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
Split the cube's 8 corners into two sets so every edge always connects a corner from one set to a corner from the other (a bipartite split).
Step 1: Understand what is being asked.
The cube has 8 corners: a, b, c, d, e, f, g, h. We need to color every corner so that any two corners joined by an edge get different colors. We want the smallest number of colors that makes this possible.
Step 2: Split the corners into two groups.
Look at the cube as two squares, the top face and the bottom face, joined by four vertical edges.
Put a, d, f, h in Group 1 and b, c, e, g in Group 2 (this is the usual alternate-corner split of a cube, where every edge joins a corner from Group 1 to a corner from Group 2, never two corners from the same group).
Check a few edges: a-b, a-c, a-e all join a Group 1 corner to a Group 2 corner, and this pattern holds for every edge of the cube.
Step 3: Color each group with one color.
Give every corner in Group 1 color X and every corner in Group 2 color Y. Since every edge always connects one corner from Group 1 to one from Group 2, no edge ever joins two corners of the same color.
So 2 colors are enough to color the whole cube correctly.
Step 4: Confirm that 1 color is not enough.
Any two corners joined by an edge cannot share a color, so we need at least 2 colors. Using only 1 color would force every edge to join two same-colored corners, which breaks the rule. Options (A) 8 and (B) 4 use far more colors than needed, and (C) 3 is also more than the minimum, since the whole cube splits cleanly into just two groups.
Final Answer:
The minimum number of colors required is 2, which is option (D).
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