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Get a numer decimal part as Integer using only math

Edit: This has to do with how computers handle floating point operations, a fact that every programmer faces once in a lifetime. I didn't understand this correctly when I asked the question.
I know the simplest way to start dealing with this would be:
val floatNumber: Float = 123.456f
val decimalPart = floatNumber - floatNumber.toInt() //This would be 0.456 (I don't care about precision as this is not the main objective of my question)
Now in a real world with a pen and a piece of paper, if I want to "convert" the decimal part 0.456 to integer, I just need to multiply 0.456 * 1000, and I get the desired result, which is 456 (an integer number).
Many proposed solutions suggest splitting the number as string and extracting the decimal part this way, but I need the solution to be obtained mathematically, not using strings.
Given a number, with an unknown number of decimals (convert to string and counting chars after . or , is not acceptable), I need to "extract" it's decimal part as an integer using only math.
Read questions like this with no luck:
How to get the decimal part of a float?
How to extract fractional digits of double/BigDecimal
If someone knows a kotlin language solution, it would be great. I will post this question also on the math platform just in case.
How do I get whole and fractional parts from double in JSP/Java?
Update:
Is there a "mathematical" way to "calculate" how many decimals a number has? (It is obvious when you convert to string and count the chars, but I need to avoid using strings) It would be great cause calculating: decimal (0.456) * 10 * number of decimals(3) will produce the desired result.
Update 2
This is not my use-case, but I guess it will clarify the idea:
Suppose you want to calculate a constant(such as PI), and want to return an integer with at most 50 digits of the decimal part of the constant. The constant doesn't have to be necessarily infinite (can be for example 0.5, in which case "5" will be returned)

I would just multiply the fractional number by 10 (or move the decimal point to the right) until it has no fractional part left:
public static long fractionalDigitsLong(BigDecimal value) {
BigDecimal fractional = value.remainder(BigDecimal.ONE);
long digits;
do {
fractional = fractional.movePointRight(1); // or multiply(BigDecimal.TEN)
digits = fractional.longValue();
} while (fractional.compareTo(BigDecimal.valueOf(digits)) != 0);
return digits;
}
Note 1: using BigDecimal to avoid floating point precision problems
Note 2: using compareTo since equals also compares the scale ("0.0" not equals "0.00")
(sure the BigDecimal already knows the size of the fractional part, just the value returned by scale())
Complement:
If using BigDecimal the whole problem can be compressed to:
public static BigInteger fractionalDigits(BigDecimal value) {
return value.remainder(BigDecimal.ONE).stripTrailingZeros().unscaledValue();
}
stripping zeros can be suppressed if desired

I am not sure if it counts against you on this specific problem if you use some String converters with a method(). That is one way to get the proper answer. I know that you stated you couldn't use String, but would you be able to use Strings within a Custom made method? That could get you the answer that you need with precision. Here is the class that could help us convert the number:
class NumConvert{
String theNum;
public NumConvert(String theNum) {
this.theNum = theNum;
}
public int convert() {
String a = String.valueOf(theNum);
String[] b = a.split("\\.");
String b2 = b[1];
int zeros = b2.length();
String num = "1";
for(int x = 0; x < zeros; x++) {
num += "0";
}
float c = Float.parseFloat(theNum);
int multiply = Integer.parseInt(num);
float answer = c - (int)c;
int integerForm = (int)(answer * multiply);
return integerForm;
}
}
Then within your main class:
public class ChapterOneBasics {
public static void main(String[] args) throws java.io.IOException{
NumConvert n = new NumConvert("123.456");
NumConvert q = new NumConvert("123.45600128");
System.out.println(q.convert());
System.out.println(n.convert());
}
}
output:
45600128
456

Float or Double are imprecise, just an approximation - without precision. Hence 12.345 is somewhere between 12.3449... and 12.3450... .
This means that 12.340 cannot be distinghuished from 12.34. The "decimal part" would be 34 divided by 100.
Also 12.01 would have a "decimal part" 1 divided by 100, and too 12.1 would have 1 divided by 10.
So a complete algorith would be (using java):
int[] decimalsAndDivider(double x) {
int decimalPart = 0;
int divider = 1;
final double EPS = 0.001;
for (;;) {
double error = x - (int)x;
if (-EPS < error && error < EPS) {
break;
}
x *= 10;
decimalPart = 10 * decimalPart + ((int)(x + EPS) % 10);
divider *= 10;
}
return new int[] { decimalPart, divider };
}

I posted the below solution yesterday after testing it for a while, and later found that it does not always work due to problems regarding precision of floats, doubles and bigdecimals. My conclusion is that this problem is unsolvable if you want infinite precision:
So I re-post the code just for reference:
fun getDecimalCounter(d: Double): Int {
var temp = d
var tempInt = Math.floor(d)
var counter = 0
while ((temp - tempInt) > 0.0 ) {
temp *= 10
tempInt = Math.floor(temp)
counter++
}
return counter
}
fun main(args: Array <String> ) {
var d = 3.14159
if (d < 0) d = -d
val decimalCounter = getDecimalCounter(d)
val decimalPart = (d - Math.floor(d))
var decimalPartInt = Math.round(decimalPart * 10.0.pow(decimalCounter))
while (decimalPartInt % 10 == 0L) {
decimalPartInt /= 10
}
println(decimalPartInt)
}
I dropped floats because of lesser precision and used doubles.
The final rounding is also necessary due to precision.

Why does my code return 1 always?

double formula1, formula2;
int plus;
int VALUE = 10000;
private void processFormula2()
{
for (int k = 0; k <= VALUE; k++) {
if (k % 2 != 0) {
if (plus % 2 == 0) {
double math = 1/k;
formula2 += math;
System.out.println("Getting Formula: "+ formula2);
plus++;
} else {
formula2 -= 1/k;
plus++;
}
// System.out.println("Term: " + formula2);
}
}
}
I am trying to get my formula to print out the result of Pi based off this formula that my teacher gave us. But for some reason it just returns 1.0, not really sure why. Any help or suggestions would be appreciated :)

Here's the problem:
double math = 1/k;
and
formula2 -= 1/k;
k is an int variable, so the JVM won't never return a decimal number in this statement. It will take only two possible values: 0 (if k > 1) or 1 (if k == 1) because the JVM performs the division before promoting the result to double.
Try this:
formula2 -= 1/(double)k;
Take a look at Numeric Promotions

Firstly, there are multiple errors with variable declaration.
double math = 1/k;will not truly work in Java due to how integer division is handled. You must either cast '1' to a double like double math = (double)1/k; or specify that you are using mixed mode arithmetic by using double math = 1.0/k;. This is also a problem for your formula2 variable (Along with you should always initialize your variables like formula1, formula2, and plus). You must also do the same thing with formula2 -= 1/k;.
Secondly, we have no idea what you are setting those variables to in the first place, nor do we have any test cases to compare to.

Round long from 1004L to 1000L (or 1006L to 1010L)

Suppose I have Long someLong = 1004L. What efficient method can I use to round this down to 1000L? Note that I do not actually know that someLong == 1004L so I can't simply do someLong -= 4L;. I need a generalizable method. I also want the ability to round down to each 5 instead of each 10, for example a function to round to 1005L (since if we're rounding by 5's then it'll round up instead of down).
More examples .. It could be that I have 1926L and I want to round to 5 meaning I need 1925L. Or I need to round to 10 meaning I need 1930L.

This is very simple.
If you want to round always down:
Your required formula is:
someLong-someLong%10
It is because someLong%10 is the remainder of someLong divided by 10. If you get this from the original number, you get the downrounded value, which you wanted.
The generalization is also simple: you can use 100, or even 13, if you want.
If you want to rounding in another direction (for example, rounding always up or always to the middle), then first to add something to this number, and then round always down.
If you want to round always up:
Then first you need to first add 9, then round always down.
someLong+9-(someLong+9)%10
If you want to round always to the middle:
...also you want to round to the nearest neightbor. Then you first add the half of the required interval, then round always down. For example, for 10 it is:
someLong+5-(someLong+5)%10

If you want to round a value towards the nearest multiple of step using the semantics of BigDecimal.ROUND_HALF_UP (if exactly halfway between two steps, round up), the necessary calculations are:
val += step/2;
val -= val%step;

Try this:
double a=1002l;
double b=a/10;
a=Math.round(b)*10;
System.out.println("Double round of value : "+a);

A generic function to round to the nearest multiple of k would be (works for positives only):
public static long round(long toRound, long k) {
long times = toRound / k;
long reminder = toRound % k;
if (reminder < k / 2) {
return k * times;
} else {
return k * (times + 1);
}
}
And a branchless variant (reminder < k / 2 => (2 * reminder / k) < 1:
public static long round(long toRound, long k) {
long times = toRound / k;
long reminder = toRound % k;
return k * (times + ((2 * reminder) / k));
}

The following example reachs what you need:
public static void main(String[] args) {
Long n = 1004L;
Long n2 = 1005L;
n = round(n);
n2 = round(n2);
System.out.println(n);
System.out.println(n2);
}
private static Long round(Long n) {
if (n%10 <=4) {
return n -=n%10;
} else {
return n += (10-n%10);
}
}

myFloor(long n, int m) {
return n - (n % m);
}
myRound(long n, int m) {
int i = (n % m) >= (m / 2) ? m : 0;
return n + i - (n % m);
}
so m could be 10 , 5 , ...

