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| 1 | +import java.util.InputMismatchException; |
| 2 | +import java.util.Scanner; |
| 3 | + |
| 4 | +/** |
| 5 | + * Class for finding the lowest base in which a given integer is a palindrome. |
| 6 | + * Includes auxiliary methods for converting between bases and reversing strings. |
| 7 | + * |
| 8 | + * NOTE: There is potential for error, see note at line 63. |
| 9 | + * |
| 10 | + * @author RollandMichael |
| 11 | + * @version 2017.09.28 |
| 12 | + * |
| 13 | + */ |
| 14 | +public class LowestBasePalindrome { |
| 15 | + |
| 16 | + public static void main(String[] args) { |
| 17 | + Scanner in = new Scanner(System.in); |
| 18 | + int n=0; |
| 19 | + while (true) { |
| 20 | + try { |
| 21 | + System.out.print("Enter number: "); |
| 22 | + n = in.nextInt(); |
| 23 | + break; |
| 24 | + } catch (InputMismatchException e) { |
| 25 | + System.out.println("Invalid input!"); |
| 26 | + in.next(); |
| 27 | + } |
| 28 | + } |
| 29 | + System.out.println(n+" is a palindrome in base "+lowestBasePalindrome(n)); |
| 30 | + System.out.println(base2base(Integer.toString(n),10, lowestBasePalindrome(n))); |
| 31 | + } |
| 32 | + |
| 33 | + /** |
| 34 | + * Given a number in base 10, returns the lowest base in which the |
| 35 | + * number is represented by a palindrome (read the same left-to-right |
| 36 | + * and right-to-left). |
| 37 | + * @param num A number in base 10. |
| 38 | + * @return The lowest base in which num is a palindrome. |
| 39 | + */ |
| 40 | + public static int lowestBasePalindrome(int num) { |
| 41 | + int base, num2=num; |
| 42 | + int digit; |
| 43 | + char digitC; |
| 44 | + boolean foundBase=false; |
| 45 | + String newNum = ""; |
| 46 | + String digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; |
| 47 | + |
| 48 | + while (!foundBase) { |
| 49 | + // Try from bases 2 to num (any number n in base n is 1) |
| 50 | + for (base=2; base<num2; base++) { |
| 51 | + newNum=""; |
| 52 | + while(num>0) { |
| 53 | + // Obtain the first digit of n in the current base, |
| 54 | + // which is equivalent to the integer remainder of (n/base). |
| 55 | + // The next digit is obtained by dividing n by the base and |
| 56 | + // continuing the process of getting the remainder. This is done |
| 57 | + // until n is <=0 and the number in the new base is obtained. |
| 58 | + digit = (num % base); |
| 59 | + num/=base; |
| 60 | + // If the digit isn't in the set of [0-9][A-Z] (beyond base 36), its character |
| 61 | + // form is just its value in ASCII. |
| 62 | + |
| 63 | + // NOTE: This may cause problems, as the capital letters are ASCII values |
| 64 | + // 65-90. It may cause false positives when one digit is, for instance 10 and assigned |
| 65 | + // 'A' from the character array and the other is 65 and also assigned 'A'. |
| 66 | + |
| 67 | + // Regardless, the character is added to the representation of n |
| 68 | + // in the current base. |
| 69 | + if (digit>=digits.length()) { |
| 70 | + digitC=(char)(digit); |
| 71 | + newNum+=digitC; |
| 72 | + continue; |
| 73 | + } |
| 74 | + newNum+=digits.charAt(digit); |
| 75 | + } |
| 76 | + // Num is assigned back its original value for the next iteration. |
| 77 | + num=num2; |
| 78 | + // Auxiliary method reverses the number. |
| 79 | + String reverse = reverse(newNum); |
| 80 | + // If the number is read the same as its reverse, then it is a palindrome. |
| 81 | + // The current base is returned. |
| 82 | + if (reverse.equals(newNum)) { |
| 83 | + foundBase=true; |
| 84 | + return base; |
| 85 | + } |
| 86 | + } |
| 87 | + } |
| 88 | + // If all else fails, n is always a palindrome in base n-1. ("11") |
| 89 | + return num-1; |
| 90 | + } |
| 91 | + |
| 92 | + private static String reverse(String str) { |
| 93 | + String reverse = ""; |
| 94 | + for(int i=str.length()-1; i>=0; i--) { |
| 95 | + reverse += str.charAt(i); |
| 96 | + } |
| 97 | + return reverse; |
| 98 | + } |
| 99 | + |
| 100 | + private static String base2base(String n, int b1, int b2) { |
| 101 | + // Declare variables: decimal value of n, |
| 102 | + // character of base b1, character of base b2, |
| 103 | + // and the string that will be returned. |
| 104 | + int decimalValue = 0, charB2; |
| 105 | + char charB1; |
| 106 | + String output=""; |
| 107 | + // Go through every character of n |
| 108 | + for (int i=0; i<n.length(); i++) { |
| 109 | + // store the character in charB1 |
| 110 | + charB1 = n.charAt(i); |
| 111 | + // if it is a non-number, convert it to a decimal value >9 and store it in charB2 |
| 112 | + if (charB1 >= 'A' && charB1 <= 'Z') |
| 113 | + charB2 = 10 + (charB1 - 'A'); |
| 114 | + // Else, store the integer value in charB2 |
| 115 | + else |
| 116 | + charB2 = charB1 - '0'; |
| 117 | + // Convert the digit to decimal and add it to the |
| 118 | + // decimalValue of n |
| 119 | + decimalValue = decimalValue * b1 + charB2; |
| 120 | + } |
| 121 | + |
| 122 | + // Converting the decimal value to base b2: |
| 123 | + // A number is converted from decimal to another base |
| 124 | + // by continuously dividing by the base and recording |
| 125 | + // the remainder until the quotient is zero. The number in the |
| 126 | + // new base is the remainders, with the last remainder |
| 127 | + // being the left-most digit. |
| 128 | + |
| 129 | + // While the quotient is NOT zero: |
| 130 | + while (decimalValue != 0) { |
| 131 | + // If the remainder is a digit < 10, simply add it to |
| 132 | + // the left side of the new number. |
| 133 | + if (decimalValue % b2 < 10) |
| 134 | + output = Integer.toString(decimalValue % b2) + output; |
| 135 | + // If the remainder is >= 10, add a character with the |
| 136 | + // corresponding value to the new number. (A = 10, B = 11, C = 12, ...) |
| 137 | + else |
| 138 | + output = (char)((decimalValue % b2)+55) + output; |
| 139 | + // Divide by the new base again |
| 140 | + decimalValue /= b2; |
| 141 | + } |
| 142 | + return output; |
| 143 | + } |
| 144 | +} |
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