-
Notifications
You must be signed in to change notification settings - Fork 0
/
scc.py
286 lines (238 loc) · 8.92 KB
/
scc.py
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
def strongly_connected_components_path(vertices, edges):
"""
Find the strongly connected components of a directed graph.
Uses a recursive linear-time algorithm described by Gabow [1]_ to find all
strongly connected components of a directed graph.
Parameters
----------
vertices : iterable
A sequence or other iterable of vertices. Each vertex should be
hashable.
edges : mapping
Dictionary (or mapping) that maps each vertex v to an iterable of the
vertices w that are linked to v by a directed edge (v, w).
Returns
-------
components : iterator
An iterator that yields sets of vertices. Each set produced gives the
vertices of one strongly connected component.
Raises
------
RuntimeError
If the graph is deep enough that the algorithm exceeds Python's
recursion limit.
Notes
-----
The algorithm has running time proportional to the total number of vertices
and edges. It's practical to use this algorithm on graphs with hundreds of
thousands of vertices and edges.
The algorithm is recursive. Deep graphs may cause Python to exceed its
recursion limit.
`vertices` will be iterated over exactly once, and `edges[v]` will be
iterated over exactly once for each vertex `v`. `edges[v]` is permitted to
specify the same vertex multiple times, and it's permissible for `edges[v]`
to include `v` itself. (In graph-theoretic terms, loops and multiple edges
are permitted.)
References
----------
.. [1] Harold N. Gabow, "Path-based depth-first search for strong and
biconnected components," Inf. Process. Lett. 74 (2000) 107--114.
.. [2] Robert E. Tarjan, "Depth-first search and linear graph algorithms,"
SIAM J.Comput. 1 (2) (1972) 146--160.
Examples
--------
Example from Gabow's paper [1]_.
>>> vertices = [1, 2, 3, 4, 5, 6]
>>> edges = {1: [2, 3], 2: [3, 4], 3: [], 4: [3, 5], 5: [2, 6], 6: [3, 4]}
>>> for scc in strongly_connected_components_path(vertices, edges):
... print(scc)
...
set([3])
set([2, 4, 5, 6])
set([1])
Example from Tarjan's paper [2]_.
>>> vertices = [1, 2, 3, 4, 5, 6, 7, 8]
>>> edges = {1: [2], 2: [3, 8], 3: [4, 7], 4: [5],
... 5: [3, 6], 6: [], 7: [4, 6], 8: [1, 7]}
>>> for scc in strongly_connected_components_path(vertices, edges):
... print(scc)
...
set([6])
set([3, 4, 5, 7])
set([8, 1, 2])
"""
identified = set()
stack = []
index = {}
boundaries = []
def dfs(v):
index[v] = len(stack)
stack.append(v)
boundaries.append(index[v])
for w in edges[v]:
if w not in index:
# For Python >= 3.3, replace with "yield from dfs(w)"
for scc in dfs(w):
yield scc
elif w not in identified:
while index[w] < boundaries[-1]:
boundaries.pop()
if boundaries[-1] == index[v]:
boundaries.pop()
scc = set(stack[index[v]:])
del stack[index[v]:]
identified.update(scc)
yield scc
for v in vertices:
if v not in index:
# For Python >= 3.3, replace with "yield from dfs(v)"
for scc in dfs(v):
yield scc
def strongly_connected_components_tree(vertices, edges):
"""
Find the strongly connected components of a directed graph.
Uses a recursive linear-time algorithm described by Tarjan [2]_ to find all
strongly connected components of a directed graph.
Parameters
----------
vertices : iterable
A sequence or other iterable of vertices. Each vertex should be
hashable.
edges : mapping
Dictionary (or mapping) that maps each vertex v to an iterable of the
vertices w that are linked to v by a directed edge (v, w).
Returns
-------
components : iterator
An iterator that yields sets of vertices. Each set produced gives the
vertices of one strongly connected component.
Raises
------
RuntimeError
If the graph is deep enough that the algorithm exceeds Python's
recursion limit.
Notes
-----
The algorithm has running time proportional to the total number of vertices
and edges. It's practical to use this algorithm on graphs with hundreds of
thousands of vertices and edges.
The algorithm is recursive. Deep graphs may cause Python to exceed its
recursion limit.
