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java binary search max int value

Posted on 2014-02-09
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Last Modified: 2014-02-13
Hi,

I have a function which attempts to evaluate whether a number is a perfect square or not.

It runs a unidirectional binary search and falls back to binary search when the product exceeds the input.

It works and runs O(logn) for numbers up to 2^30 (1,073,741,824).
For numbers equal or greater than 2^31 it throws exception as expected.

However, for numbers in between 2^30 and 2^31-1 it just stalls. no error. Anyone know why?

public class PerfectSquare{
	public static void main(String[] args){
		System.out.println(isPerfectSquare(Integer.parseInt(args[0])));
		// MAX int value: 2147483647 = 2^31-1
		// But, the maximum value I am able to work with is 2^30, larger numbers below MAX do not throw error but stall.
	}
	public static boolean isPerfectSquare(int input){
		for(int i = 1; i < input; i*=2){
			int product = i*i;
			if (product == input){
				System.out.println("power of 2: "+i);
				return true;
			}else if(product > input){
				int lowerBound = i/2;
				int upperBound = i;
				return binarySearch(lowerBound, upperBound, input);
			}
		}
		return false;
	}
	public static boolean binarySearch(int low, int high, int input){
		int mid, product, ans = 0;
		
		while (low <= high) {
			mid = (low + high) / 2;
			System.out.println("low: "+low+", high:" +high+", middle: "+mid);
			product = mid*mid;
			if(product == input){
				System.out.println("other square: "+mid);
				ans = 1;
				break;
			} else if (product > input) {
				high = mid - 1;
			} else  {
				low = mid + 1;
			}
		}
		return ans == 1 ? true : false;
	}
}

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Question by:Kyle Hamilton
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krakatoa earned 250 total points
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public class PerfectSquare{
	public static void main(String[] args){
		System.out.println(isPerfectSquare(Integer.parseInt(args[0])));
		// MAX int value: 2147483647 = 2^31-1
		// But, the maximum value I am able to work with is 2^30, larger numbers below MAX do not throw error but stall.
	}
	public static boolean isPerfectSquare(long input){
		for(long i = 1; i < input; i*=2){
			long product = i*i;
			if (product == input){
				System.out.println("power of 2: "+i/4);
				return true;
			}else if(product > input){
				long lowerBound = i/2;
				long upperBound = i;
				return binarySearch(lowerBound, upperBound, input);
			}
		}
		return false;
	}
	public static boolean binarySearch(long low, long high, long input){
		long mid, product, ans = 0;
		
		while (low <= high) {
			mid = (low + high) / 2;
			System.out.println("low: "+low/4+", high:" +high/4+", middle: "+mid/4);
			product = mid*mid;
			if(product == input){
				System.out.println("other square: "+mid/4);
				ans = 1;
				break;
			} else if (product > input) {
				high = mid - 1;
			} else  {
				low = mid + 1;
			}
		}
		return ans == 1 ? true : false;
	}
}

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by:Kyle Hamilton
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great, that works, but I was hoping for an explanation :)

I don't understand why long works and int doesn't. long is 32bit, and int is 64bit, so it seems the opposite should be true. clearly, I'm missing something.

why did you divide everything by four in the logging? what clue did that provide?

thanks.
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by:krakatoa
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I don't understand why long works and int doesn't. long is 32bit, and int is 64bit,

I thought it was the other way 'round, myself.
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by:Kyle Hamilton
Comment Utility
You're totally right. I've been staring at it for too long.

sorry
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by:Kyle Hamilton
Comment Utility
but the question does remain though. if INT is supposed to be able to handle up to 2^31-1, then why does it freak out at n > 2^30 ??

something to do with overflow??
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by:krakatoa
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If you do " product = i * i"  - on product being an int - it isn't going to last very long when you input int MAXVALUE as a start value.
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Assisted Solution

by:CPColin
CPColin earned 250 total points
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If input is greater than 2^30, your loop variable i is going to try to go from 2^30 to 2^31. Since an int can't hold that value, it does indeed overflow and wrap around to become a negative number. If you print out the value of i at the start of that loop on line 8, you'll see what's going on.
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by:Kyle Hamilton
Comment Utility
Hmmm.

