COMPSCI 575/MATH 513: Combinatorics and Graph Theory
Final Exam, Fall 2018
David Mix Barrington
17 December 2018
Directions:
- Answer the problems on the exam pages.
- There are six problems (most with multiple parts)
for 120 total points plus 10 extra credit.
Actual scale was A = 105, C = 67.5.
- If you need extra space use the back of a page.
- No books, notes, calculators, or collaboration.
- Numerical answers may be given as expressions involving
arithmetic operations (including exponentiation and factorial)
or the functions P(n, k) and C(n, k). In general a summation
is a complete answer to a counting problem, but a recurrence
or a generating function is not.
Q1: 20 points
Q2: 20+10 points
Q3: 40 points
Q4: 20 points
Q5: 20 points
Total: 120+10 points
The undirected graph G has node set {a, b, c, d, e, f| and the following
set of nine edges:
{(a, b), (a, c), (b, c), (c, d), (c, e), (c, f), (d, e), (d, f), (e, f)}
- Question 1 (20):
Identify each of these statements as true or false. The first five
deal with the graph H defined in Question 5. No explanation is
needed
or wanted, and there is no partial credit or penalty for guessing (2
points each).
- (a) G is a planar graph and can be drawn in the plane
with five regions (counting the exterior region).
- (b) The graph G has no Euler path, and it is not possible to
add one edge to it to make an undirected graph with an Euler path.
- (c) There does not exist a spanning tree of G in which
vertex c has degree 1.
- (d) Let A be the adjacency matrix of G, viewed as an
integer matrix. Then the matrix I + A + A2
has at least one zero entry.
- (e) The chromatic polynomial of G is divisible by
the linear polynomial r - 4.
- (f) Let f(x) and g(x) be the exponential
generating functions for the sequences
{an} and {bn}
respectively. Define the sequence {cn}
so that f(x)g(x) is the exponential generating
function of {cn}. Then for any r,
cr is the sum for k from 0 to r of
C(r, k)akbr-k.
- (g) Every solution to the recurrence an =
3an-a - 3an-2 + an-3
is a polynomial in n.
- (h) The rook polynomial of a 2 by 2 square is
1 + 4r + 2r2, and the number of permutations
of {1, 2, 3, 4} that do not contain any of the pairs (2, 2),
(2, 3), (3, 2), or (3, 3) is given by 1(0!) + 4(1!) + 2(2!)
= 9.
- (i) Every one-stalk game of Red-Blue Hackenbush has a value
that is a "number" in Conway's terminology.
- (j) It is not the case that every one-stalk game of
Red-Blue Hackenbush has a value that is a real number.
- Question 2 (20):
In this question we will determine the number of ways to color each
vertex of G (the undirected graph defined above) black or white, where
two colorings are considered tot be the same if there is an automorphism
of G taking one to another. (Recall that an automorphism is a
bijection of the vertices that takes edges to edges and non-edges to
non-edges.) Note also that we are considering all two-colorings here,
not just "proper colorings" as in the definition of chromatic number of
the chromatic polynomial.
- (a, 5) How many ways are there to color each vertex black or
white if we do not consider the automorshisms? (This is
a very easy question.)
- (b, 5) There are exacctly twelve automorphisms of G --
describe them and argue that there aren't any others.
- (c, 10) Determine the number of two-colorings of G up to
symmetry. One way to do this is to find and apply the
cycle index polynomial of the action of the
automorphism group of G. (If you do this, keep in mind
that if an automorphism fixes a vertex, it creates a cycle
of size one.) It's also posible to determine the number
of colorings by an ad hoc argument -- this is fine
if your reasoning is clear.
- (d, 10 XC) Determine the pattern inventory
of the colorings counted in part (c). This is a
polynomial in b and w that tells how many of the
colorings have each number of black and white vertices.
This can be done easily from the cycle index polynomial,
but there is also an ad hoc argument you could
make if you explain it clearly. For full credit, your
answer should be expressed as a sum of monomials.
