Consider the cube shown below with its 8 corners labelled a, b, c, d, e, f, g, and h. The figure is representative. All corners are to be colored such that any two corners that are connected by an edge must be of different colors. The minimum number of colors required to achieve this is ________
Show Hint
A cube graph has no odd cycles, since every face is a 4-sided square, so it is bipartite. A bipartite graph with at least one edge always needs exactly 2 colors.
Step 1: Model the problem as a graph coloring problem.
Each corner of the cube is a vertex, and each edge of the cube (the line joining two corners that are physically connected) is a graph edge. Two vertices joined by an edge must get different colors. This is exactly the standard proper vertex coloring problem, and we want the smallest number of colors, called the chromatic number of the graph.
Step 2: Check whether the cube graph has any odd length cycle.
A graph needs only 2 colors if and only if it has no cycle of odd length, which is the definition of a bipartite graph. Every face of a cube is a square, so every face boundary is a cycle of length 4, which is even. Any other cycle you can trace on a cube, for example going around two faces, is built out of these even length face boundaries and also comes out even in length. So the cube graph has no odd cycle anywhere, which means it is bipartite.
Step 3: Build an explicit 2-coloring to show 2 colors are enough.
Give each corner of the cube a coordinate triple \((x,y,z)\) with \(x,y,z \in \{0,1\}\), one coordinate per direction of the cube. Two corners are joined by a cube edge exactly when their coordinates differ in one place only. Color a corner red if \(x+y+z\) is even, and blue if \(x+y+z\) is odd. Moving along any single edge changes exactly one coordinate, so it always flips the parity of \(x+y+z\). This means every edge always joins a red corner to a blue corner, never two corners of the same color, so this 2-coloring is valid for all 8 labelled corners a to h.
Step 4: Rule out 1 color.
The cube clearly has edges, for example a is joined to its neighbouring corners, so at least two corners are adjacent. Using only 1 color would force two adjacent corners to share a color, which breaks the rule. So 1 color is never enough, and 2 is the smallest number that can work.
Why the other options are wrong:
8 colors works but wastes 6 colors that are not needed, since Step 3 already gives a valid coloring with only 2. 4 colors and 3 colors both work too, since any coloring that already needs only 2 colors also succeeds with more colors, but the question asks for the minimum, and Steps 3 and 4 together show that number is exactly 2, not 3 or 4.
Final Answer:
The cube graph is bipartite, since it has no odd cycles, so the minimum number of colors needed is 2.
\[ \boxed{2} \]