That however brings up a new question.  In isPerfectSquare(), I don't expect "i" to ever get past log(input). It's supposed to go to the binarySearch after that - (which when I put in a log, it does.)
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by:Kyle Hamilton
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i never gets that big. here's i logged out at the start of the loop as suggested for input ~2^30

i: 1
i: 2
i: 4
i: 8
i: 16
i: 32
i: 64
i: 128
i: 256
i: 512
i: 1024
i: 2048
i: 4096
i: 8192
i: 16384
i: 32768
low: 16384, high:32768, middle: 24576
low: 24577, high:32768, middle: 28672
low: 28673, high:32768, middle: 30720
low: 30721, high:32768, middle: 31744
low: 31745, high:32768, middle: 32256
low: 32257, high:32768, middle: 32512
low: 32257, high:32511, middle: 32384
low: 32257, high:32383, middle: 32320
low: 32321, high:32383, middle: 32352
low: 32353, high:32383, middle: 32368
low: 32353, high:32367, middle: 32360
low: 32361, high:32367, middle: 32364
low: 32365, high:32367, middle: 32366
low: 32365, high:32365, middle: 32365
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by:krakatoa
Comment Utility
But you are already trying to hold "input" as an int, when passed from the command line.
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by:Kyle Hamilton
Comment Utility
ok, nvm.

when I use 2^31-1, it overflows. here's a snippet:

i: 33554432
i: 67108864
i: 134217728
i: 268435456
i: 536870912
i: 1073741824
i: -2147483648
i: 0
i: 0
i: 0
i: 0
i: 0
i: 0
i: 0


the loop is supposed to exit when i*i is greater than input. that's where it goes wrong. When i=2^15, then i*i is still less than 2^31, but when i = (2^15)*2, then i*i > 2^32

thanks guys. really appreciate it.
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by:Kyle Hamilton
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thank you. :)
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by:krakatoa
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then i*i > 2^32

Yeah . . . I put that in my comment, then swapped it out for 'product'. Lol. Oops. ;)
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by:Kyle Hamilton
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conclusion: my algorithm is crappy, since it doesn't error out and worse yet goes into infinite loop :O

suggestions are most welcome.. if you're so inlclined.
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by:krakatoa
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Thanks. But would have been nice to give CPColin some points, because he was forensic about the overflow. And as I said, I fluffed my chance at the full answer. ;)
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by:Kyle Hamilton
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for what it's worth, this fixes it:

for(int i = 1; i < input && i > 0; i*=2){
}
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by:krakatoa
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conclusion: my algorithm is crappy,

probably recognising things like that is indicative of being able to take it all further - heuristics is the best way.
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by:krakatoa
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See - you beat me to it. QED. ;)
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by:Kyle Hamilton
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goddam points. I meant to split them evenly, because both of you helped me figure out what was going on. grr. I really appreciate the help, I really do.

CPColin, sorry mate, can I go back and re-assign points?
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by:dpearson
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Yes "i" should never get very large.

The problem with the original code is that whenever you do multiplications on an integer value, if that multiplied result overflows the capacity of an "int" you'll get 0 or a negative number.

You had this problem in at least 2 places in the original code:
      int product = i*i;    // i*i can overflow

...
      product = mid*mid ; // can overlow

If you simply switch to longs then the method works fine for me:

		public static boolean isPerfectSquare(int input){
			for(int i = 1; i < input; i*=2){
				long product = (long)i*i;
				System.out.println("Product " + product);
				if (product == input){
					System.out.println("power of 2: "+i);
					return true;
				}else if(product > (long)input){
					int lowerBound = i/2;
					int upperBound = i;
					return binarySearch(lowerBound, upperBound, input);
				}
			}
			return false;
		}
		public static boolean binarySearch(long low, long high, long input){
			long mid, product, ans = 0;

			while (low <= high) {
				mid = (low + high) / 2;
				System.out.println("low: "+low+", high:" +high+", middle: "+mid);
				product = mid*mid;
				if(product == input){
					System.out.println("other square: "+mid);
					ans = 1;
					break;
				} else if (product > input) {
					high = mid - 1;
				} else  {
					low = mid + 1;
				}
			}
			return ans == 1 ? true : false;
		}

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Then given:

            int input = 32769 * 32769 ;  // More than 2^30
            boolean isPerfect = PerfectSquare.isPerfectSquare(input) ;

returns true (as it should) and does it in O(log n) time (as it should).

