Develop Reference

How to check if a ≤ b with an epsilon precision

So I have two doubles, a and b, and I need to check if a ≤ b with the precision of a given epsilon.
So to check if a == b I need to do this (a and b are doubles, EPSILON is a final double, in my case 0.001):
Math.abs(a - b) < EPSILON
My idea as an answer to my own question is:
a < EPSILON + b
My problem is that with a precision of epsilon, what would be the difference between just less than and less than or equal in the final result, maybe someone has a better way to write it?

You don't want to write, a < EPSILON+b, because if b is large, then you might have b == EPSILON+b, and then a < EPSILON+b would fail if a and b were exactly equal.
(a-b) < EPSILON works.
When comparing numbers with "epsilon precision" you consider numbers to be the same if they are within EPSILON, so "a < b, with epsilon precision" is actually:
(a-b) <= -EPSILON

If you can use BigDecimal, then use it, else:
/**
*#param precision number of decimal digits
*/
public static boolean areEqualDouble(double a, double b, int precision) {
return Math.abs(a - b) <= Math.pow(10, -precision);
}

UPDATE: this post is wrong, see comments. #matttimmens' answer seem legit. (I still have to check corner cases)
UPDATE 2 I checked and learned. I thought I'd find corner cases like with integers. In Java, (Integer.MIN_VALUE == ((-Integer.MIN_VALUE))) is true. This is because integer numbers (as in: non-decimal, non-floating point) have asymmetrical ranges:
first positive number is 0, first negative number is -1, and thus
byte: MAX_VALUE is 127, while MIN_VALUE is -128, so MAX_VALUE != -MIN_VALUE and MIN_VALUE != MAX_VALUE
also goes for short, int, long
This, mixed together with silent integer overflow, creates problems when working in the extreme corners, especially the off-by-one problem:
final int a = Integer.MIN_VALUE;
final int b = Integer.MAX_VALUE;
final int eps = 1;
System.out.println("a-b = " + (a - b));
System.out.println((a - b) < eps);
a-b results in 1, and 1 < eps is false, though we clearly see that a is a lot smaller than b.
This is one corner case where the algorithm would fail on integers. Mind: the OP was about doubles.
However, floating-point numbers in Java (float, double), or rather, operations on them, compensate, and do not allow for overflows to happen:
(Double.MAX_VALUE + 1) == Double.MAX_VALUE holds true
(Double.MAX_VALUE + Double.MAX_VALUE) == Double.POSITIVE_INFINITY holds true
(-Double.MAX_VALUE + -Double.MAX_VALUE) == Double.NEGATIVE_INFINITY holds true
So before, with integers, the +1 or -1 silent integer overflows caused a problem.
So when adding a +1 (or -1 or any small number) to a really large double variable, the change gets lost to rounding. Thus the corner cases play no role here, and #matttimmens' solution (a-b)<e is as close to the truth as we can get for floating-point variables, rounding left aside.
My old, wrong post:
To counter #matttimmens' reply, run that code:
final double a = Double.MIN_VALUE;
final double b = Double.MIN_VALUE + 0;
final double c = Double.MIN_VALUE + 10;
final double e = 0.001;
System.out.println("Matches 1: " + ((a - b) < e));
System.out.println("Matches 2: " + ((b - a) < e));
System.out.println("Matches 3: " + ((a - c) < e));
System.out.println("Matches 4: " + ((c - a) < e));
Match 3 gives a false result, thus his answer is wrong.
Funny thing is that he thought about vary big numbers, but left the very small numbers unattended.

How to Correctly Round a Square Root Function?

