1. 22.1-2 Give an adjacency-list representation for a complete binary tree on 7 vertices.
Give an equivalent adjacency-matrix representation. Assume that vertices are numbered
from 1 to 7 as in a binary heap.
1
2
1
2
4
3
3
5
7
6
4
5
6
/
/
/
7
/
2
3/
4
6
5/
7/
5
0
1
0
0
0
0
0
6
0
0
1
0
0
0
0
Adjacency Matrix:
1
2
3
4
5
6
7
1
0
0
0
0
0
0
0
2
1
0
0
0
0
0
0
3
1
0
0
0
0
0
0
4
0
1
0
0
0
0
0
7
0
0
1
0
0
0
0
2. 22.1-5 The square of a directed graph G = (V, E) is the graph G2 = (V, E2) such that
(u,w)  E2 if and only if for some v  V, both (u,v)  E and (v, w)  E. That is, G2 contains
an edge between u and w whenever G contains a path with exactly two edges between u and
w. Describe efficient algorithms for computing G2 from G for both the adjacency-list and
adjacency-matrix representations of G. Analyze the running times of your algorithms.
G2 for an adjacency matrix: - Computing G2 may be done in V3 time by matrix multiplication:
for i = 1 to V
for j = 1 to V {
G2[i][j] = 0;
for k = 1 to V
if (g[i][k] == 1 && g[k][j] == 1) {
G2[i][j] == 1;
break;
}
}
G2 for an adjacency list:
Procedure G-Square (V[G], E[G])
V[G2]  V[G]
for each u  V[G]
for each v  Adj[u]
for each w  Adj[v]
E[G2]  {(u, w)}  E[G2]
3
Run time = O(V )
3. 22.2-1 Show the d and  values that result from running breadth-first search on the
directed graph of Figure 22.2(a), using vertex 3 as the source.
5/3/4
-/-/1
2
3 1/0/-
4
5
6
4/2/5
2/1/3
3/1/3
4. 22.2-3 What is the running time of BFS if its input graph is represented by an adjacency
matrix and the algorithm is modified to handle this form of input?
Each vertex can be explored once and its adjacent vertices must be determined too. This takes
(V2) time.
5. Do a DFS on figure 22.6 (p.548). Classify each edge based on the DFS tree you determine.
1/16
2/7
s
3/6
T
q
w
4/5
r
T
F
B
T
v
T
17/20
B
t
8/15
T
T
C
y
9/12 x
T
T
C
u
18/19
13/14
B
z
10/11
6. Find the strongly connected components in figure 22.6.
From 5., the first DFS gives the list R U Q T Y X Z S U W (reverse order of turning black)
5/10
r
B
C
T
17/18
T
7/8
C
16/19
u
3/4
C
y
11/14 x
w
C
C
t
T
B
v
q
C
15/20
s
1/2
T
6/9
T
B
z
12/13
Strongly connected components are {s, w, v}, {q, y, t}, {x, z}, {r}, {u}
7. 22.4-1 Show the ordering of vertices produced by TOPOLOGICAL-SORT when it is
run on the dag of Figure 22.8, under the assumption of Exercise 22.3-2.
1/20
21/26
m
22/25
T
n
T
C
T
C
C
q
C
C
r
T
p
C
s
2/5
T
C
o
6/19
F
27/28
23/24
T
C
T
C
t
3/4
u
v
7/8
15/16
w
11/14
10/17
T
x
T
T
T
y
9/18
z
12/13
p n o s m r y v x w z u q t
8. 22.5-1 How can the number of strongly connected components of a graph change if a
new edge is added?
The number of strongly connected components can be reduced.
9. 22.5-3 Professor Deaver claims that the algorithm for strongly connected components
can be simplified by using the original (instead of the transpose) graph in the second depthfirst search and scanning the vertices in order of increasing finishing times. Is the professor
correct?
Consider
1
0
2
For this algorithm, the first DFS will give a list 1 2 0 for the second DFS. All vertices will be
incorrectly reported to be in the same SCC.
For the CLRS algorithm, the first DFS will give a list 0 2 1 for the second DFS. After reversing
edges, the correct SCCs {0, 1} and {2} will be reported.
10. 22.5-4 Prove that for any directed graph G, we have ((GT)SCC)T = GSCC. That is, the
transpose of the component graph of GT is the same as the component graph of G.
Since the strongly connected relationship is an equivalence relation, G and GT will always have
the same strongly connected components. If two vertices X and Y are not strongly connected,
then there is a unique path from X to Y in G iff there is a unique path from Y to X in GT.
