Hello all,
Let us assume n = 2. I want to print like this:
0 0
0 1
1 0
1 1
Let us assume n = 3. I want to print like this:
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
In the same way I want to print for any value of "n". Is there any algorithm for this?
thanks and regards,
dakamath
Yes, it is complex. But fortunately we have "Integer.toBinaryString" which serves my current immediate requirement!
Original requirement is something like this:
I have "n" variables, each having 2 different values. I need to have all possible combinations of these variables.
thanks,
dakamath
>In the same way I want to print for any value of "n". Is there any algorithm for this?
We all know the algorithm for this. We learned it in grade school.
Think about how a gradeschool student would add 999 + 3 on a piece of paper. 9+3 = 12, carry the 1. 9+1=10, carry the 1 ...
The only difference is you're adding in base 2. And you're always adding 1. Pretty easy.
To sum up:
1. Add value.
2. Carry exess into next column.
> You obviously think what you said is simple to
> implement.
Not trivial, but simple.
>Why don't you post your code for a binary
> adder and then compare it to the code I have posted...
Here is a general purpose base-n adder. So long as N<Integer.MAX_VALUE/3;
It's not tailored to be a solution to a specific problem. Just an example of how to implement such a thing.
For example, to iterate through binary values
Use base 2 and add {0, ... 0, 1} repeadedly.
Optimizations (such as caller supplying and reusing result array) are easy enough to do for a customized implementation.
public class Adder {
private int base;
public Adder(int base)
{
this.base = base;
}
public int[] add(int[] n1, int[] n2)
{
if (n1.length!=n2.length)
throw new IllegalArgumentException("n1 and n2 must be same length");
int[] result = new int[n1.length +1 ]; //result may need extra digit.
int carry = 0;
for (int i=result.length-1; i>=1; i--)
{
result[i] = carry;
//arrays are right aligned, "Most Significant Int" first (at 0)
//we must shift to account for the result being 1 larger
//hence the [i-1]
result[i]+=n1[i-1];//
result[i]+=n2[i-1];// Algorithm: add the values
//
carry = result[i]/base;// carry the extra
result[i]%=base;//
}
result[0] = carry;
return result;
}
}
Don't forget you need this code too!
public static void main(String[] args) {
Adder adder = new Adder(2);
int[] n1 = new int[4];
int[] n2 = new int[] { 0, 0, 0, 1};
for (int i = 0; i < Math.pow(2, n1.length); i++) {
for (int j = 0; j < n1.length; j++)
System.out.print(n1[j]);
System.out.println();
int[] result = adder.add(n1, n2);
for (int j = 0; j < n1.length; j++)
n1[j] = result[j+1];
}
}