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 "", line 1, in ? File "", 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 "", line 1, in ? File "", line 3, in f StopIteration >>> g.next() # once stopped, can't be resumed Traceback (most recent call last): File "", line 1, in ? StopIteration "raise StopIteration" stops the generator too: >>> def f(): ... yield 1 ... return ... yield 2 # never reached ... >>> g = f() >>> g.next() 1 >>> g.next() Traceback (most recent call last): File "", line 1, in ? StopIteration >>> g.next() Traceback (most recent call last): File "", 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: 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 "", line 1, in ? File "", line 2, in g File "", 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 "", 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] A generator can't be resumed while it's already running. >>> def g(): ... i = me.next() ... yield i >>> me = g() >>> me.next() Traceback (most recent call last): ... File "", line 2, in g ValueError: generator already executing 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) >>> i = g() >>> type(i) >>> dir(i) ['gi_frame', 'gi_running', 'next'] >>> print i.next.__doc__ next() -- get 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) >>> i.gi_running = 42 Traceback (most recent call last): ... TypeError: object has read-only attributes >>> 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 ... ... def clear(self): ... self.__dict__.clear() >>> 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 clear(self): ... self.__dict__.clear() >>> 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 (, line 2) >>> def f(): ... yield 1 ... return 22 Traceback (most recent call last): ... SyntaxError: 'return' with argument inside generator (, 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 (, 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 (, 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 (, 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()) >>> def f(): ... if "": ... yield None >>> type(f()) >>> 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()) >>> def f(): ... if 0: ... def g(): ... yield 1 ... >>> type(f()) >>> def f(): ... if 0: ... class C: ... def __init__(self): ... yield 1 ... def f(self): ... yield 2 >>> type(f()) >>> def f(): ... if 0: ... return ... if 0: ... yield 2 >>> type(f()) >>> 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 (, 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. def flat_conjoin(gs): # rename to conjoin to run tests with this instead n = len(gs) values = [None] * n iters = [None] * n i = 0 while i >= 0: # Need a fresh iterator. if i >= n: yield values # Backtrack. i -= 1 else: iters[i] = gs[i]().next # Need next value from current iterator. while i >= 0: try: values[i] = iters[i]() except StopIteration: # Backtrack. i -= 1 else: # Start fresh at next level. i += 1 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!), so goes on at some length class Knights: def __init__(self, n, hard=0): self.n = n def coords2index(i, j): return i*n + j offsets = [( 1, 2), ( 2, 1), ( 2, -1), ( 1, -2), (-1, -2), (-2, -1), (-2, 1), (-1, 2)] succs = [] for i in range(n): for j in range(n): s = [coords2index(i+io, j+jo) for io, jo in offsets if 0 <= i+io < n and 0 <= j+jo < n] succs.append(s) del s del offsets free = [0] * n**2 # 0 if occupied, 1 if visited nexits = free[:] # number of free successors def decrexits(i0): # 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]: if free[i]: e = nexits[i] - 1 nexits[i] = e if e == 0: ne0 += 1 elif e == 1: ne1 += 1 return ne0 == 0 and ne1 < 2 def increxits(i0): for i in succs[i0]: if free[i]: nexits[i] += 1 # Generate the first move. def first(): if n < 1: return # Initialize board structures. for i in xrange(n**2): free[i] = 1 nexits[i] = len(succs[i]) # Since we're looking for a cycle, it doesn't matter where we # start. Starting in a corner makes the 2nd move easy. corner = coords2index(0, 0) free[corner] = 0 decrexits(corner) self.lastij = corner yield corner increxits(corner) free[corner] = 1 # Generate the second moves. def second(): corner = coords2index(0, 0) assert self.lastij == corner # i.e., we started in the corner if n < 3: return assert nexits[corner] == len(succs[corner]) == 2 assert coords2index(1, 2) in succs[corner] assert coords2index(2, 1) in succs[corner] # Only two choices. Whichever we pick, the other must be the # square picked on move n**2, 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, final = coords2index(i, j), coords2index(3-i, 3-j) assert free[this] and free[final] self.final = final nexits[final] += 1 # it has an exit back to 0,0 free[this] = 0 decrexits(this) self.lastij = this yield this increxits(this) free[this] = 1 nexits[final] -= 1 # Generate moves 3 thru n**2-1. def advance(): # If some successor has only one exit, must take it. # Else favor successors with fewer exits. candidates = [] for i in succs[self.lastij]: if free[i]: e = nexits[i] assert e > 0, "else decrexits() 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 decrexits(i): free[i] = 0 self.lastij = i yield i free[i] = 1 increxits(i) # Generate moves 3 thru n**2-1. Alternative version using a # stronger (but more expensive) heuristic to order successors. # Since the # of backtracking levels is n**2, a poor move early on # can take eons to undo. Smallest n for which this matters a lot is # n==52. def advance_hard(midpoint=(n-1)/2.0): # 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]: if free[i]: e = nexits[i] assert e > 0, "else decrexits() pruning flawed" if e == 1: candidates = [(e, 0, i)] break i1, j1 = divmod(i, n) d = (i1 - midpoint)**2 + (j1 - midpoint)**2 candidates.append((e, -d, i)) else: candidates.sort() for e, d, i in candidates: if i != self.final: if decrexits(i): free[i] = 0 self.lastij = i yield i free[i] = 1 increxits(i) # Generate the last move. def last(): assert self.final in succs[self.lastij] yield self.final if n <= 1: self.rowgenerators = [first] else: self.rowgenerators = [first, second] + \ [hard and advance_hard or advance] * (n**2 - 3) + \ [last] # Generate solutions. def solve(self): for x in conjoin(self.rowgenerators): yield x def printsolution(self, x): n = self.n assert len(x) == n**2 w = len(str(n**2 + 1)) format = "%" + str(w) + "d" squares = [[None] * n for i in range(n)] k = 1 for i in x: i1, j1 = divmod(i, n) squares[i1][j1] = format % k k += 1 sep = "+" + ("-" * w + "+") * n print sep for i in range(n): 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) >>> 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: # Temporary block to help track down leaks. So far, the blame # fell mostly on doctest. Later: the only leaks remaining are # in fun_tests, and only if you comment out the two LazyList.clear() # calls. for i in range(10000): doctest.master = None doctest.testmod(test_generators) else: doctest.testmod(test_generators) # This part isn't needed for regrtest, but for running the test directly. if __name__ == "__main__": test_main()