A similar property will also hold for the component graph and transpose of the component graph,
i.e. If two vertices X and Y are not strongly connected in G, then here is a unique path from
SCC(G,X) to SCC(G,Y) in (G)SCC iff there is a unique path from SCC(GT,Y) to SCC(GT,X) in
(GT)SCC).
11. 23.1-1
Let (u, v) be a minimum-weight edge in a graph G. Show that (u, v) belongs to some
minimum spanning tree of G.
Let A be a subset of some MST T such that (u, v)  A. To choose an edge to be added to A, all
the edges on the cut are considered and an edge with lowest weight is selected. Since (u, v) is the
minimum weight edge in the graph G, it gets selected on some cut.
12. 24.3-1 Run Dijkstra’s algorithm on the directed graph of Figure 24.2, first using vertex
s as the source and then using vertex z as the source. In the style of Figure 24.6, show the d
and  values and the vertices in set S after each iteration of the while loop.
Black vertices are in the set S. The black, thick arrows are the values of  and the values of d are
included are listed inside each node.
T
A.
6
∞
3
2
T
∞
4
1
5
2
6
∞
Y
3
0
2
Y
S 5
2
6
5
Y
7
3
Z
X
6
2
2
7
6
5
Z
9
4
1
S 5
Y
E.
∞
3
3
0
11
7
T
9
4
1
2
6
5
D.
∞
4
1
3
2
6
3
Z
6
X
S 5
X
3
3
7
∞
T
C.
0
B.
3
0
S
X
11
3
Z
F.
T
3
0
6
3
2
4
1
S 5
6
5
Y
3
X
T
9
3
2
3
7
0
2
X
6
4
1
S 5
11
5
Z
Y
9
6
3
2
7
11
Z
Using Vertex Z as the source.
T
A.
X
6
∞
3
T
∞
2
4
1
5
2
7
6
∞
Y
3
3
6
2
4
1
2
7
6
∞
Z
Y
C.
7
5
S
0
X
∞
3
∞
S
B.
0
3
Z
D.
T
3
3
S
6
6
2
4
1
X
T
7
6
2
7
5
6
8
Y
3
3
3
S
2
X
6
7
4
1
2
7
5
0
8
Z
Y
6
3
0
Z
T
E.
X
6
6
3
2
7
4
1
2
7
5
S
6
8
Y
G.
3
Z
6
3
2
2
7
5
6
8
Y
3
0
Z
6
6
2
7
4
1
2
7
5
Y
7
4
1
X
6
8
X
6
3
S
0
T
3
T
3
3
S
F.
3
0
Z
13. a. Determine the transitive closure of the following Boolean matrix by using Warshall’s
algorithm.
1 0 1
1 0
0 0 1
1 0
1 0 0
0 0
1 0 1
1 1
0 0 0
1 0
T(1) =
T(2)=
1
0
1
1
0
0
0
0
0
0
1
1
1
1
0
1
1
1
1
1
0
0
0
1
0
1
1
1
1
0
0
0
0
0
0
1
1
1
1
0
1
1
1
1
1
0
0
0
1
0
1
1
1
1
1
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
T(3)=
T(5)=
1
0
1
1
0
0
0
0
0
0
1
1
1
1
0
1
1
1
1
1
0
0
0
1
0
1
1
1
1
1
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
T(4)=
b. Convert the matrix to indicate successors and use the version of Warshall’s algorithm
that allows path tracing.
0
2
0
0
3
-1
-1
-1
-1
-1
2
2
0
2
3
3
3
0
3
3
3
3
0
4
3
14. 26.1-2
Prove that for any vertex v other than the source or sink, the total positive flow entering v
must equal the total positive flow leaving v.
11/16
10
s
8/13
12/12
v1
1/14
v3
15/20
7/7
4/9
v2
t
4/4
v4
11/14
Capacity constraint: For all u, v  V, we require f(u, v) <= c(u, v)
In the above figure it can be seen that for each edge f(u, v) <= c(u, v) for edge (s, v1), f(u, v1) =
11 and c(u, v) = 16. 11 <= 16. Similarly for other edges.
Skew Symmetry: For all u, v  V, we require f(u, v) <= -f(u, v)
It can be seen for each edge (u, v) for flow (u, v) signifies a negative flow –f(v, u) from (v, u)
Flow conservation: For all u, v  V – {s, t}, we require Σ f(u, v) = 0
It can be seen for each vertex v excluding s and t sum of the flows into it is equal to the flow out
of it and hence flow is conserved. For vertex v1, f(s, v1) + f(v2, v1) = f(v1, v3). 11 + 1 = 12.
