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  #31  
Old 06-18-2008, 11:00 PM
cbk1994 cbk1994 is offline
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Quote:
Originally Posted by Stryke View Post
Would this work?

PHP Code:
function onCreated()
{
  
this.start timevar2;
  
this.incompatability = {
                         {
23,46},
                         {
29,44},
                         {
45,96},
                         {
11,77},
                         {
2,32},
                         {
1,12},
                         {
13,69},
                         };
  
this.houselist NULL;
  for(
temp.i=0temp.i<100temp.i++)
  {
    
this.intA 0;
    
this.intB 0;
    while(!
this.intA && !this.intB)
    {
      
this.intA int(random(0,400));
      
this.intB int(random(0,400));
      for(
temp.j=0temp.j<100temp.j++)
      {
        if(
this.intA == this.houselist[temp.j][0])
          
this.int 0;
        else if(
this.intA == this.houselist[temp.j][1])
          
this.int 0;
        else if(
this.intB == this.houselist[temp.j][0])
          
this.int 0;
        else if(
this.intB == this.houselist[temp.j][1])
          
this.int 0;
        if(
this.intA == this.incompatibility[temp.j])
          if(
this.intB == this.incompatibility[temp.j])
            
this.int 0;
        else if(
this.intB == this.incompatibility[temp.j])
          if(
this.intA == this.incompatibility[temp.j])
            
this.int 0;
      }
      if(
this.intA == this.intB)
        
this.int 0;
    }
    
this.houselist.add({this.intA,this.intB});
  }
  
this.end timevar2;
}
savelog2("houseList.txt",this.houselist);
echo(
"Time: " this.end this.start); 
Time: 0.026829004

Damn that's slow compared to others lol
I didn't try it, but, like said, anything using random will not work well with larger numbers.
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  #32  
Old 06-18-2008, 11:00 PM
Dan Dan is offline
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I think it's slow because of the logging you're doing. But yeah, quite some loops also.
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  #33  
Old 06-21-2008, 06:36 AM
Stryke Stryke is offline
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Oh it's only 50? :P

Time remaining: 0.007151126

Still crappy. :P I took out the log too.
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  #34  
Old 06-21-2008, 06:52 PM
Tolnaftate2004 Tolnaftate2004 is offline
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We can assume that the algorithm would eventually need to be used to find an appropriate room assignment for a group of students for which there is only one solution (i.e. worst case scenario). Because of this, the algorithm would need to "know" that student A needs to pair with student B rather than student C. Hence why random is a horrible idea.
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  #35  
Old 07-19-2008, 09:26 PM
WhiteDragon WhiteDragon is offline
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PHP Code:
//#CLIENTSIDE
function onWeaponFired() {
  
// Bad match array
  
temp.bad = {
             {
91,50},
             {
60,230},
             {
132,399},
             {
52,293},
             {
22,25},
             {
83,293},
             {
83,82}
             };
  
// Generate students array
  
temp.students = new[400];
  for (
temp.0temp.400temp.i++)
    
temp.students[temp.i] = temp.i;



  
// START of exercise
  
temp.start timevar2;

  
// Shuffle students array
  
temp.temp.students.size();
  for (
temp.0temp.temp.1temp.i++) {
    
temp.temp.random(0,32767) / (32767 / (temp.temp.i) + 1);
    
temp.temp.students[temp.j];
    
temp.students[temp.j] = temp.students[temp.i];
    
temp.students[temp.i] = temp.t;
  }
  
  
// Heapsort the bad match array
  
temp.bad.sortascending();
  
  
temp.pairs = new[50];
  
temp.0;
  while (
temp.50) {
    
temp.pair = {temp.students[temp.x], temp.students[temp.x+1]};

    
// Binary search for the current pair in the bad pairs
    
if (binarysearcharray(temp.badtemp.pair) == -1) {
      
temp.pairs[temp.i] = temp.pair;
      
temp.x++;
    }
    
temp.x++;
    
temp.i++;
  }
  
  
  
// END of exercise
  
temp.time timevar2 temp.start;
    
  
player.chat temp.pairs;
  echo (
temp.time);
}

// Modified binary search
function binarysearcharray(temp.atemp.v) {
  
temp.low 0;
  
temp.high temp.a.size()-1;
  while (
temp.low <= temp.high) {
    
temp.mid = (temp.low temp.high) /2;
    if (
temp.a[temp.mid][0] > temp.v[0] || temp.a[temp.mid][0] == temp.v[0] && temp.a[temp.mid][1] > temp.v[1]) 
      
temp.high temp.mid 1;
    else if (
temp.a[temp.mid][0] < temp.v[0] || temp.a[temp.mid][0] == temp.v[0] && temp.a[temp.mid][1] < temp.v[1])
      
temp.low temp.mid +1;
    else
      return 
temp.mid;
  }
  return -
1;

Average running time of 0.00326736083, but it's a rather small "bad pair" set.

