Department of Electronics Engineering
National Chiao Tung University
Algorithms, Fall 2010
October 3, 2010
Iris Hui-Ru Jiang
DEE3504:
Sample Solutions of Homework #3
Exercises:
1. (10 pt) 3.1
2.
(30 pt) 3.6
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Department of Electronics Engineering
National Chiao Tung University
Algorithms, Fall 2010
3.
October 3, 2010
Iris Hui-Ru Jiang
(30 pt) Consider the given procedure. Initially, time is 0. The execution interval of DFS(u)
is defined as the interval from begin[u] to end[u]. Line 2 visits u’s neighbor in ascending
order of node index.
(a) Execute DFS(1) on the given graph. Draw the diagram to indicate the execution
interval for each node along time-line.
(b) Based on your drawing, please outline the relationship between any two execution
intervals.
DFS(u)
0. begin[u] = ++time
1. mark u as explored and add u to R
2. for each edge (u, v) incident to u do
3.
if (v is not marked as explored) then
4.
recursively invoke DFS(v)
5. end[u] = ++time
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Department of Electronics Engineering
National Chiao Tung University
Algorithms, Fall 2010
October 3, 2010
Iris Hui-Ru Jiang
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Department of Electronics Engineering
National Chiao Tung University
Algorithms, Fall 2010
4.
October 3, 2010
Iris Hui-Ru Jiang
(30 pt) Based on the procedure of testing strong connectivity of a graph, modify it to
generate all strong components of a given directed graph. Please use DFS if you need to
traverse this graph. Prove your procedure is correct and can be run in linear time.
TestSC(G)
Grev: reverse the direction of every edge in G
1. pick any node s in G
(u, v) in Grev if (v, u) in G
2. R = BFS(s, G)
3. Rrev = BFS(s, Grev)
4. if (R = V = Rrev) then return true else false
a
b
c
d
e
f
g
h
Strongly-Connected-Components(G)
1. call DFS(G) to compute finishing times f[u] for each vertex u
2. compute GT
3. call DFS(GT), but in the main loop of DFS, consider the vertices in order of decreasing
f[u] (as computed in line 1)
4.
output the vertices of each tree in the depth-first forest of step 3 as a separate strongly
connected component
Theorem 3: Strongly-connected-component(G) correctly computes the strongly connected
components of a directed graph G.
Proof: Induction on # of depth-first trees found in line 3 that the vertices of each tree form an
SCC.
 Hypothesis: the first k trees produced by line 3 are SCCs.
 Basis: k=0, trivial.
 Inductive step: k holds, consider (k+1)st tree.
 If u is the root of this tree and in SCC C, f[u] = f[C]>f[C’] for any other unvisited C’
 When the search visits u, other vertices in C are white and thus descendants of u.
 Any edges in GT leaving C are to other SCCs already been visited.
 Thus, no vertex in any SCC other than C will be a descendant of u during depth-first
search on GT.

 The vertices of the depth-first tree in GT rooted at u form exactly one SCC.
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