Download MC302—GRAPH THEORY—SOLUTIONS TO HOMEWORK #1—9

MC302—GRAPH THEORY—SOLUTIONS TO HOMEWORK #1—9/19/13
68 points + 6 extra credit points
1. [CH] p. 13, #1.3.3.
a. In each case, for the two graphs you say are isomorphic, justify it by labeling their
vertices so that they become identical; in other words, label the vertices in the two
graphs with numbers 1, 2, …, n, so that corresponding vertices get the same number.
b. In addition, provide justification why the third graph is not isomorphic to the other two,
i.e., give some “isomorphic property” that one graph has but not the other (recall that
an isomorphic property is one that two isomorphic graphs must either both have or both
not have)
Solution to #1 (15 pts):
• 1.3.3 (a) The first two are isomorphic; the third is not because
it has a triangle but the others do not.
• 1.3.3 (b) The second two are isomorphic; the first is not
because its degree-3 vertices do not form a cycle.
• 1.3.3 (c) The second and third are isomorphic; the first is not
because it has 3 degree-2 vertices while the others each have
only 2 degree-2 vertices.
2. [CH] p. 13, #1.3.4. In addition to listing all the graphs, give a brief explanation why (a) there are
no others, and (b) why no two graphs on the list are isomorphic.
Solution to #2 (5 pts): If the sizes of the two bipartite sets are and , then both and must be
positive, and 2 ≤ + ≤ 7. The graph , has vertices of degree and vertices of degree .
Since , ≅ , , listing all , with 1 ≤ ≤ and 2 ≤ + ≤ 7 will list all possible graphs up to
isomorphism. There are twelve such graphs:
, , , , , , , , , , , , , , , , , , , , , , , .
No two of these three graphs are isomorphic because their degree sequences are all different.
3. Show that, in any group of two or more people, there are always two with exactly the same
number of friends inside the group. Do this as follows:
a. First, restate this as an “if-then” statement about graphs and graph properties, i.e.,
“If G is any simple graph, then …,”
b. Now prove the claim about graphs in order to answer the original question. Hint: Show
that if G is a simple graph with n vertices, then there are fewer possible values for the
degree of a vertex than there are vertices, so some value must be repeated.
Solution to #3 (10 pts: a-4, b-6):
a. We let the people be represented as vertices of a graph G, with an edge indicating that the
two people corresponding to its endpoints are friends. Then this sentence translates as follows:
If G is any simple graph with two or more vertices, then G must have two vertices with the same
degree.
b. Suppose G has n vertices, where ≥ 2. The smallest possible degree is 0 (a person with no
friends), and the largest is − 1 (a person who is friends with everyone else). Thus there are n
possible values (0, 1, . . . , − 1) for the degrees of vertices in G. But notice that 0 and − 1
cannot both be degrees, because if one vertex has no neighbors, then no vertex is adjacent to
every other one, and vice-versa. Thus the possible values are either 0, 1, . . . , − 2 or
1, 2, . . . , − 1. In either case there are only − 1 values that must be assigned to n vertices,
so some value must be used twice.
4. In each part use the Havel-Hakimi Theorem (even if there is some other, shorter way) to
determine whether the sequence is graphic. If the answer is yes, use the “reverse” process
described on Slide 8 of the slides from 9/5/13 to generate a graph for the original sequence from
a graph for the shorter sequence. Label each vertex in the final graph with its degree, and if
you'd like, redraw the graph in a more aesthetically pleasing way (without changing the graph,
of course).
a. (5, 5, 4, 2, 1, 1, 1, 1)
b. (6, 6, 5, 5, 3, 3, 2, 1, 1)
Solution to #4 (10 pts: a-4, b-6):
a. (5, 5, 4, 2, 1, 1, 1, 1) → (4, 3, 1, 0, 0, 1, 1) = (4, 3, 1, 1, 1, 0, 0) → (2, 0, 0, 0, 0, 0). This last
sequence is not graphic, because if some vertex has degree 2, then each of its neighbors
has degree 1 or more, but all the other numbers are 0.
b. (6, 6, 5, 5, 3, 3, 2, 1, 1) → (5, 4, 4, 2, 2, 1, 1, 1) → (3, 3, 1, 1, 0, 1, 1) = (3, 3, 1, 1, 1, 1, 0) →
(2, 0, 0, 1, 1, 0). The last sequence is easily seen to be graphic. So we use the reverse
procedure to construct a graph with the original degree sequence.
(2, 0, 0, 1, 1, 0).