I am currently working on a Java math library which will include a variety of correctly rounded functions (i.e. sqrt, cbrt, exp, sin, gamma, and ln). I have already used the Babylonian method to write a square root algorithm that is correct to within 1 ulp of the correct answer. However, I cannot figure out how to properly calculate which way the number should be rounded to represent the best possible approximation to the actual square root of the input. Answers containing principles which can be extended to other functions would be preferred, but I have heard that sqrt is a simpler case than many transcendental functions, and specialized solutions would also be much appreciated.
Also, here is a cleaned-up version of my code as of this question's original submission:
public static double sqrt(double x) {
long bits = Double.doubleToLongBits(x);
// NaN and non-zero negatives:
if (Double.isNaN(x) || x < 0) return Double.NaN;
// +-0 and 1:
if (x == 0d || x == 1d) return x;
// Halving the exponent to come up with a good initial guess:
long exp = bits << 1;
exp = (exp - 0x7fe0000000000000L >> 1) + 0x7fe0000000000000L >>> 1 & 0x7ff0000000000000L;
double guess = Double.longBitsToDouble(bits & 0x800fffffffffffffL | exp);
double nextUp, nextDown, guessSq, nextUpSq, nextDownSq;
// Main loop:
while (true) {
guessSq = guess * guess;
if (guessSq == x) return guess;
nextUp = Math.nextUp(guess);
nextUpSq = nextUp * nextUp;
if (nextUpSq == x) return nextUp;
if (guessSq < x && x < nextUpSq) {
double z = x / nextUp;
if (z * nextUp > x) z = Math.nextDown(z);
return z < nextUp ? nextUp : guess;
}
nextDown = Math.nextDown(guess);
nextDownSq = nextDown * nextDown;
if (nextDownSq == x) return nextDown;
if (nextDownSq < x && x < guessSq) {
double z = x / guess;
if (z * guess > x) z = Math.nextDown(z);
return z < guess ? guess : nextDown;
}
// Babylonian method:
guess = 0.5 * (guess + x / guess);
}
}
As you can see, I was using division as a test. However, I believe that requires the division to round towards 0, which obviously doesn't happen in Java.

By the Taylor theorem, the square root function is locally approximated by a linear function, of slope 1/2√x, which is positive. So you can relate the error to the error in the square, x - (√x)², where √x is understood to be the approximate root. Then you round in the direction that minimizes this error.
Anyway, the computation of x - (√x)² is subjected to catastrophic cancellation and you may need extended accuracy to compute it reliably. Not sure the benefit is worth the effort.

Java integer division doesn't give floor for negative numbers

I was trying to use java's integer division, and it supposedly takes the floor. However, it rounds towards zero instead of the floor.
public class Main {
public static void main(String[] args) {
System.out.println(-1 / 100); // should be -1, but is 0
System.out.println(Math.floor(-1d/100d)); // correct
}
}
The problem is that I do not want to convert to a double/float because it needs to be efficient. I'm trying to solve this with a method, floorDivide(long a, long b). What I have is:
static long floorDivide(long a, long b) {
if (a < 0) {
// what do I put here?
}
return a / b;
}
How can I do this without a double/float?

floorDiv() from Java.Math that does exactly what you want.
static long floorDiv(long x, long y)
Returns the largest (closest to positive infinity) long value that is less than or equal to the algebraic quotient.

Take the absolute value, divide it, multiply it by -1.
Weird bug.

You can use
int i = (int) Math.round(doubleValToRound);
It will return a double value that you can cast into an int without lost of precission and without performance problems (casts haven't a great computational cost)
Also it's equivalent to
int a = (int) (doubleValToRound + 0.5);
//in your case
System.out.println((int) ((-1 / 100) + 0.5));
With this last one you won't have to enter into tedious and unnecesary "if" instructions. Like a good suit, its valid for every moment and has a higher portability for other languages.

This is ugly, but meets the requirement to not use a double/float. Really you should just cast it to a double.
The logic here is take the floor of a negative result from the integer division if it doesn't divide evenly.
static long floorDivide(long a, long b)
{
if(a % b != 0 && ((a < 0 && b > 0) || (a > 0 && b < 0)))
{
return (a / b - 1);
}
else
{
return (a / b);
}
}

Just divide the two integers. then add -1 to the result (in case the absolute value of both numerator and denominator are not same). For example -3/3 gives you -1, the right answer without adding -1 to the division.

Since a bit late, but you need to convert your parameters to long or double
int result = (int) Math.floor( (double) -1 / 5 );
// result == -1
This worked for me elegantly.

I would use floorDiv() for a general case, as Frank Harper suggested.
Note, however, that when the divisor is a power of 2, the division is often substituted by a right shift by an appropriate number of bits, i.e.
x / d
is the same as
x >> p
when p = 0,1,...,30 (or 0,1,...,62 for longs), d = 2p and x is non-negative. This is not only more effective than ordinary division but gives the right result (in mathematical sense) when x is negative.

