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  #16  
Old 11-20-2007, 01:52 AM
Inverness Inverness is offline
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Quote:
Originally Posted by Skyld View Post
PHP Code:
temp.list = new [0]; 
Quote:
Originally Posted by Graal v5.006 Info
You can now create empty arrays simplier by doing "var = {};" (it was displaying a syntax error before)
...
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  #17  
Old 11-20-2007, 02:33 AM
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Quote:
Originally Posted by Googi View Post
I'm pretty seriously fourth or fifth rate at scripting.
Where did this rank come from?
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  #18  
Old 11-20-2007, 03:09 AM
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Quote:
Originally Posted by Twinny View Post
Where did this rank come from?
I think it should be pretty clear that I'm just using "fourth or fifth" as a synonym for "low" rather than an accurate expression of tier rank.
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  #19  
Old 11-20-2007, 05:25 AM
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Quote:
Originally Posted by Skyld View Post
PHP Code:
temp.list = new [0]; 
PHP Code:
temp.list = {}; 
Support for that was recently added my Stefan. I use it as it is a few characters shorter and in all honesty, it really doesn't matter .

Quote:
Originally Posted by Kristi View Post
Zero, that is incredibly inefficient. We want something better then the naive algorithm
Well you made it in your example that they have to be in order of the first array so that was the only real algorithm that would really work. So far, I haven't seen anything better in this thread that actually works.
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  #20  
Old 11-20-2007, 07:52 AM
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Quote:
Originally Posted by Googi View Post
PHP Code:
  if (temp.array1.size || temp.array2.size 2) { 

PHP Code:
for (temp.0temp.temp.array1.size 3temp.i++) { 
wouldn't work at all
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  #21  
Old 11-20-2007, 08:05 AM
Tolnaftate2004 Tolnaftate2004 is offline
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PHP Code:
function S3(arrA,arrB) {
  
temp.intersect.clear();
  
temp.sets.clear();
  if (
arrA.size() > arrB.size()) temp.= {arrB,arrA};
  else                           
temp.= {arrA,arrB};

  for (
elmttemp.c[0]) {
    if (
elmt in temp.c[1])
      
temp.intersect.add(elmt);
  }

  
temp.len temp.intersect.size();
  for (
temp.i=0temp.i<temp.len-2temp.i++) {
    for (
temp.j=temp.i+1temp.j<temp.len-1temp.j++) {
      for (
temp.k=temp.j+1temp.ktemp.lentemp.k++) {
        
temp.sets.add({temp.intersect[temp.i],temp.intersect[temp.j],temp.intersect[temp.k]});
      }
    }
  }
  return 
temp.sets;

This has O(N+N!/((N-3)!*3!)), N ≤ min{a.size(),b.size()}.
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Last edited by Tolnaftate2004; 11-21-2007 at 03:34 AM..
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  #22  
Old 11-20-2007, 11:46 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
PHP Code:
function S3(arrA,arrB) {
  
temp.intersect.clear();
  
temp.sets.clear();
  if (
arrA.size() > arrB.size()) temp.= {arrB,arrA};
  else                           
temp.= {arrA,arrB};

  for (
elmttemp.c[0]) {
    if (
elmt in temp.c[0])
      
temp.intersect.add(elmt);
  }

  
temp.len temp.intersect.size();
  for (
temp.i=0temp.i<temp.len-2temp.i++) {
    for (
temp.j=temp.i+1temp.j<temp.len-1temp.j++) {
      for (
temp.k=temp.j+1temp.ktemp.lentemp.k++) {
        
temp.sets.add({temp.intersect[temp.i],temp.intersect[temp.j],temp.intersect[temp.k]});
      }
    }
  }
  return 
temp.sets;

This has O(N+N!/((N-3)!*3!)), N ≤ min{a.size(),b.size()}.
^_^ this is an open discussion. Please explain why this is better then the naive algorithm
Also, in your big O, did you include the search time for the "in" operator in your intersect portion? I can see a few ways to optimize that.
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Last edited by Kristi; 11-20-2007 at 12:00 PM..
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  #23  
Old 11-20-2007, 10:42 PM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Kristi View Post
Also, in your big O, did you include the search time for the "in" operator in your intersect portion? I can see a few ways to optimize that.
A good point: O(N1*N2 + N!/((N-3)!*3!))