2
1
3
4
6
5
7
→ (, 3, 1, 1, 1, 1, 0)
2
1
3
4
6
5
7
2
1
3
4
= (3, 3, 1, 1, 0, 1, 1)
→ (, 4, 4, 2, 2, 1, 1, 1)
6
5
8
7
4
1
9
2
3
→ (6, 6, 5, 5, 3, 3, 2, 1, 1)
8
5
6
5. [CH] p. 23, 1.5.2 and 1.5.3. For 1.5.2, justify briefly in each case why the graph is or is not selfcomplementary. The book gives a hint for 1.5.3 (b); here is a slightly expanded version of that
hint. First, let e be the number of edges in the self-complementary graph G. Use the book's hint
to find a formula for e in terms of n. Then use this formula to show that either n is a multiple of 4
(i.e., e = 4t for some t), or n is a multiple of 4 plus 1, (i.e., e = 4t+1 for some t).
Solution to #5 (28 pts: 1.5.2--6 pts, 1.5.3(a)--6 pts, 1.5.3(b)--6 pts, 1.5.3(c)--10 pts):
1.5.2: The edges of the original graph are shown as gray dotted lines; the edges of the complement
(i.e., all the edges not in the original graph are solid red lines).
1.5.3(a) First graph: no (different degree sequences); second yes (both 5-cycles); third no (different
degree sequences).
1.5.3(b) If G is self-complementary, then G and "̅ have the same number of edges, which together
comprise all the edges of $ . Thus, if G has % edges, then % + % = 2% = ( − 1)/2. It follows that
% = ( − 1)/4. Since % is an integer, this implies that ( − 1) is divisible by 4. But and − 1
differ by 1, so one of them is odd and the other is even. Therefore the even value must be divisible
by 4. If is even, then it is divisible by 4, so = 4', for some integer '. If − 1 is even, then it is
divisible by 4, so − 1 = 4', i.e., = 4' + 1, for some integer '.
1.5.3(c). From the previous problem we know that a self-complementary graph " has ( − 1)/4
edges. It is also useful to notice that " cannot have a vertex of degree − 1, because in the
complement that vertex would have degree 0. But since " and "̅ are isomorphic, that would mean
they both have vertices of degree − 1 and 0, which we know from an earlier problem is
impossible. By the same reasoning " cannot have a vertex of degree 0. With these facts to cut
down on the possibilities, let’s now look at = 4 and = 5.
• = 4: By what we just said, a self-complementary graph " must have 3 edges and no
vertices of degree 3 or 0. By trial and error, we see that the only way we can place 3 edges
on 4 vertices with no vertices of degree 3 or 0 is to have the path of length 4, and that graph
is self-complementary:
•
= 5: A self-complementary graph on 5 vertices must have 5 edges and no vertices of
degree 4 or 0. There are several ways to place the edges, so we first assume that some
vertex has degree 3, as indicated in the first picture below, with the degree-3 vertex labeled.
There is one vertex that so far has no incident edges, but it cannot have degree 0, so there
must be an edge between it and one of the other 3 vertices, as in picture 2. There is one
more edge to place, and 5 ways to place it (but only 3 ways up to isomorphism). By trying
out each way and looking at the complement, we find one self-complementary graph (up to
isomorphism). That graph and its complement are shown in the 3rd and 4th pictures.
•
3
3
•
3
3
3
3
3
3
3
3
3
3
This takes care of the case when there is a degree-3 vertex, so now suppose there isn’t a
degree 3 vertex. Then there can’t be a degree-1 vertex either, because that vertex would
have degree-3 in the complement. So all the vertices must have degree 2, and the only way
to do that is with a 5-cycle, shown below, which is self-complementary. Thus, up to
isomorphism, there are two self-complementary graphs on 5 vertices.
Extra credit: We saw in class that the Degree-Sum Theorem is true for both simple graphs and
multigraphs, provided that a loop edge counts 2 toward the degree of the vertex it has at both
ends. However, the same is not the case for the Havel-Hakimi Theorem: there are some
sequences that are not graphic, but which are the degree sequences of multigraphs. Prove the
following result: A finite sequence of nonnegative integers is the degree sequence of some
multigraph if and only if its sum is even. Note that you must prove necessity and sufficiency:
a. For necessity: Assume that a finite sequence S is the degree sequence of some
multigraph, and prove that the sum of the numbers in S must be even.
b. For sufficiency: Assume that a finite sequence S of nonnegative numbers has an even
sum. Prove that there is some multigraph whose degree sequence is S.
Solution to extra credit problem (6 pts: a-2, b-4):
a. We observed in class that the Degree-Sum Theorem holds for multigraphs, so the sum of the
degrees must be even.
b. If the sum is even, then there is an even number of odd values. We create a multigraph G with
this degree sequence as follows: For each even value k, G has a distinct vertex with k/2 incident
loops; then this vertex has degree k. Divide the odd values into sets of two. For each such pair
(k, m), add two adjacent vertices. Give one of the vertices (k-1)/2 loops and the other (m-1)/2 loops.
Then the first vertex has degree k and the second has degree m. This process is illustrated below
for the sequence (7, 5, 4, 3, 2, 1), in which I paired 7 with 5 and 3 with 1. The vertices are labeled with
their degrees: