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Com S 511: Homework #2
Due on Friday, September 18, 2009
Yuheng Long
1
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Com S 511 : Homework #2
Problem 1
Problem 1
(a) It is suﬃce to show that the ﬂow satisﬁes the ﬂow capacity constraint. Let
c
k
(
e
) be the capacity of
edge e in
G
k
. Clearly, 2
c
k

1
(
e
)
≤
c
k
(
e
)
≤
2
c
k

1
(
e
) + 1. Because,
f
*
k

1
is the maximum ﬂow in G, for
each edge e,
f
*
k

1
(
e
)
≤
c
k

1
(
e
). That is, 2
f
*
k

1
(
e
)
≤
2
c
k

1
(
e
)
≤
c
k
(
e
).
(b) Let (A,B) be a minumum cut in
G
k

1
, and
c
0
k

1
be its capacity. Then we have
v
(
f
*
k

1
) =
c
0
k

1
=
∑
e out of A
c
k

1
(
e
).
For any edge e, we have
c
k
(
e
)
≤
2
c
k

1
(
e
) + 1, thus
∑
e out of A
c
k
(
e
)
≤
2
∑
e out of A
(
c
k

1
(
e
) + 1)
≤
2
v
(
f
*
k

1
) +
∑
e out of A
1.
Therefore,
v
(
f
*
k
)
≤
∑
e out of A
c
k
(
e
)
≤
2
v
(
f
*
k

1
) +
m
and
v
(
f
*
k
)

2
v
(
f
*
k

1
)
≤
m
.
(c) We have
O
(
lg C
) rounds to compute
G
1
,G
2
,...,G
B
. In each round, we begin with 2
f
*
k

1
and by (b), it
could have at most m augmentations. In each augmentation, we use BFS/DFS to search a path, which
is O(m+n) (3.13). Also,
m
≥
n/
2. Thus the overall running time is
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