# vb.net code to generate barcode It prints the same values as shown in Table 9.2. in Java Make GTIN - 13 in Java It prints the same values as shown in Table 9.2.

It prints the same values as shown in Table 9.2.
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TRACING A RECURSIVE CALL Hand tracing the execution of a method usually helps clarify it. EXAMPLE 9.7 Tracing the Recursive Factorial Function
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Here is a trace of the call f(5) to the recursive factorial function defined in Example 9.2:
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Figure 9.1 Tracing the recursive factorial function
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The call originates in the main() function, passing 5 to the f() function. There, the value of the parameter n is 5, so it calls f(4), passing 4 to the f() function. There the value of the parameter n is 4, so it calls f(3), passing 3 to the f() function. This process continues (recursively) until the call f(1) is made from within the call f(2). There, the value of the parameter n is 1, so it returns 1 immediately, without making any more calls. Then the call f(2) returns 2*1 = 2 to the call f(3). Then the call f(3) returns 3*2 = 6 to the call f(4). Then the call f(4) returns 4*6 = 24 to the call f(5). Finally, the call f(5) returns the value 120 to main().
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The trace in Example 9.7 shows that the call f(n) to the recursive implementation of the factorial function will generate n 1 recursive calls. This is clearly very inefficient compared to the iterative implementation shown in Example 9.3. EXAMPLE 9.8 Tracing the Recursive Fibonacci Function
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The Fibonacci function (Example 9.6) is more heavily recursive than the factorial function (Example 9.2) because it includes two recursive calls. The consequences can be seen from the trace of the call
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RECURSION
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[CHAP. 9
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fib(5), shown in Figure 9.2 on page 168. The call originates in the main() function, passing 5 to the fib() function. There, the value of the parameter n is 5, so it calls fib(4) and fib(3), passing 4 and 3,
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respectively. Each of these calls then makes two more recursive calls, continuing down to the basis calls f(1) and f(0). Each of these basis calls returns 1. The recursive calls then return the sum of the two values returned to them, ultimately resulting in the value 8 being returned to main().
Figure 9.2 Tracing the recursive Fibonacci function
THE RECURSIVE BINARY SEARCH The nonrecursive binary search algorithm is given on page 31. It uses the divide-and-conquer strategy, each time splitting the sequence in half and continuing the search on one half. This is naturally recursive. EXAMPLE 9.9 The Recursive Binary Search
Here is the recursive binary search algorithm: (Precondition: s = {s0, s1, . . ., sn 1} is a sorted sequence of n ordinal values of the same type as x.) (Postcondition: either the index i is returned where si = x, or 1 is returned.) 1. If the sequence is empty, return 1. 2. Let si be the middle element of the sequence. 3. If si = x, return its index i . 4. If si < x, apply the algorithm on the subsequence that lies above si . 5. Apply the algorithm on the subsequence of s that lies below si . It is implemented in Example 9.10.
The recursive binary search runs in O(lgn) time. The running time is proportional to the number of recursive calls made. Each call processes a subsequence that is half as long as the previous one. So the number of recursive calls is the same as the number of times that n can be divided in two, namely lg n.
CHAP. 9]
RECURSION
EXAMPLE 9.10 Testing the Recursive Binary Search
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
public class TestBinarySearch { public static void main(String[] args) { int[] a = {22, 33, 44, 55, 66, 77, 88, 99}; print(a); System.out.println("search(a, 44): " + search(a, 44)); System.out.println("search(a, 50): " + search(a, 50)); System.out.println("search(a, 77): " + search(a, 77)); System.out.println("search(a, 100): " + search(a, 100)); } public static void print(int[] a) { System.out.printf("{%d", a); for (int i = 1; i < a.length; i++) { System.out.printf(", %d", a[i]); } System.out.println("}"); } public static int search(int[] a, int x) { return search(a, 0, a.length-1, x); } public static int search(int[] a, int lo, int hi, int x) { // PRECONDITION: a <= a <= ... <= a[a.length-1]; // POSTCONDITIONS: returns i; // if i >= 0, then a[i] == x; otherwise i == -1; if (lo > hi) { return -1; // basis } int i = (lo + hi)/2; if (a[i] == x) { return i; } else if (a[i] < x) { return search(a, i+1, hi, x); } else { return search(a, lo, i-1, x); } }
The output is:
{22, 33, 44, 55, 66, 77, 88, 99} search(a, 44): 2 search(a, 50): -1 search(a, 77): 5 search(a, 100): -1 The search() method returns the index of the target x: search(a, 44) returns 2 because a = 44 and search(a, 77) returns 5 because a = 77. The method returns 1 when the target is not in the array: search(a, 50) returns 1 because 50 is not in the array.
BINOMIAL COEFFICIENTS The binomial coefficients are the coefficients that result from the expansion of a binomial expression of the form (x + 1)n. For example, (x + 1)6 = x 6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1 The seven coefficients generated here are 1, 6, 15, 20, 15, 6, and 1.