Download MATH CIRCLE: Graph Theory 2011.05.02 Activity: problem solving

MATH CIRCLE: Graph Theory
2011.05.02
Activity: problem solving competition.
1. Divide the class into two groups and choose a teacher (or two) to be the judge(s)
2. Hand out the worksheet with problems and solutions
3. Give at least 20 minutes for the two groups to understand the problems and solutions
4. One group gets to go first (it’s up to you how you’d like to determine which group goes
first)
5. At least three members of one group come up to the board and explain a problem (and
its solution) of their choice entirely. They get about 5–10 minutes to do this.
6. When the group finishes, the other group gets a chance for a rebuttal if they think the
first group did not explain it well. It is then up to the judge how the points for that
problem gets distributed amongst the groups based on the explanation given.
7. Repeat for the other group...
8. Tally up the points 5 minutes before end of class and congratulate the winners!
Graph Theory
MATCH CIRCLE
2011.05.02
1. (20 pts.) Suppose we have four colors: red, blue, green, yellow and we denote these colors
by numbers such that
1=
2=
3=
4=
red
blue
green
yellow.
How many ways can you color the complete graph below if no two adjacent vertices can be
colored with the same color? Now, suppose we represent the number of colors by the variable
x. Can you come up with a polynomial that counts the number of ways to color the same
graph?
Solution.
First, we pick a vertex. We have four colors to choose from. The next vertex we pick will
have only three color choices. The last vertex is adjacent to the other two, so this one will
only have two color choices. Putting this together we have 4(3)(2) = 24 ways to color the
complete graph with four colors.
If we represent four as the variable x, then we get the polynomial
x(x − 1)(x − 2).
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Graph Theory
MATCH CIRCLE
2011.05.02
2. (10 pts.) A signed graph is a regular graph we’ve seen before, except now the edges are
either a positive or a negative. Let us represent the vertices to be people, a negative edge to
be a hostile relationship between two people and a positive edge to be a friendly relationship
between two people. Draw a signed graph to show the famous quote “an enemy of my enemy
is my friend”.
Solution.
3. (15 pts) A path graph P is a graph that starts at a vertex and takes a “path” and end
at a terminal vertex where when we trace this with a pencil, we don’t lift our pencil. For
example, the graph below is a path graph from vertex a to vertex d. Show at least 5 paths
of G other than the one shown. You can get 5 more bonus points if you come up with 10
paths.
b
b
a
a
G
c
d
P
c
d
Figure 1: A path P = abcd from vertex a to vertex d in G.
Solution.
abc
abdc
bcd
acd
acbd
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Graph Theory
MATCH CIRCLE
2011.05.02
4. (50 pts.)Suppose we have the graph below. If we represent the X and Y vertices as axes
on the xy-plane and denote available colors by numbers (see number 1), draw the xy-plane
representation to show the number of ways to color the graph below with the same condition
that adjacent vertices cannot be colored with the same color.
X
Y
Solution.
Since the X vertex and the Y vertex cannot be colored the same, the (x, y) pair where x = y
is not included in our xy-plane graph. For example, the point (2, 2) means we color X with
color 2 and Y with color 2 and this is against the graph coloring rules. The point (3, 5) is
okay because this means we color X with color 3 and Y with color 5.
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Graph Theory
MATCH CIRCLE
2011.05.02
5. (10 pts.) The Wolf, Cabbage, Goat and Farmer Problem
A farmer is bringing a wolf, a cabbage and a goat to market. The farmer arrives with all
three at one side of a river that they need to cross. The farmer has a boat which can accommodate only one of the three (it’s a big cabbage.) If the wolf is left along with the goat, the
wolf will eat the goat. And, if the goat is left alone with the cabbage, the goat will eat the
cabbage. The wolf can be left alone with the cabbage because wolves don’t like cabbages.
How can all three get across the river intact? Represent this as a graph.
Solution.
Denote the four principals by W , C, G, and F . According to the rules of the problem, the
following combinations are legal on either side of the river
W CGF W C W
W CF GF C
W GF G
CGF
∅ (empty set)
Each label in the river represents a passenger in the boat besides the farmer.
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