Generate a random binary number with a variable proportion of '1' bits

I need a function to generate random integers. (assume Java long type for now, but this will be extended to BigInteger or BitSet later.)
The tricky part is there is a parameter P that specifies the (independent) probability of any bit in the result being 1.
If P = 0.5 then we can just use the standard random number generator. Some other values of P are also easy to implement. Here's an incomplete example:
Random random = new Random();
// ...
long nextLong(float p) {
if (p == 0.0f) return 0L;
else if (p == 1.0f) return -1L;
else if (p == 0.5f) return random.nextLong();
else if (p == 0.25f) return nextLong(0.5f) & nextLong(0.5f);
else if (p == 0.75f) return nextLong(0.5f) | nextLong(0.5f);
else if (p == 0.375f) return nextLong(0.5f) & nextLong(0.75f); // etc
else {
// What goes here??
String message = String.format("P=%f not implemented yet!", p);
throw new IllegalArgumentException(message);
}
}
Is there a way to generalise this for any value of P between 0.0 and 1.0?

First a little ugly math that you're already using in your code.
Define x and y are bits with probability of being 1 of X = p(x=1), Y = p(y=1) respectively.
Then we have that
p( x & y = 1) = X Y
p( x | y = 1) = 1 - (1-X) (1-Y)
p( x ^ y = 1) = X (1 - Y) + Y (1 - X)
Now if we let Y = 1/2 we get
P( x & y ) = X/2
P( x | y ) = (X+1)/2
Now set the RHS to the probability we want and we have two cases that we can solve for X
X = 2 p // if we use &
X = 2 p - 1 // if we use |
Next we assume we can use this again to obtain X in terms of another variable Z...
And then we keep iterating until we've done "enough".
Thats a bit unclear but consider p = 0.375
0.375 * 2 = 0.75 < 1.0 so our first operation is &
0.75 * 2 = 1.5 > 1.0 so our second operation is |
0.5 is something we know so we stop.
Thus we can get a variable with p=0.375 by X1 & (X2 | X3 )
The problem is that for most variables this will not terminate. e.g.
0.333 *2 = 0.666 < 1.0 so our first operation is &
0.666 *2 = 1.333 > 1.0 so our second operation is |
0.333 *2 = 0.666 < 1.0 so our third operation is &
etc...
so p=0.333 can be generated by
X1 & ( X2 | (X3 & (X4 | ( ... ) ) ) )
Now I suspect that taking enough terms in the series will give you enough accuracy, and this can be written as a recursive function. However there might be a better way that that too... I think the order of the operations is related to the binary representation of p, I'm just not sure exactly how... and dont have time to think about it deeper.
Anyway heres some untested C++ code that does this. You should be able to javaify it easily.
uint bitsWithProbability( float p )
{
return bitsWithProbabilityHelper( p, 0.001, 0, 10 );
}
uint bitsWithProbabilityHelper( float p, float tol, int cur_depth, int max_depth )
{
uint X = randbits();
if( cur_depth >= max_depth) return X;
if( p<0.5-tol)
{
return X & bitsWithProbabilityHelper( 2*p, 0.001, cur_depth+1, max_depth );
}
if(p>0.5+tol)
{
return X | bitsWithProbabilityHelper( 2*p-1, 0.001, cur_depth+1, max_depth );
}
return X;
}

Distribute proportional number of bits throughuot the number.
Pseudocode:
long generateNumber( double probability ){
int bitCount = 64 * probability;
byte[] data = new byte[64]; // 0-filled
long indexes = getRandomLong();
for 0 to bitCount-1 {
do {
// distribute this bit to some postition with 0.
int index = indexes & 64;
indexes >> 6;
if( indexes == 0 ) indexes = getRandomLong();
} while ( data[index] == 0 );
data[index] = 1;
}
return bytesToLong( data );
}
I hope you get what I mean. Perhaps the byte[] could be replaced with a long and bit operations to make it faster.

Here's how I solved it in the end.
Generate an integer N between 0..16, following the binomial distribution. This gives the number of '1' bits in the 16-bit partial result.
Randomly generate an index into a lookup table that contains 16-bit integers containing the desired number of '1' bits.
Repeat 4 times to get four 16-bit integers.
Splice these four 16-bit integers together to get a 64-bit integer.
This was partly inspired by Ondra Žižka's answer.
The benefit is that it reduces the number of calls to Random.nextLong() to 8 calls per 64 bits of output.
For comparison, rolling for each individual bit would require 64 calls. Bitwise AND/OR uses between 2 and 32 calls depending on the value of P
Of course calculating binomial probabilities is just as expensive, so those go in another lookup table.
It's a lot of code, but it's paying off in terms of performance.
Update - merged this with the bitwise AND/OR solution. It now uses that method if it guesses it will be more efficient (in terms of calls to Random.next().)

Use a random generator that generates a uniform float number r between 0 and 1. If r>p then set the bit to 0, otherwise set it to 1

If you're looking to apply some distribution where with probability P you get a 1 and with probability 1-P you get a 0 at any particular bit your best bet is simply to generate each bit independently with probability P of being a 1 (that sounds like a recursive definition, I know).
Here's a solution, I'll walk through it below:
public class MyRandomBitGenerator
{
Random pgen = new Random();
// assumed p is well conditioned (0 < p < 1)
public boolean nextBitIsOne(double p){
return pgen.nextDouble() < p ? true : false;
}
// assumed p is well conditioned (0 < p < 1)
public long nextLong(double p){
long nxt = 0;
for(int i = 0; i < 64; i++){
if(nextBitIsOne(p)){
nxt += 1 << i;
}
}
return nxt;
}
}
Basically, we first determine how to generate a value of 1 with probability P: pgen.nextDouble() generates a number between 0 and 1 with uniform probability, by asking if it's less than p we're sampling this distribution such that we expect to see p 1s as we call this function infinitely.

Here's another variant of Michael Anderson's answer
To avoid recursion, we process the bits of P iteratively from right-to-left instead of recursively from left-to-right. This would be tricky in floating-point representation so we extract the exponent/mantissa fields from the binary representation instead.
class BitsWithProbabilityHelper {
public BitsWithProbabilityHelper(float prob, Random rnd) {
if (Float.isNaN(prob)) throw new IllegalArgumentException();
this.rnd = rnd;
if (prob <= 0f) {
zero = true;
return;
}
// Decode IEEE float
int probBits = Float.floatToIntBits(prob);
mantissa = probBits & 0x7FFFFF;
exponent = probBits >>> 23;
// Restore the implicit leading 1 (except for denormals)
if (exponent > 0) mantissa |= 0x800000;
exponent -= 150;
// Force mantissa to be odd
int ntz = Integer.numberOfTrailingZeros(mantissa);
mantissa >>= ntz;
exponent += ntz;
}
/** Determine how many random words we need from the system RNG to
* generate one output word with probability P.
**/
public int iterationCount() {
return - exponent;
}
/** Generate a random number with the desired probability */
public long nextLong() {
if (zero) return 0L;
long acc = -1L;
int shiftReg = mantissa - 1;
for (int bit = exponent; bit < 0; ++ bit) {
if ((shiftReg & 1) == 0) {
acc &= rnd.nextLong();
} else {
acc |= rnd.nextLong();
}
shiftReg >>= 1;
}
return acc;
}
/** Value of <code>prob</code>, represented as m * 2**e where m is always odd. */
private int exponent;
private int mantissa;
/** Random data source */
private final Random rnd;
/** Zero flag (special case) */
private boolean zero;
}

Suppose the size of bit array is L. If L=1, the chance that the 1st bit is 1 will be P, and that being 0 will be 1-P. For L=2, the probability of getting a 00 is (1-P)2, a 01 or 10 is P(1-P) each and 11 is P2. Extending this logic, we can first determine the first bit by comparing a random number with P, then scale the random number such that we can again get anything between 0 to 1. A sample javascript code:
function getRandomBitArray(maxBits,probabilityOf1) {
var randomSeed = Math.random();
bitArray = new Array();
for(var currentBit=0;currentBit<maxBits;currentBit++){
if(randomSeed<probabilityOf1){
//fill 0 at current bit
bitArray.push(0);
//scale the sample space of the random no from [0,1)
//to [0.probabilityOf1)
randomSeed=randomSeed/probabilityOf1;
}
else{
//fill 1 at current bit
bitArray.push(1);
//scale the sample space to [probabilityOf1,1)
randomSeed = (randomSeed-probabilityOf1)/(1-probabilityOf1);
}
}
}
EDIT:
This code does generate completely random bits. I will try to explain the algorithm better.
Each bit string has a certain probability of occurring. Suppose a string has a probability of occurrence p; we want to choose that string if our random number falls is some interval of length p. The starting point of the interval must be fixed, but its value will not make much difference. Suppose we have chosen upto k bits correctly. Then, for the next bit, we divide the interval corresponding to this k-length bit-string into two parts of sizes in the ratio P:1-P (here P is the probability of getting a 1). We say that the next bit will be 1 if the random number is in the first part, 0 if it is in the second part. This ensure that the probabilities of strings of length k+1 also remain correct.
Java code:
public ArrayList<Boolean> getRandomBitArray(int maxBits, double probabilityOf1) {
double randomSeed = Math.random();
ArrayList<Boolean> bitArray = new ArrayList<Boolean>();
for(int currentBit=0;currentBit<maxBits;currentBit++){
if(randomSeed<probabilityOf1){
//fill 0 at current bit
bitArray.add(false);
//scale the sample space of the random no from [0,1)
//to [0.probabilityOf1)
randomSeed=randomSeed/probabilityOf1;
}
else{
//fill 1 at current bit
bitArray.add(true);
//scale the sample space to [probabilityOf1,1)
randomSeed = (randomSeed-probabilityOf1)/(1-probabilityOf1);
}
}
return bitArray;
}

Fastest way to determine if an integer's square root is an integer

I'm looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer):
I've done it the easy way, by using the built-in Math.sqrt()
function, but I'm wondering if there is a way to do it faster by
restricting yourself to integer-only domain.
Maintaining a lookup table is impractical (since there are about
231.5 integers whose square is less than 263).
Here is the very simple and straightforward way I'm doing it now:
public final static boolean isPerfectSquare(long n)
{
if (n < 0)
return false;
long tst = (long)(Math.sqrt(n) + 0.5);
return tst*tst == n;
}
Note: I'm using this function in many Project Euler problems. So no one else will ever have to maintain this code. And this kind of micro-optimization could actually make a difference, since part of the challenge is to do every algorithm in less than a minute, and this function will need to be called millions of times in some problems.
I've tried the different solutions to the problem:
After exhaustive testing, I found that adding 0.5 to the result of Math.sqrt() is not necessary, at least not on my machine.
The fast inverse square root was faster, but it gave incorrect results for n >= 410881. However, as suggested by BobbyShaftoe, we can use the FISR hack for n < 410881.
Newton's method was a good bit slower than Math.sqrt(). This is probably because Math.sqrt() uses something similar to Newton's Method, but implemented in the hardware so it's much faster than in Java. Also, Newton's Method still required use of doubles.
A modified Newton's method, which used a few tricks so that only integer math was involved, required some hacks to avoid overflow (I want this function to work with all positive 64-bit signed integers), and it was still slower than Math.sqrt().
Binary chop was even slower. This makes sense because the binary chop will on average require 16 passes to find the square root of a 64-bit number.
According to John's tests, using or statements is faster in C++ than using a switch, but in Java and C# there appears to be no difference between or and switch.
I also tried making a lookup table (as a private static array of 64 boolean values). Then instead of either switch or or statement, I would just say if(lookup[(int)(n&0x3F)]) { test } else return false;. To my surprise, this was (just slightly) slower. This is because array bounds are checked in Java.

