summaryrefslogtreecommitdiffstats
path: root/Lib/test/test_generators.py
blob: 2476a275f9d94650ef2d2efa2fad0e241c1fb97b (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009
1010
1011
1012
1013
1014
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071
1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
1117
1118
1119
1120
1121
1122
1123
1124
1125
1126
1127
1128
1129
1130
1131
1132
1133
1134
1135
1136
1137
1138
1139
1140
1141
1142
1143
1144
1145
1146
1147
1148
1149
1150
1151
1152
1153
1154
1155
1156
1157
1158
1159
1160
1161
1162
1163
1164
1165
1166
1167
1168
1169
1170
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
1192
1193
1194
1195
1196
1197
1198
1199
1200
1201
1202
1203
1204
1205
1206
1207
1208
1209
1210
1211
1212
1213
1214
1215
1216
1217
1218
1219
1220
1221
1222
1223
1224
1225
1226
1227
1228
1229
1230
1231
1232
1233
1234
1235
1236
1237
1238
1239
1240
1241
1242
1243
1244
1245
1246
1247
1248
1249
1250
1251
1252
1253
1254
1255
1256
1257
1258
1259
1260
1261
1262
1263
1264
1265
1266
1267
1268
1269
1270
1271
1272
1273
1274
1275
1276
1277
1278
1279
1280
1281
1282
1283
1284
1285
1286
1287
1288
1289
1290
1291
1292
1293
1294
1295
1296
1297
1298
1299
1300
1301
1302
1303
1304
1305
1306
1307
1308
1309
1310
1311
1312
1313
1314
1315
1316
1317
1318
1319
1320
1321
1322
1323
1324
1325
1326
1327
1328
1329
1330
1331
1332
1333
1334
1335
1336
1337
1338
1339
1340
1341
1342
1343
1344
1345
1346
1347
1348
1349
1350
1351
1352
1353
1354
1355
1356
1357
1358
1359
1360
1361
1362
1363
1364
1365
1366
1367
1368
1369
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: 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]

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 "<string>", 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)
<type 'function'>
>>> i = g()
>>> type(i)
<type 'generator'>

XXX dir(object) *generally* doesn't return useful stuff in descr-branch.
>>> dir(i)
[]

Was hoping to see this instead:
['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 'None'>

>>> def f():
...     if 0:
...         class C:
...             def __init__(self):
...                 yield 1
...             def f(self):
...                 yield 2
>>> type(f())
<type 'None'>

>>> def f():
...     if 0:
...         return
...     if 0:
...         yield 2
>>> type(f())
<type 'generator'>


>>> def f():
...     if 0:
...         lambda x:  x        # shouldn't trigger here
...         return              # or here
...         def f(i):
...             return 2*i      # or here
...         if 0:
...             return 3        # but *this* sucks (line 8)
...     if 0:
...         yield 2             # because it's a generator
Traceback (most recent call last):
SyntaxError: 'return' with argument inside generator (<string>, line 8)
"""

# conjoin is a simple backtracking generator, named in honor of Icon's
# "conjunction" control structure.  Pass a list of no-argument functions
# that return iterable objects.  Easiest to explain by example:  assume the
# function list [x, y, z] is passed.  Then conjoin acts like:
#
# def g():
#     values = [None] * 3
#     for values[0] in x():
#         for values[1] in y():
#             for values[2] in z():
#                 yield values
#
# So some 3-lists of values *may* be generated, each time we successfully
# get into the innermost loop.  If an iterator fails (is exhausted) before
# then, it "backtracks" to get the next value from the nearest enclosing
# iterator (the one "to the left"), and starts all over again at the next
# slot (pumps a fresh iterator).  Of course this is most useful when the
# iterators have side-effects, so that which values *can* be generated at
# each slot depend on the values iterated at previous slots.

def conjoin(gs):

    values = [None] * len(gs)

    def gen(i, values=values):
        if i >= len(gs):
            yield values
        else:
            for values[i] in gs[i]():
                for x in gen(i+1):
                    yield x

    for x in gen(0):
        yield x

# That works fine, but recursing a level and checking i against len(gs) for
# each item produced is inefficient.  By doing manual loop unrolling across
# generator boundaries, it's possible to eliminate most of that overhead.
# This isn't worth the bother *in general* for generators, but conjoin() is
# a core building block for some CPU-intensive generator applications.

