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vb.net code to generate barcode Figure 10.3 A leaftoroot path in Java
Figure 10.3 A leaftoroot path EAN13 Reader In Java Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications. Draw EAN13 In Java Using Barcode encoder for Java Control to generate, create European Article Number 13 image in Java applications. TREES
Scan EAN / UCC  13 In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. Bar Code Encoder In Java Using Barcode encoder for Java Control to generate, create barcode image in Java applications. [CHAP. 10
Decode Barcode In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Create European Article Number 13 In Visual C# Using Barcode generator for VS .NET Control to generate, create EAN13 image in .NET applications. The degree of a node is the number of its children. In Example 10.2, b has degree 1, d has degree 0, and h has degree 5. The order of a tree is the maximum degree among all of its nodes. A tree is said to be full if all of its internal nodes have the same degree and all of its leaves are at the same level. The tree shown in Figure 10.4 is a full Figure 10.4 A full tree tree of degree 3. Note that it has a total of 40 nodes. dh + 1 1 Theorem 10.2 The full tree of order d and height h has  nodes. d 1 For a proof, see Problem 10.1 on page 194. Corollary 10.1 The height of a full tree of order d and size n is h = log d (nd n + 1) 1. dh + 1 1 Corollary 10.2 The number of nodes in any tree of height h is at most  where d is d 1 the maximum degree among its nodes. DECISION TREES A decision tree is a tree diagram that summarizes all the different possible stages of a process that solves a problem by means of a sequence of decisions. Each internal node is labeled with a question, each arc is labeled with an answer to its question, and each leaf node is labeled with the solution to the problem. EXAMPLE 10.3 Finding the Counterfeit Coin GS1  13 Creation In VS .NET Using Barcode creation for ASP.NET Control to generate, create EAN13 image in ASP.NET applications. Encode EAN13 Supplement 5 In Visual Studio .NET Using Barcode generation for .NET Control to generate, create EAN13 image in .NET framework applications. Five coins that appear identical are to be tested to determine which one of them is counterfeit. The only feature that distinguishes the counterfeit coin is that it weighs less than the legitimate coins. The only available test is to weigh one subset of the coins against another. How should the subsets be chosen to find the counterfeit Drawing EAN 13 In VB.NET Using Barcode generator for Visual Studio .NET Control to generate, create EAN / UCC  13 image in Visual Studio .NET applications. 1D Printer In Java Using Barcode creator for Java Control to generate, create Linear Barcode image in Java applications. Figure 10.5 A decision tree
Painting UCC  12 In Java Using Barcode generator for Java Control to generate, create UPCA image in Java applications. Code 128A Creator In Java Using Barcode printer for Java Control to generate, create USS Code 128 image in Java applications. CHAP. 10] British Royal Mail 4State Customer Code Drawer In Java Using Barcode generator for Java Control to generate, create British Royal Mail 4State Customer Code image in Java applications. 1D Barcode Creator In VB.NET Using Barcode generation for .NET Control to generate, create 1D image in VS .NET applications. TREES
EAN 128 Drawer In Java Using Barcode generator for Android Control to generate, create EAN / UCC  14 image in Android applications. Making UCC.EAN  128 In Java Using Barcode generator for BIRT reports Control to generate, create EAN 128 image in BIRT reports applications. In the decision tree shown in Figure 10.5, the symbol means to compare the weights of the two operands. So, for example, {a, b} {d, e} means to weight coins a and b against coins d and e. Decode Barcode In Visual Basic .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in Visual Studio .NET applications. Create Barcode In .NET Framework Using Barcode encoder for ASP.NET Control to generate, create bar code image in ASP.NET applications. TRANSITION DIAGRAMS A transition diagram is a tree or graph (see 15) whose internal nodes represent different states or situations that may occur during a multistage process. As in a decision tree, each leaf represents a different outcome from the process. Each branch is labeled with the conditional probability that the resulting child event will occur, given that the parent event has occurred. EXAMPLE 10.4 The Game of Craps 2D Barcode Generation In Visual Basic .NET Using Barcode maker for Visual Studio .NET Control to generate, create 2D Barcode image in .NET applications. Encode DataMatrix In VB.NET Using Barcode generator for VS .NET Control to generate, create Data Matrix image in Visual Studio .NET applications. The game of craps is a dice game played by two players, X and Y. First X tosses the pair of dice. If the sum of the dice is 7 or 11, X wins the game. If the sum is 2, 3, or 12, Y wins. Otherwise, the sum is designated as the point, to be matched by another toss. So if neither player has won on the first toss, then the dice are tossed repeatedly until either the point comes up or a 7 comes up. If a 7 comes up first, Y wins. Otherwise, X wins when the point comes up. The transition diagram shown in Figure 10.6 models the game of craps. Figure 10.6 A decision tree for the game of craps
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[CHAP. 10
When a pair of dice is tossed, there are 36 possible outcomes (6 outcomes on the first die, and 6 outcomes on the second for each outcome on the first die). Of those 36 outcomes, 1 will produce a sum of 2 (1 + 1), 2 will produce a sum of 3 (1 + 2 or 2 + 1), and 1 will produce a sum of 12 (6 + 6). So there are a total of 4 chances out of 36 of the event 2, 3, or 12 happening. That s a probability of 4/36 = 1/9. Similarly, there are 6 ways that a sum of 7 will occur and 2 ways that a sum of 11 will occur, so the probability of the event 7 or 11 is 8/36 = 2/9. The other probabilities on the first level of the tree are computed similarly. To see how the probabilities are computed for the second level of the tree, consider the case where the point is 4. If the next toss comes up 4, X wins. If it comes up 7, Y wins. Otherwise, that step is repeated. The transition diagram shown in Figure 10.7 summarizes those three possibilities. The probabilities 1/12, 1/6, and 3/4 are computed as shown in the transition diagram in Figure 10.7: Figure 10.7 The game of craps P(4) = 3/36 = 1/12 P(7) = 6/36 = 1/3 P(2,3,5,6,8,9,10,11, or 12) = 27/36 = 3/4 So once the point 4 has been established on the first toss, X has a probability of 1/12 of winning on the second toss and a probability of 3/4 of getting to the third toss. So once the point 4 has been established on the first toss, X has a probability of (3/4)(1/12) of winning on the third toss and a probability of (3/4)(3/4) of getting to the fourth toss. Similarly, once the point 4 has been established on the first toss, X has a probability of (3/4)(1/12) + (3/4)(3/4)(1/12) of winning on the fourth toss, and so on. Summing these partial probabilities, we find that once the point 4 has been established on the first toss, the probability that X wins on any toss thereafter is 1 3 1 3 21 3 31 3 41 3 51  P 4 =  +   +   +   +   +   + 12 4 12 4 12 4 12 4 12 4 12 = 3 1 4 1 12 = 1 4 1 = 3 112 This calculation applies the formula for geometric series. (See page 323.) If the probability is 1/3 that X wins once the point 4 has been established on the first toss, the probability that Y wins at that point must be 2/3. The other probabilities at the second level are computed similarly. Now we can calculate the probability that X wins the game from the main transition diagram: 1 5 5 1 1 2 1 P =  +  P 4 +  P 5 +  P 6 +  P8 +  P9 +  P 10 9 36 36 9 12 9 12 2 1 1 1 2 5 5 5 5 1 2 1 1       =  +   +   +   +   +   +  9 12 3 9 5 36 11 36 11 9 5 12 3 244 = 495 = 0.4929 So the probability that X wins is 49.29 percent, and the probability that Y wins is 50.71 percent. CHAP. 10]

