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 Negative integer division  (Read 3972 times) 0 Members and 1 Guest are viewing this topic.
loom_weaver

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 « Posted 2011-11-21 00:32:35 »

Here's a puzzler for you.

Given the following coordinates (same for x or y-axis):
 1 `...|-300 ... -201|-200 ... -101|-100 ... -1|0 ... 99|100 ... 199|200 ... 299|...`

Can you come up with a simple function that given an int, returns the anchor point?  The anchor point is the left-hand number in the range and is a multiple of 100.  For example for any value between -200 ... -101 return -200.  For 100 ... 199 return 100.

The biggest problem for me is that integer division of negative values seems to be different from language to language!  (-1 / 100) gives me -1 in python but 0 in Java.

Otherwise I'd use f(x) = x / 100 * 100
Riven
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 « Reply #1 - Posted 2011-11-21 00:38:45 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

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SimonH
 « Reply #2 - Posted 2011-11-21 00:55:05 »

Another option would be to limit the play area to +ve coordinates only.
Integer.MAX_VALUE*Integer.MAX_VALUE is a quarter the size but it's still very, very big!

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theagentd

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 « Reply #3 - Posted 2011-11-21 01:17:13 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

... or just
 1 `return (int)Math.floor(x/100.0);`

It might be slower, though, but it's at least more readable.

Myomyomyo.
counterp

Senior Devvie

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 « Reply #4 - Posted 2011-11-21 02:00:02 »

 1  2  3 `   public static final int anchor(int in) {      return (in / 100 + (in >> 31)) * 100;   }`

or

 1  2  3 `   public static final int anchor(int in) {      return (in / 100 - (in >>> 31)) * 100;   }`

depending on if you want -100 to be -200 or -100
Riven
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 « Reply #5 - Posted 2011-11-21 02:18:31 »

 1  2  3 `   public static final int anchor(int in) {      return (in / 100 + (in >> 31)) * 100;   }`

or

 1  2  3 `   public static final int anchor(int in) {      return (in / 100 - (in >>> 31)) * 100;   }`

depending on if you want -100 to be -200 or -100
both these examples are incorrect (partly because they are identical), loom_weaver is very clear about the spec. you're off by one for all negative multiples of 100. there is a reason i used 99 as opposed to 100 in my solution. both -1 and -100 should result in -100.

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loom_weaver

JGO Coder

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 « Reply #6 - Posted 2011-11-21 02:36:40 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

f(-250) = -400.  Should be -300.

I think this is the best I can do.  I was hoping for a non-branching version:

 1  2  3  4  5  6  7 `int anchor(int x) {    if (x < 0) {        return -((-x + 99) / 100 * 100)    } else {        return x / 100 * 100    }}`
Riven
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 « Reply #7 - Posted 2011-11-21 02:42:12 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

f(-250) = -400.  Should be -300.
I don't know how you tested that, but it simply returns -300, as it should.

 1  2  3  4  5  6  7  8  9 `   public static final int anchor(int x)   {      return (x - ( x<0 ? 99 : 0 )) / 100 * 100;   }   public static void main(String[] args)   {      System.out.println(anchor(-250)); // "-300"   }`

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loom_weaver

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 « Reply #8 - Posted 2011-11-21 02:54:41 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

f(-250) = -400.  Should be -300.
I don't know how you tested that, but it simply returns -300, as it should.

Oops... was testing in Python again.  Argh.  Oh well my lesson learned is be careful with division involving -ve integers.

In the old C standard it did not define what direction to round to (-ve INFINITY or 0) if dividing with a negative int.  In later standards, it rounds towards zero.

http://stackoverflow.com/questions/828092/python-style-integer-division-modulus-in-c
http://stackoverflow.com/questions/319880/integer-division-rounding-with-negatives-in-c
Riven
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 « Reply #9 - Posted 2011-11-21 02:57:02 »

As you were hoping for one, here is a non-branching solution:

 1  2  3  4 `   public static final int anchor(int x)   {      return (x - ((x >>> 31) * 99)) / 100 * 100;   }`

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Riven
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 « Reply #10 - Posted 2011-11-21 02:59:27 »

If you want the offset (relative to your anchor), in a non-branching way:

 1  2  3  4 `   public static final int offset(int x)   {      return (x%100+100)%100;   }`

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loom_weaver

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 « Reply #11 - Posted 2011-11-21 03:03:48 »

Dang... taking the high-order bit and multiplying by 99.

Thanks for the formulas guys!
Riven
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 « Reply #12 - Posted 2011-11-21 03:06:10 »

it you really want to get it 'fast', you can take out the multiply...
 1 `((x >>> 31) * 99) == ((x >> 31) & 99)`

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ra4king

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 « Reply #13 - Posted 2011-11-21 05:40:59 »

 1 `return (x - ( x<0 ? 99 : 0 )) / 100 * 100;`

... or just
 1 `return (int)Math.floor(x/100.0);`

It might be slower, though, but it's at least more readable.
Your example is completely incorrect

@Riven
That last one returns true or false, not a value. What is its purpose?
Hehe looked at it a bit more and realized that you mean to substitute the right-hand side for the left-hand side. Wow that is quite clever and much much faster!

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