def conjoin(gs):

    n = len(gs)
    values = [None] * n

    # Do one loop nest at time recursively, until the # of loop nests
    # remaining is divisible by 3.

    def gen(i, values=values):
        if i >= n:
            yield values

        elif (n-i) % 3:
            ip1 = i+1
            for values[i] in gs[i]():
                for x in gen(ip1):
                    yield x

        else:
            for x in _gen3(i):
                yield x

    # Do three loop nests at a time, recursing only if at least three more
    # remain.  Don't call directly:  this is an internal optimization for
    # gen's use.

    def _gen3(i, values=values):
        assert i < n and (n-i) % 3 == 0
        ip1, ip2, ip3 = i+1, i+2, i+3
        g, g1, g2 = gs[i : ip3]

        if ip3 >= n:
            # These are the last three, so we can yield values directly.
            for values[i] in g():
                for values[ip1] in g1():
                    for values[ip2] in g2():
                        yield values

        else:
            # At least 6 loop nests remain; peel off 3 and recurse for the
            # rest.
            for values[i] in g():
                for values[ip1] in g1():
                    for values[ip2] in g2():
                        for x in _gen3(ip3):
                            yield x

    for x in gen(0):
        yield x

# And one more approach:  For backtracking apps like the Knight's Tour
# solver below, the number of backtracking levels can be enormous (one
# level per square, for the Knight's Tour, so that e.g. a 100x100 board
# needs 10,000 levels).  In such cases Python is likely to run out of
# stack space due to recursion.  So here's a recursion-free version of
# conjoin too.
# NOTE WELL:  This allows large problems to be solved with only trivial
# demands on stack space.  Without explicitly resumable generators, this is
# much harder to achieve.  OTOH, this is much slower (up to a factor of 2)
# than the fancy unrolled recursive conjoin.

def flat_conjoin(gs):  # rename to conjoin to run tests with this instead
    n = len(gs)
    values = [None] * n
    iters  = [None] * n
    _StopIteration = StopIteration  # make local because caught a *lot*
    i = 0
    while 1:
        # Descend.
        try:
            while i < n:
                it = iters[i] = gs[i]().next
                values[i] = it()
                i += 1
        except _StopIteration:
            pass
        else:
            assert i == n
            yield values

        # Backtrack until an older iterator can be resumed.
        i -= 1
        while i >= 0:
            try:
                values[i] = iters[i]()
                # Success!  Start fresh at next level.
                i += 1
                break
            except _StopIteration:
                # Continue backtracking.
                i -= 1
        else:
            assert i < 0
            break

# A conjoin-based N-Queens solver.

class Queens:
    def __init__(self, n):
        self.n = n
        rangen = range(n)

        # Assign a unique int to each column and diagonal.
        # columns:  n of those, range(n).
        # NW-SE diagonals: 2n-1 of these, i-j unique and invariant along
        # each, smallest i-j is 0-(n-1) = 1-n, so add n-1 to shift to 0-
        # based.
        # NE-SW diagonals: 2n-1 of these, i+j unique and invariant along
        # each, smallest i+j is 0, largest is 2n-2.

        # For each square, compute a bit vector of the columns and
        # diagonals it covers, and for each row compute a function that
        # generates the possiblities for the columns in that row.
        self.rowgenerators = []
        for i in rangen:
            rowuses = [(1L << j) |                  # column ordinal
                       (1L << (n + i-j + n-1)) |    # NW-SE ordinal
                       (1L << (n + 2*n-1 + i+j))    # NE-SW ordinal
                            for j in rangen]

            def rowgen(rowuses=rowuses):
                for j in rangen:
                    uses = rowuses[j]
                    if uses & self.used == 0:
                        self.used |= uses
                        yield j
                        self.used &= ~uses

            self.rowgenerators.append(rowgen)

    # Generate solutions.
    def solve(self):
        self.used = 0
        for row2col in conjoin(self.rowgenerators):
            yield row2col

    def printsolution(self, row2col):
        n = self.n
        assert n == len(row2col)
        sep = "+" + "-+" * n
        print sep
        for i in range(n):
            squares = [" " for j in range(n)]
            squares[row2col[i]] = "Q"
            print "|" + "|".join(squares) + "|"
            print sep

