diff --git a/Misc/LowestBasePalindrome.java b/Misc/LowestBasePalindrome.java new file mode 100644 index 000000000000..fce2d2168465 --- /dev/null +++ b/Misc/LowestBasePalindrome.java @@ -0,0 +1,144 @@ +import java.util.InputMismatchException; +import java.util.Scanner; + +/** + * Class for finding the lowest base in which a given integer is a palindrome. + * Includes auxiliary methods for converting between bases and reversing strings. + * + * NOTE: There is potential for error, see note at line 63. + * + * @author RollandMichael + * @version 2017.09.28 + * + */ +public class LowestBasePalindrome { + + public static void main(String[] args) { + Scanner in = new Scanner(System.in); + int n=0; + while (true) { + try { + System.out.print("Enter number: "); + n = in.nextInt(); + break; + } catch (InputMismatchException e) { + System.out.println("Invalid input!"); + in.next(); + } + } + System.out.println(n+" is a palindrome in base "+lowestBasePalindrome(n)); + System.out.println(base2base(Integer.toString(n),10, lowestBasePalindrome(n))); + } + + /** + * Given a number in base 10, returns the lowest base in which the + * number is represented by a palindrome (read the same left-to-right + * and right-to-left). + * @param num A number in base 10. + * @return The lowest base in which num is a palindrome. + */ + public static int lowestBasePalindrome(int num) { + int base, num2=num; + int digit; + char digitC; + boolean foundBase=false; + String newNum = ""; + String digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; + + while (!foundBase) { + // Try from bases 2 to num (any number n in base n is 1) + for (base=2; base0) { + // Obtain the first digit of n in the current base, + // which is equivalent to the integer remainder of (n/base). + // The next digit is obtained by dividing n by the base and + // continuing the process of getting the remainder. This is done + // until n is <=0 and the number in the new base is obtained. + digit = (num % base); + num/=base; + // If the digit isn't in the set of [0-9][A-Z] (beyond base 36), its character + // form is just its value in ASCII. + + // NOTE: This may cause problems, as the capital letters are ASCII values + // 65-90. It may cause false positives when one digit is, for instance 10 and assigned + // 'A' from the character array and the other is 65 and also assigned 'A'. + + // Regardless, the character is added to the representation of n + // in the current base. + if (digit>=digits.length()) { + digitC=(char)(digit); + newNum+=digitC; + continue; + } + newNum+=digits.charAt(digit); + } + // Num is assigned back its original value for the next iteration. + num=num2; + // Auxiliary method reverses the number. + String reverse = reverse(newNum); + // If the number is read the same as its reverse, then it is a palindrome. + // The current base is returned. + if (reverse.equals(newNum)) { + foundBase=true; + return base; + } + } + } + // If all else fails, n is always a palindrome in base n-1. ("11") + return num-1; + } + + private static String reverse(String str) { + String reverse = ""; + for(int i=str.length()-1; i>=0; i--) { + reverse += str.charAt(i); + } + return reverse; + } + + private static String base2base(String n, int b1, int b2) { + // Declare variables: decimal value of n, + // character of base b1, character of base b2, + // and the string that will be returned. + int decimalValue = 0, charB2; + char charB1; + String output=""; + // Go through every character of n + for (int i=0; i9 and store it in charB2 + if (charB1 >= 'A' && charB1 <= 'Z') + charB2 = 10 + (charB1 - 'A'); + // Else, store the integer value in charB2 + else + charB2 = charB1 - '0'; + // Convert the digit to decimal and add it to the + // decimalValue of n + decimalValue = decimalValue * b1 + charB2; + } + + // Converting the decimal value to base b2: + // A number is converted from decimal to another base + // by continuously dividing by the base and recording + // the remainder until the quotient is zero. The number in the + // new base is the remainders, with the last remainder + // being the left-most digit. + + // While the quotient is NOT zero: + while (decimalValue != 0) { + // If the remainder is a digit < 10, simply add it to + // the left side of the new number. + if (decimalValue % b2 < 10) + output = Integer.toString(decimalValue % b2) + output; + // If the remainder is >= 10, add a character with the + // corresponding value to the new number. (A = 10, B = 11, C = 12, ...) + else + output = (char)((decimalValue % b2)+55) + output; + // Divide by the new base again + decimalValue /= b2; + } + return output; + } +} pFad - Phonifier reborn

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