I figured out a method that works ~35% faster than your 6bits+Carmack+sqrt code, at least with my CPU (x86) and programming language (C/C++). Your results may vary, especially because I don't know how the Java factor will play out.
My approach is threefold:
First, filter out obvious answers. This includes negative numbers and looking at the last 4 bits. (I found looking at the last six didn't help.) I also answer yes for 0. (In reading the code below, note that my input is int64 x.)
if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )
return false;
if( x == 0 )
return true;
Next, check if it's a square modulo 255 = 3 * 5 * 17. Because that's a product of three distinct primes, only about 1/8 of the residues mod 255 are squares. However, in my experience, calling the modulo operator (%) costs more than the benefit one gets, so I use bit tricks involving 255 = 2^8-1 to compute the residue. (For better or worse, I am not using the trick of reading individual bytes out of a word, only bitwise-and and shifts.)
int64 y = x;
y = (y & 4294967295LL) + (y >> 32);
y = (y & 65535) + (y >> 16);
y = (y & 255) + ((y >> 8) & 255) + (y >> 16);
// At this point, y is between 0 and 511. More code can reduce it farther.
To actually check if the residue is a square, I look up the answer in a precomputed table.
if( bad255[y] )
return false;
// However, I just use a table of size 512
Finally, try to compute the square root using a method similar to Hensel's lemma. (I don't think it's applicable directly, but it works with some modifications.) Before doing that, I divide out all powers of 2 with a binary search:
if((x & 4294967295LL) == 0)
x >>= 32;
if((x & 65535) == 0)
x >>= 16;
if((x & 255) == 0)
x >>= 8;
if((x & 15) == 0)
x >>= 4;
if((x & 3) == 0)
x >>= 2;
At this point, for our number to be a square, it must be 1 mod 8.
if((x & 7) != 1)
return false;
The basic structure of Hensel's lemma is the following. (Note: untested code; if it doesn't work, try t=2 or 8.)
int64 t = 4, r = 1;
t <<= 1; r += ((x - r * r) & t) >> 1;
t <<= 1; r += ((x - r * r) & t) >> 1;
t <<= 1; r += ((x - r * r) & t) >> 1;
// Repeat until t is 2^33 or so. Use a loop if you want.
The idea is that at each iteration, you add one bit onto r, the "current" square root of x; each square root is accurate modulo a larger and larger power of 2, namely t/2. At the end, r and t/2-r will be square roots of x modulo t/2. (Note that if r is a square root of x, then so is -r. This is true even modulo numbers, but beware, modulo some numbers, things can have even more than 2 square roots; notably, this includes powers of 2.) Because our actual square root is less than 2^32, at that point we can actually just check if r or t/2-r are real square roots. In my actual code, I use the following modified loop:
int64 r, t, z;
r = start[(x >> 3) & 1023];
do {
z = x - r * r;
if( z == 0 )
return true;
if( z < 0 )
return false;
t = z & (-z);
r += (z & t) >> 1;
if( r > (t >> 1) )
r = t - r;
} while( t <= (1LL << 33) );
The speedup here is obtained in three ways: precomputed start value (equivalent to ~10 iterations of the loop), earlier exit of the loop, and skipping some t values. For the last part, I look at z = r - x * x, and set t to be the largest power of 2 dividing z with a bit trick. This allows me to skip t values that wouldn't have affected the value of r anyway. The precomputed start value in my case picks out the "smallest positive" square root modulo 8192.
Even if this code doesn't work faster for you, I hope you enjoy some of the ideas it contains. Complete, tested code follows, including the precomputed tables.
typedef signed long long int int64;
int start[1024] =
{1,3,1769,5,1937,1741,7,1451,479,157,9,91,945,659,1817,11,
1983,707,1321,1211,1071,13,1479,405,415,1501,1609,741,15,339,1703,203,
129,1411,873,1669,17,1715,1145,1835,351,1251,887,1573,975,19,1127,395,
1855,1981,425,453,1105,653,327,21,287,93,713,1691,1935,301,551,587,
257,1277,23,763,1903,1075,1799,1877,223,1437,1783,859,1201,621,25,779,
1727,573,471,1979,815,1293,825,363,159,1315,183,27,241,941,601,971,
385,131,919,901,273,435,647,1493,95,29,1417,805,719,1261,1177,1163,
1599,835,1367,315,1361,1933,1977,747,31,1373,1079,1637,1679,1581,1753,1355,
513,1539,1815,1531,1647,205,505,1109,33,1379,521,1627,1457,1901,1767,1547,
1471,1853,1833,1349,559,1523,967,1131,97,35,1975,795,497,1875,1191,1739,
641,1149,1385,133,529,845,1657,725,161,1309,375,37,463,1555,615,1931,
1343,445,937,1083,1617,883,185,1515,225,1443,1225,869,1423,1235,39,1973,
769,259,489,1797,1391,1485,1287,341,289,99,1271,1701,1713,915,537,1781,
1215,963,41,581,303,243,1337,1899,353,1245,329,1563,753,595,1113,1589,
897,1667,407,635,785,1971,135,43,417,1507,1929,731,207,275,1689,1397,
1087,1725,855,1851,1873,397,1607,1813,481,163,567,101,1167,45,1831,1205,
1025,1021,1303,1029,1135,1331,1017,427,545,1181,1033,933,1969,365,1255,1013,
959,317,1751,187,47,1037,455,1429,609,1571,1463,1765,1009,685,679,821,
1153,387,1897,1403,1041,691,1927,811,673,227,137,1499,49,1005,103,629,
831,1091,1449,1477,1967,1677,697,1045,737,1117,1737,667,911,1325,473,437,
1281,1795,1001,261,879,51,775,1195,801,1635,759,165,1871,1645,1049,245,
703,1597,553,955,209,1779,1849,661,865,291,841,997,1265,1965,1625,53,
1409,893,105,1925,1297,589,377,1579,929,1053,1655,1829,305,1811,1895,139,
575,189,343,709,1711,1139,1095,277,993,1699,55,1435,655,1491,1319,331,
1537,515,791,507,623,1229,1529,1963,1057,355,1545,603,1615,1171,743,523,
447,1219,1239,1723,465,499,57,107,1121,989,951,229,1521,851,167,715,
1665,1923,1687,1157,1553,1869,1415,1749,1185,1763,649,1061,561,531,409,907,
319,1469,1961,59,1455,141,1209,491,1249,419,1847,1893,399,211,985,1099,
1793,765,1513,1275,367,1587,263,1365,1313,925,247,1371,1359,109,1561,1291,
191,61,1065,1605,721,781,1735,875,1377,1827,1353,539,1777,429,1959,1483,
1921,643,617,389,1809,947,889,981,1441,483,1143,293,817,749,1383,1675,
63,1347,169,827,1199,1421,583,1259,1505,861,457,1125,143,1069,807,1867,
2047,2045,279,2043,111,307,2041,597,1569,1891,2039,1957,1103,1389,231,2037,
65,1341,727,837,977,2035,569,1643,1633,547,439,1307,2033,1709,345,1845,
1919,637,1175,379,2031,333,903,213,1697,797,1161,475,1073,2029,921,1653,
193,67,1623,1595,943,1395,1721,2027,1761,1955,1335,357,113,1747,1497,1461,
1791,771,2025,1285,145,973,249,171,1825,611,265,1189,847,1427,2023,1269,
321,1475,1577,69,1233,755,1223,1685,1889,733,1865,2021,1807,1107,1447,1077,
1663,1917,1129,1147,1775,1613,1401,555,1953,2019,631,1243,1329,787,871,885,
449,1213,681,1733,687,115,71,1301,2017,675,969,411,369,467,295,693,
1535,509,233,517,401,1843,1543,939,2015,669,1527,421,591,147,281,501,
577,195,215,699,1489,525,1081,917,1951,2013,73,1253,1551,173,857,309,
1407,899,663,1915,1519,1203,391,1323,1887,739,1673,2011,1585,493,1433,117,
705,1603,1111,965,431,1165,1863,533,1823,605,823,1179,625,813,2009,75,
1279,1789,1559,251,657,563,761,1707,1759,1949,777,347,335,1133,1511,267,
833,1085,2007,1467,1745,1805,711,149,1695,803,1719,485,1295,1453,935,459,
1151,381,1641,1413,1263,77,1913,2005,1631,541,119,1317,1841,1773,359,651,
961,323,1193,197,175,1651,441,235,1567,1885,1481,1947,881,2003,217,843,
1023,1027,745,1019,913,717,1031,1621,1503,867,1015,1115,79,1683,793,1035,
1089,1731,297,1861,2001,1011,1593,619,1439,477,585,283,1039,1363,1369,1227,
895,1661,151,645,1007,1357,121,1237,1375,1821,1911,549,1999,1043,1945,1419,
1217,957,599,571,81,371,1351,1003,1311,931,311,1381,1137,723,1575,1611,
767,253,1047,1787,1169,1997,1273,853,1247,413,1289,1883,177,403,999,1803,
1345,451,1495,1093,1839,269,199,1387,1183,1757,1207,1051,783,83,423,1995,
639,1155,1943,123,751,1459,1671,469,1119,995,393,219,1743,237,153,1909,
1473,1859,1705,1339,337,909,953,1771,1055,349,1993,613,1393,557,729,1717,
511,1533,1257,1541,1425,819,519,85,991,1693,503,1445,433,877,1305,1525,
1601,829,809,325,1583,1549,1991,1941,927,1059,1097,1819,527,1197,1881,1333,
383,125,361,891,495,179,633,299,863,285,1399,987,1487,1517,1639,1141,
1729,579,87,1989,593,1907,839,1557,799,1629,201,155,1649,1837,1063,949,
255,1283,535,773,1681,461,1785,683,735,1123,1801,677,689,1939,487,757,
1857,1987,983,443,1327,1267,313,1173,671,221,695,1509,271,1619,89,565,
127,1405,1431,1659,239,1101,1159,1067,607,1565,905,1755,1231,1299,665,373,
1985,701,1879,1221,849,627,1465,789,543,1187,1591,923,1905,979,1241,181};
bool bad255[512] =
{0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,
1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1,
0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1,
1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,
1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1,
1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,
1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,
1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,
0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,
1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1,
0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1,
1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1,
1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1,
1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,
1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,
1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,
0,0};
inline bool square( int64 x ) {
// Quickfail
if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )
return false;
if( x == 0 )
return true;
// Check mod 255 = 3 * 5 * 17, for fun
int64 y = x;
y = (y & 4294967295LL) + (y >> 32);
y = (y & 65535) + (y >> 16);
y = (y & 255) + ((y >> 8) & 255) + (y >> 16);
if( bad255[y] )
return false;
// Divide out powers of 4 using binary search
if((x & 4294967295LL) == 0)
x >>= 32;
if((x & 65535) == 0)
x >>= 16;
if((x & 255) == 0)
x >>= 8;
if((x & 15) == 0)
x >>= 4;
if((x & 3) == 0)
x >>= 2;
if((x & 7) != 1)
return false;
// Compute sqrt using something like Hensel's lemma
int64 r, t, z;
r = start[(x >> 3) & 1023];
do {
z = x - r * r;
if( z == 0 )
return true;
if( z < 0 )
return false;
t = z & (-z);
r += (z & t) >> 1;
if( r > (t >> 1) )
r = t - r;
} while( t <= (1LL << 33) );
return false;
}

I'm pretty late to the party, but I hope to provide a better answer; shorter and (assuming my benchmark is correct) also much faster.
long goodMask; // 0xC840C04048404040 computed below
{
for (int i=0; i<64; ++i) goodMask |= Long.MIN_VALUE >>> (i*i);
}
public boolean isSquare(long x) {
// This tests if the 6 least significant bits are right.
// Moving the to be tested bit to the highest position saves us masking.
if (goodMask << x >= 0) return false;
final int numberOfTrailingZeros = Long.numberOfTrailingZeros(x);
// Each square ends with an even number of zeros.
if ((numberOfTrailingZeros & 1) != 0) return false;
x >>= numberOfTrailingZeros;
// Now x is either 0 or odd.
// In binary each odd square ends with 001.
// Postpone the sign test until now; handle zero in the branch.
if ((x&7) != 1 | x <= 0) return x == 0;
// Do it in the classical way.
// The correctness is not trivial as the conversion from long to double is lossy!
final long tst = (long) Math.sqrt(x);
return tst * tst == x;
}
The first test catches most non-squares quickly. It uses a 64-item table packed in a long, so there's no array access cost (indirection and bounds checks). For a uniformly random long, there's a 81.25% probability of ending here.
The second test catches all numbers having an odd number of twos in their factorization. The method Long.numberOfTrailingZeros is very fast as it gets JIT-ed into a single i86 instruction.
After dropping the trailing zeros, the third test handles numbers ending with 011, 101, or 111 in binary, which are no perfect squares. It also cares about negative numbers and also handles 0.
The final test falls back to double arithmetic. As double has only 53 bits mantissa,
the conversion from long to double includes rounding for big values. Nonetheless, the test is correct (unless the proof is wrong).
Trying to incorporate the mod255 idea wasn't successful.

You'll have to do some benchmarking. The best algorithm will depend on the distribution of your inputs.
Your algorithm may be nearly optimal, but you might want to do a quick check to rule out some possibilities before calling your square root routine. For example, look at the last digit of your number in hex by doing a bit-wise "and." Perfect squares can only end in 0, 1, 4, or 9 in base 16, So for 75% of your inputs (assuming they are uniformly distributed) you can avoid a call to the square root in exchange for some very fast bit twiddling.
Kip benchmarked the following code implementing the hex trick. When testing numbers 1 through 100,000,000, this code ran twice as fast as the original.
public final static boolean isPerfectSquare(long n)
{
if (n < 0)
return false;
switch((int)(n & 0xF))
{
case 0: case 1: case 4: case 9:
long tst = (long)Math.sqrt(n);
return tst*tst == n;
default:
return false;
}
}
When I tested the analogous code in C++, it actually ran slower than the original. However, when I eliminated the switch statement, the hex trick once again make the code twice as fast.
int isPerfectSquare(int n)
{
int h = n & 0xF; // h is the last hex "digit"
if (h > 9)
return 0;
// Use lazy evaluation to jump out of the if statement as soon as possible
if (h != 2 && h != 3 && h != 5 && h != 6 && h != 7 && h != 8)
{
int t = (int) floor( sqrt((double) n) + 0.5 );
return t*t == n;
}
return 0;
}
Eliminating the switch statement had little effect on the C# code.

