1.
4%,4%
(a)
C58  56
(b) C06  C16 *1! C26 * 2!  37
2.
7%
path 1
1
2
3
k
……
k’
2k
1'
2'
…
3'
Join the odd-degree vertices in pairs with k new edges.
We can generate an Euler circuit with all even-degree vertices.
Therefore, we can create an Euler path (take path 1 as an example, remove the added edge 1-1’,
it’s an Euler path starting from 1 and ending at 1’)
Rearrange the Euler path as following:
2
2'
3
3'
Remove 2-2’, 3-3’, …, k-k’ away, the resulting k simple paths form edge set E(G).
3.
7%
Assume G has n vertices, and G is regular of degree k.
The complement G’ must have n vertices, and G’ is also regular of degree (n-1)-k.
n≧2k+2 implies (n-1)-k=n-(k+1)≧n-n/2=n/2
According the theorem, if the graph G’ has no loops or parallel edges
if |V(G’)|=n≧3
if deg(v)≧n/2
ok
ok
ok
G’ is Hamiltonian.
4.
10%
Assume there exist two different minimum spanning trees, S and T.
There must be some edges in S but not in T.
Consider the edge u  S \ T.
Then T{u} forms a cycle C.
In the same cycle, there must exist another edge v such that v  T \ S.
Since |u| ≠ |v|, C contains only one Euler path with minimum weight.
(It’s either without u or without v, depends on which one has smaller weight.)
This contradicts to the assumption of co-existence of S and T.
5.
7%
Any algorithm results a maximal spanning tree is fine.
6.
7%
Euler circuit requires both m and n to be even.
Hamilton circuit requires this bipartite graph have all vertices with degree 2.
∴K2,2.
7.
4%,3%
(a)
(b)
3
2
1
d
e
f
h
g
7
6
h
5
4
8.
4%,4%
(a)
The variables x1, x2, … are wff’s and the constant 2 is a wff.
S1=x12^ is a wff.
If Sn is a wff, so are xn+12^ and Sn+1=Snxn+12^+
So all Sn’s are wff’s by induction.
By (B)
By (R)
(b)
Sn=x12+x22+…+xn2
9.
7%
x1=0.49 x2=0.26 x3=0.12 x4=0.04 x5=0.04 x6=0.03 x7=0.02
0
1
0.49
0.26
0.12
0.02
x1=0
x2=11
x3=110
x4=10111
0.03 0.04
x5=10110
0.04
x6=10101
10. 4%,6%
(a)
q
r
z
s
y
t
x
w
v
(b) No, s is the root without any children.
u
x7=10100
11. 8%
Assume no loops or parallel edges.
Consider a longest path v1, v2, …, vn with distinct vertices.
Since each vertex has degree at least 2, there exists another edge at vn.
Adjoin it to the path to get a closed path and since all vertices distinct, it’s a cycle.
12. 8%
Assume G1={v1, …, vn, E(G1)} and G2={u1, …, un, E(G2)}
Let G1’ be the complement of G1, and G2’ be the complement of G2.
(→)
Since G1 and G2 are isomorphic, then all vertices and edges of G1 will have a one-to-one
correspondence to G2.
To check the isomorphism of G1’ and G2’, we need to check the one-to-one
correspondence with respect to all the vertices and edges between G1’ and G2’.
By definition, G1’ contains the same vertex set as G1 (same as G2 and G2’).
Therefore, use the same mapping between G1 and G2, we can map all vertices of G1’ to
G2’.
{vertex ok, now we need to check edge}
By definition, G1’ has edges which don’t exist in G1 (same as G2 and G2’).
To build the edge correspondence:
If an edge (vi, vj)G1 and the corresponding edge (ui, uj)G2, then an edge (vi,
vj)G1’, and an edge (ui, uj)G2’.
Mapping edge (vi, vj) in G1’ to edge (ui, uj) in G2’.
If an edge (vi, vj)G1 and the corresponding edge (ui, uj)G2, then an edge (vi,
vj)G1’, and an edge (ui, uj)G2’.
(←)
Apply G1’ to G1 and G2’ to G2. The rest is the same as above.
13. 4%,4%
The n-cube Qn is defined recursively. Q0 is just a vertex. Qn+1 is gotten by taking 2 copies of
Qn and joining each vertex v of Qn with its copy v’.
Q0
Q1
Q2
Q3
Q4 (hypercube)
edge(Q0)=0 vertex(Q0)=1
vertex(Qn) = 2*vertex(Qn-1)
edge(Qn) = 2*edge(Qn-1)+vertex(Qn-1)
(a) n
(b) 2n-1*n
14. 7%
5
7
5
6
20
8