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