15. 26.2-1
In Figure 26.1(b), which is the flow across the cut ({s, v2, v4}, {v1, v3, t})? What is the
capacity of this cut?
12/12
v1
v3
11/16
15/20
10
s
8/13
1/14
v2
7/7
4/9
v4
t
4/4
11/14
Net flow across the cut is f(s, v1) + f(v2,v1) + f(v2,v3) + f(v3,v4) + f(v4,t) = 11 + 1 + -4 + 7 + 4
= 19 and its capacity is = 16 + 14 +7 + 4 = 41
16
26.2-2
12
v1
v3
10
s
4
12 / 12
v1
20
16
v3
12/ 20
12 / 16
7
t
9
10
s
13
4
t
7
9
13
v2
4
v4
v2
14
14
12
v1
4
v4
v3
12 / 12
v1
8
12
v3
12 / 20
12 / 16
4
12
10
s
4
7
t
9
13
v2
4
7
v2
4 / 14
12
v3
4
12 / 12
v1
8
12
v3
19 / 20
12 / 16
12
10
s
4 /4
v4
14
v1
t
9
4 / 13
4
v4
10
s
4
7
9
9
s
t
4
7 /7
v2
v2
4
v4
v4
11 / 14
4
12
v1
v3
1
12
4
19
10
4
7
s
t
9
2
3
11
t
9
11 / 13
10
4
10
v2
4
v4
11
17 26.3-1
2
1
s
1
1
1
1
1
1
3
1
4 1
1
5
1
6
1
1
7
1
1
1
1
8
9
1
Augmenting path
t
S  1  6  t
4 /4
Flow N/W:
2
1
s
1/ 1
1
1/ 1
1
1
1
3
1
4
1
6
1
7
1
1
1
1/ 1
1
t
1
8
1
1
9
5
Residual N/W:
1
1
2
1
s
1
1
1
3
1
4 1
1
1
1
6
1
1
7
1
1
1
1
8
1
2
1
s
1
1
1
3
1
4 1
1
1
1
6
1
1
7
1
1
1
Augmenting path
t
1
8
1
9
5
Final Residual graph:
1
1
s
2
1
1
1
1
1
3
4 1
1
5
1
1
6
1
1
7
1
1
1
S  2  8  t
1
9
5
1
Augmenting path
t
1
8
9
1
t
S  3  7 t
18 15.2-1
Finding the optimal parenthesization of a matrix-chain product whose sequence of dimensions is
<5, 10, 3, 12, 5, 50, 6>
The m table
0
150 330 405 1655 2010
0
0
360 330 2430 1950
0
0
0
180 930
1770
0
0
0
0
3000 1860
0
0
0
0
0
1500
0
0
0
0
0
0
The s table
0
1
0
0
0
0
0
0
0
0
0
0
2
2
0
0
0
0
2
2
3
0
0
0
4
2
4
4
0
0
2
2
4
4
5
0
The optimal parenthesization
(A1 * A2) * ((A3 *A4) * (A5 * A6)))
19 15.2-2
MATRIX-CHAIN-MULTIPLY (A, s, i, j)
{
if i =j
C  Ai
else
A  MATRIX-CHAIN-MULTIPLY (A, s, i, s[i, j])
B  MATRIX-CHAIN-MULTIPLY (A, s, s[i,j]+1, j)
C MATRIX-MULTIPLY (A, B)
return (C)
}
20 15.2-4
We can find the sum by making a note of the access pattern for the m table
Example: For n = 5
1
2
3
4
5
1
4
X
X
X
X
2
3
4
X
X
X
3
2
3
4
X
X
4
1
2
3
4
X
5
0
1
2
3
4
X – Don’t Care; The numbers in the cells indicate the number of accesses to the cell
It can be seen that
R(i,j) =
i(i-1)
=
= n/6 ((n+1) (2n+1)) – n/2 (n+1)
= (n3 – n) /3
21 15.4-1
x = <1,0,0,1,0,1,0,1>
y = <0,1,0,1,1,0,1,1,0>
0
1
2
3
4
5
6
7
8
9
yj
0
1
0
1
1
0
1
1
0
0 xi
0
0
0
0
0
0
0
0
0
0
1 1
0
0
1
1
1
1
1
1
1
2 0
0
1
1
2
2
2
2
2
2
3 0
0
1
1
2
2
2
3
3
3
4 1
0
1
2
2
3
3
3
4
4
5 0
0
1
2
3
3
3
4
4
4
5
6 1
0
1
2
3
4
4
4
5
7 0
0
1
2
3
4
4
5
5
5
8 1
0
1
2
3
4
5
5
6
6
1
2
3
4
5
5
6
6
22 15.4-2
Print LCS without the b table
print_lcs (i,j)
{
int i,j;
if ((i = = 0) || (j = = 0))
return;
if(x[i] = = y [j])
{
print_lcs (i -1, j -1);
printf(“%d”, x[i]);
}
else if (c[i-1][j] >= c[i][j])
{
print_lcs (i -1, j);
}
else
{
print_lcs (i, j-1);
}
return;
}
/* i, j are the lengths of the two lists */