Recursive function taking ages to run

I profiled my code and found that my program spent roughly 85% of the time executing this particular recursive function. The function aims to calculate the probability of reaching a set of states in a markov chain, given an initial position (x,y).
private static boolean condition(int n){
int i = 0;
while ( n >= i){
if( n == i*4 || n == (i*4 - 1))
return true;
i++;
}
return false;
}
public static double recursiveVal(int x, int y, double A, double B){
if(x> 6 && (x- 2 >= y)){ return 1;}
if(y> 6 && (y- 2 >= x)){ return 0;}
if(x> 5 && y> 5 && x== y){ return (A*(1-B) / (1 -(A*B) - ((1-A)*(1-B))));}
if(condition(x+ y)){
return (recursiveVal(x+1, y,A,B)*A + recursiveVal(x, y+1,A,B)*(1-A));
}
else{
return (recursiveVal(x+1, y,A,B)*(1-B) + recursiveVal(x,y+1,A,B)*B);
}
}
I was once told that 99% of recursive functions could be replaced by a while loop. I'm having trouble doing this though. Does anyone know how I could improve the execution time or rewrite this as a iterative loop?
Thanks

You could try to use a technique called memoization which basically caches previously computed results for recursive calls.
Wikipedia article on memoization.
As a side note, I recommend reformatting your code a bit. Here is a simplified version of yoru code.
private static boolean condition(int n){
for (int i = 0; i <= n; i++)
if(n == i*4 || n == (i * 4 - 1))
return true;
return false;
}
public static double recursiveVal(int x, int y, double A, double B){
if (x > 6 && (x - 2 >= y))
return 1;
if (y > 6 && (y - 2 >= x))
return 0;
if(x > 5 && y > 5 && x == y)
return A*(1-B) / (1 -(A*B) - ((1-A)*(1-B)));
double val1 = recursiveVal(x+1, y, A, B);
double val2 = recursiveVal(x, y+1, A, B);
return condition(x + y)
? A * val1 + val2 * (1-A)
: (1-B) * val1 + B * val2;
}

If you want to refactor a recursive function to an iterative one the steps are as follows:
1) Determine the base case of the Recursion. Base case, when reached, causes Recursion to end. Every Recursion must have a defined base case. In addition, each recursive call must make a progress towards the base case (otherwise recursive calls would be performed infinitely).
2) Implement a loop that will iterate until the base case is reached.
3) Make a progress towards the base case. Send the new arguments to the top of the loop instead to the recursive method.

Example of how trying to understand what it is doing can make the code faster/simpler...
This method is trying to determine if 'n' is a multiple of 4 or n+1 is a multiple of 4. This can be written much shorter as.
private static boolean condition(int n){
return (n+1) & 3 <= 1;
}

To convert this to an iterative form, note that you are computing a function on two (discrete) variables. You can use a table to store the values of the function, and fill in the table in a specific order so that you have already computed values you need by the time you need them. (in this case, from higher values of x and y).
In this case, the boundary cases (corresponding to the base cases in the original recursion) are:
f(7, y..5), f(8, 6)
f(x..5, 7), f(6, 8)
f(7, 7)
Then we fill in f(7, 6) and f(6, 7) and then proceed "downwards" - i.e.
f(6, 6), f(5, 6) ... f(x, 6), f(6, 5), f(5, 5) ... f(x, 5) ... f(x, y).
Note that what looks like function call syntax corresponds to a table lookup (it really converts to array syntax).

Binary Search to Compute Square root (Java)

I need help writing a program that uses binary search to recursively compute a square root (rounded down to the nearest integer) of an input non-negative integer.
This is what I have so far:
import java.util.Scanner;
public class Sqrt {
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
System.out.print("Enter A Valid Integer: ");
int value = console.nextInt();
calculateSquareRoot(value);
}
public static int calculateSquareRoot(int value) {
while (value > 0) {
double sqrt = (int) Math.sqrt(value);
System.out.println(sqrt);
}
return -1;
}
}
The fact that it has to use binary search to compute the square root is the part that is confusing me. If anyone has any suggestions on how to do this, it would be greatly appreciated. Thank you

Teh codez:
def sqrt(n):
low = 0
high = n+1
while high-low > 1:
mid = (low+high) / 2
if mid*mid <= n:
low = mid
else:
high = mid
return low
To understand it, just think of the loop invariant, namely:
lowlow <= n < highhigh
If you understand this code, writing a recursive version should be trivial.