I had originally intended to do something along these lines:
PHP Code:
  for (elmttemp.c[0]) {
    if (
elmt in temp.c[1])
      
temp.intersect.add(elmt);
  } 
becomes
PHP Code:
  while (temp.c[0].size()) {
    
temp.elmt temp.c[0][0];
    if (
temp.elmt in temp.c[1]) {
      
temp.intersect.add(temp.elmt);
      
temp.c[0].remove(elmt);
      
temp.c[1].remove(elmt);
    } else 
temp.c[0].delete(0);
  } 
Still O(N1*N2), but often less.
It just slipped my mind. Silly me.
Also, first thing should be if (arrA.size() > 2 && arrB.size() > 2)...

As for the naive algorithm, It has O(N2^3*N1!/((N1-3)!*3!)) (?), I think.
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Last edited by Tolnaftate2004; 11-21-2007 at 03:34 AM..
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  #24  
Old 11-21-2007, 12:41 AM
Googi Googi is offline
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Quote:
Originally Posted by xAndrewx View Post
wouldn't work at all
Those are syntax errors, they mean that the script wouldn't work if actually used, they don't mean that the method I'm trying to use is nonsensical.
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  #25  
Old 11-21-2007, 02:52 AM
Novo Novo is offline
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PHP Code:
function getSets()
{
  return {{
2,3,5}, {2,3,6}, {2,3,10}, {2,5,6}, {2,5,10}, {2,6,10}, {3,5,6}, {3,5,10}, {3,6,10}, {5,6,10}};

I win.
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  #26  
Old 11-21-2007, 02:53 AM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Novo View Post
PHP Code:
function getSets()
{
  return {{
2,3,5} {2,3,6} {2,3,10} {2,5,6} {2,5,10} {2,6,10} {3,5,6} {3,5,10} {3,6,10} {5,6,10}};

I win.
Not quite. :P
e: nice ninja edit.
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  #27  
Old 11-21-2007, 03:14 AM
Crono Crono is offline
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Using my leet skillz I have created the ultimate:

PHP Code:
//Returns array containing sets of 3 numbers both arrays have ing common
function ex1(arr1arr2) {
  
temp.result = {};
  
temp.pos 0;
  while (
temp.pos temp.arr1.size() - 2) {  //Make sure there's 3 elements left
    
temp.check = {temp.arr1[temp.pos], temp.arr1[temp.pos 1], temp.arr1[temp.pos 2]};
    if (
temp.result.pos(temp.check) < && searchArr(temp.arr2temp.check)) {
      
temp.result.add(temp.check); 
      
temp.pos += 3;  //Set is found, jump over whole set
    
}
    else {
      
temp.pos++;  //Set not found, go to next number
    
}
  }
  return 
temp.result;
}

function 
ArrayComp(array1,array2) {
  if (
temp.array1.size || temp.array2.size 2) {
    
temp.checkarray temp.array2;
    
temp.checkarray.delete(temp.array2.size 1);
    
temp.checkarray.delete(temp.array2.size 2);
    
temp.failarray = new [0];
    for (
temp.0temp.temp.array1.size 3temp.i++) {
      if (
temp.failarray.index(temp.array1[temp.i]) == -1) {
        
temp.var = temp.checkarray.index(temp.array1[temp.i]);
        if (
temp.var == -1) {
          