I was thinking about the horrible times I've spent in Numerical Analysis course.
And then I remember, there was this function circling around the 'net from the Quake Source code:
float Q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // wtf?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
#ifndef Q3_VM
#ifdef __linux__
assert( !isnan(y) ); // bk010122 - FPE?
#endif
#endif
return y;
}
Which basically calculates a square root, using Newton's approximation function (cant remember the exact name).
It should be usable and might even be faster, it's from one of the phenomenal id software's game!
It's written in C++ but it should not be too hard to reuse the same technique in Java once you get the idea:
I originally found it at: http://www.codemaestro.com/reviews/9
Newton's method explained at wikipedia: http://en.wikipedia.org/wiki/Newton%27s_method
You can follow the link for more explanation of how it works, but if you don't care much, then this is roughly what I remember from reading the blog and from taking the Numerical Analysis course:
the * (long*) &y is basically a fast convert-to-long function so integer operations can be applied on the raw bytes.
the 0x5f3759df - (i >> 1); line is a pre-calculated seed value for the approximation function.
the * (float*) &i converts the value back to floating point.
the y = y * ( threehalfs - ( x2 * y * y ) ) line bascially iterates the value over the function again.
The approximation function gives more precise values the more you iterate the function over the result. In Quake's case, one iteration is "good enough", but if it wasn't for you... then you could add as much iteration as you need.
This should be faster because it reduces the number of division operations done in naive square rooting down to a simple divide by 2 (actually a * 0.5F multiply operation) and replace it with a few fixed number of multiplication operations instead.

I'm not sure if it would be faster, or even accurate, but you could use John Carmack's Magical Square Root, algorithm to solve the square root faster. You could probably easily test this for all possible 32 bit integers, and validate that you actually got correct results, as it's only an appoximation. However, now that I think about it, using doubles is approximating also, so I'm not sure how that would come into play.

If you do a binary chop to try to find the "right" square root, you can fairly easily detect if the value you've got is close enough to tell:
(n+1)^2 = n^2 + 2n + 1
(n-1)^2 = n^2 - 2n + 1
So having calculated n^2, the options are:
n^2 = target: done, return true
n^2 + 2n + 1 > target > n^2 : you're close, but it's not perfect: return false
n^2 - 2n + 1 < target < n^2 : ditto
target < n^2 - 2n + 1 : binary chop on a lower n
target > n^2 + 2n + 1 : binary chop on a higher n
(Sorry, this uses n as your current guess, and target for the parameter. Apologise for the confusion!)
I don't know whether this will be faster or not, but it's worth a try.
EDIT: The binary chop doesn't have to take in the whole range of integers, either (2^x)^2 = 2^(2x), so once you've found the top set bit in your target (which can be done with a bit-twiddling trick; I forget exactly how) you can quickly get a range of potential answers. Mind you, a naive binary chop is still only going to take up to 31 or 32 iterations.

I ran my own analysis of several of the algorithms in this thread and came up with some new results. You can see those old results in the edit history of this answer, but they're not accurate, as I made a mistake, and wasted time analyzing several algorithms which aren't close. However, pulling lessons from several different answers, I now have two algorithms that crush the "winner" of this thread. Here's the core thing I do differently than everyone else:
// This is faster because a number is divisible by 2^4 or more only 6% of the time
// and more than that a vanishingly small percentage.
while((x & 0x3) == 0) x >>= 2;
// This is effectively the same as the switch-case statement used in the original
// answer.
if((x & 0x7) != 1) return false;
However, this simple line, which most of the time adds one or two very fast instructions, greatly simplifies the switch-case statement into one if statement. However, it can add to the runtime if many of the tested numbers have significant power-of-two factors.
The algorithms below are as follows:
Internet - Kip's posted answer
Durron - My modified answer using the one-pass answer as a base
DurronTwo - My modified answer using the two-pass answer (by #JohnnyHeggheim), with some other slight modifications.
Here is a sample runtime if the numbers are generated using Math.abs(java.util.Random.nextLong())
0% Scenario{vm=java, trial=0, benchmark=Internet} 39673.40 ns; ?=378.78 ns # 3 trials
33% Scenario{vm=java, trial=0, benchmark=Durron} 37785.75 ns; ?=478.86 ns # 10 trials
67% Scenario{vm=java, trial=0, benchmark=DurronTwo} 35978.10 ns; ?=734.10 ns # 10 trials
benchmark us linear runtime
Internet 39.7 ==============================
Durron 37.8 ============================
DurronTwo 36.0 ===========================
vm: java
trial: 0
And here is a sample runtime if it's run on the first million longs only:
0% Scenario{vm=java, trial=0, benchmark=Internet} 2933380.84 ns; ?=56939.84 ns # 10 trials
33% Scenario{vm=java, trial=0, benchmark=Durron} 2243266.81 ns; ?=50537.62 ns # 10 trials
67% Scenario{vm=java, trial=0, benchmark=DurronTwo} 3159227.68 ns; ?=10766.22 ns # 3 trials
benchmark ms linear runtime
Internet 2.93 ===========================
Durron 2.24 =====================
DurronTwo 3.16 ==============================
vm: java
trial: 0
As you can see, DurronTwo does better for large inputs, because it gets to use the magic trick very very often, but gets clobbered compared to the first algorithm and Math.sqrt because the numbers are so much smaller. Meanwhile, the simpler Durron is a huge winner because it never has to divide by 4 many many times in the first million numbers.
Here's Durron:
public final static boolean isPerfectSquareDurron(long n) {
if(n < 0) return false;
if(n == 0) return true;
long x = n;
// This is faster because a number is divisible by 16 only 6% of the time
// and more than that a vanishingly small percentage.
while((x & 0x3) == 0) x >>= 2;
// This is effectively the same as the switch-case statement used in the original
// answer.
if((x & 0x7) == 1) {
long sqrt;
if(x < 410881L)
{
int i;
float x2, y;
x2 = x * 0.5F;
y = x;
i = Float.floatToRawIntBits(y);
i = 0x5f3759df - ( i >> 1 );
y = Float.intBitsToFloat(i);
y = y * ( 1.5F - ( x2 * y * y ) );
sqrt = (long)(1.0F/y);
} else {
sqrt = (long) Math.sqrt(x);
}
return sqrt*sqrt == x;
}
return false;
}
And DurronTwo
public final static boolean isPerfectSquareDurronTwo(long n) {
if(n < 0) return false;
// Needed to prevent infinite loop
if(n == 0) return true;
long x = n;
while((x & 0x3) == 0) x >>= 2;
if((x & 0x7) == 1) {
long sqrt;
if (x < 41529141369L) {
int i;
float x2, y;
x2 = x * 0.5F;
y = x;
i = Float.floatToRawIntBits(y);
//using the magic number from
//http://www.lomont.org/Math/Papers/2003/InvSqrt.pdf
//since it more accurate
i = 0x5f375a86 - (i >> 1);
y = Float.intBitsToFloat(i);
y = y * (1.5F - (x2 * y * y));
y = y * (1.5F - (x2 * y * y)); //Newton iteration, more accurate
sqrt = (long) ((1.0F/y) + 0.2);
} else {
//Carmack hack gives incorrect answer for n >= 41529141369.
sqrt = (long) Math.sqrt(x);
}
return sqrt*sqrt == x;
}
return false;
}
And my benchmark harness: (Requires Google caliper 0.1-rc5)
public class SquareRootBenchmark {
public static class Benchmark1 extends SimpleBenchmark {
private static final int ARRAY_SIZE = 10000;
long[] trials = new long[ARRAY_SIZE];
#Override
protected void setUp() throws Exception {
Random r = new Random();
for (int i = 0; i < ARRAY_SIZE; i++) {
trials[i] = Math.abs(r.nextLong());
}
}
public int timeInternet(int reps) {
int trues = 0;
for(int i = 0; i < reps; i++) {
for(int j = 0; j < ARRAY_SIZE; j++) {
if(SquareRootAlgs.isPerfectSquareInternet(trials[j])) trues++;
}
}
return trues;
}
public int timeDurron(int reps) {
int trues = 0;
for(int i = 0; i < reps; i++) {
for(int j = 0; j < ARRAY_SIZE; j++) {
if(SquareRootAlgs.isPerfectSquareDurron(trials[j])) trues++;
}
}
return trues;
}
public int timeDurronTwo(int reps) {
int trues = 0;
for(int i = 0; i < reps; i++) {
for(int j = 0; j < ARRAY_SIZE; j++) {
if(SquareRootAlgs.isPerfectSquareDurronTwo(trials[j])) trues++;
}
}
return trues;
}
}
public static void main(String... args) {
Runner.main(Benchmark1.class, args);
}
}
UPDATE: I've made a new algorithm that is faster in some scenarios, slower in others, I've gotten different benchmarks based on different inputs. If we calculate modulo 0xFFFFFF = 3 x 3 x 5 x 7 x 13 x 17 x 241, we can eliminate 97.82% of numbers that cannot be squares. This can be (sort of) done in one line, with 5 bitwise operations:
if (!goodLookupSquares[(int) ((n & 0xFFFFFFl) + ((n >> 24) & 0xFFFFFFl) + (n >> 48))]) return false;
The resulting index is either 1) the residue, 2) the residue + 0xFFFFFF, or 3) the residue + 0x1FFFFFE. Of course, we need to have a lookup table for residues modulo 0xFFFFFF, which is about a 3mb file (in this case stored as ascii text decimal numbers, not optimal but clearly improvable with a ByteBuffer and so forth. But since that is precalculation it doesn't matter so much. You can find the file here (or generate it yourself):
public final static boolean isPerfectSquareDurronThree(long n) {
if(n < 0) return false;
if(n == 0) return true;
long x = n;
while((x & 0x3) == 0) x >>= 2;
if((x & 0x7) == 1) {
if (!goodLookupSquares[(int) ((n & 0xFFFFFFl) + ((n >> 24) & 0xFFFFFFl) + (n >> 48))]) return false;
long sqrt;
if(x < 410881L)
{
int i;
float x2, y;
x2 = x * 0.5F;
y = x;
i = Float.floatToRawIntBits(y);
i = 0x5f3759df - ( i >> 1 );
y = Float.intBitsToFloat(i);
y = y * ( 1.5F - ( x2 * y * y ) );
sqrt = (long)(1.0F/y);
} else {
sqrt = (long) Math.sqrt(x);
}
return sqrt*sqrt == x;
}
return false;
}
I load it into a boolean array like this:
private static boolean[] goodLookupSquares = null;
public static void initGoodLookupSquares() throws Exception {
Scanner s = new Scanner(new File("24residues_squares.txt"));
goodLookupSquares = new boolean[0x1FFFFFE];
while(s.hasNextLine()) {
int residue = Integer.valueOf(s.nextLine());
goodLookupSquares[residue] = true;
goodLookupSquares[residue + 0xFFFFFF] = true;
goodLookupSquares[residue + 0x1FFFFFE] = true;
}
s.close();
}
Example runtime. It beat Durron (version one) in every trial I ran.
0% Scenario{vm=java, trial=0, benchmark=Internet} 40665.77 ns; ?=566.71 ns # 10 trials
33% Scenario{vm=java, trial=0, benchmark=Durron} 38397.60 ns; ?=784.30 ns # 10 trials
67% Scenario{vm=java, trial=0, benchmark=DurronThree} 36171.46 ns; ?=693.02 ns # 10 trials
benchmark us linear runtime
Internet 40.7 ==============================
Durron 38.4 ============================
DurronThree 36.2 ==========================
vm: java
trial: 0