You can use this java method (Iterative)
public class Solution {
// basic idea is using binary search
public int sqrt(int x) {
if(x == 0 || x == 1) {
return x;
}
int start = 1, end = x / 2;
while(start <= end) {
int mid = start + (end - start) / 2;
if(mid == x / mid) {
return mid;
}
if(mid < x / mid) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return start - 1;
}
}
You can drive your own recursive method

Essentially the idea is that you can use binary search to get closer to the answer.
For example, say you are given 14 as an input. Then, you are sure that the square root of 14 is between 0 and 14. So, 0 and 14 are your current "boundaries". You bisect these two end points and obtain the mid point: 7. Then you try 7 as a candidate - If the square of 7 is greater than 14, then you have a new boundary (0,7); otherwise you would have a new boundary (7,14).
You keep repeating this bisection until you are "close enough" to the answer, for example you have a number square of which is within 14-0.01 and 14+0.01 - then you declare that as the answer.
OK, that much hint should be good enough for HW. Don't forget to cite StackOverflow.

I'm assuming this is homework so I'm only going to give a hint.
To conduct a binary search, you pick a point as close as possible the median of possible correct values. So the question becomes what is a typical median value for a square root, that is either constant or can be computed via multiplication. Obviously using an arbitrary constant will not work for most inputs, so you need to arrive at your guess by multiplying the input by a constant.
As for what that constant C to multiply by should be, that should be chosen based on what values you expect as input. For example, if you expect your inputs to be around 250,000, then:
C * 250,000 ~= sqrt(250,000)
C = sqrt(250,000) / 250,000
C = 500 / 250,000
C = 1 / 500

