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Thread: Vampire Numbers
 06292010, 03:31 PM #1
Vampire Numbers
This is an exercise in the book "Thinking in Java" and it says I need to find all the 4 digit Vampire Numbers
I have indeed finished the exercise and produced working code:
Java Code:import java.util.Arrays; class Vampire { public static void main(String[] args) { for(int i=11; i<100; i++) { for(int j=i; j<100; j++) { int k = i * j; String kStr = Integer.toString(k); String checkStr = Integer.toString(i) + Integer.toString(j); //if(kStr.length() != 4) break; char[] kChar = kStr.toCharArray(); char[] checkChar = checkStr.toCharArray(); Arrays.sort(kChar); Arrays.sort(checkChar); boolean isVampire = Arrays.equals(kChar, checkChar); if(isVampire) { System.out.println(i + " * " + j + " = " + k); } } } } }
 06292010, 04:17 PM #2
so 
i = 11
j = 11
k = 121
kStr = 121
checkStr = 1111
if(kStr.length() != 4) break; (Right now that would be false) Whats the point of this line?
shouldn't it be
Java Code:if(kStr.length() != 4){ break; } else { char[] kChar = kStr.toCharArray(); char[] checkChar = checkStr.toCharArray(); Arrays.sort(kChar); Arrays.sort(checkChar); boolean isVampire = Arrays.equals(kChar, checkChar); if(isVampire) { System.out.println(i + " * " + j + " = " + k); }
:rolleyes: ~ Sno ~ :rolleyes:
'~ B.S. Computer Science ~'
 06292010, 04:18 PM #3
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If you break from the inner loop if the product k doesn't have four digits you may stop the search too early, e.g. 15x15 has only three digits (225); if you break out of the loop you'll never find 15x93 which is a Vampire number. You can break out of the loop when k has more than four digits (which is never going to happen).
kind regards,
Jos
edit: you can start at j=Math.max(1000/i, i)Last edited by JosAH; 06292010 at 04:22 PM.
 06292010, 04:23 PM #4
 06292010, 04:25 PM #5
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 06292010, 04:36 PM #6
Well the way he had it,
I was taught, that if you did not have brackets, it would only effect the line that it is on, and the one below,
two lines below it wouldn't consider it to be part of the if statement.
so I figured his if statement was only affecting the break, and not dealing with anything else.:rolleyes: ~ Sno ~ :rolleyes:
'~ B.S. Computer Science ~'
 06292010, 04:44 PM #7
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 06292010, 04:51 PM #8
Good thinking! That will certainly eliminate all the 3 Digit results
 06292010, 06:09 PM #9
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 07022010, 12:53 PM #14
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 07022010, 03:39 PM #15Member
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Here it is:
Java Code:public class E10_Vampire { public static void main(String[] args) { int[] startDigit = new int[4]; int[] productDigit = new int[4]; for(int num1 = 10; num1 <= 99; num1++) for(int num2 = num1; num2 <= 99; num2++) { // Pete Hartley's theoretical result: // If x·y is a vampire number then // x·y == x+y (mod 9) if((num1 * num2) % 9 != (num1 + num2) % 9) continue; int product = num1 * num2; startDigit[0] = num1 / 10; startDigit[1] = num1 % 10; startDigit[2] = num2 / 10; startDigit[3] = num2 % 10; productDigit[0] = product / 1000; productDigit[1] = (product % 1000) / 100; productDigit[2] = product % 1000 % 100 / 10; productDigit[3] = product % 1000 % 100 % 10; int count = 0; for(int x = 0; x < 4; x++) for(int y = 0; y < 4; y++) { if(productDigit[x] == startDigit[y]) { count++; productDigit[x] = 1; startDigit[y] = 2; if(count == 4) System.out.println(num1 + " * " + num2 + " : " + product); } } } } } /* Output: 15 * 93 : 1395 21 * 60 : 1260 21 * 87 : 1827 27 * 81 : 2187 30 * 51 : 1530 35 * 41 : 1435 80 * 86 : 6880 *///:~
 07022010, 03:54 PM #16
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 08032011, 12:26 PM #19
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