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Extending Gaylord's Problem

  • To: mathgroup at
  • Subject: [mg5003] Extending Gaylord's Problem
  • From: Jack Goldberg <jackgold at>
  • Date: Sat, 19 Oct 1996 02:25:49 -0400
  • Sender: owner-wri-mathgroup at

Hi Group;

style for this problem:  

place this entry in a new list  Lnew  and remove it from  L1.  Repeat 
until all entries of   L1    have been removed.  At this point   Lnew 
is a permutation of  L1.    (As pointed out by others, a tie 
breaking scheme is needed to make the problem uniquely solvable.)

to me that there is an extension whose solution might be of some interest.  
Here it is. 

new list   Lnew   by taking the maximum of the first entries of the lists 
L1,..., Lm.  The winning element is removed from the list in which it 
appeared and the second element of that list becomes its new first 
element.  In the event of a tie, choose the element from the list with 
the lowest subscript.  Repeat until  Lnew  contains  r  entries.  (r  is 
set in advance and cannot be more than the total number of entries in 
all of the lists.)  When all the elements of a list are gone, the 
selection proceeds without that list.  It seems to me that this game 
has a unique solution, but programming it appears somewhat of a 
challange.  The lengths of the lists need not be equal, but they can 
always be arranged so by adding sufficiently many  -Infinity  at the 
end of the shorter lists. 

Simply let  L2 = Reverse[L1]   and   r = Length[L1].

Jack Goldberg  

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