Is there any other problem that still needs to be solved?

Doug
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by:krakatoa
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If you simply switch to longs then the method works fine for me:

Doug - there is a funny echo in the room  somehow.

We said that at the top. ;)
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by:krakatoa
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class inttest{

static int i=0;

public static void main(String[] args){

System.out.println(Integer.parseInt(args[0]));

i=Integer.parseInt(args[0])+1;

System.out.println(i);

}


}

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by:Kyle Hamilton
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well, then if I use long, and put in a max long value, I'll run into the same problem.

and in fact my "pseudo" fix, is no good either, as the function will return false for inputs between 2^30 and 2^31-1 that are possible perfect squares (in the case if INT, that is, and larger values for long)
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by:Kyle Hamilton
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actually, long does something different doesn't it? it loops around itself when it overflows...
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by:krakatoa
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If I haven't upset Doug, he might help further, but my money says that whatever type you use, it's finite and will overrun sometime.
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by:Kyle Hamilton
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yeah, bottom line is, this is a bad approach, and i will have nightmares trying to figure out a better one, but that is the price of learning.

I'm pretty sure this problem has been solved before, haha.

cheers.
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by:krakatoa
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For what it's worth :

class inttest{

static int i=0;
static long l =0;

public static void main(String[] args){

if (Integer.parseInt(args[0])<2147483647) {args[0]="2147483647";}

System.out.println(Integer.parseInt(args[0]));

i=Integer.parseInt(args[0])+1;

System.out.println(i);
l= (long)i;
try{
while (l<Long.MAX_VALUE&&l>Long.MIN_VALUE){
l *= 2;

System.out.println(l);
}
}catch(Exception e){}

}


}

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ciao.
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by:dpearson
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If I haven't upset Doug, he might help further, but my money says that whatever type you use, it's finite and will overrun sometime.

Yeah that's a pretty sound statement :)  (And I'm pretty hard to upset - I saw the 'long' at the top, just wasn't sure what was still unsolved here so I collapsed it together).

Although if you really want to handle arbitrary sized numbers for this algorithm you could use a BigInteger instead of primitive types (int, long etc.)

To state the problem that you're running into with the overflow more generally, it's that you are computing values that are larger than the upper bound for the size of value that you're using.

E.g. If you are passing in an 'int', the algorithm checks for "product > input".  If input is Integer.MAX_VALUE then it should be pretty easy to see that product will never exceed it *unless* product is stored in a larger type (like long).

This will be true for any fixed size for input.

So your options are:
a) Use BigInteger
b) Use a larger primitive type internally than the type of input (e.g. longs with int input)
c) Rewrite those tests so they are more aware of the upper bound they may exceed.
   i.e. before you do:
         product = i * i ;
   you could check:

         // Make sure product won't overflow
         if (i > Math.sqrt(Integer.MAX_VALUE))
            return false ;   // If we his this limit there will be no divisor, so we're done

         int product = i * i ;

Similarly for the product inside binarySearch() method.

Doug
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by:Kyle Hamilton
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thanks Doug. I like option b which doesn't require any additional logic.

cheers.
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by:Kyle Hamilton
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biginteger sounds like fun. what happens when you run out of memory? how do they calculate big things like distances between stars and Universes? sigh... clusters of computers and other stuff I'm totally clueless about ?? .... I'm sooo not even scratching the surface... sigh again, baby steps, baby steps....

goodnight.
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by:dpearson
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Number of atoms in universe ~= 10^82 or in binary call it ~= 2^86

So to record the number of atoms in the universe you need about 86 bits in your number or about 11 bytes.

So how big of a number could you store in say 11GB instead of 11 bytes?  Probably big enough :)

Doug
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by:CPColin
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CPColin, sorry mate, can I go back and re-assign points?

If you want, you can use the "Request Attention" button below the body of this question to request help from the moderators. I'm not too worried about it, though.
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by:Kyle Hamilton
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thanks again. i'll be back for more punishment :)
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by:krakatoa
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Spare the rod, spoil the child.

;)
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