of these things.
>>> def g():
... yield 1
...
>>> type(g)
<type 'function'>
>>> i = g()
>>> type(i)
<type 'generator'>
>>> [s for s in dir(i) if not s.startswith('_')]
['gi_frame', 'gi_running', 'next']
>>> print i.next.__doc__
x.next() -> the next value, or raise StopIteration
>>> iter(i) is i
True
>>> import types
>>> isinstance(i, types.GeneratorType)
True
And more, added later.
>>> i.gi_running
0
>>> type(i.gi_frame)
<type 'frame'>
>>> i.gi_running = 42
Traceback (most recent call last):
...
TypeError: readonly attribute
>>> def g():
... yield me.gi_running
>>> me = g()
>>> me.gi_running
0
>>> me.next()
1
>>> me.gi_running
0
A clever union-find implementation from c.l.py, due to David Eppstein.
Sent: Friday, June 29, 2001 12:16 PM
To: python-list@python.org
Subject: Re: PEP 255: Simple Generators
>>> class disjointSet:
... def __init__(self, name):
... self.name = name
... self.parent = None
... self.generator = self.generate()
...
... def generate(self):
... while not self.parent:
... yield self
... for x in self.parent.generator:
... yield x
...
... def find(self):
... return self.generator.next()
...
... def union(self, parent):
... if self.parent:
... raise ValueError("Sorry, I'm not a root!")
... self.parent = parent
...
... def __str__(self):
... return self.name
>>> names = "ABCDEFGHIJKLM"
>>> sets = [disjointSet(name) for name in names]
>>> roots = sets[:]
>>> import random
>>> gen = random.WichmannHill(42)
>>> while 1:
... for s in sets:
... print "%s->%s" % (s, s.find()),
... print
... if len(roots) > 1:
... s1 = gen.choice(roots)
... roots.remove(s1)
... s2 = gen.choice(roots)
... s1.union(s2)
... print "merged", s1, "into", s2
... else:
... break
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
merged D into G
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
merged C into F
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
merged L into A
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
merged H into E
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
merged B into E
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
merged J into G
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
merged E into G
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
merged M into G
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
merged I into K
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
merged K into A
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
merged F into A
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
merged A into G
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
"""
# Emacs turd '
# Fun tests (for sufficiently warped notions of "fun").
fun_tests = """
Build up to a recursive Sieve of Eratosthenes generator.
>>> def firstn(g, n):
... return [g.next() for i in range(n)]
>>> def intsfrom(i):
... while 1:
... yield i
... i += 1
>>> firstn(intsfrom(5), 7)
[5, 6, 7, 8, 9, 10, 11]
>>> def exclude_multiples(n, ints):
... for i in ints:
... if i % n:
... yield i
>>> firstn(exclude_multiples(3, intsfrom(1)), 6)
[1, 2, 4, 5, 7, 8]
>>> def sieve(ints):
... prime = ints.next()
... yield prime
... not_divisible_by_prime = exclude_multiples(prime, ints)
... for p in sieve(not_divisible_by_prime):
... yield p
>>> primes = sieve(intsfrom(2))
>>> firstn(primes, 20)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]
Another famous problem: generate all integers of the form
2**i * 3**j * 5**k
in increasing order, where i,j,k >= 0. Trickier than it may look at first!
Try writing it without generators, and correctly, and without generating
3 internal results for each result output.
>>> def times(n, g):
... for i in g:
... yield n * i
>>> firstn(times(10, intsfrom(1)), 10)
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
>>> def merge(g, h):
... ng = g.next()
... nh = h.next()
... while 1:
... if ng < nh:
... yield ng
... ng = g.next()
... elif ng > nh:
... yield nh
... nh = h.next()
... else:
... yield ng
... ng = g.next()
... nh = h.next()
The following works, but is doing a whale of a lot of redundant work --
it's not clear how to get the internal uses of m235 to share a single
generator. Note that me_times2 (etc) each need to see every element in the
result sequence. So this is an example where lazy lists are more natural
(you can look at the head of a lazy list any number of times).
>>> def m235():
... yield 1
... 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
Don't print "too many" of these -- the implementation above is extremely
inefficient: each call of m235() leads to 3 recursive calls, and in
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 (<doctest test.test_generators.__test__.syntax[0]>, line 2)
>>> def f():
... yield 1
... return 22
Traceback (most recent call last):
..
SyntaxError: 'return' with argument inside generator (<doctest test.test_generators.__test__.syntax[1]>, 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 (<doctest test.test_generators.__test__.syntax[2]>, 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 (<doctest test.test_generators.__test__.syntax[4]>, 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 (<doctest test.test_generators.__test__.syntax[5]>, 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 'NoneType'>
>>> def f():
... if 0:
... class C:
... def __init__(self):
... yield 1
... def f(self):
... yield 2
>>> type(f())
<type 'NoneType'>
>>> 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 (<doctest test.test_generators.__test__.syntax[22]>, line 8)
This one caused a crash (see SF bug 567538):
>>> def f():
... for i in range(3):
... try:
... continue
... finally:
... yield i
...
>>> g = f()
>>> print g.next()
0
>>> print g.next()
1
>>> print g.next()
2
>>> print g.next()
Traceback (most recent call last):
StopIteration
"""
# 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 True True
1 2 True True
2 4 True True
3 8 True True
4 16 True True
5 32 True True
6 64 True True
7 128 True True
8 256 True True
9 512 True True
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|
+---+---+---+---+---+---+---+---+---+---+
"""
weakref_tests = """\
Generators are weakly referencable:
>>> import weakref
>>> def gen():
... yield 'foo!'
...
>>> wr = weakref.ref(gen)
>>> wr() is gen
True
>>> p = weakref.proxy(gen)
Generator-iterators are weakly referencable as well:
>>> gi = gen()
>>> wr = weakref.ref(gi)
>>> wr() is gi
True
>>> p = weakref.proxy(gi)
>>> list(p)
['foo!']
"""
__test__ = {"tut": tutorial_tests,
"pep": pep_tests,
"email": email_tests,
"fun": fun_tests,
"syntax": syntax_tests,
"conjoin": conjoin_tests,
"weakref": weakref_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(verbose=None):
from test import test_support, test_generators
test_support.run_doctest(test_generators, verbose)
# This part isn't needed for regrtest, but for running the test directly.
if __name__ == "__main__":
test_main(1)
|