Should run in O(n log n log m) where n is the number of pairs and m is the number of bad pairs.

Last edited by WhiteDragon; 07-20-2008 at 09:20 PM.. Reason: Slight mistake
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  #36  
Old 07-21-2008, 07:47 AM
WhiteDragon WhiteDragon is offline
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I realized I made an error, but it wouldn't let me edit it... so here is another giant block of code:

PHP Code:
//#CLIENTSIDE
function onWeaponFired() {
  
// Bad match array
  
temp.bad = {
             {
91,50},
             {
60,230},
             {
132,399},
             {
52,293},
             {
22,25},
             {
83,293},
             {
83,82}
             };
  
// Generate students array
  
temp.students = new[400];
  for (
temp.0temp.400temp.i++)
    
temp.students[temp.i] = temp.i;



  
// START of exercise
  
temp.start timevar2;

  
// Shuffle students array
  
temp.temp.students.size();
  for (
temp.0temp.temp.1temp.i++) {
    
temp.temp.random(0,32767) / (32767 / (temp.temp.i) + 1);
    
temp.temp.students[temp.j];
    
temp.students[temp.j] = temp.students[temp.i];
    
temp.students[temp.i] = temp.t;
  }
  
  
// Heapsort the bad match array
  
temp.bad.sortascending();
  
  
temp.pairs = new[50];
  
temp.0;
  while (
temp.50) {
    
temp.pair = {temp.students[temp.x], temp.students[temp.x+1]};

    
// Binary search for the current pair in the bad pairs
    
if (binarysearcharray(temp.badtemp.pair) == -1) {
      
// Add it
      
temp.pairs[temp.i] = temp.pair;
      
temp.students.remove(temp.students[temp.x]);
      
temp.students.remove(temp.students[temp.y]);
      
temp.i++;
      
temp.0;
      
temp.0;
    } else {
      
// Look more
      
if (temp.== temp.students.size()) {
        
temp.x++;
        
temp.0;
      } else {
        
temp.y++;
      }
    }
  }
  
  
  
// END of exercise
  
temp.time timevar2 temp.start;
    
  
player.chat temp.pairs;
  echo (
temp.time);
}

// Modified binary search
function binarysearcharray(temp.atemp.v) {
  
temp.low 0;
  
temp.high temp.a.size()-1;
  while (
temp.low <= temp.high) {
    
temp.mid = (temp.low temp.high) /2;
    if (
temp.a[temp.mid][0] > temp.v[0] || temp.a[temp.mid][0] == temp.v[0] && temp.a[temp.mid][1] > temp.v[1]) 
      
temp.high temp.mid 1;
    else if (
temp.a[temp.mid][0] < temp.v[0] || temp.a[temp.mid][0] == temp.v[0] && temp.a[temp.mid][1] < temp.v[1])
      
temp.low temp.mid +1;
    else
      return 
temp.mid;
  }
  return -
1;

Worst case is actually O(n**2 log n log m), but it uses Randomization and Restriction, so it's considerably fast than the worst case.
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  #37  
Old 08-02-2008, 11:24 AM
_Z3phyr_ _Z3phyr_ is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
Today's challenge will include 2 problems:
  1. hard Suppose that you are organizing housing accommodations for a group of four hundred university students. Space is limited and only one hundred of the students will receive places in the dormitory. To complicate matters, the Dean has provided you with a list of pairs of incompatible students, and requested that no pair from this list appear in your final choice.
    The point of this exercise is to GENERATE DISCUSSION. We are not asking you to WRITE the code, merely speculate as to how you might do it. BUT IF YOU FEEL COMPELLED, write the code to accomodate arbitrary lists from the Dean.

Note From Hell Raven: the 100 students are rooming in pairs of 2. So that is 50 pairs (I do not believe this was clear...)

answer:





proof:

N and B are disjoint sets. (see above picture)

---

this is an abstract problem. idk what you guys have been doing with all of that prime number crap.

Last edited by _Z3phyr_; 08-02-2008 at 01:40 PM..
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  #38  
Old 08-02-2008, 10:35 PM
WhiteDragon WhiteDragon is offline
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This problem is NP-complete, so it's impossible to do it in polynomial time.
The runtime is actually O(n**m) if you check all conditions. The best way to attack the problem would most likely be dynamic programming.
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