Q1: 20 points Q2: 20 points Q3: 20 points Q4: 20 points Q5: 30 points Q6: 15 points Total: 125 points
Question text is in black, solutions in blue.
TRUE. The actual chromatic polynomial is r6 - 7r5 + 21r4 - 33r3 + 27r2 - 9r. This is pretty tedious to compute by the method given in Tucker. But if we look at the Inclusion/Exclusion solution, there are r6 total colorings, and r5 that fail to be valid on each of the seven edges. So the I/E formula gives us r6 - 7r5, plus all the terms for colorings that fail to be valid on two or more edges. These latter terms are all of degree at most 4.
TRUE. H has exactly two vertices of odd degree, so any Euler path must starts at one and end at the other. There is such a path because H is connected, and it is also easy to find one by inspection such as A-B-C-D-E-F-A-D.
FALSE. A spanning tree must consist of exactly five of the seven edges. There are 21 total ways to remove two of the seven edges, but not all of them yield spanning trees. (For example, if we remove (A, B) and (C, D), the resulting graph is not connected.) It turns out that there are 15 different spanning trees.
FALSE. The only Hamilton circuit is A-B-C-D-E-F-G-A or some rotation thereof. (The problem should have specified that two rotations are not considered distinct tours.) But even if we don't observe this, note that since a tour has six edges, its weight is 28 minus the weight of the edge not included. The seven choices of this edge leave tours with seven different weights.
TRUE. Because planar graphs obey the rule E ≤ 3V - 6, we cannot add six edges to make 13. But we can add, for example, (A, C) and (A, E), place these edges inside the cycle A-B-C-D-E-F-A along with (A, D), then add (B, D), (B, F), and (D, F) and place them outside the cycle.
TRUE. This is standard: the EGF for each dog without the restriction is ex, and forbidding each dog from getting 0 or 1 treat is accomplished by subtracting the r = 0 and r = 1 term. We get the EGF for distributions to the three dogs by multiplying the three individual EGF's.
TRUE. We could solve the characteristing equation of the recurrence, or just substitute A2r into the recurrence and divide it by 2r-3 to get 8A = 4A + 2A + 2A, which is true even if A = 0.
TRUE. We know that one action is trivial and the other is its own inverse. So every element under every action is either fixed or exchanged with another element. Thus every orbit is of size 1 or 2, and the cycle index polynomial only needs to deal with orbits of size 1 or 2.
FALSE. A number must have each of its left options less than each of its right options.
TRUE. The second player will always win this game. If Left moves first she must make the game negative (a Right win) and if Right moves first he must make the game positive (a Left win).
It's clear that the union spans the vertices, because any two nodes
are connected by their respective paths to x. The difficulty is
to prove that there can be no cycles in the union. This was not
as obvious as many of you thought. In principle, a path could
exist in the union because each of its edges was on the path from
x to a different y -- we need to rule this out.
If there is a cycle, there must be a node y that has two
different simple paths from x. By the unique path property, one
of these paths is longer. Look at the last edge on this longer
path that is not on the shorter path, and rename y to the node
at the end of this edge away from x, so that our edge is (w, y).
We get a contradiction
by proving that this edge cannot be in the union.
If it were in the union, it would be on the shortest path
from x to some z. But then we could get a shorter path from
x to y in the original graph by replacing the path from x to w
and the edge (w, y) by the actual shortest path from x to y.
Node set {x, a, b, c}, edges (x, a) of weight 2 and (a, b), (b, c), and (x, c) each of weight 1. The unique path property holds by inspection. The MST consists of the three weight-1 edges, since Kruskal's algorithm will take these first. But the shortest path from x to a is the edge of weight 2, rather than the path of weight 3 in the MST. So the shortest-path tree has edges (x, a), (x, c), and (b, c), and has total weight 4 while the MST has weight 3.
Same example -- the MST is clearly unique and does not have the shortest path in G from x to a.
If Whistle has one treat, the other ten may be divided into unequal sets (each of size more than 1) in three ways to give distributions of 8-2-1, 7-3-1, or 6-4-1. If Whistle has two treats, the other nine may divided into two unequal sets (each of size more than 2) in two ways to give 6-3-2 or 5-4-2. If Whistle has 3 or more treats, then Duncan has at least 4 and Cardie at least 5, so there are at least 12 total. We thus have five total distributions of 11 treats following the rules.
Let X be the number of Duncan's treats minus the number of
Whistle's, and let Y be the number of Cardie's minus the number
of Duncan's. We are looking then for triples (X, Y, W) where
all three numbers are positive and X + 2Y + 3W = r. The
generating function for this is
(x + x2 + x3 + ...) ×
(x2 + x4 + x6 + ...) ×
(x3 + x6 + x9 + ...) =
x/(1-x) × x2/(1-x2) ×
x3/(1-x3).
That's enough for this problem, but if we wanted to evaluate
coefficients we'd start by rewriting
this as x6 × 1/(1-x)3
× 1/(1+x) × 1/(1+x+x2).
Every such distribution is a permutation of the set {c, d, w}
applied to a solution for the rules of part (a) and (b). So the
number is exactly 3! = 6 times the answer to (b).
A string in Xr can only be made by appending a letter to
a string in Xr-1, and there is exactly one letter that
can be appended to each such string.