It should be much faster to use Newton's method to calculate the Integer Square Root, then square this number and check, as you do in your current solution. Newton's method is the basis for the Carmack solution mentioned in some other answers. You should be able to get a faster answer since you're only interested in the integer part of the root, allowing you to stop the approximation algorithm sooner.
Another optimization that you can try: If the Digital Root of a number doesn't end in
1, 4, 7, or 9 the number is not a perfect square. This can be used as a quick way to eliminate 60% of your inputs before applying the slower square root algorithm.

I want this function to work with all
positive 64-bit signed integers
Math.sqrt() works with doubles as input parameters, so you won't get accurate results for integers bigger than 2^53.

An integer problem deserves an integer solution. Thus
Do binary search on the (non-negative) integers to find the greatest integer t such that t**2 <= n. Then test whether r**2 = n exactly. This takes time O(log n).
If you don't know how to binary search the positive integers because the set is unbounded, it's easy. You starting by computing your increasing function f (above f(t) = t**2 - n) on powers of two. When you see it turn positive, you've found an upper bound. Then you can do standard binary search.

Just for the record, another approach is to use the prime decomposition. If every factor of the decomposition is even, then the number is a perfect square. So what you want is to see if a number can be decomposed as a product of squares of prime numbers. Of course, you don't need to obtain such a decomposition, just to see if it exists.
First build a table of squares of prime numbers which are lower than 2^32. This is far smaller than a table of all integers up to this limit.
A solution would then be like this:
boolean isPerfectSquare(long number)
{
if (number < 0) return false;
if (number < 2) return true;
for (int i = 0; ; i++)
{
long square = squareTable[i];
if (square > number) return false;
while (number % square == 0)
{
number /= square;
}
if (number == 1) return true;
}
}
I guess it's a bit cryptic. What it does is checking in every step that the square of a prime number divide the input number. If it does then it divides the number by the square as long as it is possible, to remove this square from the prime decomposition.
If by this process, we came to 1, then the input number was a decomposition of square of prime numbers. If the square becomes larger than the number itself, then there is no way this square, or any larger squares, can divide it, so the number can not be a decomposition of squares of prime numbers.
Given nowadays' sqrt done in hardware and the need to compute prime numbers here, I guess this solution is way slower. But it should give better results than solution with sqrt which won't work over 2^54, as says mrzl in his answer.

It's been pointed out that the last d digits of a perfect square can only take on certain values. The last d digits (in base b) of a number n is the same as the remainder when n is divided by bd, ie. in C notation n % pow(b, d).
This can be generalized to any modulus m, ie. n % m can be used to rule out some percentage of numbers from being perfect squares. The modulus you are currently using is 64, which allows 12, ie. 19% of remainders, as possible squares. With a little coding I found the modulus 110880, which allows only 2016, ie. 1.8% of remainders as possible squares. So depending on the cost of a modulus operation (ie. division) and a table lookup versus a square root on your machine, using this modulus might be faster.
By the way if Java has a way to store a packed array of bits for the lookup table, don't use it. 110880 32-bit words is not much RAM these days and fetching a machine word is going to be faster than fetching a single bit.

The following simplification of maaartinus's solution appears to shave a few percentage points off the runtime, but I'm not good enough at benchmarking to produce a benchmark I can trust:
long goodMask; // 0xC840C04048404040 computed below
{
for (int i=0; i<64; ++i) goodMask |= Long.MIN_VALUE >>> (i*i);
}
public boolean isSquare(long x) {
// This tests if the 6 least significant bits are right.
// Moving the to be tested bit to the highest position saves us masking.
if (goodMask << x >= 0) return false;
// Remove an even number of trailing zeros, leaving at most one.
x >>= (Long.numberOfTrailingZeros(x) & (-2);
// Repeat the test on the 6 least significant remaining bits.
if (goodMask << x >= 0 | x <= 0) return x == 0;
// Do it in the classical way.
// The correctness is not trivial as the conversion from long to double is lossy!
final long tst = (long) Math.sqrt(x);
return tst * tst == x;
}
It would be worth checking how omitting the first test,
if (goodMask << x >= 0) return false;
would affect performance.

For performance, you very often have to do some compromsies. Others have expressed various methods, however, you noted Carmack's hack was faster up to certain values of N. Then, you should check the "n" and if it is less than that number N, use Carmack's hack, else use some other method described in the answers here.

