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% modulo 4%3 = 1%3 != (Read 3227 times)
alphachapmtl
% modulo 4%3 = 1%3 !=
Oct 15th, 2007, 12:00pm  
Mathematically,
the integer sequence
(..,-5,-4,-3,-2,-1, 0, 1, 2, 3, 4, 5,..) modulo 3
becomes
(.., 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2,..)

In Processing,
(..,-5,-4,-3,-2,-1, 0, 1, 2, 3, 4, 5,..) % 3
becomes
(..,-2,-1, 0,-2,-1, 0, 1, 2, 0, 1, 2,..)

This is equivalent (and correct), but a consequence is that:

(5%3)==(2%3) is true,  but (2%3)==(-1%3) is false

It would be nice to have a non-negative version of % .

I built my own method to do this, but it would be nice to have it in the language
(maybe it could be called %%).
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davbol
Re: % modulo 4%3 = 1%3 !=
Reply #1 - Oct 15th, 2007, 5:18pm  
The confusion comes from calling "%" the "modulo" operator, which is common but technically incorrect, because it's really the "remainder" operator, such that if x/y=z and x%y=w, then z*y+w=x.  (Java/Processing doesn't have a true modular arithmetic operator - as you've already discovered)
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Eric F.
Re: % modulo 4%3 = 1%3 !=
Reply #2 - Feb 10th, 2009, 8:40am  
can you post your method?  i just tore some hairs out wondering why an equation wasn't working, and this is why...  

or is there another convention people use to wrap numbers around in both directions?
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Eric F.
Re: % modulo 4%3 = 1%3 !=
Reply #3 - Feb 10th, 2009, 9:33am  
i wrote this quickly, just curious how you did it.  mine doesn't handle all cases, just quick fix.

Code:

float mod(float a, float b) {
  if (a>=0) {
    return a%b;  
  } else {
    return a+b;
  }
}


detailed description of the particulars:
http://mindprod.com/jgloss/modulus.html
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alphachapmtl
Re: % modulo 4%3 = 1%3 !=
Reply #4 - Feb 10th, 2009, 6:25pm  
I think your method is incorrect as written.
"a+b" should be "(a%b)+b"

float mod(float a, float b) {
 if (a>=0) {
   return a%b;  
 } else {
   return a+b; // <<-- (a%b)+b
 }
}

---
For int or long values,
I have been using those methods (below):
---
The 1st one is for integer values.
The 2nd one is for long values.
(long is not properly supported in Processing, so I had to use this modified method)
---
The 1st one could be modified for float values.
---

// return k mod m
public static int mod(int k, int m) {
 k = k % m;
 if(k < 0) k+=m;
 return(k);
}//mod()

// return k mod m
public static long mod(long k, long m) {
 boolean minus=(k<0);
 if (minus) k=-k;
 k = k-((k/m)*m);
 if (minus) k=m-k;
 return(k);
}//mod()
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Eric F.
Re: % modulo 4%3 = 1%3 !=
Reply #5 - Feb 10th, 2009, 7:39pm  
ah yes, good catch.  thanks.  
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PhiLho 
Re: % modulo 4%3 = 1%3 !=
Reply #6 - Feb 10th, 2009, 9:46pm  
alphachapmtl wrote on Feb 10th, 2009, 6:25pm:
"a+b" should be "(a%b)+b"

Not really, as the result is 'b' when a%b is 0!
Your function is correct, though.
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alphachapmtl
Re: % modulo 4%3 = 1%3 !=
Reply #7 - Feb 11th, 2009, 12:02am  
Hey, you`re right!(didn`t check that case...)
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alphachapmtl
Re: % modulo 4%3 = 1%3 !=
Reply #8 - Feb 11th, 2009, 12:06am  
So, for float values, better use:

// return k mod m
public static float mod(float k, float m) {
 k = k % m;
 if(k < 0) k+=m;
 return(k);
}//mod()

or

// return a mod b
public static float mod(float a, float b) {
 a = a % b;
 if(a < 0) a+=b;
 return(a);
}//mod()
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