Players (2) take turns, each playing m rounds where they are offense

and each playing m rounds where they are defense. A player's grand

score is the sum of the player's scores for rounds where they played

offense.

Both players on a round *take turns* adding one number at a time to a

list.

The number should be from 1 to n (where n is, say, 20) and must not

have

been already put in the list by either player.

After n turns the list is complete and represents a permutation of the

integers 1 through n.

Scoring (by the offensive player) is as follows.

Write below the list of integers (if the integer list was made from

left to right) a list of (n-1) U's and D's, placing each letter between

and below each pair of neighboring integers, a U for up if the

right-most element of the pair above the letter is greater than the

left-most element, or put a D for down if the right-most element of the

pair above the letter is less than the left-most element of the pair.

(See example below.)

The offensive player gets a point for each element in the *longest*

sequence of U's and D's that occurs somewhere else in the U/D list.

And it is possible that a particular U/D sub-sequence shares specific

elements with its duplicate.

For example:

If we have the 10-integer game:

1, 7, 3, 5, 2, 9, 10, 4, 6, 8,

(Player 1 added the 1, 3, 2, 10, and 6 to the list;

Player 2 added the 7, 5, 9, 4, and 8 to the list)

we would have the U/D sequence:

U, D, U, D, U, U, D, U, U.

Now the sub-sequence U, D, U, U occurs twice in the sequence

(with the pair sharing one of their U's).

So the offensive player gets 4 points.

If the finished game looked like:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,

then we would have the U/D sequence of only U's (9 in number),

U, U, U, U, U, U, U, U, U.

And what would the score be here?

8 points, because the subsequence which equals another subsequence

is made up of 8 U's, and the two subsequences each share the middle 7

U's.

So, as I have implied, the two equal subsequences can share any

number of common elements but not share every element --

they must terminate at different locations in the U/D sequence.

thanks,

Leroy Quet

## Sunday, September 21, 2008

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