This is the fastest Java implementation I could come up with, using a combination of techniques suggested by others in this thread.
Mod-256 test
Inexact mod-3465 test (avoids integer division at the cost of some false positives)
Floating-point square root, round and compare with input value
I also experimented with these modifications but they did not help performance:
Additional mod-255 test
Dividing the input value by powers of 4
Fast Inverse Square Root (to work for high values of N it needs 3 iterations, enough to make it slower than the hardware square root function.)
public class SquareTester {
public static boolean isPerfectSquare(long n) {
if (n < 0) {
return false;
} else {
switch ((byte) n) {
case -128: case -127: case -124: case -119: case -112:
case -111: case -103: case -95: case -92: case -87:
case -79: case -71: case -64: case -63: case -60:
case -55: case -47: case -39: case -31: case -28:
case -23: case -15: case -7: case 0: case 1:
case 4: case 9: case 16: case 17: case 25:
case 33: case 36: case 41: case 49: case 57:
case 64: case 65: case 68: case 73: case 81:
case 89: case 97: case 100: case 105: case 113:
case 121:
long i = (n * INV3465) >>> 52;
if (! good3465[(int) i]) {
return false;
} else {
long r = round(Math.sqrt(n));
return r*r == n;
}
default:
return false;
}
}
}
private static int round(double x) {
return (int) Double.doubleToRawLongBits(x + (double) (1L << 52));
}
/** 3465<sup>-1</sup> modulo 2<sup>64</sup> */
private static final long INV3465 = 0x8ffed161732e78b9L;
private static final boolean[] good3465 =
new boolean[0x1000];
static {
for (int r = 0; r < 3465; ++ r) {
int i = (int) ((r * r * INV3465) >>> 52);
good3465[i] = good3465[i+1] = true;
}
}
}

You should get rid of the 2-power part of N right from the start.
2nd Edit
The magical expression for m below should be
m = N - (N & (N-1));
and not as written
End of 2nd edit
m = N & (N-1); // the lawest bit of N
N /= m;
byte = N & 0x0F;
if ((m % 2) || (byte !=1 && byte !=9))
return false;
1st Edit:
Minor improvement:
m = N & (N-1); // the lawest bit of N
N /= m;
if ((m % 2) || (N & 0x07 != 1))
return false;
End of 1st edit
Now continue as usual. This way, by the time you get to the floating point part, you already got rid of all the numbers whose 2-power part is odd (about half), and then you only consider 1/8 of whats left. I.e. you run the floating point part on 6% of the numbers.

Project Euler is mentioned in the tags and many of the problems in it require checking numbers >> 2^64. Most of the optimizations mentioned above don't work easily when you are working with an 80 byte buffer.
I used java BigInteger and a slightly modified version of Newton's method, one that works better with integers. The problem was that exact squares n^2 converged to (n-1) instead of n because n^2-1 = (n-1)(n+1) and the final error was just one step below the final divisor and the algorithm terminated. It was easy to fix by adding one to the original argument before computing the error. (Add two for cube roots, etc.)
One nice attribute of this algorithm is that you can immediately tell if the number is a perfect square - the final error (not correction) in Newton's method will be zero. A simple modification also lets you quickly calculate floor(sqrt(x)) instead of the closest integer. This is handy with several Euler problems.

The sqrt call is not perfectly accurate, as has been mentioned, but it's interesting and instructive that it doesn't blow away the other answers in terms of speed. After all, the sequence of assembly language instructions for a sqrt is tiny. Intel has a hardware instruction, which isn't used by Java I believe because it doesn't conform to IEEE.
So why is it slow? Because Java is actually calling a C routine through JNI, and it's actually slower to do so than to call a Java subroutine, which itself is slower than doing it inline. This is very annoying, and Java should have come up with a better solution, ie building in floating point library calls if necessary. Oh well.
In C++, I suspect all the complex alternatives would lose on speed, but I haven't checked them all.
What I did, and what Java people will find usefull, is a simple hack, an extension of the special case testing suggested by A. Rex. Use a single long value as a bit array, which isn't bounds checked. That way, you have 64 bit boolean lookup.
typedef unsigned long long UVLONG
UVLONG pp1,pp2;
void init2() {
for (int i = 0; i < 64; i++) {
for (int j = 0; j < 64; j++)
if (isPerfectSquare(i * 64 + j)) {
pp1 |= (1 << j);
pp2 |= (1 << i);
break;
}
}
cout << "pp1=" << pp1 << "," << pp2 << "\n";
}
inline bool isPerfectSquare5(UVLONG x) {
return pp1 & (1 << (x & 0x3F)) ? isPerfectSquare(x) : false;
}
The routine isPerfectSquare5 runs in about 1/3 the time on my core2 duo machine. I suspect that further tweaks along the same lines could reduce the time further on average, but every time you check, you are trading off more testing for more eliminating, so you can't go too much farther on that road.
Certainly, rather than having a separate test for negative, you could check the high 6 bits the same way.
Note that all I'm doing is eliminating possible squares, but when I have a potential case I have to call the original, inlined isPerfectSquare.
The init2 routine is called once to initialize the static values of pp1 and pp2.
Note that in my implementation in C++, I'm using unsigned long long, so since you're signed, you'd have to use the >>> operator.
There is no intrinsic need to bounds check the array, but Java's optimizer has to figure this stuff out pretty quickly, so I don't blame them for that.

I like the idea to use an almost correct method on some of the input. Here is a version with a higher "offset". The code seems to work and passes my simple test case.
Just replace your:
if(n < 410881L){...}
code with this one:
if (n < 11043908100L) {
//John Carmack hack, converted to Java.
// See: http://www.codemaestro.com/reviews/9
int i;
float x2, y;
x2 = n * 0.5F;
y = n;
i = Float.floatToRawIntBits(y);
//using the magic number from
//http://www.lomont.org/Math/Papers/2003/InvSqrt.pdf
//since it more accurate
i = 0x5f375a86 - (i >> 1);
y = Float.intBitsToFloat(i);
y = y * (1.5F - (x2 * y * y));
y = y * (1.5F - (x2 * y * y)); //Newton iteration, more accurate
sqrt = Math.round(1.0F / y);
} else {
//Carmack hack gives incorrect answer for n >= 11043908100.
sqrt = (long) Math.sqrt(n);
}

Considering for general bit length (though I have used specific type here), I tried to design simplistic algo as below. Simple and obvious check for 0,1,2 or <0 is required initially.
Following is simple in sense that it doesn't try to use any existing maths functions. Most of the operator can be replaced with bit-wise operators. I haven't tested with any bench mark data though. I'm neither expert at maths or computer algorithm design in particular, I would love to see you pointing out problem. I know there is lots of improvement chances there.
int main()
{
unsigned int c1=0 ,c2 = 0;
unsigned int x = 0;
unsigned int p = 0;
int k1 = 0;
scanf("%d",&p);
if(p % 2 == 0) {
x = p/2;
}
else {
x = (p/2) +1;
}
while(x)
{
if((x*x) > p) {
c1 = x;
x = x/2;
}else {
c2 = x;
break;
}
}
if((p%2) != 0)
c2++;
while(c2 < c1)
{
if((c2 * c2 ) == p) {
k1 = 1;
break;
}
c2++;
}
if(k1)
printf("\n Perfect square for %d", c2);
else
printf("\n Not perfect but nearest to :%d :", c2);
return 0;
}

This a rework from decimal to binary of the old Marchant calculator algorithm (sorry, I don't have a reference), in Ruby, adapted specifically for this question:
def isexactsqrt(v)
value = v.abs
residue = value
root = 0
onebit = 1
onebit <<= 8 while (onebit < residue)
onebit >>= 2 while (onebit > residue)
while (onebit > 0)
x = root + onebit
if (residue >= x) then
residue -= x
root = x + onebit
end
root >>= 1
onebit >>= 2
end
return (residue == 0)
end
Here's a workup of something similar (there may be coding style/smells or clunky O/O - it's the algorithm that counts, and C++ is not my home language). In this case, we're looking for residue == 0:
#include <iostream>
using namespace std;
typedef unsigned long long int llint;
class ISqrt { // Integer Square Root
llint value; // Integer whose square root is required
llint root; // Result: floor(sqrt(value))
llint residue; // Result: value-root*root
llint onebit, x; // Working bit, working value
public:
ISqrt(llint v = 2) { // Constructor
Root(v); // Take the root
};
llint Root(llint r) { // Resets and calculates new square root
value = r; // Store input
residue = value; // Initialise for subtracting down
root = 0; // Clear root accumulator
onebit = 1; // Calculate start value of counter
onebit <<= (8*sizeof(llint)-2); // Set up counter bit as greatest odd power of 2
while (onebit > residue) {onebit >>= 2; }; // Shift down until just < value
while (onebit > 0) {
x = root ^ onebit; // Will check root+1bit (root bit corresponding to onebit is always zero)
if (residue >= x) { // Room to subtract?
residue -= x; // Yes - deduct from residue
root = x + onebit; // and step root
};
root >>= 1;
onebit >>= 2;
};
return root;
};
llint Residue() { // Returns residue from last calculation
return residue;
};
};
int main() {
llint big, i, q, r, v, delta;
big = 0; big = (big-1); // Kludge for "big number"
ISqrt b; // Make q sqrt generator
for ( i = big; i > 0 ; i /= 7 ) { // for several numbers
q = b.Root(i); // Get the square root
r = b.Residue(); // Get the residue
v = q*q+r; // Recalc original value
delta = v-i; // And diff, hopefully 0
cout << i << ": " << q << " ++ " << r << " V: " << v << " Delta: " << delta << "\n";
};
return 0;
};

I checked all of the possible results when the last n bits of a square is observed. By successively examining more bits, up to 5/6th of inputs can be eliminated. I actually designed this to implement Fermat's Factorization algorithm, and it is very fast there.
public static boolean isSquare(final long val) {
if ((val & 2) == 2 || (val & 7) == 5) {
return false;
}
if ((val & 11) == 8 || (val & 31) == 20) {
return false;
}
if ((val & 47) == 32 || (val & 127) == 80) {
return false;
}
if ((val & 191) == 128 || (val & 511) == 320) {
return false;
}
// if((val & a == b) || (val & c == d){
// return false;
// }
if (!modSq[(int) (val % modSq.length)]) {
return false;
}
final long root = (long) Math.sqrt(val);
return root * root == val;
}
The last bit of pseudocode can be used to extend the tests to eliminate more values. The tests above are for k = 0, 1, 2, 3
a is of the form (3 << 2k) - 1
b is of the form (2 << 2k)
c is of the form (2 << 2k + 2) - 1
d is of the form (2 << 2k - 1) * 10
It first tests whether it has a square residual with moduli of power of two, then it tests based on a final modulus, then it uses the Math.sqrt to do a final test. I came up with the idea from the top post, and attempted to extend upon it. I appreciate any comments or suggestions.
Update: Using the test by a modulus, (modSq) and a modulus base of 44352, my test runs in 96% of the time of the one in the OP's update for numbers up to 1,000,000,000.

