# Tricky Double

Other languages: 中文 Español Português

This text is a translation of the original article by my friend, Maksim Pelevin. If you are comfortable with reading Russian, you might find more interesting articles in his blog.

You already know this:

The `double`

data type is not very precise

Let’s reinforce this understanding by solving a problem on CodeWars.

## Problem statement

The task as described on CodeWars is as follows:

Your task is to construct a building, which will be a pile of *n* cubes.
The cube at the bottom will have a volume of *n^3*,
the cube above will have volume of *(n-1)^3*
and so on until the top which will have a volume of *1^3*.

You are given the total volume *m* of the building.
Being given *m*, can you find the number *n* of cubes you will have to build?

The parameter of the function `findNb`

will be an integer `m`

and you have to return the integer `n`

such as
*n^3 + (n-1)^3 + … + 1^3 = m* if such a *n* exists or `-1`

if there is no such *n*.

## Naive solution

The most straightforward solution using loops would be something like:

```
public static long findNb(long m) {
long n = 0, totalVolume = 0;
while (totalVolume < m) {
totalVolume += ++n * n * n;
}
return totalVolume == m ? n : -1;
}
```

## Mathematical solution

Back in university, I learned about mathematical series and the methods to calculate their sums, so here is my take on a mathematical solution.

I no longer recall the formulas, but thankfully, Wolfram Alpha is here to help. To get the formula, let’s run the following query: n^3, n = 1 to k.

The results tell us that:

Let’s substitute `m`

for the left side of the equation,
then apply the square root property
and multiply both sides of the equation by two.
After simplifying, we get:

### Write the code

Here is a solution for the equation above, written in Java:

```
public static long findNb(long m) {
double res = Math.sqrt(m);
double d = 1 + 8 * res;
double n = (Math.sqrt(d) - 1) / 2;
return n - Math.floor(n) == 0 ? (long) n : -1;
}
```

### Results

It turns out that this code returns an incorrect result for a subset of test data.
This inconcistency is visible, for example, when
you pass `1646842954019151682L`

as the input.

While calculating a square root in Java, you may get a result that isn’t wholly accurate. Let’s take a look:

```
System.out.println(String.format("%.30f", Math.sqrt(1646842954019151682L)));
// output 1283293791.000000000000000000000000000000
// correct 1283293791.00000000038962239473657673134031176401122912...
```

Obviously, the output and the correct answer are different. Fin!

## Conclusion

The most technically correct and potentially fastest solution is not always the best one to choose. This may be no surprise to a skilled developer, but those new to programming might puzzle over the cause of the error for quite a while.

If you’re wondering how to fix the problem with the mathematical solution,
one approach is to use `BigDecimal`

. This code passes all the tests:

```
public static long findNb(long m) {
MathContext mc = new MathContext(64);
BigDecimal res = BigDecimal.valueOf(m).sqrt(mc);
BigDecimal d = res.multiply(BigDecimal.valueOf(8)).add(BigDecimal.ONE);
BigDecimal n = d.sqrt(mc).subtract(BigDecimal.ONE).divide(BigDecimal.valueOf(2));
return Objects.equals(n, n.setScale(0, RoundingMode.FLOOR)) ? n.longValue() : -1;
}
```

… and a little bonus: can you guess which of the following `true`

,
and which is `false`

?

`(1646842954019151681L == 1646842954019151682D)`

`(1646842954019151681L == (long) 1646842954019151682D)`