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Index › Programming Questions & Help › Programs › math problem
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math problem (Read 587 times)
distro
math problem
May 12th, 2009, 10:05am  
Hello
I've been working on this:

int x=1;
int y =1;
int z=0;
for(int i=0;i<=7;i=i+1){
 
x=x*-1;
 y=y*x;

 if(i<4){
   z=-1;
 }
 else{
   z=1;
 }
 println("x="+x+"  y="+y+"  z="+z);
}

This program defines "x","y",and "z" coordinates for the vertices from a 3D cube. I think the way "x" and "y" coordinates are calculated is ok, but i can't figure out how to calculate "z" without using "if"and"else". Although the result is correct, i think the way isn't.

Can you help me please?
Thanks.
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PhiLho 
Re: math problem
Reply #1 - May 12th, 2009, 11:48am  
Anyway giving the correct result is "right", as long as it isn't incredibly inefficient! Smiley

That said, you can make your code a bit more Java-esque, using some shortcuts:
Code:
int x = 1;
int y = 1;
int z = 1;
for (int i = 0; i < 8; i++) {
 x *= -1;
 y *= x;
 z = i < 4 ? -1 : 1;
 println("x="+x+"  y="+y+"  z="+z);
}

The ? : is actually a conditional in disguise.
If you really don't want a conditional, you can compute z as follow:

z = 1 - 2 * (i / 4);

Not going the readability road, though! Smiley
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distro
Re: math problem
Reply #2 - May 12th, 2009, 12:59pm  
thanks
I think that solves my problem
Wink
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NoahBuddy
Re: math problem
Reply #3 - May 12th, 2009, 1:06pm  
As PhiLho suggested (in other words) "if it isn't broke, don't fix it."

BUT... Cheesy

I kinda like puzzles.
This was a bit of a challenge to do without comparisons.

It won't work in all cases.
Also, this is not good practice as it does not test for a div/zero condition.

If your goal was to confuse someone trying to take your code, it could throw them off. See http://en.wikipedia.org/wiki/Obfuscated_code

Code:
int x=1;
int y =1;
int z=0;
for(int i=-7;i<=8;i=i+2){//odd start value to skip over zero
  x=x*-1;
  y=y*x;
  
  z = i / abs(i);

  println("x="+x+"  y="+y+"  z="+z);
}


I think PhiLho's is easier to understand. Smiley
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davbol
Re: math problem
Reply #4 - May 12th, 2009, 1:16pm  
not that it matters, as problem already solved, but all three coords can be expressed as f(i) if you wish:
Quote:
  for (int i=0; i<8; i++) {
   int x = (i&1) * 2 - 1;
   int y = (i&2) - 1;
   int z = (i&4) / 2 - 1;
 }
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PhiLho 
Re: math problem
Reply #5 - May 12th, 2009, 1:36pm  
davbol wrote on May 12th, 2009, 1:16pm:
not that it matters, as problem already solved

Hey, if we see the question as a puzzle, all answers are good! :D
I like your solution, I saw there was a binary enumeration, but I didn't thought of this way.
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