# ¿Cómo verificar si el número dado es divisible de 15 de la manera más rápida?

Division in processor takes much time, so I want to ask how to check in fastest way if number is divisible of some other number, in my case I need to check if number is divisible by 15.

Also I've been looking through web and found diversión ways to check if number is divisible of some number, but I'm looking for fast option.

NOTA: as division takes much time I'm looking for answer without `/` y `%`.

preguntado el 09 de septiembre de 13 a las 21:09

`a % 15 == 0`: trust the compiler and hardware to do it efficiently. -

@awashburn modulus is not the fastest option in some cases -

@user2623967 hence the word confianza. The compiler most likely knows better then you how to optimize for the local system. -

posible duplicado de Comprueba si un número es divisible por 3 (my answer has solution for N=15 too) -

@awashburn: "trust the hardware and compiler to do it efficiently" is unfortunately not true as some compiler don't implement non-division solutions when while others - like gcc - do. When they don't you're stuck with inefficient code and you have to resort to manually implementing the code. -

## 6 Respuestas

Obligatory answer for other learners who might come looking for an answer.

`if (number % n == 0)`

In la mayor cases, you can always do this, trusting the smart modern compilers.

This doesn't mean you get discouraged from learning fun ways though. Check out these links.

Pruebas de divisibilidad rápidas (por 2,3,4,5, .., 16)?

Pequeños trucos de Twiddling

contestado el 23 de mayo de 17 a las 13:05

"Obligatory answer?" Learners want to learn, the really interested ones expand the envelope. "Trusting the smart modern compilers?" Yeah, right. In your example I can trust the smart modern compiler to use a 60+ clock cycle divide instruction which is what compilers always do when the divisor is variable. Silly anser. - Olof Forshell

And what if the "fun ways" are the better ways, the faster ways, the more efficient ways? Then they should replace the current ways. - Olof Forshell

@OlofForshell When divisor is variable, there is not much else you can do anyway. - Erbureth

@Erbureth: there are almost always things that can be done. That they may be more or less feasible is another issue. If the programmer knows the range of values for n that might lead to openings. Certainly for the {2, 3, 4 ... 16} set there are generic possibilities which might clobber performance for the {2, 4, 8, 16} set but speed up the others. Usually it's about trading space for time and if you're in a jam, what choice do you have other than to try? Or if you want to explore the possibilities, learning something all by yourself - not being told how something should be done? - Olof Forshell

Multiplication takes less time then division, so you can try this:

``````inline bool divisible15(unsigned int x)
{
//286331153 = (2^32 - 1) / 15
//4008636143 = (2^32) - 286331153
return x * 4008636143u <= 286331153u;
}
``````

This way works because `2^32-1` (max 32-bit value) is divisible of 15, however if you take, for example 7, it would look like working, but wouldn't work in all cases.

EDIT: Ver este, it proves that this solution (en algunos compiladores) is faster then module.

EDIT: Aquí is the explanation and the generalization.

respondido 13 mar '18, 17:03

There are working solutions for 7 and other non-power-of-2 values. gcc implements these solutions when the divisor is constant. I got a real fright the first time I saw the generated assembly. But it works and it's fast. - Olof Forshell

@OlofForshell I used 7 as an example number which do not work in this case, I know it is other ways to check if value is divisible, but in 15 case, this is probably fastest. - ST3

7 may be tested in the same manner but using different constants. For the moment the multiplication method is the fastest but who says that won't ever change? - Olof Forshell

@user2623967: I spent some time looking for this (homepage.cs.uiowa.edu/~jones/bcd/divide.html) link which explains a more structured approach and you can also check two publications here (gmplib.org/~tege) concerning division by invariant (constant) integers. - Olof Forshell

Your new idea works with 15. It doesn't work with 30, which last time I checked is a multiple of 15. It also gives a false positive for 31. It's just testing if the number is one less than a multiple of 16. - HierroMensan

Sólo tiene que utilizar `i % 15 == 0`

1. Since the compiler can easily see that 15 will never change it can feel free to make any optimization it wants on the mod operation. It is a compiler writers job to make this kind of optimization if they haven't thought of a better way to do it you won't.

2. For example it is very easy to check if a number is divisible by 2 because you just check the first bit. Compiler writers know this and you could write the code yourself to check the first bit but especially a mature compiler will have people thinking about and working on these things for years. This type of optimization is a very simple one to make as it only requires changing an instruction or 2, optimizations like better register allocation are much harder to achieve

3. One more thing to consider is that your compiler was written for the system that it is on, you code on the other hand is the same everywhere, if you write some strange code that may be just as fast on one system (probably still not faster) but on another system which has a special hardware optimization your code may lose by an order of magnitude. Since you wrote some esoteric code to check for the divisibility the compiler is unlikely to realize it can optimize to a single hardware op, writing the obvious thing to do makes life better for you and easier for the compiler.

