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				August 6th, 2003, 09:43 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		No, that's not right.  
 
LGM, I didn't mean any offense. I looked at your initial post and didn't think it solved teh problme, but after looking at it again I see it does. Can you explain the math your program is doing to find the solutions? That is what I am trying to figure out. 
 
Geoschmo 
		
	
		
		
		
		
		
		
			
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				August 6th, 2003, 09:47 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Only solutions are 3, 7, 15, 31,... 
 
Other numbers of players leave a three some with two players and no one to match them with.  Someone demonstrated this earlier with 9 players. 
		
	
		
		
		
		
		
		
		
		
		
	
		
		
	
	
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				August 6th, 2003, 09:49 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Sorry slick, this is way over my head.  
"The general formula for  C(n,r) = n!/[r!(n-r)!]"
 
What do n and r represent in the formula, and what are the exclamation points for?
 
And is C(n,r) the total numebr of players needed in the tourney, or the number of games played by each person in the tourney, or something else alltogether?      
		
	
		
		
		
		
		
		
			
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				August 6th, 2003, 09:50 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		
	Quote: 
	
	
		
			
				Originally posted by LGM: 
Only solutions are 3, 7, 15, 31,... 
 
Other numbers of players leave a three some with two players and no one to match them with.  Someone demonstrated this earlier with 9 players.
			
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 Right, but other then working them out by hand, or having you plug it into yoru black box and give me the result, can you explain how you know this? What I am trying to learn is not the answer, but how the answer is derived. 
 
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				August 6th, 2003, 09:55 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Here is the code.  Crudely done by brute force in a rather esoteric flavor of basic: 
 
PRIMES = '2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61  ,67,71,73,79,83,89,97,101,103,109,113,127,131,137' 
SWAP ',' WITH @AM IN PRIMES 
ILIMIT = DCOUNT(PRIMES,@AM) 
PRINT ILIMIT 
FOR I = 4 TO ILIMIT 
  NBR = I 
  COMBOS = '' 
  COMBOX = '' 
  IDX = 1 
  FOR J = 1 TO I 
    FOR K = J+1 TO I 
      FOR L = K+1 TO I 
        GOSUB FIND.DUPLICATE 
        IF NOT(FOUND.DUPLICATE) THEN 
          COMBOS<-1> = PRIMES*PRIMES*PRIMES 
          COMBOX<-1> = J:',':K:',':L 
        END 
      NEXT L 
    NEXT K 
  NEXT J 
  GOSUB CHECK.SOLUTION 
NEXT I 
PRINT 'DONE' 
STOP 
* 
SOLVED:* 
  PRINT '---------------------' 
  PRINT 'SOLVED' 
  PRINT NBR 
  FOR M = DCOUNT(COMBOX,@AM) TO 1 STEP -1 
    PRINT COMBOX 
  NEXT M 
  INPUT CH,1_ 
RETURN 
* 
CHECK.SOLUTION: * 
  FOR J = 1 TO I 
    FOR K = J + 1 TO I 
      NBR.MATCHES = 0 
      FOR M = DCOUNT(COMBOS,@AM) TO 1 STEP -1 
        A.COMBO = COMBOS 
        IF MOD(A.COMBO,PRIMES)=0 AND MOD(A.COMBO,PRIMES)=0 THEN 
          NBR.MATCHES = NBR.MATCHES + 1 
          IF NBR.MATCHES > 1 THEN RETURN 
        END 
      NEXT M 
      IF NBR.MATCHES = 0 THEN RETURN 
    NEXT K 
   NEXT J 
   GOSUB SOLVED 
RETURN 
* 
FIND.DUPLICATE: * 
  FOUND.DUPLICATE = 0 
  FOR M = DCOUNT(COMBOS,@AM) TO 1 STEP -1 
    A.COMBO = COMBOS 
    J.MOD = MOD(A.COMBO,PRIMES) 
    K.MOD = MOD(A.COMBO,PRIMES) 
    L.MOD = MOD(A.COMBO,PRIMES) 
    IF J.MOD = 0 THEN 
      IF K.MOD=0 OR L.MOD=0 THEN 
        FOUND.DUPLICATE = 1 
      END  
    END ELSE IF K.MOD=0 AND L.MOD=0 THEN 
      FOUND.DUPLICATE = 1 
    END 
  NEXT M 
RETURN 
		
	
		
		
		
		
		
		
		
		
		
	
		
			
			
			
			
				 
			
			
			
			
            
			
			
				
			
			
			
		 
		
	
	
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				August 6th, 2003, 10:01 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Ok, so bascially you are having the couputer run through all the possible options fast and see if a number works. Ok, that does the job but isn't really what I was looking for. Mayeb there is no simple mathematical formula for what I am looking for.  
 
Basically I can take x players and group them into games of three players each and see fairly quickly that 7 players will work, but 5 will not. What I was hoping to find was a formula that I could plug the numbers in and be able to tell if a combination of x and y would work without having to go through all teh grouping by hand. 
 
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				August 6th, 2003, 10:06 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		By the way LGM, what is the reason for using the prime numbers in the program? I'm not a programmer so I can't really make heads or tails of your code there, not that it wasn't apprecieated.     
		
	
		
		
		
		
		
		
			
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				August 6th, 2003, 10:07 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Try X raised to the Yth power and subtract one.  Just a theory.  I am not sure if that works for 4 or 5 man games yet. X is Number of players per game minus 1. Incidentally, that does not work for 2 players games as any number works for 'round robin' leagues. 
		
	
		
		
		
		
		
		
		
		
		
	
		
		
	
	
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				August 6th, 2003, 10:09 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		Prime numbers are a nice way to test if something is a member of a set.  Products of prime numbers is a common way of encoding sets because you can divide by a given prime.  If the remainder is zero, it is an element of the set. 
		
	
		
		
		
		
		
		
		
		
		
	
		
		
	
	
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				August 6th, 2003, 10:12 PM
			
			
			
		  
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				Re: Math problem
			 
             
			
		
		
		
		You could use 21 players (3 sets of 7) with a championship game from the winners of each set. 
 
Or you could use 49 players (7 sets of 7) and have a championship round of the top 7 playing each of the top 7. 
  
 [ August 06, 2003, 21:13: Message edited by: LGM ] 
		
	
		
		
		
		
		
		
		
		
		
	
		
		
	
	
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