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// recursive part
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Here is a simple test driver for the factorial method:
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public static void main(String[] args) { for (int n=0; n<10; n++) { System.out.println("f("+n+") = "+f(n)); } }
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It prints the same values as shown in Table 9.1.
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EXAMPLE 9.3 Iterative Implementation of the Factorial Function
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The factorial function is also easy to implement iteratively:
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public static int f(int n) { int f = 1; for (int i = 2; i <= n; i++) { f *= i; } return f; }
Note that the function header is identical to that used in Example 9.2; only the body is different. This allows us to use the same test driver for both implementations. The output should be the same.
BASIS AND RECURSIVE PARTS To work correctly, every recursive function must have a basis and a recursive part. The basis is what stops the recursion. The recursive part is where the function calls itself. EXAMPLE 9.4 The Basis and Recursive Parts of the Factorial Function
In the Java method that implements the factorial function in Example 9.2, the basis and the recursive parts are labeled with comments. The recursive part invokes the method, passing a smaller value of n. So starting with a positive value like 5, the values on the successive invocations will be 4, 3, 2, 1, and 0. When 0 is passed, the basis executes, thereby stopping the recursion and beginning the chain of returns, returning 1, 1, 2, 6, 24, and finally 120.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377
EXAMPLE 9.5 The Fibonacci Numbers
The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . Each number after the second is the sum of the two preceding numbers. This is a naturally recursive definition: Fn = 0, if n = 0 1, if n = 1 F n 1 + F n 2 , if n > 1
The first 15 values of the Fibonacci sequence are shown in Table 9.2. The first two values, F0 and F1, are defined by the first two parts of the definition: F0 = 0 (for n = 0) and F1 = 1 (for n = 1). These two parts form the basis of the recursion. All the other values are defined by the recursive part of the definition:
Table 9.2 Fibonacci numbers
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RECURSION
For n = 2, F2 = Fn = Fn 1 + Fn 2 = F(2) 1 + F(2) 2 = F1 + F0 = 1 + 0 = 1. For n = 3, F3 = Fn = Fn 1 + Fn 2 = F(3) 1 + F(3) 2 = F2 + F1 = 1 + 1 = 2. For n = 4, F4 = Fn = Fn 1 + Fn 2 = F(4) 1 + F(4) 2 = F3 + F2 = 2 + 1 = 3. For n = 5, F5 = Fn = Fn 1 + Fn 2 = F(5) 1 + F(5) 2 = F4 + F3 = 3 + 2 = 5. For n = 6, F6 = Fn = Fn 1 + Fn 2 = F(6) 1 + F(6) 2 = F5 + F4 = 5 + 3 = 8. For n = 7, F7 = Fn = Fn 1 + Fn 2 = F(7) 1 + F(7) 2 = F6 + F5 = 8 + 5 = 13.
EXAMPLE 9.6 Recursive Implementation of the Fibonacci Function
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public static int fib(int n) { if (n < 2) { return n; // basis } return fib(n-1) + fib(n-2); // recursive part }
Here is a simple test driver for the Fibonacci method:
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public static void main(String[] args) { for (int n = 0; n < 16; n++) { System.out.println("fib(" + n + ") = " + fib(n)); } }
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