temp.failarray.add(temp.array1[temp.i]);
        }
        else {
          
temp.dupecheck temp.checkarray;
          while (
temp.var != -1) {
            if (
temp.array1[temp.1] == temp.array2[temp.var + 1]) {
              if (
temp.array1[temp.2] == temp.array2[temp.var + 2]) {
                
temp.returnvar++;
              }
            }
            
temp.dupecheck.replace(temp.var,"x");
            
temp.var = temp.dupecheck.index(temp.array1[temp.i]);
          }
        }
      }
    }
  }
  return 
temp.returnvar;
}  

function 
ex1(a1a2) {
  
temp.out = {};
  
temp.0;
  
temp.sub 0;

  for (
temp.0a.size() - 2i++) {
    if (!(
a[iin b)) continue;
    for (
temp.1a.size() - 1j++) {
      if (!(
a[jin b)) continue;
      for (
temp.1a.size(); k++) {
        if (
a[kin b) list.add({a[i], a[j], a[k]});

function 
S3(arrA,arrB) {
  
temp.intersect.clear();
  
temp.sets.clear();
  if (
arrA.size() > arrB.size()) temp.= {arrB,arrA};
  else                           
temp.= {arrA,arrB};

function 
getSets()
{
  return {{
2,3,5}, {2,3,6}, {2,3,10}, {2,5,6}, {2,5,10}, {2,6,10}, {3,5,6}, {3,5,10}, {3,6,10}, {5,6,10}};

i SO did not copy paste bits from everyone else's scripts
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  #28  
Old 11-21-2007, 04:22 AM
coreys coreys is offline
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Rather long, don't you think? XD
But I don't even have the competence to try.
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  #29  
Old 11-29-2007, 08:47 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
A good point: O(N1*N2 + N!/((N-3)!*3!))

I had originally intended to do something along these lines:
PHP Code:
  for (elmttemp.c[0]) {
    if (
elmt in temp.c[1])
      
temp.intersect.add(elmt);
  } 
becomes
PHP Code:
  while (temp.c[0].size()) {
    
temp.elmt temp.c[0][0];
    if (
temp.elmt in temp.c[1]) {
      
temp.intersect.add(temp.elmt);
      
temp.c[0].remove(elmt);
      
temp.c[1].remove(elmt);
    } else 
temp.c[0].delete(0);
  } 
Still O(N1*N2), but often less.
It just slipped my mind. Silly me.
Also, first thing should be if (arrA.size() > 2 && arrB.size() > 2)...

As for the naive algorithm, It has O(N2^3*N1!/((N1-3)!*3!)) (?), I think.
Deleting elements from the array and etc is actually going to make it take longer to process.

Anyway, hashes would optimize finding the intersect, but graal has no support for it. However, we can dynamically name variables, so we can emulate it well enough for this situation.

code snippet:
PHP Code:
for(etemp.array1)
   
temp.b.("v"@e) = true;

for(
etemp.array2)
   if(
temp.b.("v"@e))
     
temp.intersect.add(e); 
with larger datasets, this gets exponentially faster. with 5000-10000 element arrays, its literally 120x faster then searching through the arrays for intersects.

Also, it would be slightly faster to make an array with the max possible size first, so you're not resizing the array (and thus possibly moving it to a new location in memory constantly.)
PHP Code:
  temp.intersect = new[smallerarray.size()];

  for(
etemp.largerarray)
   
temp.b.("v"@e) = true;

  for(
esmallerarray)
   if(
temp.b.("v"@e)) {
     
temp.intersect[temp.j] = e;
     
temp.j++;
   } 
This does in fact improve performance, although it's not terribly noticeable.

I'll optimize finding the sets from the intersects later. Lets let this soak in first.
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Last edited by Kristi; 11-29-2007 at 09:08 PM..
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  #30  
Old 11-29-2007, 08:53 AM
Tolnaftate2004 Tolnaftate2004 is offline
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The maximum size of the intersect should be the size of the small array as it cannot possibly have more elements in common than there are in the set.
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