       Extending Gaylord's Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg5003] Extending Gaylord's Problem
• From: Jack Goldberg <jackgold at math.lsa.umich.edu>
• Date: Sat, 19 Oct 1996 02:25:49 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```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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