Here is a divide and conquer solution.
If the square root of a natural number (number) is a natural number (solution), you can easily determine a range for solution based on the number of digits of number:
number has 1 digit: solution in range = 1 - 4
number has 2 digits: solution in range = 3 - 10
number has 3 digits: solution in range = 10 - 40
number has 4 digits: solution in range = 30 - 100
number has 5 digits: solution in range = 100 - 400
Notice the repetition?
You can use this range in a binary search approach to see if there is a solution for which:
number == solution * solution
Here is the code
Here is my class SquareRootChecker
public class SquareRootChecker {
private long number;
private long initialLow;
private long initialHigh;
public SquareRootChecker(long number) {
this.number = number;
initialLow = 1;
initialHigh = 4;
if (Long.toString(number).length() % 2 == 0) {
initialLow = 3;
initialHigh = 10;
}
for (long i = 0; i < Long.toString(number).length() / 2; i++) {
initialLow *= 10;
initialHigh *= 10;
}
if (Long.toString(number).length() % 2 == 0) {
initialLow /= 10;
initialHigh /=10;
}
}
public boolean checkSquareRoot() {
return findSquareRoot(initialLow, initialHigh, number);
}
private boolean findSquareRoot(long low, long high, long number) {
long check = low + (high - low) / 2;
if (high >= low) {
if (number == check * check) {
return true;
}
else if (number < check * check) {
high = check - 1;
return findSquareRoot(low, high, number);
}
else {
low = check + 1;
return findSquareRoot(low, high, number);
}
}
return false;
}
}
And here is an example on how to use it.
long number = 1234567;
long square = number * number;
SquareRootChecker squareRootChecker = new SquareRootChecker(square);
System.out.println(square + ": " + squareRootChecker.checkSquareRoot()); //Prints "1524155677489: true"
long notSquare = square + 1;
squareRootChecker = new SquareRootChecker(notSquare);
System.out.println(notSquare + ": " + squareRootChecker.checkSquareRoot()); //Prints "1524155677490: false"

Newton's Method with integer arithmetic
If you wish to avoid non-integer operations you could use the method below. It basically uses Newton's Method modified for integer arithmetic.
/**
* Test if the given number is a perfect square.
* #param n Must be greater than 0 and less
* than Long.MAX_VALUE.
* #return <code>true</code> if n is a perfect
* square, or <code>false</code> otherwise.
*/
public static boolean isSquare(long n)
{
long x1 = n;
long x2 = 1L;
while (x1 > x2)
{
x1 = (x1 + x2) / 2L;
x2 = n / x1;
}
return x1 == x2 && n % x1 == 0L;
}
This implementation can not compete with solutions that use Math.sqrt. However, its performance can be improved by using the filtering mechanisms described in some of the other posts.

Square Root of a number, given that the number is a perfect square.
The complexity is log(n)
/**
* Calculate square root if the given number is a perfect square.
*
* Approach: Sum of n odd numbers is equals to the square root of n*n, given
* that n is a perfect square.
*
* #param number
* #return squareRoot
*/
public static int calculateSquareRoot(int number) {
int sum=1;
int count =1;
int squareRoot=1;
while(sum<number) {
count+=2;
sum+=count;
squareRoot++;
}
return squareRoot;
}

Here is the simplest and most concise way, although I do not know how it compares in terms of CPU cycles. This works great if you only wish to know if the root is a whole number. If you really care if it is an integer, you can also figure that out. Here is a simple (and pure) function:
private static final MathContext precision = new MathContext(20);
private static final Function<Long, Boolean> isRootWhole = (n) -> {
long digit = n % 10;
if (digit == 2 || digit == 3 || digit == 7 || digit == 8) {
return false;
}
return new BigDecimal(n).sqrt(precision).scale() == 0;
};
If you do not need micro-optimization, this answer is better in terms of simplicity and maintainability. If you will be calculating negative numbers, you will need to handle that accordingly, and send the absolute value into the function. I have included a minor optimization because no perfect squares have a tens digit of 2, 3, 7, or 8 due to quadratic residues mod 10.
On my CPU, a run of this algorithm on 0 - 10,000,000 took an average of 1000 - 1100 nanoseconds per calculation.
If you are performing a lesser number of calculations, the earlier calculations take a bit longer.
I had a negative comment that my previous edit did not work for large numbers. The OP mentioned Longs, and the largest perfect square that is a Long is 9223372030926249001, so this method works for all Longs.

This question got me wondering, so I did some simple coding and I'm presenting it here because I think it's interesting, relevant, but I don't know how useful. There's a simple algorithm
a_n+1 = (a_n + x/a_n)/2
for calculating square roots, but it's meant to be used for decimals. I wondered what would happen if I just coded the same algorithm using integer maths. Would it even converge on the right answer? I didn't know, so I wrote a program...
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <math.h>
_Bool isperfectsquare(uint64_t x, uint64_t *isqrtx) {
// NOTE: isqrtx approximate for non-squares. (benchmarked at 162ns 3GHz i5)
uint32_t i;
uint64_t ai;
ai = 1 + ((x & 0xffff000000000000) >> 32) + ((x & 0xffff00000000) >> 24) + ((x & 0xffff0000) >> 16);
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = (ai + x/ai)/2;
ai = ai & 0xffffffff;
if (isqrtx != NULL) isqrtx[0] = ai;
return ai*ai == x;
}
void main() {
uint64_t x, isqrtx;
uint64_t i;
for (i=1; i<0x100000000; i++) {
if (!isperfectsquare(i*i, &isqrtx)) {
printf("Failed at %li", i);
exit(1);
}
}
printf("All OK.\n");
}
So, it turns out that 12 iterations of the formula is enough to give correct results for all 64 bit unsigned longs that are perfect squares, and of course, non-squares will return false.
simon#simon-Inspiron-N5040:~$ time ./isqrt.bin
All OK.
real 11m37.096s
user 11m35.053s
sys 0m0.272s
So 697s/2^32 is approx 162ns. As it is, the function will have the same runtime for all inputs. Some of the measures detailed elsewhere in the discussion could speed it up for non-squares by checking the last four bits etc. Hope someone finds this interesting as I did.

If speed is a concern, why not partition off the most commonly used set of inputs and their values to a lookup table and then do whatever optimized magic algorithm you have come up with for the exceptional cases?

"I'm looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer)."
The answers are impressive, but I failed to see a simple check :
check whether the first number on the right of the long it a member of the set (0,1,4,5,6,9) . If it is not, then it cannot possibly be a 'perfect square' .
eg.
4567 - cannot be a perfect square.

It ought to be possible to pack the 'cannot be a perfect square if the last X digits are N' much more efficiently than that! I'll use java 32 bit ints, and produce enough data to check the last 16 bits of the number - that's 2048 hexadecimal int values.
...
Ok. Either I have run into some number theory that is a little beyond me, or there is a bug in my code. In any case, here is the code:
public static void main(String[] args) {
final int BITS = 16;
BitSet foo = new BitSet();
for(int i = 0; i< (1<<BITS); i++) {
int sq = (i*i);
sq = sq & ((1<<BITS)-1);
foo.set(sq);
}
System.out.println("int[] mayBeASquare = {");
for(int i = 0; i< 1<<(BITS-5); i++) {
int kk = 0;
for(int j = 0; j<32; j++) {
if(foo.get((i << 5) | j)) {
kk |= 1<<j;
}
}
System.out.print("0x" + Integer.toHexString(kk) + ", ");
if(i%8 == 7) System.out.println();
}
System.out.println("};");
}
and here are the results:
(ed: elided for poor performance in prettify.js; view revision history to see.)