Hi Neil.

17-May-04 17:10:04, Neil Heller wrote to Jasen Betts

 NH> JB>I get the feeling that you want the number of races that will
 NH> suffice JB>to determine the 5 fastest in all cases.

 NH> JB>in my first attempt I can do it in 26 races, but I don't think
 NH> that's JB>anything special. there is probably away to do this with
 NH> far fewer JB>than 26 races.

 NH> JB>if acurate relative spacing is also available (by a nose,
 NH> JB>two lengths begind etc) JB>is available the answer is again 6
 NH> races.

 NH> 6 or 26?

 NH> Relative spacing, I think, is the answer.  Using that, the field
 NH> of 25 could be easily cut down to 10-15 after just 6 races.

if it can be measured accurately it is as good as a stopwatch.. (and
therefore probably against the rules too)

 NH> Still, I think that 2 more races would be necessary in the close
 NH> cases.

after thinking some more on the matter,

any any horse that has finished behind 5 other horses isn't elegible for
the top 5

immagine a 5x5 grid  race the rows against each other and sort the members
of each row according to order     (5 races)

then do the same with the coplumns (5 more races)

this will puss the fastest horses to the top and left edge of the grid:

possible arrangements of the top 5:

     * * * * *     * * * * -     * * * - -     * * * - -
     - - - - -     * - - - -     * * - - -     * - - - -
     - - - - -     - - - - -     - - - - -     * - - - -
     - - - - -     - - - - -     - - - - -     - - - - -
     - - - - -     - - - - -     - - - - -     - - - - -

     * * - - -     * * - - -     * - - - -
     * * - - -     * - - - -     * - - - -
     * - - - -     * - - - -     * - - - -
     - - - - -     * - - - -     * - - - -
     - - - - -     - - - - -     * - - - -

basically now (after 10 races) there's field of 9 horses of which you need
to find the
fastest 4 (since we know X is the fastest horse)

     X a b c d
     f e . . .
     g . . . .
     h . . . .
     i . . . .

now... race  a b e h i and take the two fastest,
and then race f g c d and the third place getter from
the last race and take the two fastest

that'll do it

12 races.

is there  a better solution?

 -=> Bye <=-

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