most popular name prior to Instant Insanity). One version of the puzzle is currently being marketed by Winning Moves Games USA. Solution Given the already colored cubes and the four distinct colors are (Red, Green, Blue, White), we will try to generate a graph which gives a clear picture of all the positions of colors in all the cubes. The resultant graph will contain four vertices one for each color and we will number each edge from one through four (one number for each cube). If an edge connects two vertices (Red and Green) and the number of the edge is three, then it means that the third cube has Red and Green faces opposite to each other. To find a solution to this problem we need the arrangement of four faces of each of the cubes. To represent the information of two opposite faces of all the four cubes we need a directed subgraph instead of an undirected one because two directions can only represent two opposite faces, but not whether a face should be at the front or at the back. So if we have two directed subgraphs, we can actually represent all the four faces (which matter) of all the four cubes. First directed graph will represent the front and back faces. Second directed graph will represent the left and right faces. We cannot randomly select any two subgraphs - so what are the criteria for selecting? We need to choose graphs such that: the two subgraphs have no edges in common, because if there is an edge which is common that means at least one cube has the pair of opposite faces of exactly the same color, that is, if a cube has Red and Blue as its front and back faces, then the same is true for its left and right faces. a subgraph contains only one edge from each cube, because the sub graph has to account for all the cubes and one edge can completely represent a pair of opposite faces. a subgraph can contain only vertices of degree two, because a degree of two means a color can only be present at faces of two cubes. Easy way to understand is that there are eight faces to be equally divided into four colors. So, two per color. After understanding these restrictions if we try to derive the two sub graphs, we may