import java.util.Arrays;
/**
* Created with IntelliJ IDEA.
* Print the N-bit binary reflected Gray code using recursion.
*
* Gray Code of 3:
* 000
* 001
* 011
* 010
* 110
* 111
* 101
* 100
*/
public class GrayCode {
public static void main(String[] args) {
int N = 3;
grayCode(N, new boolean[N]);
grayCode(N);
}
public static void grayCode(int n, boolean[] show) {
if (n == 0) {
showCode(show);
} else {
show[n - 1] = false;
grayCode(n - 1, show);
show[n - 1] = true;
yargCode(n - 1, show);
}
}
// append reverse of order n gray code
public static void yargCode(int n, boolean[] show) {
if (n == 0) {
showCode(show);
} else {
show[n - 1] = true;
grayCode(n - 1, show);
show[n - 1] = false;
yargCode(n - 1, show);
}
}
public static void showCode(boolean[] show) {
String code = "";
for (boolean yes : show) {
if (yes)
code = "1" + code;
else
code = "0" + code;
}
System.out.println(code);
// System.out.println(Integer.parseInt(code, 2));
}
/*
** The BinaryToGray method is base on the bit of gray cod loop occur: 0,1,1,0
** Hence, we could calculate each bits of gray code by mod 4 to find the matching value
**
** However, after search on the website, the convert formula is just simply as (i>>1) ^ i.
** The ith bit is equal to ith bit plus (i+1)th bit with no carry.
*/
public static String[] grayCode(int n) {
int[] code = new int[1 > 1) ^ i);
codeString[i] = toBinaryString((i >> 1) ^ i);
}
System.out.println(Arrays.toString(codeString));
System.out.println(Arrays.toString(code));
return codeString;
}
public static String toBinaryString(int integer) {
StringBuilder builder = new StringBuilder();
int temp;
while (integer >= 0) { // 0 case
temp = integer;
integer = (temp >> 1);
builder.append(String.valueOf(temp % 2));
// if insert at 0, no need to reverse
// builder.insert(0, String.valueOf(temp % 2));
if(integer ==0)
break;
}
return builder.reverse().toString();
}
}