vb.net code to generate barcode int lg(int if (n == return } return 1 } n) { 1) { 0; + lg(n/2); // basis // recursion in Java

Encoding EAN-13 Supplement 5 in Java int lg(int if (n == return } return 1 } n) { 1) { 0; + lg(n/2); // basis // recursion

int lg(int if (n == return } return 1 } n) { 1) { 0; + lg(n/2); // basis // recursion
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CHAP. 9]
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boolean isPalindrome(String s) { int len = s.length(); if (len < 2) { return true; } else if (s.charAt(0) != s.charAt(len-1)) return false; } else if (len == 2) { return true; } else { return isPalindrome(s.substring(1,len-1)); } }
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// basis // basis
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// basis // recursion
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String binary(int n) { String s; if (n%2 == 0) { s = "0"; } else { s = "1"; } if (n < 2) { return s; } return binary(n/2) + s; }
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// basis // recursion
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A recursive function that converts decimal to hexadecimal:
String hexadecimal(int n) { if (n < 16) { return Integer.toString(n%16) } return hexadecimal(n/16) + s; // recursion }
A recursive function that prints permutations:
void print(String str) { print("",str); } void print(String left, String right) { int n = right.length(); if (n == 0) { return; } else if (n == 1) System.out.println(left+right); return; } StringBuilder buf = new StringBuilder(right); for (int i = 0; i < n; i++) { char temp = s.charAt(i); s.setCharAt(i, s.charAt(0)); s.setCharAt(0, temp); print(left+temp, s.substring(1, n)); } }
RECURSION
Iterative implementation of the Fibonacci function:
int fib(int n) { if (n < 2) { return n; } int f0 = 0, f1 = 1, f = f0+f1; for (int i = 2; i < n; i++) { f0 = f1; f1 = f; f = f0 + f1; } return f; }
[CHAP. 9
The Ackermann function:
int ackermann(int m, int n) { if (m == 0) { return n + 1; } else if (n == 0) { return ackermann(m - 1, 1); } else { return ackermann(m - 1, ackermann(m, n - 1)); } // basis // basis // recursion
Consider the relationship c(8,3) = 56 = 35 + 21 = c(7,3) + c(7,2) from the expansion of (x + 1)8: (x + 1)8= (x + 1)(x + 1)7 = (x + 1)(x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1) = x8 + 7x7 + 21x6 + 35x5 + 35x4 + 21x3 + 7x2 + x + x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1 8 + 8x7 + 28x6 + 56x5 + 70x4 + 56x3 + 28x2 + 7x + 1 =x The coefficient c(8,3) is for the x5 term, which is 35x5 + 21x5 = 56x5. The sum 35x5 + 21x5 came from (x)(35x4) and (1)(21x5). So those coefficients are 35 = c(7,3) and 21 = c(7,2). The general proof is based upon the same argument: c(n,k) is the coefficient of the term xk in the expansion of (x + 1)n. Since (x + 1)n = (x + 1)(x + 1)n 1, that term comes from the sum (x)(c(n 1, k 1) x k 1) + (1)(c(n 1, k)x k) = (c(n 1, k 1) + c(n 1, k))x k Therefore c(n, k) = c(n 1, k 1) + c(n 1, k). Figure 9.7 shows the trace of the call gcd(616, 231):
Figure 9.7 Tracing the Euclidean algorithm
Iterative implementation of the Euclidean algorithm:
int gcd(int m, int n) { while (m != n) { if (m < n) { n -= m; } else { m -= n; } } return n; } // INVARIANT: gcd(m,n)
CHAP. 9]
RECURSION
Recursive implementation of the Euclidean algorithm using the remainder operator:
int gcd(int m, int n) { if (m == 0) { return n; } else if (n == 0) { return m; } else if (m < n) { return gcd(m, n%m); } else { return gcd(m%n, n); } } // basis // basis // recursion // recursion
Iterative implementation of the Euclidean algorithm using the remainder operator:
int gcd(int m, int n) { while (n > 0) { // INVARIANT: gcd(m,n) int r = m%n; m = n; n = r; } return m; }
To prove that the recursive implementation of the Fibonacci function is correct, first verify the basis. The calls fib(0) and fib(1) return the correct values 0 and 1 because of the first line
if (n < 2) { return n; }
Next, we assume that the function returns the correct value for all integers less than some n > 1. Then the second line
return fib(n-1) + fib(n-2);
will return the correct value because (by the inductive hypothesis) the calls fib(n-1) and fib(n-2) return the correct values for Fn 1 and Fn 2, respectively, and Fn = Fn 1 + Fn 2 by definition. Note that the basis here requires the verification of the first two steps in the sequence because the recurrence relation Fn = Fn 1 + Fn 2 applies only for n > 1. 9.21 If n = 1, then the basis executes, returning a[0] which is the maximum element because it is the only element. If n > 1, then the function correctly computes the maximum m of the first n 1 elements (by the inductive hypothesis). If the condition (a[n-1] > m) is true, then that element a[n-1] is returned, and it is the largest because it is larger than m, which is larger than all the others. On the other hand, if the condition (a[n-1] > m) is false, then m is returned, and that is the largest because it is not smaller than a[m-1], and it is the largest among all the others. If n = 1, then the basis executes, returning a[0] which is the maximum element because it is the only element. If n > 1, then the function correctly computes the maxima m1 and m2 of the first and second halves of the array (by the inductive hypothesis). One of these two numbers is the correct maximum for the entire array. The larger is returned. If n = 1, then the basis executes, returning 0, which is the number of times n can be divided by 2. If n > 1, then the function correctly computes the number of times n/2 can be divided by 2 (by the inductive hypothesis). This is 1 less than the number of times n can be divided by 2, so the value returned, 1 + lg(n/2), is correct. First, we prove the conjecture that the call print(left, right) will print n! distinct strings, all having the same prefix string left, where n = right.length(). If n = 1, the method prints left+right and returns; that is 1! (distinct) string. Assume that when right.length() = n 1, the call print(left,right) prints (n 1)! distinct strings all having the same left prefix string. Then, when right.length() = n, the for loop makes n calls of the form print(left+temp,ss), where temp is a distinct character and ss = s.substring(1,n). Since the length of s.substring(1,n) is n 1, each of those calls will print (n 1)! distinct strings all having the same left+temp prefix string.
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