Re: Extending Gaylord's Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg5030] Re: Extending Gaylord's Problem
• From: espen.haslund at fys.uio.no (Espen Haslund)
• Date: Sat, 19 Oct 1996 16:40:33 -0400
• Organization: Universitet i Oslo
• Sender: owner-wri-mathgroup at wolfram.com

```In article <542t6a\$p9f at dragonfly.wolfram.com>, jackgold at math.lsa.umich.edu says...
>
>Hi Group;
....
>        Suppose we have  m  lists labelled  L1, L2, ..., Lm.  We form a
>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.
>
>        Incidentally, Gaylord's Problem is a special case with  m = 2.
>Simply let  L2 = Reverse[L1]   and   r = Length[L1].
>
>Jack Goldberg
>
>
Hi Jack;

I think maybe this one meets your specifications.
It is more or less an extension of Allan Hayes
solution to Gaylord's Problem.

takeFirst[data_, n_:-1] :=
Module[{r, k, i, ii, big, lis, ll},
ll = Map[Length, data];
If[n == -1, r = Plus @@ ll, r = n];
ii = Table[1, {Length[data]}];
Table[
big = -Infinity;
Do[
lis = data[[j]];
If[(i = ii[[j]]) <= ll[[j]],
If[lis[[i]] >= big,
big = lis[[i]]; k = j
]
],
{j, Length[data], 1, -1}];
ii[[k]] = ii[[k]] + 1;
big,
{r}]
];

(* data -- {L1, L2, ..., Ln} *)

-Espen

```

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