I need to be able to create a boolean array of one combination and run it through a program to see if it works. If not then I dispose of it and go to the next combination. My issue is that I don't know how to create this array because n can be equal anywhere from 1-1000. So I was planning on using Integer.toBinaryString but that won't work due to its too big when it gets to past 32. Any help would be greatful.
Thanks!
The "accepted answer" states that
Tested and this will work for high values of n, such as 10000 and so on.
But this is incorrect.
public static void main(String[] args) {
final int n = 3;
for (int i = 0; i < Math.pow(2, n); i++) {
String bin = Integer.toBinaryString(i);
while (bin.length() < n)
bin = "0" + bin;
char[] chars = bin.toCharArray();
boolean[] boolArray = new boolean[n];
for (int j = 0; j < chars.length; j++) {
boolArray[j] = chars[j] == '0' ? true : false;
}
System.out.println(Arrays.toString(boolArray));
}
}
When n > 31 it will loop forever repeating the first 2^31 combinations since i will overflow and will never reach Math.pow(2, n). You can easily test this with
public static void main2(String[] args){
int n = 32;
for (int i = 0; i < Math.pow(2, n); i++){
if (i == Integer.MIN_VALUE) {
// i overflows
System.out.println("i exceeded Integer.MAX_VALUE");
}
}
}
Code above will indefinitely print i exceeded Integer.MAX_VALUE
However this can easily be corrected using BigInteger or a similar data structure for looping. The code below will work for n <= Integer.MAX_VALUE
public static void main(String[] args) {
final int n = 32;
BigInteger bi = BigInteger.ZERO;
BigDecimal rows = new BigDecimal(Math.pow(2, n));
while (bi.compareTo(rows.toBigInteger()) < 0) {
String bin = bi.toString(2);//Integer.toBinaryString(i);
while (bin.length() < n)
bin = "0" + bin;
char[] chars = bin.toCharArray();
boolean[] boolArray = new boolean[n];
for (int j = 0; j < chars.length; j++) {
boolArray[j] = chars[j] == '0' ? true : false;
}
System.out.println(Arrays.toString(boolArray));
bi = bi.add(BigInteger.ONE);
}
}
boolean[] boolArray = new boolean[n]; long mask = 1; for (int j = 0; j < n; j++) { boolArray[j] = (i & mask) > 0; mask <<= 1; } (Oh, do note that I used longs instead of ints or bigintegers, this means it will work for n <= 63, also the speed comparison was compared to the example using integers and strings, so there will be an even better speedup compared to the BigInteger version (but of course there will be the n limit in this case)I've found the answer to your problem on another SO question, and I've adapted it for you:
public class Foo {
public static void main(String[] args) {
final int n = 3;
for (int i = 0; i < Math.pow(2, n); i++) {
String bin = Integer.toBinaryString(i);
while (bin.length() < n)
bin = "0" + bin;
char[] chars = bin.toCharArray();
boolean[] boolArray = new boolean[n];
for (int j = 0; j < chars.length; j++) {
boolArray[j] = chars[j] == '0' ? true : false;
}
System.out.println(Arrays.toString(boolArray));
}
}
}
Will produce:
[true, true, true]
[true, true, false]
[true, false, true]
[true, false, false]
[false, true, true]
[false, true, false]
[false, false, true]
[false, false, false]
Tested and this will work for high values of n, such as 10000 and so on.
OutOfMemoryError's when I try higher numbers for n. In that case, you'd simply have to create 2^n arrays with a size of [n]. I've updated my answer.
for (int i = 0; i < Math.pow(2, n); i++) when n > 31 will loop forever as the integer i will overflow as assertTrue(Integer.MAX_VALUE + 1 == Integer.MIN_VALUE) See my answer for rectified implementationI know that the there is a Java tag. I just want to add my swift code converted from Java to the answer.
let SIZE = 4
let max = (pow(2, SIZE) as NSDecimalNumber).intValue;
for i in 0..<max {
var bin = String(i, radix: 2)
while (bin.count < SIZE){
bin = "0" + bin
}
var boolArray = [Bool]();
var count = 0
for ch in bin {
boolArray.append(ch == "0")
count = count + 1
}
print(boolArray)
}
