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cpython/Lib/test/test_generators.py
Tim Peters a0a6222509 Teach regrtest how to pass on doctest failure msgs. This is done via a 
horridly inefficient hack in regrtest's Compare class, but it's about as
clean as can be:  regrtest has to set up the Compare instance before
importing a test module, and by the time the module *is* imported it's too
late to change that decision.  The good news is that the more tests we
convert to unittest and doctest, the less the inefficiency here matters.
Even now there are few tests with large expected-output files (the new
cost here is a Python-level call per .write() when there's an expected-
output file).
2001-09-09 06:12:01 +00:00

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from __future__ import generators
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 + `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 leaves 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
1
>>> import types
>>> isinstance(i, types.GeneratorType)
1
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
>>> random.seed(42)
>>> while 1:
... for s in sets:
... print "%s->%s" % (s, s.find()),
... print
... if len(roots) > 1:
... s1 = random.choice(roots)
... roots.remove(s1)
... s2 = random.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
"""
# 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 (<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 '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 (<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(verbose=None):
import doctest, test_support, test_generators
if 0: # change to 1 to run forever (to check for leaks)
while 1:
doctest.master = None
test_support.run_doctest(test_generators, verbose)
print ".",
else:
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)
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