4. Since you haven't actually checked that the speed matters writing code the strange way will make the code very difficult to read for the next person and more error prone ( La optimización temprana es la raíz de todo mal)

5. It still works whether the input is 16, 32, or 64 bits since it doesn't rely on an bit manipulation.

6. Even if the compiler writer hasn't implemented it, it is clearly possible for someone to implement it (even yourself)

Respondido el 09 de Septiembre de 13 a las 22:09

5. It still works whether the input is 16, 32, or 64 bits (or any other size) - adam rosenfield

1. What if the 15 is not constant? The compiler has little hope of optimizing it, but if the possible values all admit techniques parallel to this one, a solution with a table indexed by the modulus could be used. 3. True in general, but unlikely to apply to this case. 4. Easily solved with a comment. - R .. GitHub DEJA DE AYUDAR A ICE

@R.. what's ur point, the 15 is constant in this case, obviously a mod table only works if the input range is small, are you criticizing my answer BTW I can't tell - Aaronman

Just providing some reasons it podría be reasonable to manually optimize this like OP asked for. - R .. GitHub DEJA DE AYUDAR A ICE

I have problems respecting anyone who throws around the "premature optimiztion ..." quote without stating that there are precise conditions (3/97%) in it, i e when it holds true and when it doesn't. I feel that for an ongoing, commercial project the 3% condition is valid - though many project managers would disagree. If you are exploring the subject for fun, pleasure and recreation why should you be bothered with the expression at all? - Olof Forshell

On a reasonably modern process, dividing by 15 shouldn't be that terrible. The AMD optimisation guide defines it based on the quotient (the value that is being divided), and it takes 8 + bit position of the most significant bit of the quotient. So if your numbers have the 63rd bit set, you end up with 71 cycles - which is quite a long instruction, of course. But for a 32-bit number with a few zeros in the top bits, we're talking 30-40 cycles. If the number fits in a 16-bit value, we the maximum is 23 cycles.

To get the remainder is one more clockcycle on top of that.

If you are doing this ALL the time, of course, you may find that this time is quite long, but I'm not sure there is a trivial way to avoid it.

Like others have said, compiler may be able to replace it with something better. But 15 does not, to my knowledge have an obvious fast solution (if you have 16 instead of 15, then we can use the trick of `x & 15`).

If it's a limited range, you could build a table [`vector<bool>` for example, which will store 1 bit per entry], but you'll quite soon run into the problem that the non-cached memory access takes just as long as a divide operation...

There are some interesting ways to figure out if a number divides by 3, 5 and so on by summing the digits, but unfortunately, those only work based on decimal digits, which involves a long sequence of divides.

Respondido el 09 de Septiembre de 13 a las 22:09

There is a trivial way to avoid it which is called "division by invariant integers using multiplication." user2623967's answer - an hour ahead of your's - points to this possible solution. It's been around for a long time and is implemented in some compilers, one of which is gcc. You need to do some reading up I think. - Olof Forshell

Here's another approach, which is probably slower than others, but uses only addition, bitwise-and and shift:

``````int divisible15(unsigned int x) {
if (x==0) return 1;
x = (x & 0x0f0f0f0f) + ((x&0xf0f0f0f0)>>4);
x = (x & 0x00ff00ff) + ((x&0xff00ff00)>>8);
x = (x & 0x0000ffff) + ((x&0xffff0000)>>16);
x = (x & 0x0f) + ((x&0xf0)>>4);
return x==15;
}
``````

The idea is that divisibility by 15, in base 16, is like divisibility by 9 in base 10 - the sum of digits must be divisible by 15.
So the code sums up all hex digits (similarly to the way you count bits), and the sum must equal 15 (except for 0).

Respondido 08 Oct 13, 08:10

You could shift before masking so each line would use the same constant twice. This would probably help on RISC architectures where large immediate constants have to be constructed in registers with multiple instructions (unlike x86, where any instruction can have a 32bit immediate). - pedro cordes

@PeterCordes, I guess so. Something like `x = (x & 0x0f0f0f0f) + ((x>>4) & 0x0f0f0f0f);` - ugoren

Well, it's very easy to do in your head if you have the hex representation. Just sum all the digits, until you have a single digit. If the answer is '0xf', it's divisible by 15.

Ejemplo `0x3a98` : 3 + 0xa + 9 + 8 = 0x1e = 1 + 0xe = 0xf, so that's divisible by 15.

This works for all factors on X-1, where X is the base used to represent the number. (For smaller factors, the final digit must be divisible by the factor).

Don't expect this to be fast in code, though.

Respondido el 29 de enero de 16 a las 12:01

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