vb.net print barcode zebra Figure 10.3 A leaf-to-root path in Java

Encode Data Matrix ECC200 in Java Figure 10.3 A leaf-to-root path

Figure 10.3 A leaf-to-root path
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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
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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
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Figure 10.5 A decision tree
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operands. So, for example, {a, b} TRANSITION DIAGRAMS
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In the decision tree shown in Figure 10.5, the symbol means to compare the weights of the two {d, e} means to weight coins a and b against coins d and e.
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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
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
TREES
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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 1----12
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.
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