A string in Yr can be made in two ways. We could
take a string in Xr-1 and append either of the two
letters that do not occur in it, or take a string in
Yr-1 and append one of the two letters that do occur
in it.
A string in Zr may be made by appending any of the
three letters to a string in Zr-1, or by appending the
one missing letter to a string in Yr-1.
The initial conditions are x1 = 3, y1 = 0,
and z1 = 0, since all three strings of length 1 are in
X1.
Since xr = xr-1 and x1 = 3,
we have easily by induction that xr = 3 for all
positive r.
The second recurrence now becomes yr =
2(yr-1 + 3), with initial condition y1 =
0, which has solution yr =
3(2r - 2). We could infer this solution from small
examples or test all possible solutions of the form
A2r + B. (The solution to the
homogeneous part is 2r.)
The third recurrence now becomes zr =
3(Zr-1 - 2r-1 + 1). The solution to the
homogenous part is 3r, so we are looking for an
overall solution of the form zr =
A3r + B2r + C for the initial
condition z1 = 0 and z2 = 0.
(The latter can be computed from the r = 1 conditions using
the recurrence.) Substituting r = 1 and 2,
we get
3A + 2B + C = 0 and 9A + 4B + C = 0, which yields
6A + 2B = 0 or B = -3A. Now the recurrence becomes
A3r - 3A2r + C =
3(A3r-1 - 3A2r-1 + C - 2r-1
+ 2). The 3r terms cancel to give
-6A2r-1 + C = (-9A+3)2r-1 + 3C + 6.
Equating terms, we get C = 3C + 6 (so that C = -3) and
-6A = -9A + 3 (so that A = 1. This makes our final solution
zr = 3r - 3(2r) + 3.
The total number of strings is 3r. The size of each of the Ai's is 2r, the size of each intersection of two Ai's is 1r = 1, and the intersection of all three sets is empty. So we have 3r - 3(2r) + 3 for zr, the number of strings not in any of the three Ai's.
Any isomorphism must take degree-3 nodes to degree-3 nodes, so it must either leave A and D fixed or map them to each other. If it leaves A and D fixed, it must take neighbors of A to neighbors of A, and similarly for D. It must thus either fix B and F (in which case it must also fix C and E to maintain the edges to B and F), making it the identity, or exchange B and F (and thus exchange C and E). So there are two isomporphisms that fix A and D. If it exchanges A and D, then B and F must be mapped to C and E, and vice versa. If we map B to C we must map F to E, and if we map B to E we must map F to C. So there are two isomorphisms that exchange A and D, and thus four in all.
In cycle notation, the four automorphisms are the identity, (B F)(C E), (A D)(B C)(E F), and (A D)(B E)(C F). Thus the cycle index polynomial is (1/4)(x16 + x12x22 + 2x23).
We can just substitute 2 for x1 and x2 in the cycle index polynomial to get (1/4)(64 + 16 + 16) or 24. Using Burnside's Lemma directly, we can see that there are 64 colorings, of which the identity fixes all, the other automorphism fixing A and D fixes 16, and the two that exchange A and D fix 8 each, so that again we have (1/4)(64 + 16 + 16) = 24 classes.
Applying the Polya theorem, we have (1/4)((b+w)6 + (b+w)2(b2+w2)2 + 2(b2+w2)3). This simplifies to b6 + 2b5w + 6b4w2 + 6b3w3 + 6b2w4 + 2bw5 + w6, but all you need for part (e) is that the coefficient of b2w4 is 6.
If A and D are both black, then the other four vertices must be white, and there is only one choice. If one of A and D is white and the other black, we must choose which of the degree-1 vertices is black. There are two choices: either it is a neighbor of the black degree-3 vertex or it isn't. If A and D are both white, we must choose two of B, C, E, and F to be black. There are three choices of how to do this: the black nodes could be neighbors, neighbors of the same degree-3 node, or opposite one another. This gives 1 + 2 + 3 = 6 classes of colorings in all, matching the coefficient found in (d). Overall there are C(6, 2) = 15 ways to color the graph with four white and two black before we consider automorphism. One has A and D both black, eight have one white and the other black (four in each class), and six have both white (two in each class).
The second player has a winning strategy. If the first player takes one stone, the second player takes the stone diagonally opposite. Then the first player must take one stone, and the second player wins by taking the other. If the first player begins by taking two stones, the second player wins immediately by taking the other two.
There are two positions with two stones (adjacent or opposite) and
one each with zero, one, three, and four stones. The kernel
includes three of these six positions: zero stones, two stones
opposite, and four stones. By inspection, no move from any kernel
position is to another kernal position. From three stones, there
is a move to the kernel by taking one to leave two stones opposite.
From two stones adjacent or from one stone, there is a move to the
kernal by taking all the stones.
Many people got this wrong because they missed the winning move
from the three-stone position, which comes from not reading the
rules carefully.
Zero stones: no moves, Grundy number is 0
One stone: only move is to 0, Grundy number is 1
Two adjacent: moves to 0 or 1, Grundy number is 2
Two opposite: only move is to 1, Grundy number is 0
Three stones: moves to 0, 1, or 2, Grundy number is 3
Four stones: moves to 2 or 3, Grundy number is 0
Last modified 11 January 2017