# A conjoin-based Knight's Tour solver.  This is pretty sophisticated
# (e.g., when used with flat_conjoin above, and passing hard=1 to the
# constructor, a 200x200 Knight's Tour was found quickly -- note that we're
# creating 10s of thousands of generators then!), and is lengthy.

class Knights:
    def __init__(self, m, n, hard=0):
        self.m, self.n = m, n

        # solve() will set up succs[i] to be a list of square #i's
        # successors.
        succs = self.succs = []

        # Remove i0 from each of its successor's successor lists, i.e.
        # successors can't go back to i0 again.  Return 0 if we can
        # detect this makes a solution impossible, else return 1.

        def remove_from_successors(i0, len=len):
            # If we remove all exits from a free square, we're dead:
            # even if we move to it next, we can't leave it again.
            # If we create a square with one exit, we must visit it next;
            # else somebody else will have to visit it, and since there's
            # only one adjacent, there won't be a way to leave it again.
            # Finelly, if we create more than one free square with a
            # single exit, we can only move to one of them next, leaving
            # the other one a dead end.
            ne0 = ne1 = 0
            for i in succs[i0]:
                s = succs[i]
                s.remove(i0)
                e = len(s)
                if e == 0:
                    ne0 += 1
                elif e == 1:
                    ne1 += 1
            return ne0 == 0 and ne1 < 2

        # Put i0 back in each of its successor's successor lists.

        def add_to_successors(i0):
            for i in succs[i0]:
                succs[i].append(i0)

        # Generate the first move.
        def first():
            if m < 1 or n < 1:
                return

            # Since we're looking for a cycle, it doesn't matter where we
            # start.  Starting in a corner makes the 2nd move easy.
            corner = self.coords2index(0, 0)
            remove_from_successors(corner)
            self.lastij = corner
            yield corner
            add_to_successors(corner)

        # Generate the second moves.
        def second():
            corner = self.coords2index(0, 0)
            assert self.lastij == corner  # i.e., we started in the corner
            if m < 3 or n < 3:
                return
            assert len(succs[corner]) == 2
            assert self.coords2index(1, 2) in succs[corner]
            assert self.coords2index(2, 1) in succs[corner]
            # Only two choices.  Whichever we pick, the other must be the
            # square picked on move m*n, as it's the only way to get back
            # to (0, 0).  Save its index in self.final so that moves before
            # the last know it must be kept free.
            for i, j in (1, 2), (2, 1):
                this  = self.coords2index(i, j)
                final = self.coords2index(3-i, 3-j)
                self.final = final

                remove_from_successors(this)
                succs[final].append(corner)
                self.lastij = this
                yield this
                succs[final].remove(corner)
                add_to_successors(this)

        # Generate moves 3 thru m*n-1.
        def advance(len=len):
            # If some successor has only one exit, must take it.
            # Else favor successors with fewer exits.
            candidates = []
            for i in succs[self.lastij]:
                e = len(succs[i])
                assert e > 0, "else remove_from_successors() pruning flawed"
                if e == 1:
                    candidates = [(e, i)]
                    break
                candidates.append((e, i))
            else:
                candidates.sort()

            for e, i in candidates:
                if i != self.final:
                    if remove_from_successors(i):
                        self.lastij = i
                        yield i
                    add_to_successors(i)

        # Generate moves 3 thru m*n-1.  Alternative version using a
        # stronger (but more expensive) heuristic to order successors.
        # Since the # of backtracking levels is m*n, a poor move early on
        # can take eons to undo.  Smallest square board for which this
        # matters a lot is 52x52.
        def advance_hard(vmid=(m-1)/2.0, hmid=(n-1)/2.0, len=len):
            # If some successor has only one exit, must take it.
            # Else favor successors with fewer exits.
            # Break ties via max distance from board centerpoint (favor
            # corners and edges whenever possible).
            candidates = []
            for i in succs[self.lastij]:
                e = len(succs[i])
                assert e > 0, "else remove_from_successors() pruning flawed"
                if e == 1:
                    candidates = [(e, 0, i)]
                    break
                i1, j1 = self.index2coords(i)
                d = (i1 - vmid)**2 + (j1 - hmid)**2
                candidates.append((e, -d, i))
            else:
                candidates.sort()

            for e, d, i in candidates:
                if i != self.final:
                    if remove_from_successors(i):
                        self.lastij = i
                        yield i
                    add_to_successors(i)

