Map Coloring Worksheet: Key
There are many ways to do this but it can be done correctly with 4 colors--but not fewer.
The goal of a map coloring problem is to color a map so that regions sharing a common border have different colors. (Regions that meet only in a point may share a common color.)
In the language of mathematics, we are assuming that each region is a closed curve. Simply put, this means that for a region it is not possible to move from inside the region to outside the region without crossing the border of the region and it is always possible to move from a point within a region to any other point within the region without crossing the border of the region. The chromatic number of a map is the mimimum number of colors required to color that map.
If v is the largest degree (or valence) of any vertex in a connected graph, G, AND G is not a complete graph nor a circuit of odd length, then the chromatic number of G is less than or equal to v.
There are many correct ways of coloring both maps but, the S.E. Missouri map has a chromatic number of 4 while the S.E. United States map has a chromatic number of 3.
Brook's Theorem assures us that we can color the first graph with 3 or fewer colors. It turns out that you do need three colors to color this graph.
For the second map (the pie chart with 7 sectors), you need 3 colors. In general, any pie chart with an even number of sectors can be colored with 2 colors while any pie chart with an odd number of sectors (but more than one) requires 3 colors.
Your graphs are correct if the region representing each state was transformed into a vertex and any two vertices are connected by an edge if and only if they share a border. The chromatic number of each graph is 3.
There are many correct answers. One answer is AREA 1: Eagle, Panda; AREA 2: Rhino, Elephant, Tiger; AREA 3: Giraffe, Zebra, Oryx. Your answer is correct if you have animals in exactly 3 areas and no two animals in the same area are represented in the graph by vertices joined by an edge.
There are many correct answers. Your graph is correct if each of the 10 species was transformed into a vertex and any two vertices are connected by an edge if and only if the two species are incompatible (there is an X in the appropriate table cell).
Brook's Theorem guarantees that you will need 6 or fewer
(Of course, the Four Color Theorem guarantees that we actually need no more than 4.)
In fact, this situation does require 4 tanks. One posible solution is Tank A: species 1, 6, 10; Tank B: species 2, 5, 7; Tank C: species 3, 8; Tank D: species 4, 9. There are many other correct solutions.
Additional Exercises (solutions)
COMING SOON!!! (maybe)