I see two important computing concepts in your question. The first is binary search, the second is recursion. Since this is homework, here is a contribution towards understanding a binary search, recursion and how to think about them.
Think of binary search as dividing the solution "space" in half, keeping the half the solution is in and doing that in succession so that the process converges on the solution. The key concepts for doing this are that you need to engineer a solution "space" that has the following properties:
1) can be subdivided, usually in half or at least two pieces
2) of the two pieces after subdivision, there is a way to determine which half has the solution so that the process can be repeated on only one half.
Recursion involves a function (method in O-O speak) invoking itself. Recursion works really well for a process that converges to a conclusion. It either recurses forever or until you run out of some resource, usually memory, and it fatally stops. The two key concepts for recursion are:
1) convergence through some invariance (more on invariance below).
2) termination condition (one that recognizes sufficient convergence).
Now, for your square root routine. The requirements for the routine are:
1) Integer input.
2) Integer square-root approximation that gives the floor integer closest to the actual square root.
3) Use recursion.
4) Use binary search.
It helps to know some mathematics about square roots for this. Elementary calculus and analytical geometry concepts are helpful too. Lets do some reasoning.
We have an arbitrary positive integer x. We want its root y. If we choose some test value for y, we can see if it is the root of x if y * y = x. If y is too big, y * y > x. if y is too small, y * y < x. We also know that 0 <= root <= x and that square-roots of 0 and 1 are trivially zero and 1. Since we are looking for largest integer where y * y <= x (i.e. a floor value) we'll have to account for that too.
Here is some mathematical reasoning to help. We know that x = y * y where y is the square root of x. That means: y = x/y.
Hmmm... what happens if y is to large to be the square root of x? Then: x < y * y and: x/y < y which means x/y is also too small to be the square root of x. So we know that, for y too large, x/y < square-root of x < y. So, lets find a new y, say y1, between x/y and y as a new test value. The average of x/y and y will do. y1 = (x/y0 + y0)/2 will give a y1 that is closer to the square root of x than y0 if y0 is too large.
Does this converge? Well, in mathematics using positive real numbers, the average will always be above the value but getting closer each iteration. This satisfies the condition that we successively divide the solution "space" into two parts and know which of the two to keep. In this case, we successively calculate new values below previous ones and below which the answer still lies, allowing us to discard all values above the new one. We stop when we reach a condition where no more new values above the answer exist. Using computers, however, results in binary approximations of real numbers. With integers, there is truncation in division. This may affect the convergence beneficially or adversely. In addition, your answer is supposed to be the largest integer smaller than or equal to the square root. It's wise to take a look at the kind of convergence we will get.
Because of integer division turncation, y1 = (x/y0 + y0)/2 will converge until successive iterations reach an integer root or a floor value for (i.e. the largest integer less than) the root. This is ideal. If we start with a proposed value for the root that has to be larger than the root, say x itself, the first value for yn where yn * yn <= x is the desired result.
The simple answer is that, when we start with y0 > y, the first new yn that is less than or equal to y, then y - yn < 1. That is, yn is now the floor value for which we've been looking and we now have a termination condition that exactly satisfies the conditions for the required answer.
Here are basic iterative and recursive solutions. The solutions don't incude safety features to ensure negative values are not input for x. The one major concern is to avoid dividing by zero in case someone wants to find the square root of 0. Since that is a trivial answer, both the recursive and iterative methods return 0 before division by zero can take place. Both the recursive and iterative solutions work with the trivial cases for finding the square roots of 0 and of 1.
There is another analysis that always has to be done with int and long arithmetic in Java. A major concern is integer overflow since Java does nothing about int or long overflow. Overflow results in twos-complement values (look that up elsewhere) that can lead to bogus results and Java does not throw exceptions with int or long overflow.
In this case, it is easy to avoid arithmetic that could result in an internal overflow with large values of x. If we create a termination condition such as y0 * y0 < x we risk overflow if x is greater than the square root of Integer.MAX_VALUE since y0 * y0, an intermediate value, will immediately exceed the maximum int value. However, we can rearrange the termination condition to y0 < x / y0. We still have a problem with the calculations: ((x / y0) + y0) / 2) if x and y0 are Integer.MAX_VALUE since it wll attempt Integer.MAX_VALUE + 1. However, we can always start with a value less than x that is guaranteed to be > y. x / 2 works for all values of x > 1. Since the square root of x where x is either 0 or 1 is simply x, we can easily test for those values and simply return the correct and trivial value. You can construct code to prevent using values < 0 or values > Integer.MAX_VALUE. The same can be applied if we use long instead of int. Welcome to computing in the real world!
public static int intSqRootRecursive (int x) {
// square roots of 0 and 1 are trivial and x / 2 for
// the y0 parameter will cause a divide-by-zero exception
if (x == 0 || x == 1) {
return x;
}
// starting with x / 2 avoids overflow issues
return intSqRootRecursive (x, x / 2);
} // end intSqRootRecursive
private static int intSqRootRecursive(int x, int y0) {
// square roots of 0 and 1 are trivial
// y0 == 0 will cause a divide-by-zero exception
if (x == 0 || x == 1) {
return x;
} // end if
if (y0 > x / y0) {
int y1 = ((x / y0) + y0) / 2;
return intSqRootRecursive(x, y1);
} else {
return y0;
} // end if...else
} // end intSqRootRecursive
public static int intSqRootIterative(int x) {
// square roots of 0 and 1 are trivial and
// y == 0 will cause a divide-by-zero exception
if (x == 0 || x == 1) {
return x;
} // end if
int y;
// starting with y = x / 2 avoids overflow issues
for (y = x / 2; y > x / y; y = ((x / y) + y) / 2);
return y;
} // end intSqRootIterative
You can test the recursive solution to find out how many instances will result on the frame stack, but you will see that it converges very fast. It's interesting to see that the iterative solution is much smaller and faster than the recursive one, something that is often not the case and is why recursion gets used where it can be predicted that stack resources are sufficient for the recursion depth.

Here is the recursive solution in Java using binary search :
public class FindSquareRoot {
public static void main(String[] args) {
int inputNumber = 50;
System.out.println(findSquareRoot(1, inputNumber, inputNumber));
}
public static int findSquareRoot(int left, int right, int inputNumber){
// base condition
if (inputNumber ==0 || inputNumber == 1){
return inputNumber;
}
int mid = (left + right)/2;
// if square of mid value is less or equal to input value and
// square of mid+1 is less than input value. We found the answer.
if (mid*mid <= inputNumber && (mid+1)*(mid+1) > inputNumber){
return mid;
}
// if input number is greater than square of mid, we need
// to find in right hand side of mid else in left hand side.
if (mid*mid < inputNumber){
return findSquareRoot(mid+1, right, inputNumber);
}
else{
return findSquareRoot(left, mid-1, inputNumber);
}
}
}

Iterative binary solution:
public static double sqrt(int n) {
double low = 0;
double high = n;
double mid = (high - low) / 2;
while (Math.abs((mid * mid) - n) > 0.000000000001) {
if ((mid * mid) > n) {
high = mid;
mid = (high - low) / 2;
} else{
low = mid;
mid = mid + ((high - low) / 2);
}
}
return mid;
}

edst solution is good, but there is a mistake in line 11:
mid = (high - low) / 2;
should be
mid = low + (high - low) / 2;