- Question 3 (40):
This problem concerns the chromatic polynomial of an
undirected graph, which is a polynomial in r that gives the number of
proper colorings of the vertices of the graph using r
colors.
(A proper coloring is one where no edge connects two vertices of
the same color.)
- (a, 10)
Determine the chromatic polynomial of G. You may use any
method, but explain your reasoning. (The general algorithm
given in Tucker may not be the fastest way to get the polynomial
for this particular graph.) You may find the results of parts
(c) and/or (d) useful.
- (b, 10)
We stated in lecture that the 3-colorability problem,
of deciding whether a given undirected graph has a proper
coloring with three colors, is NP-complete. Assuming
this fact, prove that if there is a polynomial-time
algorithm that takes an undirected graph as input and
computes its chromatic polynomial, then P = NP.
- (c, 10) Let tn(r) be the chromatic
polynomial of a tree with n nodes and let
kn(r) be the chromatic polynomial of the
complete graph Kn. Find and justify two
recurrences (each with a base case for n = 1) that
compute tn in terms of tn-1(r) and
kr in terms of kn-1(r). Solve
these recurrences to get closed forms for
tn(r) and kn(r). (A consequence
of the correct argument here is that all n-node trees
have the same chromatic polynomial.)
- (d, 10). In this part we will compute tn(r)
in another way, using inclusion-exclusion. Consider an
arbitrary n-node tree T. For each edge e of the tree, let
Ce be the set of colorings that are not proper
because the two endpoints of e have the same color. Determine
the size of Ce and, for each k from 1 to the number
of edges in T, determine the size of any intersection of any k
of the Ce's. (You should argue that all of the
intersections for a given k have the same size.) Using
inclusion-exclusion, determine the number of colorings of T that
are proper, i.e., that are not members of the
union of the Ce's. Argue that this number is the
same polynomial that you computed in part (c).
- Question 4 (20):
In this problem we will yet again distribute r identical treats
to three distinguished dogs, this time Cardie, Duncan, and
Guinness. Our goal will be to find a generating function for
the number sn of distributions in which Cardie gets
strictly more treats than Duncan, Cardie gets strictly more than
Guinness, and each dog gets at least one treat.
- (a, 5) Find the explicit values of s7 and
s8, by any method. (Listing the solutions without
justification is fine if you are correct.)
- (b, 10) We also define tr to be the
total number of distibutions of r treats to the three dogs,
with only the condition that each dog gets at least one
treat. We define ur to be the number where each
dog gets at least one treat, and Cardie and Duncan tie for
the largest number of treats (possibly as part of a
three-way tie), and vr to be the number where
all three dogs get the same (positive) number of treats.
Determine generating functions for tr,
ur, and vr, each as ratios of polynomials.
- (c, 5) Use Inclusion/Exclusion to express the
generating function for sr in terms of
those for tr, ur, and
vr, and thus compute the generating
function of sr as a ratio of
polynomials. You need not simplify. (Hint: Cardie
gets strictly the most treats in exactly one-third
of the distributions in which there is no tie for
the most treats.)
- Question 5 (20): In this problem we have a two-player
game played with a single chess king on an eight-by-eight
chessboard. We will call the top left (northwest) square (0,
0), the top right (northeast) square (0, 7), the southwwst
square (7, 0), and the southeast square (7, 7). The king starts
on (7, 7) and the game ends when one player wins by moving it to
(0, 0). On each turn a player may move the king one square
north, west, or northwest. (A chess king could normally move
one square in any of the eight directions, but we disallow this.)
- (a, 10) Recall that the kernel of such a game
is the set of positions from which the second player wins
under optimal play. (Thus (0, 0) is in the kernel, since
if the game starts there the first player has no more and
immediately loses.) Determine the kernel of this game and
hence which player (first or second) wins from each
possible starting position.
- (b, 10) Recall that the Grundy number of a
position in such a game is the size of a one-pile Nim
game that is equal (in Conway's terminology) to
the game starting at that position. Find the Grundy
number of (7, 7) in this game. (This will probably
involve finding the Grundy numbers of most or all of the
other positions.)
Last modified 4 January 2019