`vertices` will be iterated over exactly once, and `edges[v]` will be
iterated over exactly once for each vertex `v`. `edges[v]` is permitted to
specify the same vertex multiple times, and it's permissible for `edges[v]`
to include `v` itself. (In graph-theoretic terms, loops and multiple edges
are permitted.)
References
----------
.. [1] Harold N. Gabow, "Path-based depth-first search for strong and
biconnected components," Inf. Process. Lett. 74 (2000) 107--114.
.. [2] Robert E. Tarjan, "Depth-first search and linear graph algorithms,"
SIAM J.Comput. 1 (2) (1972) 146--160.
Examples
--------
Example from Gabow's paper [1]_.
>>> vertices = [1, 2, 3, 4, 5, 6]
>>> edges = {1: [2, 3], 2: [3, 4], 3: [], 4: [3, 5], 5: [2, 6], 6: [3, 4]}
>>> for scc in strongly_connected_components_tree(vertices, edges):
... print(scc)
...
set([3])
set([2, 4, 5, 6])
set([1])
Example from Tarjan's paper [2]_.
>>> vertices = [1, 2, 3, 4, 5, 6, 7, 8]
>>> edges = {1: [2], 2: [3, 8], 3: [4, 7], 4: [5],
... 5: [3, 6], 6: [], 7: [4, 6], 8: [1, 7]}
>>> for scc in strongly_connected_components_tree(vertices, edges):
... print(scc)
...
set([6])
set([3, 4, 5, 7])
set([8, 1, 2])
"""
identified = set()
stack = []
index = {}
lowlink = {}
def dfs(v):
index[v] = len(stack)
stack.append(v)
lowlink[v] = index[v]
for w in edges[v]:
if w not in index:
# For Python >= 3.3, replace with "yield from dfs(w)"
for scc in dfs(w):
yield scc
lowlink[v] = min(lowlink[v], lowlink[w])
elif w not in identified:
lowlink[v] = min(lowlink[v], lowlink[w])
if lowlink[v] == index[v]:
scc = set(stack[index[v]:])
del stack[index[v]:]
identified.update(scc)
yield scc
for v in vertices:
if v not in index:
# For Python >= 3.3, replace with "yield from dfs(v)"
for scc in dfs(v):
yield scc
def strongly_connected_components_iterative(vertices, edges):
"""
This is a non-recursive version of strongly_connected_components_path.
See the docstring of that function for more details.
Examples
--------
Example from Gabow's paper [1]_.
>>> vertices = [1, 2, 3, 4, 5, 6]
>>> edges = {1: [2, 3], 2: [3, 4], 3: [], 4: [3, 5], 5: [2, 6], 6: [3, 4]}
>>> for scc in strongly_connected_components_iterative(vertices, edges):
... print(scc)
...
set([3])
set([2, 4, 5, 6])
set([1])
Example from Tarjan's paper [2]_.
>>> vertices = [1, 2, 3, 4, 5, 6, 7, 8]
>>> edges = {1: [2], 2: [3, 8], 3: [4, 7], 4: [5],
... 5: [3, 6], 6: [], 7: [4, 6], 8: [1, 7]}
>>> for scc in strongly_connected_components_iterative(vertices, edges):
... print(scc)
...
set([6])
set([3, 4, 5, 7])
set([8, 1, 2])
"""
identified = set()
stack = []
index = {}
boundaries = []
for v in vertices:
if v not in index:
to_do = [('VISIT', v)]
while to_do:
operation_type, v = to_do.pop()
if operation_type == 'VISIT':
index[v] = len(stack)
stack.append(v)
boundaries.append(index[v])
to_do.append(('POSTVISIT', v))
# We reverse to keep the search order identical to that of
# the recursive code; the reversal is not necessary for
# correctness, and can be omitted.
to_do.extend(
reversed([('VISITEDGE', w) for w in edges[v]]))
elif operation_type == 'VISITEDGE':
if v not in index:
to_do.append(('VISIT', v))
elif v not in identified:
while index[v] < boundaries[-1]:
boundaries.pop()
else:
# operation_type == 'POSTVISIT'
if boundaries[-1] == index[v]:
boundaries.pop()
scc = set(stack[index[v]:])
del stack[index[v]:]
identified.update(scc)
yield scc