/* x, y are the two lists whose LCS is to be found */
/* c is the c table as in the algorithm */
23 15.4-5
Algorithm



a[1 ..n] is the input sequence
length[1..n] contains the length of the monotonically increasing subsequences up
to a[i]= { i = 1 ..n}
lms is the length of the longest monotonically increasing subsequence
for i = 2 to n do
Begin
for j = 1 to i -1 do
Begin
Search for the j such that length[j] is the largest and a[i] can be
included in the subsequence it represents.
End
length[i] = length[j] + 1
if lms < length[i] then lms = length[i]
End
24 15-1
Bitonic TSP
Points P0… Pn-1 are sorted by increasing X- Coordinate
C(i, j) = Cost of achieving optimal pair of paths such that are paths ends with Pi, the other with Pj
( i < j)
Base Case
C(0,1) = dist (0, 1)
General Case
C(i-1, i) = min {C(j, i-1) + dist(j, i)}
0 j< i-1
C(i, j) = C(i, j-1) + dist(j-1, j) where i <j-1
Final solution
min {C(i, n-1) + dist(i,n-1)}
0 i< n-1
25 16.1-1
/* f [1…n] contains finishing times (sorted) of activities
s [1…n] contains the starting times of those activities
m[1…n] contains the number of activities from 1 .. i that can be scheduled mi in the
problem
fm[1…n] indicates the finishing times of the tasks scheduled in each of m[1…n] */
Begin
m[1] = 1
fm[1] = f[1]
for i = 2 to n do
Begin
if( fm[i-1]  s[i] then
Begin
m[i] = m[i-1] + 1
fm[i] = f[i]
End
else
Begin
fm[i] = fm[i-1]
m[i] = m[i-1]
End
End i
End
26 16.1-3
n  length [s]
for i  1 to n
A[i]  {  }
//each A[i] (lecture Hall) has a set of activities
LIST_INSERT( L, i );
k0
while L  
do k  k + 1
i  head [L]
for j  i + 1 to tail[L]
do if sj  fi
then A[k]  A[k] U { j }
ij
LIST_DELETE( L, j )
return L // the final value of ‘k’ has the number of lecture halls
27 16.2-4
The greedy strategy would be to fill up gas at the last moment i.e., Travel to the farthest gas
station that can be reached from the current gas station (without falling short)
28 16.2-5
Sort the points in ascending order of their k values
The greedy strategy would be to enclose the leftmost unenclosed point and all points that lie
within a unit distance of this point. The next interval will begin at the closest point to the right of
this interval
29 16.3-2
0
54
1
0
h : 21
1
33
0
g : 13
1
20
0
f:8
1
12
0
e :5
7
d :3
1
0
0
c :2
b :1
Generalization:
code =
k-1 1’s followed by a ‘0’, if k  n-1
k 1’s, k = n
1
4
2
1
a :1
30
String ababbabbababbababbabb
Pattern ababbabbababbababbabb
Fail link table 1
0 a -1
1b0
2a0
3b1
4b2
5a0
6b1
7b2
8a0
9b1
10 a 2
11 b 3
12 b 4
13 a 5
14 b 6
15 a 7
16 b 3
17 b 4
18 a 5
19 b 6
20 b 7
Fail link table 2
0 a -1
1b0
2 a -1
3b0
4b2
5 a -1
6b0
7b2
8 a -1
9b0
10 a -1
11 b 0
12 b 2
13 a -1
14 b 0
15 a 7
16 b 0
17 b 2
18 a -1
19 b 0
20 b 2