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MATH 145 SAMPLE PROBLEMS FOR EXAM 2

Problems

(1) Let an denote the number of strings of n digits from {0, 1, 2}, containing no consecutive 1s or 2s. Find
a1, a2, a3. Derive a second order homogeneous linear recurrence for an.

(2) Let dn be the derangement sequence. Using the asymptotic fact that converges rapidly to 1/e , find
a good approximation to the probability that a random shuffling of a 52-card deck will leave at most two
cards in their original position.

(3) Use bubble sort to arrange the list of numbers 7,4,6,3 in increasing order . List all the intermediate stages
of this process.

(4) In determining the generating function for the Catalan numbers, we came up with the quadratic equation
, where f(x) is regarded as the unknown . The quadratic formula then gave us
, an ambiguous result. Explain how to make the correct choice for f(x).

(5) Show that if you remove any edge from a tree with at least two vertices , you create a disconnected graph.

(6) A certain graph G has n red vertices, n white vertices, and n blue vertices. What is the maximum
number of edges it can have if no edge has both of its end vertices the same color?

(7) A spanning tree T for a graph G is a tour if each vertex of the tree has degree ≤ 2. Give an example of
a graph with no tour. Show that Kn has a tour.

(8) What is the total number of simple graphs you can make with three vertices? How about 4 vertices? Is
there a way of counting the number of graphs on n + 1 vertices, once you know the number for n vertices?

9 Let , be the simple graph with vertices where the edges are

Is bipartite? How about planar?

Solutions

(1) Let an denote the number of strings of n digits from {0, 1, 2}, containing no consecutive 1s or 2s. Find
a1, a2, a3. Derive a second order homogeneous linear recurrence for an.

a1 = 3, a2 = 7, a3 = 17. Call a string good if it contains no consecutive 1s or 2s. Each
string of length n + 1 is formed by adding a digit to the end of a string of length n. Let
be the number of good strings of length n that end in 0, 1, 2, respectively. Then
Now it is clear that ; goodness is neither ruined nor created by
affixing a 0. The only way to create a good (n + 1)-length string ending in a 1 is for the n-
string you affix the 1 to to be good and to end in either 0 or 2. This tells us
Likewise, . Clearly, by symmetry (switch 1s and 2s in any given string),
. Let this number be . Then

This gives us the initial value recurrence ,

(2) Let dn be the derangement sequence. Using the asymptotic fact that converges rapidly to 1/e , find
a good approximation to the probability that a random shuffling of a 52-card deck will leave at most two
cards in their original position.

The probability P is exactly

To get the desired approximation, we replace each by , obtaining

(3) Use bubble sort to arrange the list of numbers 7,4,6,3 in increasing order. List all the intermediate stages
of this process.

(4) In determining the generating function for the Catalan numbers, we came up with the quadratic equation
, where f(x) is regarded as the unknown. The quadratic formula then gave us
, an ambiguous result. Explain how to make the correct choice for f(x).

Let . Then, as x -> 0, the denominator converges to 0, while the numerator
converges to 2. Then . Since every power series in x is defined at x = 0,
g cannot be the generating function for . The other possibility is . As
x -> 0, both numerator and denominator converge to 0; so we may use rule to
obtain

(5) Show that if you remove any edge from a tree with at least two vertices, you create a disconnected graph.

Suppose T be the tree in question; say T has p≥2 vertices. Then the number of edges is
p − 1. Let the resulting graph be G. Then it has p vertices and p − 2 edges, so cannot be a
tree. However, since removal of edges does not create cycles, G can have no cycles; hence, if
it were connected, it would have to be a tree. We conclude that G cannot be connected.

(6) A certain graph G has n red vertices, n white vertices, and n blue vertices. What is the maximum
number of edges it can have if no edge has both of its end vertices the same color?

Let the vertex set V be the disjoint union , where |R| = |W| = |B| = n. Then
there are n^2 possible edges between vertices in R and vertices in W; likewise for R and B ;
likewise for W and B. No two edge sets have any edges in common ; so the total possible
number of edges is 3n^2.

(7) A spanning tree T for a graph G is a tour if each vertex of the tree has degree ≤ 2. Give an example of
a graph with no tour. Show that has a tour.

The triod has no tour. In fact, any tree with a vertex of degree ≥ 3 has no tour. As for
, list its vertices as . Then there is a unique edge from for each
1 ≤ i < n, and this path gives us a tour.

(8) What is the total number of simple graphs you can make with three vertices? How about 4 vertices? Is
there a way of counting the number of graphs on n + 1 vertices, once you know the number for n vertices?

Let be the number of simple graphs on the vertices . Then ; since
there are no loops, there is just one vertex and no edges. With two vertices, they’re either
adjacent or they aren’t; so . For three vertices, you have either no edges (1 way), one
edge (3 ways), two edges (3 ways), or three edges (1 way). Hence . Suppose we want
to count the number of simple graphs on vertex set . We may specify any
such graph by determining the adjacency structure on vertices and
then pick a set of vertices from to form an edge with . Thus
. ( Solving this recurrence gives us . Another way to get this result is
to line up the pairs of vertices ( ways) and to decide which of these pairs are to receive
an edge (2 ways for each). This also gives us .)

9 Let , be the simple graph with vertices , where the edges are

Is bipartite? How about planar?

The graph is not bipartite because it has a cycle of odd length 3. It is
planar because it contains no subdivision of . (This latter fact may be determined
by induction.)

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