The exact number is in Java Add QR in Java The exact number is

The exact number is use applet qr integrating topaint qr barcode on java SQL server (64 x 43) = 41,472.. 280 9 Graphs Four colored cubes and a graphical representation. Problem; you do not even ha ve to possess the actual cubes of "Instant Insanity" in order to play. The graph in Fig 9.5 describes the four cubes of "Instant Insanity.

" Could it also be used to describe a stacking FRONT W. BACK Cube 4 Cube 3 Cube 2 -. - 1. uUDe Cube I in Fig 9.6 has its b lue face at the front and white face at the back. This orientation can be shown in the graph by emphasizing the edge labeled I joining vertices B and W.

Similarly, the green-white, red-white, and white-blue front-back faces can be shown by emphasizing three other edges in the graph. The stacking shown on the left of Fig 9.6 determines the subgraph on the right.

This stacking does not solve the puzzle, however, because there are too many white faces at the back; moreover, any rotation of any of the cubes through 180 always results in more than one white face at the front or the back. This is more easily seen by observing that in the subgraph, there are too many edges joined to vertex W. On the other hand, there are exactly two edges in the subgraph (I and 4) joined to vertex B, corresponding to the fact that there is just one blue face on the front (part of the B-W pair of faces of cube 1) and just one blue face at the back (part of the W-B pair of faces of cube 4).

When the cubes are stacked with their fronts and backs correct (that is, with four different colors on the front and four different colors on the back), the corresponding subgraph will . contain all four vertices R, B, W, G; . consist of four edges, one from each cube; .

have exactly two edges or one circle meeting at each vertex.. 9.1 A Gentle Introduction Figure 9.7 shows two subgra phs, each of which corresponds to a stacking of cubes in which the fronts and backs are correct..

Two acceptable subgraphs which are not edge disjoint. Draw an arrangement of fron ts and backs (like that on the left in Fig 9.6), which is represented by the subgraph on the left in Fig 9.7.

All four colors should appear on the front and on the back. As anyone who has tried this puzzle knows, getting the front and back of the column correct is easy. It is next to get the sides correct that provides the fun( ).

Graphically, we require a second subgraph of the type described previously to represent a correct stacking of the sides. Moreover, since a given edge cannot represent both front-back and side-side at the same time, the second subgraph must be edge disjoint from the first: No edge can appear in both subgraphs. The subgraphs shown in Fig 9.

7, for instance, are not edge disjoint; they have edges 3 and 4 in common. On the other hand, the subgraphs shown in Fig 9.8 are indeed edge disjoint and so provide us with a solution to "Instant Insanity," as shown in Fig 9.

9.. Two acceptable edge disjoint subgraphs. FRONT Cube 4 Cube 3 Cube 2 Cube 1 Figure 9.9 B G R W BACK R W G B RIGHT W B G R LEFT B R W G A solution to the game of " Instant Insanity.". 282 9 Graphs 77= -. While our game of "Instant QR-Code for Java Insanity" had a unique solution, if we try to make other games by assigning colors to the faces of four cubes in other ways, it is not hard to find games where there are several winning configurations or where there is no solution at all. Also, whereas the subgraphs for our game had certain special properties, such as connectedness3 and a lack of edge crossovers, in general, be on the lookout for other types of subgraphs, three of which are depicted in Fig 9.10.

. W Figure 9.10 Possible subgraphs for other cube games. Newspapers and magazines of ten contain mathematical teasers whose solutions can be helped by drawing simple graphs. We close this section with an example of one such puzzle and include another in the exercises for this section. PROBLEM 1.

You and your buddy return home after a semester at college and are greeted at the airport by your mothers and your buddy"s two sisters. Not uncharacteristically, there is a certain amount of hugging! Later, the other five people tell you the number of hugs they got and, curiously, these numbers are all different. Assume that you and your buddy did not hug each other, your mothers did not hug each other, and your buddy"s sisters did not hug each other.

Assume also that the same two people hugged at most once. How many people did you hug How many people hugged your buddy Solution. The conditions on who did not hug whom dictate that no person hugged more than four people.

Since the other five hugged different numbers of people, this set of numbers must be {0, 1, 2, 3, 4}. Next, it is important to realize that the person who hugged four others could not be You Al BI your buddy. Why The graph to the right should help.

The four people other than you 0 : and your buddy are labeled Al, A2 , BI, B2 , and an edge indicates a hug. Think of A 1 and Buddy A2 B2 A2 as sisters (or mothers), and B1 and B2 as mothers (or sisters). Remembering that you and your buddy did not hug, the graph shows that if your buddy had four hugs, it would be impossible for one of the group Buddy, Al, A2 , Bl, B2 to have reported no hugs.

Thus, your buddy could not have had four hugs.. A graph is connected if any QR Code JIS X 0510 for Java two vertices are joined by a sequence of edges. We shall introduce the notion of connectedness formally in Section 10.1.

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