        # Generate the last move.
        def last():
            assert self.final in succs[self.lastij]
            yield self.final

        if m*n < 4:
            self.squaregenerators = [first]
        else:
            self.squaregenerators = [first, second] + \
                [hard and advance_hard or advance] * (m*n - 3) + \
                [last]

    def coords2index(self, i, j):
        assert 0 <= i < self.m
        assert 0 <= j < self.n
        return i * self.n + j

    def index2coords(self, index):
        assert 0 <= index < self.m * self.n
        return divmod(index, self.n)

    def _init_board(self):
        succs = self.succs
        del succs[:]
        m, n = self.m, self.n
        c2i = self.coords2index

        offsets = [( 1,  2), ( 2,  1), ( 2, -1), ( 1, -2),
                   (-1, -2), (-2, -1), (-2,  1), (-1,  2)]
        rangen = range(n)
        for i in range(m):
            for j in rangen:
                s = [c2i(i+io, j+jo) for io, jo in offsets
                                     if 0 <= i+io < m and
                                        0 <= j+jo < n]
                succs.append(s)

    # Generate solutions.
    def solve(self):
        self._init_board()
        for x in conjoin(self.squaregenerators):
            yield x

    def printsolution(self, x):
        m, n = self.m, self.n
        assert len(x) == m*n
        w = len(str(m*n))
        format = "%" + str(w) + "d"

        squares = [[None] * n for i in range(m)]
        k = 1
        for i in x:
            i1, j1 = self.index2coords(i)
            squares[i1][j1] = format % k
            k += 1

        sep = "+" + ("-" * w + "+") * n
        print sep
        for i in range(m):
            row = squares[i]
            print "|" + "|".join(row) + "|"
            print sep

conjoin_tests = """

Generate the 3-bit binary numbers in order.  This illustrates dumbest-
possible use of conjoin, just to generate the full cross-product.

>>> for c in conjoin([lambda: iter((0, 1))] * 3):
...     print c
[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]

For efficiency in typical backtracking apps, conjoin() yields the same list
object each time.  So if you want to save away a full account of its
generated sequence, you need to copy its results.

>>> def gencopy(iterator):
...     for x in iterator:
...         yield x[:]

>>> for n in range(10):
...     all = list(gencopy(conjoin([lambda: iter((0, 1))] * n)))
...     print n, len(all), all[0] == [0] * n, all[-1] == [1] * n
0 1 1 1
1 2 1 1
2 4 1 1
3 8 1 1
4 16 1 1
5 32 1 1
6 64 1 1
7 128 1 1
8 256 1 1
9 512 1 1

And run an 8-queens solver.

>>> q = Queens(8)
>>> LIMIT = 2
>>> count = 0
>>> for row2col in q.solve():
...     count += 1
...     if count <= LIMIT:
...         print "Solution", count
...         q.printsolution(row2col)
Solution 1
+-+-+-+-+-+-+-+-+
|Q| | | | | | | |
+-+-+-+-+-+-+-+-+
| | | | |Q| | | |
+-+-+-+-+-+-+-+-+
| | | | | | | |Q|
+-+-+-+-+-+-+-+-+
| | | | | |Q| | |
+-+-+-+-+-+-+-+-+
| | |Q| | | | | |
+-+-+-+-+-+-+-+-+
| | | | | | |Q| |
+-+-+-+-+-+-+-+-+
| |Q| | | | | | |
+-+-+-+-+-+-+-+-+
| | | |Q| | | | |
+-+-+-+-+-+-+-+-+
Solution 2
+-+-+-+-+-+-+-+-+
|Q| | | | | | | |
+-+-+-+-+-+-+-+-+
| | | | | |Q| | |
+-+-+-+-+-+-+-+-+
| | | | | | | |Q|
+-+-+-+-+-+-+-+-+
| | |Q| | | | | |
+-+-+-+-+-+-+-+-+
| | | | | | |Q| |
+-+-+-+-+-+-+-+-+
| | | |Q| | | | |
+-+-+-+-+-+-+-+-+
| |Q| | | | | | |
+-+-+-+-+-+-+-+-+
| | | | |Q| | | |
+-+-+-+-+-+-+-+-+

>>> print count, "solutions in all."
92 solutions in all.

And run a Knight's Tour on a 10x10 board.  Note that there are about
20,000 solutions even on a 6x6 board, so don't dare run this to exhaustion.

>>> k = Knights(10, 10)
>>> LIMIT = 2
>>> count = 0
>>> for x in k.solve():
...     count += 1
...     if count <= LIMIT:
...         print "Solution", count
...         k.printsolution(x)
...     else:
...         break
Solution 1
+---+---+---+---+---+---+---+---+---+---+
|  1| 58| 27| 34|  3| 40| 29| 10|  5|  8|
+---+---+---+---+---+---+---+---+---+---+
| 26| 35|  2| 57| 28| 33|  4|  7| 30| 11|
+---+---+---+---+---+---+---+---+---+---+
| 59|100| 73| 36| 41| 56| 39| 32|  9|  6|
+---+---+---+---+---+---+---+---+---+---+
| 74| 25| 60| 55| 72| 37| 42| 49| 12| 31|
+---+---+---+---+---+---+---+---+---+---+
| 61| 86| 99| 76| 63| 52| 47| 38| 43| 50|
+---+---+---+---+---+---+---+---+---+---+
| 24| 75| 62| 85| 54| 71| 64| 51| 48| 13|
+---+---+---+---+---+---+---+---+---+---+
| 87| 98| 91| 80| 77| 84| 53| 46| 65| 44|
+---+---+---+---+---+---+---+---+---+---+
| 90| 23| 88| 95| 70| 79| 68| 83| 14| 17|
+---+---+---+---+---+---+---+---+---+---+
| 97| 92| 21| 78| 81| 94| 19| 16| 45| 66|
+---+---+---+---+---+---+---+---+---+---+
| 22| 89| 96| 93| 20| 69| 82| 67| 18| 15|
+---+---+---+---+---+---+---+---+---+---+
Solution 2
+---+---+---+---+---+---+---+---+---+---+
|  1| 58| 27| 34|  3| 40| 29| 10|  5|  8|
+---+---+---+---+---+---+---+---+---+---+
| 26| 35|  2| 57| 28| 33|  4|  7| 30| 11|
+---+---+---+---+---+---+---+---+---+---+
| 59|100| 73| 36| 41| 56| 39| 32|  9|  6|
+---+---+---+---+---+---+---+---+---+---+
| 74| 25| 60| 55| 72| 37| 42| 49| 12| 31|
+---+---+---+---+---+---+---+---+---+---+
| 61| 86| 99| 76| 63| 52| 47| 38| 43| 50|
+---+---+---+---+---+---+---+---+---+---+
| 24| 75| 62| 85| 54| 71| 64| 51| 48| 13|
+---+---+---+---+---+---+---+---+---+---+
| 87| 98| 89| 80| 77| 84| 53| 46| 65| 44|
+---+---+---+---+---+---+---+---+---+---+
| 90| 23| 92| 95| 70| 79| 68| 83| 14| 17|
+---+---+---+---+---+---+---+---+---+---+
| 97| 88| 21| 78| 81| 94| 19| 16| 45| 66|
+---+---+---+---+---+---+---+---+---+---+
| 22| 91| 96| 93| 20| 69| 82| 67| 18| 15|
+---+---+---+---+---+---+---+---+---+---+
"""

__test__ = {"tut":      tutorial_tests,
            "pep":      pep_tests,
            "email":    email_tests,
            "fun":      fun_tests,
            "syntax":   syntax_tests,
            "conjoin":  conjoin_tests}

# Magic test name that regrtest.py invokes *after* importing this module.
# This worms around a bootstrap problem.
# Note that doctest and regrtest both look in sys.argv for a "-v" argument,
# so this works as expected in both ways of running regrtest.
def test_main():
    import doctest, test_generators
    if 0:   # change to 1 to run forever (to check for leaks)
        while 1:
            doctest.master = None
            doctest.testmod(test_generators)
            print ".",
    else:
        doctest.testmod(test_generators)

# This part isn't needed for regrtest, but for running the test directly.
if __name__ == "__main__":
    test_main()